Quant Class Sessions Guide - Removed

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By Sandeep Gupta | GMAT 800/800, Harvard Final Admit
Quant Session 1
Arithmetic + Word Problems
PERCENTAGES
To express a% as a fraction divide it by 100
a % = a/100
To express a fraction as a percent multiply it by 100
a/b = [(a/b) ⨯ 100] %
To express percentage as a decimal we remove the symbol % and shift the decimal point by two
places to the left. For example
10% can be expressed as 0.1.
6.5% = 0.065 etc.
To express decimal as a percentage we shift the decimal point by two places to the right and write
the number obtained with the symbol % or simply we multiply the decimal with 100. Similarly, 0.7 =
70%.
Increase % = [ Increase / Original value] ⨯ 100%
Decrease % = [ Decrease / Original value] ⨯ 100%
Change % = [ Change / Original value] ⨯ 100%
In increase %, the denominator is smaller, whereas in decrease %, the denominator is larger.

GENERAL CONCEPTS IN PERCENTAGES:
Let’s start with a number X (= 1 X)
X increased by 10% would become X + 0.1 X = 1.1 X

X increased by 1% would become X + 0.01 X = 1.01 X
X increased by 0.1% would become X + 0.001 X = 1.001 X
X decreased by 10% would become X – 0.1 X = 0.9 X
X decreased by 1% would become X – 0.01 X = 0.99 X
X decreased by 0.1% would become X – 0.001 X = 0.999 X
X increased by 200% would become X + 2X = 3X
X decreased by 300% would become X – 3X = –2X
Also, let us remember that:
2 = 200% (or 100% increase)

3 = 300% (or 200% increase)
3.26 = 326% (means 226% increase)
Fourfold (4 times) = 400 % of original = 300% increase
10 times means 1000% of the original means 900% increase
0.6 means 60% of the original means 40% decrease
0.31 times means 31% of the original means 69% decrease etc.
1/2 = 50%
3/2 = 1 + 1/2 = 100 + 50 = 150%
5/2 = 2 + 1/2 = 200 + 50 = 250% etc.
2/3 = 1 – 1/3 = 100 – 33.33 = 66.66%
4/3 = 1 + 1/3 = 100 + 33.33 = 133.33%,
5/3 = 1 + 2/3 = 100 + 66.66 % = 166.66%
7/3 = 2 + 1/3 = 200 + 33.33 = 233.33%
8/3 = 2 + 2/3 = 200 + 66.66 = 3 – 1/3 = 300 – 33.33 = 266.66%
1/4 = 25%
3/4 = 75%
5/4 = (1 + 1/4) = 125% (= 25% increase)
7/4 = (1 + 3/4 = 2 – 1/4) = 175% (= 75% increase)
9/4 = (2 + 1/4) = 225% (= 125% increase)
11/4 = 275% = (175% increase)

1/5 = 20%
2/5 = 40%
3/5 = 60%
4/5 = 80%
6/5 = (1 + 1/5) = 120%
7/5 = (1 + 2/5) = 140% etc.
1/6 = 16.66%
5/6 = 83.33%
7/6 (1 + 1/6) = 116.66%
11/6 = 183.33%
1/8 = 12.5%
3/8 = 37.5%
5/8 = 62.5%
7/8 = 87.5%
9/8 = (1 + 1/8) = 112.5%
11/8 = (1 + 3/8) = 137.5%
13/8 = 162.5%
15/8 = 187.5%
1/9 = 11.11%
2/9 = 22.22%
4/9 = 44.44%
5/9 = 55.55%
7/9 = 77.77%
8/9 = 88.88%
10/9 = (1 + 1/9) = 111.11%
11/9 = (1 + 2/9) = 122.22% etc.

If the present population of a town is p and if there is an increase of X% per annum. Then:
(i) population after n years = p [1 + (X/100)]n
(ii) population n years ago = p / [1 + (X/100)]n
If the population of a town (or value of a machine) decreases at R% per annum, then:
(i) population (or value of machine) after n years = p [1 – (R/100)]n
(ii) population (or value of machine) n years ago = p / [1 –(R/100)]n
Profit % = (Profit /CP) x 100%
Loss % = (Loss/CP) x 100 %
In problems on DISCOUNT, remember the following:
Marked price is the price listed on the article (called list price).
Discount is calculated on Marked price and NOT on Cost price.
So, Marked Price – Discount = Sale Price. Also Cost Price + Profit = Sale Price.
Solved Examples
1. A child spends 30% of his pocket money, and has Rs 126 left. How much had he at first?
Sol.
Let the pocket money be X

70% of pocket money = 126
Or (70/100) ⨯ X = 126
X = 180 Rs.
2. When the cost of petroleum increases by 40%, a man reduces his annual consumption by 20%. Find
the percentage change in his annual expenditure on petroleum.
Sol.
Expenditure = Consumption ⨯ Price
First expenditure: Suppose 10 liters of petroleum at 10 units of money per liter, then total expenditure
= 10 ⨯ 10 units of money = 100 units of money
Second expenditure: Now, 8 liters of petroleum at 14 units of money per liter, total expenditure =
8 ⨯ 14 units of money = 112 units
So, the expenditure increases by 12%

3. The number of students in a school increases at a certain rate per cent. The number at present is 1323
and the number two years ago was 1200; find the rate per cent of the increase.
Sol.
By formula, we have 1200 ⨯ (1 + R/100)² = 1323 so R = 5%

1323 / 1200 = 441/400
(1 + R/100)2 = 1323/1200 = 441/400
Take square root
(1 + R/100) = 21/20
Subtract 1
R/100 = 1/20
So, R = 100/20 = 5%
4. A trader marks his goods at 50 percent above cost price and allows discount of 20% percent for cash
payment. What profit percent does he make?
Sol.
If the CP is 100, Marked Price = 150 But discount to the cash purchaser = 20% of Rs 150 = Rs 30
Now the reduced price he gets from the cash purchaser = 150 – 30 = 120
So, profit percent = 20%
5. A reduction of 20% in the price of apples could enable a man to get 120 more for Rs 1,440. Find the
first price of one apple.
Sol.
We have 1440 = X ⨯ Y .................(1)
X = no. of apples, Y = price of one apple.
Now 1440 = (X + 120) ⨯ 0.8Y ......(2)
Equate the two (as both are 1440)
X ⨯ Y = (X + 120) ⨯ 0.8Y
Y cancels
X = 480
Substitute X = 480 in (1) Y = Rs 3

6. A man’s working hours a day were increased by 25%, and his wages per hour were increased by 20%.
By how much percent were his daily earnings increased?
Sol.
wages per hour ⨯ number of hours = earnings

let each value be 10, so that earnings = 100
10 increased by 20% is 12
10 increased by 25% is 12.5
10 ⨯ 10 = 100
Now: 12⨯12.5 = 150 From 100 to 150 → 50% increase
OR
Let initially X = number of hours and Y = wages/hour
Later these become 1.25 X and 1.2 X respectively. Daily earnings initially = X ⨯ Y
Now Daily earnings = 1.25X ⨯ 1.2Y = 1.5 XY Hence 50% increase.
7. A shopkeeper allows a discount of 15% on the marked price. How much above the C.P. must he mark
his goods to make a profit of 19%?
Sol.
Let CP = 100, Profit = 19, SP = 100 + 19 = 119
Now marked price should be such that Marked price reduced by 15% is equal to 119
So, 85% of M.P. = 119 or MP = 119 ⨯ 100/85 = Rs 140
Answer = 40% above the C.P.
8. The production of a firm increases from 340 MT to 500 MT. What is the percent increase?
Sol. 160/340 = 8/17 = 47.05%.
9. The production of a firm decreases from 500 MT to 340 MT. What is the percent decrease?
Sol. 160/500 = 32%.
Note the answers to the above 2 questions are different.
10. The production of a firm increases by 20%, 25% and 50% in 3 successive years over the previous
year. If the production is 160 MT in the first year, find the production at the end of 3 years.
Sol. 160 × 1.2 × 1.25 × 1.5 = 360

11. The production of a firm decreases by 20% in the first year, then decreases by 25% in the next year
and then increases by 50% the next year and then increases by 10% in the next year. All percentage
changes being consecutive (over the previous year). If at the end of the changes, the value is 990 MT,
what was the value initially?
Sol. A × 0.8 × 0.75 × 1.5 × 1.1 = 990 or A = 1000
12. Which is bigger: 0.004% of 25000 or 25000% of 0.004?
Sol. Both are equal. A% of B = B % of A.
13. If price decreases by 25%, by what % should consumption increase so that the expenditure does
not increase?
Sol. Expenditure = price ⨯ consumption … assume values
100 = 100 ⨯ 1
100 = 75 ⨯ x
x = 100/75 = 4/3 = 1.33
1 becomes 1.33 → 33.33% increase
14. If speed increases by 33.33%, what is the percent reduction in the time taken to travel the same
distance?
Sol.
Distance = speed ⨯ time assume values

100 = 100 ⨯ 1
100 = 133.33 ⨯ x
x = 100/133.33 = 3/4 = 0.75
1 becomes 0.75
% decrease = 25%
15. The length of a rectangle increases by 25%. Find the percent drop in its width for area to remain
same.
Sol.
Area = L ⨯ B
100 = 1 ⨯ 100
100 = 1.25 ⨯ x
x = 100/1.25 = 80
x decreases from 100 to 80, so 20% decrease
16. If the length of a rectangle is decreased by 40% and the breadth is increased by 30%, then find the
percentage change in the area of the rectangle.
Assume values
10 ⨯ 10 = 100
6 ⨯ 13 = 78
Hence the area of the rectangle decreases by 22%
17. The edge of a cube increases by 20%. Find the % increase in its surface area and volume.
Sol.
Remember that Surface area is proportional to the square of length and the volume is proportional to
the cube of length.
Area ∝ length2
Volume ∝ length3
For area:
10 × 10 = 100
12 × 12 = 144, so area increases by 44%.
For volume
10 × 10 × 10 = 1000
12 × 12 × 12 = 1728, so 72.8% increase.
18. The side of a square decreases by 30%. Find the % decrease in its perimeter and area.
Sol. Perimeter changes by the same percent as length so 30% decrease.
For area: 10 × 10 = 100 7 × 7 = 49, so area decreases by 51%.

Questions Based on Venn Diagrams (Overlapping Sets):
On your test, you are likely to see questions with 2 or 3 variables.
In case of 2 variables, there are a maximum of four divisions possible:
Imagine that at a B-school, applicants can choose Marketing and Finance among other specializations,
where dual major is allowed. In this case, there can be only four types of sets of people possible:
1. Students taking Marketing Only

2. Students taking Finance Only
3. Students taking both Marketing and Finance
4. Students taking neither Marketing nor Finance
These types of questions are best solved by making a 2-way-matrix (table). Just remember that if one
row has “Marketing”, the other has to have “Not Marketing”. If one column has “Finance”, the other has
to have “Not Finance”.
The table can be completed by putting all the given values in the question and thus the unknown can be
found out by simple addition / subtraction of rows / columns. In case of percentage problems (where
all values are mentioned in terms of percentage), it is better to take the total as 100.
Let us solve a practical example:
50% of the apartments in a certain building have windows and hardwood floors. 25% of the
apartments without windows have hardwood floors. If 40% of the apartments do not have hardwood
floors, what percent of the apartments with windows have hardwood floors?
10% 16.66% 40% 50% 83.33%
This problem involves two sets:
Set 1: Apartments with windows / Apartments without windows
Set 2: Apartments with hardwood floors / Apartments without hardwood floors.
It is easiest to organize two-set problems by using a matrix as follows:
Windows NO TOTAL
Windows
Hardwood
Floors
NO Hardwood
Floors
TOTAL
The problem is difficult for two reasons. First, it uses percentages instead of real numbers. Second, it
involves complicated and subtle wording.
Let's attack the first difficulty by converting all of the percentages into REAL numbers. To do this, let's
say that there are 100 total apartments in the building. This is the first number we can put into our
matrix. The absolute total is placed in the lower right-hand corner of the matrix as follows:
Windows NO TOTAL
Windows
Hardwood
Floors
NO Hardwood
Floors
TOTAL 100
Next, we will attack the complex wording by reading each piece of information separately, and
filling in the matrix accordingly.
Information: 50% of the apartments in a certain building have windows and hardwood floors.
Thus, 50 of the 100 apartments have BOTH windows and hardwood floors. This number is now
added to the matrix:
Windows NO Windows TOTAL

Hardwood Floors 50
NO Hardwood
Floors
TOTAL 100
Information: 25% of the apartments without windows have hardwood floors. Here's where
the subtlety of the wording is very important. This does NOT say that 25% of ALL the apartments
have no windows and have hardwood floors. Instead it says that OF the apartments without
windows, 25% have hardwood floors. Since we do not yet know the number of apartments without
windows, let's call this number x. Thus, the number of apartments without windows and with
hardwood floors is .25x. These figures are now added to the matrix:

Hardwood Floors 50 .25x
NO Hardwood
Floors
TOTAL X 100
Information: 40% of the apartments do not have hardwood floors. Thus, 40 of the 100
apartments do not have hardwood floors. This number is put in the Total box at the end of the "No
Hardwood Floors" row of the matrix:

Hardwood Floors 50 .25x
NO Hardwood 40
Floors
TOTAL x
Before answering the question, we must complete the matrix. To do this, fill in the numbers that
yield the given totals. First, we see that there must be 60 total apartments with Hardwood Floors
(since 60 + 40 = 100) Using this information, we can solve for x by creating an equation for the first
row of the matrix:
50 + 0.25x = 60 or x = 40
Now we put these numbers in the matrix:

Hardwood Floors 50 10 60
NO Hardwood 40
Floors
TOTAL 40 100
Finally, we can fill in the rest of the matrix:

Hardwood Floors 50 10 60
NO Hardwood 10 30 40
Floors
TOTAL 60 40 100
We now return to the question: What percent of the apartments with windows have hardwood
floors?
Again, pay very careful attention to the subtle wording. The question does NOT ask for the
percentage of TOTAL apartments that have windows and hardwood floors. It asks what percent OF
the apartments with windows have hardwood floors. Since there are 60 apartments with windows,
and 50 of these have hardwood floors, the percentage is calculated as follows:
50/60 = 83.33%
Thus, the correct answer is E.
In case of 3 variables, there are a maximum of eight divisions possible. So,

a table will become very complicated. So, we will deal with such questions
by drawing 3 overlapping circles. Focus on the example below:
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum
of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing
club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for
the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
2 5 6 8 9
This is a three-set overlapping sets problem. When given three sets, a Venn diagram can be used. The
first step in constructing a Venn diagram is to identify the three sets given. In this case, we have students
signing up for the poetry club, the history club, and the writing club. The shell of the Venn diagram will
look like this:
When filling in the regions of a Venn diagram, it is important to work from inside out. If we let x
represent the number of students who sign up for all three clubs, a represent the number of students
who sign up for poetry and writing, b represent the number of students who sign up for poetry and
history, and c represent the number of students who sign up for history and writing, the Venn diagram
will look like this:
We are told that the total number of poetry club members is 22, the total number of history club
members is 27, and the total number of writing club members is 28. We can use this information to fill
in the rest of the diagram:
We can now derive an expression for the total number of students by adding up all the individual
segments of the diagram. The first bracketed item represents the students taking two or three courses.
The second bracketed item represents the number of students in only the poetry club, since it's derived
by adding in the total number of poetry students and subtracting out the poetry students in multiple
clubs. The third and fourth bracketed items represent the students in only the history or writing clubs
respectively.
59 = [a + b + c + x] + [22 – (a + b + x)] + [27 – (b + c + x)] + [28 – (a + c + x)]

59 = a + b + c + x + 22 – a – b – x + 27 – b – c – x + 28 – a – c – x
59 = 77 – 2x – a – b – c
59 = 77 – 2x – (a + b + c)
By examining the diagram, we can see that (a + b + c) represents the total number of students who sign
up for two clubs. We are told that 6 students sign up for exactly two clubs. Consequently:
59 = 77 – 2x – 6
2x = 12
x=6
So, the number of students who sign up for all three clubs is 6.
The correct answer is C.

RATIOS:
The comparison between two quantities of the same kind of unit is the Ratio of one quantity to another.
The ratio of a and b is usually written as a : b or a/b, where a is called the antecedent (numerator) and
b the consequent (denominator).
1. a : b = ka : kb where k is a constant
2. a : b = a/k : b/k
3. a : b : c = X : Y : Z is equivalent to a/X = b/Y = c/Z
4. If a/b > 1 or a > b then (a + X) / (b + X) < a/b a, b, X are natural numbers
5. If a/b < 1 or a a/b a, b, X are natural numbers
VARIATION:
Direct proportion: If two quantities X & Y are related such that any increase or decrease in ‘Y’ produces
a proportionate increase or decrease in ‘X’ or vice versa, then the two quantities are said to be in direct
proportion.
In other words, X : Y = X/Y = k (a constant)
or X = KY or Y = K’X (where K and K’ are constants)
X is directly proportional to Y is written as X ∝ Y or X = K Y
Inverse proportion: Here two quantities X & Y are related such that, any increase in X would lead to a
decrease in Y or any decrease in X would lead to an increase in Y. Thus, the quantities X & Y are said to
be inversely related and X is inversely proportional to Y is written as X ∝ 1/Y or X = k/Y or XY = k
(constant) or the product of two quantities remains constant.
Solved Examples:
1. In what ratio should tea @ 35/kg be mixed with tea @ 27/kg so that mixture may cost Rs. 30/kg?
Sol.
The average cost is 30
Let x kg of tea at Rs. 35/kg be mixed with y kg of tea at Rs. 27/kg
Total amount of tea = x + y kg
Average price = Rs. 30/kg
So, 35x + 27y = 30 (x + y)

Or 5x = 3y
Or x/y = 3/5
2. Find a : b : c if 6a = 9b = 10c.
Sol.
a/b = 9/6 = 3 : 2 = 15 : 10, b/c = 10/9 = 10 : 9. Hence a : b : c = 15 : 10 : 9.

3. A’s present age is to be B’s as 8 : 5; and 20 years ago, it was as 12 : 5. Find the present age of each.
Sol.
Let the ages be 8X and 5X
(8X – 20) / (5X – 20) = 12 / 5 Solving this, we get: X = 7
A’s age = 8 X = 56 years, B’s age = 5 X = 35 years
4. An alloy contains 24% of tin by weight. How much more tin to the nearest kg must be added to
100 kg of the alloy so that the % of tin may be doubled?
Sol.
Let X kg of tin be added to the alloy.

Tin (kg) Alloy (kg)
24 100
24 + X 100 + X
Now, Tin / Total must be 48%
So, (24 + X) / (100 + X) = 48/100
So, X = 46
Hence 46 kg of tin must be added to the alloy. Ans.
5. The expenses of a hotel consist of two parts. One part varies with the number of inmates, while
the other is constant. When the number of inmates is 200 & 250, the expenses are respectively
Rs. 1300 & Rs. 1600. Then find the expenses for 300 inmates.
Sol.
Let E = K1X + K2
E stands for expenses, X for the number of inmates.
When X = 200 & E = 1300 we have [200 K1 + K2 = 1300].
When X = 250 & E = 1600 we have [250 K1 + K2 = 1600]. Solving the equations, we have
K1 = 6 & K2 = 100
So, E = 6X + 100. Now when X = 300, E = 6 x 300 + 100 = Rs 1900 Ans.

6. Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
form a mixture containing 5 kg of wheat and 11 kg of rice?
Sol.
Let X kg of mixture be taken from A, then (16 – X) kg is taken from B
So, 2X/5 kg of wheat from A and 3(16 – X)/10 kg of wheat from B is to be taken.
Equate wheat from both sides:
We have, 2X/5 + 3(16 – X)/10 = 5 or X = 2 kg Or 2 kg from A and 14 kg from B.
7. Two vessels contain mixtures of spirit and water. In the first vessel the ratio of spirit to water is
8 : 3 and in the second vessel the ratio is 5 : 1. A 35–liter cask is filled from these vessels so as to
contain a mixture of spirit and water in the ratio of 4 : 1. How many liters are taken from the first
vessel?
Sol.
Let X liters be taken from the first vessel; then (35 – X) liters are taken from the second. In the
first vessel 8/11 of the mixture, and in the second vessel 5/6 of the mixture, is spirit
So, the spirit in the 35–liter cask is 4/5 of the mixture
Equate spirit on both sides:
So, 8/11 X + 5/6 (35 – X) = 4/5 ⨯ 35 So, X = 11
So, 11 liters are taken from the first vessel Ans.
8. A bag contains $ 600 in the form of one–dollar, 50 cents & 25–cents coins in the ratio 3 : 4 : 12.
Find the number of 25 cents coins.
Sol.
Ratio of values of coins = 3/1 : 4/2 : 12/4 = 3 : 2 : 3. Value of 25 cents coins
= Rs 600 ⨯ 3/ (3 + 2 + 3) = 225. No. of 25 cents coins = 225 x 4 = 900 Ans.
Alternate method:
Assume that the number of 1 $ coins is 3X. Then the value equation would be 3 X + 4 X (0.50) +
12 X (0.25) = 600. Find X and get answer = 12 X.
9. A mixture contains milk & water in the ratio 5 : 1. on adding 5 liters of water, the ratio of milk
and water becomes 5 : 2. Find the quantity of milk in the original mixture.
Sol.
Let the quantity of milk be 5X & that of water X. Then 5X / (X + 5) = 5/2 or X = 5. So, quantity of
milk = 5X = 25 liters
10. The ratio of the number of boys to the number of girls in a school of 546 is 4 : 3. If the number of
girls increases by 6, what must be the increase in the number of boys to make the new ratio of
boys to girls 3 : 2?
Sol.
Original no. of boys = 546 ⨯ 4/7 = 312. Original no. of girls = 78 ⨯ 3 = 234. Final no. of girls =
234 + 6 = 240 … so no. of boys reqd. to make the new ratio = 240 ⨯ 3/2 = 360 … so, the reqd.
increase in the no. of boys = 360 – 312 = 48 Ans.
11. Two numbers are in the ratio of 3 : 4. If 5 is subtracted from each, the resulting numbers are in
the ratio 2 : 3. Find the numbers.
Sol.
Let 3X and 4X be the numbers. So, (3X – 5) / (4X – 5) = 2/3 … so, 9X – 15 = 8X – 10 … or X = 5

So, the required numbers are 15 and 20 Ans.
Work / Rate:
FUNDAMENTAL CONCEPTS:
• If A alone takes X hours and B alone takes Y hours to do a piece of work, and if T is the total time
taken when they work together, then we have: 1/X + 1/Y = 1/T or T = XY/(X + Y)
So, If A and B can do a piece of work in X & Y days respectively while working alone, they will
together take XY / (X + Y) days to complete it.
It is best to take the LCM of times taken to avoid using fractions in such questions.
• If A is twice as good a workman as B, then A will take half the time B takes to finish a piece of work.
Solved examples:
1. If A and B together finish a piece of work in 10 days & B alone can finish it in 20 days. In how many
days can A alone finish the work?
2. Four men working together all day, can finish a piece of work in 11 days; but two of them having
other engagements can work only one half–time and quarter time respectively. How long will it take
them to complete the work?
3. 20 men can complete a piece of work in 10 days, but after every 4 days 5 men are called off, in what
time will the work be finished?
4. A vessel can be filled by one pipe A in 10 minutes, by a second pipe B in 15 minutes. It can be emptied
by a waste pipe C in 9 minutes. In what time will the vessel be filled if all the three were turned on
at once?
5. Three pipes A, B and C can fill a tank in 15, 20 and 30 min respectively. They were all turned on at
the same time. After 5 minutes the first two pipes were turned off. In what time will the tank be
filled?
6. A tank can be filled by two taps A and B in 12 minutes and 14 minutes respectively and can be
emptied by a third in 8 minutes. If all the taps are turned on at the same moment, what part of the
tank will remain unfilled at the end of 7 minutes?
Solutions:
1. Let X and Y be the number of days required by A and B respectively. By the standard formula,
1/X + 1/Y = 1/10
Or
XY / (X + Y) = 10
Y = 20 OR X ⨯ 20 / (X + 20) = 10 or X = 20 days Ans.
2. Each man will take 11 ⨯ 4 = 44 days to complete the work. If one man works half day/day he will
take 44 ⨯ 2 = 88 days to finish the work. Similarly, a man working quarter day/day will take 44 ⨯ 4
= 176 days to finish the work. When these people work together, they will require
We have:
1/44 + 1/44 + 1/88 + 1/176 = 1/T
Or T = 16 days.
3. Total work = 20 x 10 = 200 man-days

First 4 days' work = 20 x 4 = 80 man-days
Next 4 days' work = 15 x 4 = 60 man-days
Total 200 man-days … Hence, days required = 4 + 4 + 4 + 4 = 16
4. 1/10 + 1/15 – 1/9 = 1/T or T = 18 … so the vessel will be filled in 18 minutes Ans.
5. A, B and C can fill (1/15 + 1/20 + 1/30) or 3/20 of the tank in 1 minute
A, B and C filled (3/20 x 5) or 3/4 of the tank in 5 min.
Now A and B are turned off
(1 – 3/4) or 1/4 of the tank will be filled by C
So, C will fill 1/4 of the tank in (30 x 1/4) or 7.5 minutes
So, the tank will be filled in 7.5 + 5 or 12.5 min. Ans
6. We have (7/12) + (7/14) – 7/8 = 5/24 part filled in 7 minutes. Hence 1 – 5/24 = 19/24th of the tank
is unfilled.
Time / Speed / Distance
• If A goes from X to Y at U km/hour and comes back from Y to X at V km/hour, then Average speed
during the whole journey = 2UV / (U + V) km/hr.
• If a man changes his speed in the ratio m : n then the ratio of times taken becomes n : m.
• When two objects travel in the same direction, relative speed = difference of speeds and time to
catch / overtake = lead distance / difference of speeds
SO … Time to overtake (same directions) = Gap distance / difference of speeds
• When two objects travel in the Opposite directions, relative speed = sum of speeds and time to meet
= lead distance / sum of speeds.
SO … Time to meet (opposite directions: towards each other) = Gap distance / sum of speeds
• Average Speed = Total Distance / Total Time
• If the speed of a boat (or man) in still water be X km/hour, and the speed of the stream (or current)
be Y km/hour, then
(a) Speed of boat with the stream (or Downstream or D/S) = (X + Y) km/hour
(b) Speed of boat against the stream (or upstream or U/S) = (X – Y) km/hour
Solved Examples:
1. A policeman goes after a thief who is 176 m ahead of him. When and where will the policeman catch
the thief when they run at the rates of 11440 and 10560 meters per hour respectively?
2. If I walk at the rate of 4 kms an hour, I reach my destination 30 min too late; If I walk at the rate of
5 kms an hour I reach 30 minutes too soon. How far is my destination?
3. A man rows 18 kms down a river in 4 hours with the stream and returns while traveling upstream
in 12 hours; find his speed and also the velocity of the stream.
4. A, B and C can walk at the rates of 3, 4, 5 kms an hour. They start from X at 1, 2, 3 o’clock respectively;
when B catches up with A, B sends him back with a message to C; when will C get the message?
5. A student walks to school at the rate of 2.5 kms an hour and reaches 6 minutes too late. Next day he
increases his speed by 2 kms an hour and then reaches there 10 minutes too soon. Find the distance
of the school from his home.
6. A man can row in still water a distance of 4 kms in 20 minutes and 4 kms with the current in 16
min. How long will it take him to row the same distance against the current?
Solutions:
1. Time to catch / overtake = lead distance / difference of speeds = 176 / (11440 – 10560) = 176 / 880
= 1/5 hours = 12 minutes … so the time required to overtake the thief = 12 min. The distance from
the starting point = 11440 ⨯ 12/60 kms = 2288 meters
2. Let time taken be T hours for the distance to be covered at the normal speed (neither fast nor slow).
Then we have 4 (T + 0.5) = 5 (T – 0.5) {Note: 0.5 here is 30 min} … so T = 4.5 hours
Distance = 4 (T + 0.5) = 4 x 5 = 20 kms.
3. Speed with the stream = 18/4 = 4.5 kms an hour. Speed against the stream = 18/12 = 1.5 kms an
hour.
x + y = 4.5
x – y = 1.5
So, x = 3 and y = 1.5
4.
A starts at 1 o clock
And B starts at 2 o clock
If I consider time 0hr when A started,
after 1 hr, A is 3 km away and B has just started. After 2 hrs, A is 6km away and B is 4km away. After 3
hr, A is 9km and B is 8km
Next are, they both meet at 12km. You can also, get this by taking the LCM of 3 and 4 that is 12km
So, after 4hr, A gets the message. So at 5 o clock, he got it. Since A started at 1 o clock.
Now, C started at 3 o clock. That means it has walked for 2 hrs. So he walked 10km. Thus now, the
distance between A and C is
12–10=2km.
Now they move in opposite directions at speed 3 and 5km/hr
Distance / sum of speeds = time to meet
2/ (5 + 3) = 2/8 = 1/4 hours = 15 min
OR
In 15 mins, A walks 3/4km and C walks 5/4km
Total, 3/4+5/4=8/4=2km
Thus they meet after 15 mins
Thus C gets the letter at 5:15.

5. Let t be the usual time
We have
2.5 ⨯ (t + 1/10) = 4.5 (t – 1/6), or t = 1/2 hours.
Hence distance = 2.5 (1/2 + 1/10) = 2.5 x 6 /10 = 1.5 kms.
6. X = 4 / (20/60) = 12, X + Y = 4 ⨯ 60 /16 = 15 or Y = 3. or Time = 4 / (X – Y) = 4 x 60 /9 = 80/3 minutes

Solved examples on translating Word Problems into equations:
1. Find two consecutive odd numbers the difference of whose squares is 296.
2. A is 29 years older than B; B is 3 years older than C and D is 2 years younger than C. Two years hence
A’s age will be twice the united ages of B, C and D. Find their present ages.
3. A number consists of three consecutive digits, that in the unit’s place being the greatest of the three.
The number formed by reversing the digits exceeds the original number by 22 times the sum of the
digit. Find the number.
(A) 234 (B) 345 (C) 456 (D) 567 (E) 678
4. The crew of a boat can row at the rate of 5 miles an hour in still water. If to row 24 miles, they take
4 times as long upstream as to row the same distance down the river, find the speed at which the
river flows.
5. The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is
decreased by 3 meters. The area remains unaffected if the length is decreased by 7 m and breadth
is increased by 5 m. Find the dimensions of the rectangle.
6. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them
saves Rs. 200 per month, find their monthly incomes.
7. Find two consecutive even numbers such that 1/6th of the greater exceeds 1/10th of the smaller by
29.
8. A number consists of two digits whose sum is 12. The ten’s digit is three times the unit’s digit. What
is the number?
9. A train travelled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the
journey would have occupied 4 hours less; and had the speed been 6 miles an hour less, the journey
would have occupied 6 hours more. Find the distance.
10. A sum of money was divided equally among a certain number of persons; had there been six more,
each would have received a rupee less, and had there been four fewer, each would have received a
rupee more than he did; find the sum of money and the number of men.
Solutions:
1. Let the numbers be 2X + 1 and 2X + 3 Then (2X + 3)² – (2X + 1)² = 296 OR X = 36
Hence 2X + 1 = 2 x 36 + 1 = 73 and 2X + 3 = 75
The required numbers are 73 and 75.
[Verification. (75)² – (73)²= 5625 – 5329 = 296]
2. Let D’s age be = X years Then C’s age = (X + 2) years, B’s age = (X + 5) years and A’s age = (X + 34)
years. Two years hence A’s, B’s, C’s and D’s ages will be X + 36, X + 7, X + 4 and X + 2 years
respectively. So, we have: 2 (X + 2 + X + 4 + X + 7) = X + 36 OR X = 2 So, A’s age = 36 years; B’s age =
7 years, C’s age = 4 years; D’s age = 2 years.
3. Try option A
234 ... 2 + 3 + 4 = 9
432 – 234 = 198 = 22*9
All verified. So, A is correct
Detailed Solution:
Let the hundred’s digit be X. Then the ten’s digit = X + 1 and the unit’s digit = X + 2. The number =
100 x X + 10(X + 1) + X + 2 = 111X + 12. The number formed by reversing the digits = 100(X + 2) +
10(X + 1) + X = 111X + 210
We have 111X + 210 – 111X – 12 = 22 (X + 2 + X + 1 + X). So, we get, X = 2. Hence the required
number = 234.
4. Let X miles per hour be the speed of the river. Hence, on equating the times, we get:
Speed upstream = 5 – X
Speed downstream = 5 + X
Time taken upstream = Distance / speed = 24 / (5 – X)
Time taken downstream = Distance / speed = 24 / (5 + X)
24 / (5 – X) = 4 ⨯ 24/ (5 + X) OR X = 3 Thus, the river flows at the rate of 3 miles an hour.
5. (X + 7) (Y – 3) = XY or – 3X + 7Y – 21 = 0 .... (1)
(X – 7) (Y + 5) = XY or 5X – 7Y – 35 = 0 .... (2)
Solve the two simultaneous equations to get
Y = 15 and X = 28 Hence length = 28 m and breadth = 15 m Answer.
6. Let the monthly income of first person be Rs 9X and the monthly income of second person be Rs 7X.
Let the expenditure of first person be 4Y and the expenditure of second person be 3Y. Saving of the
first person = Rs (9X – 4Y) and solving of second person = Rs (7X – 3Y). Using the given information,
we have: 9X – 4Y = 200 .... (1) and 7X – 3Y = 200 .... (2) So, X = 200 … Hence, the monthly income of
first person = Rs. 9 ⨯ 200 = Rs. 1800 and the monthly income of second person = Rs. 7 ⨯ 200 = Rs.
1400 Ans.
7. Let the numbers be 2X and 2X + 2 Then (2X + 2)/6 – 2X/10 = 29. So, X = 21 Hence 2X = 430 and
2X + 2 = 432. The required numbers are 430 and 432. [Verification. 432/6 – 430/10 = 72 – 43 = 29]
8. Let the unit’s digit be X, Then the ten’s digit is 12 – X.
We have 3X = 12 – X or X = 3
Hence the number is 93. [Verification. 9 = 3 x 3; and 9 = 3 = 12]
9. Let us suppose that X miles per hour is the speed of the train and Y hours is the time taken for the
journey. Distance travelled = XY = (X + 6) (Y – 4) = (X – 6) (Y + 6)
Solve exactly like Q. 5 above

Generate two simultaneous equations. You will get: X = 30, Y = 24
Distance = XY = 720 miles Ans.
10. Let X be the number of persons and Rs Y be the share of each. Then by the conditions of the problem,
we have
(X + 6) (Y – 1) = XY .......(1)
(X – 4) (Y + 1) = XY .......(2).
Solve exactly like Q. 5 above

Generate two simultaneous equations. You will get: X = 24 and Y = 5
Thus, the number of people is X = 24 and the share of each is Y = Rs 5. The sum of money = 5 x Rs 24
= Rs 120
Questions based on compound interest / Rate of Increase

𝑹 𝒏
Compound Interest formula: 𝑨 = 𝑷 (𝟏 + 𝟏𝟎𝟎) CI = A – P.
𝑅 2𝑛
For half yearly calculation of the interest: 𝐴 = 𝑃 (1 + 200)
𝑅 4𝑛
For quarterly calculation of the interest: 𝐴 = 𝑃 (1 + 400)
In all these results: A = Final Amount, P = Principal (Initial Amount), R = Rate per annum, n = number
of years.
Population increase formula:

𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
𝐹𝑖𝑛𝑎𝑙 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 [𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛]𝑇𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑜𝑛𝑒 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛
Solved examples:
1. Donald plans to invest x dollars in a savings account that pays interest at an annual rate of 8%
compounded quarterly. Approximately what amount is the minimum that Donald will need to invest
to earn over $100 in interest within 6 months?
$1500 $1750 $2000 $2500 $3000
2. Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria
multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria
each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the
lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to
250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3
p.m.?
50,000 62,500 65,000 86,666 125,000
3. The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were
1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed
250,000 locusts?
6 8 10 12 14
4. A scientist is studying bacteria whose cell population doubles at constant intervals. In the last 2
hours, the population has quadrupled, increasing by 3,750 cells. How many cells will the population
contain four hours from now?
Solutions:
1. The formula for calculating compound interest is A = P(1 + r/n)nt where the variables represent the
following:
A = amount of money accumulated after t years (principal + interest)
P = principal investment
r = interest rate (annual)
n = number of times per year interest is compounded
t = number of years
In this case, x represents the unknown principal, r = 8%, n = 4 since the compounding is done quarterly,
and t = .5 since the time frame in question is half a year (6 months).
r = 8% = 8/100
So, we have:
P (1 + 8/400)2
From the options: 2500 (1 + 8/400)2 = 2601 … so the interest = 101, which is close to 100.
OR … by approximation:
8% interest over half a year, however that interest is compounded, is approximately 4% interest. So, to
compute the principal, it's actually a very simple calculation:
100 = 0.04x
2500 = x
The correct answer is D.

2. If we decide to find a constant multiple by the hour, then we can say that the population was
multiplied by a certain number three times from 1 p.m. to 4 p.m.: once from 1 to 2 p.m., again from
2 to 3 p.m., and finally from 3 to 4 p.m.
Let's call the constant multiple x.
2,000(x)(x)(x) = 250,000
2,000(x3) = 250,000
x3 = 250,000/2,000 = 125
x=5
Therefore, the population gets five times bigger each hour.
At 3 p.m., there were 2,000(5)(5) = 50,000 bacteria.
The correct answer is A.
3. A population problem is best solved with a population chart that illustrates the swarm population
at each unit of time. An example of a population chart is shown below:
Time Population
4 hours ago 1,000
2 hours ago 2,000
NOW 4,000
in 2 hours 8,000
in 4 hours 16,000
in 6 hours 32,000
in 8 hours 64,000
in 10 hours 128,000
in 12 hours 256,000
As can be seen from the chart, in 12 hours the swarm population will be equal to 256,000 locusts.
Thus, we can infer that the number of locusts will exceed 250,000 in slightly less than 12 hours.
Since we are asked for an approximate value, 12 hours provides a sufficiently close approximation
and is therefore the correct answer.
OR
28 =256 (more than 250 times) so 8 intervals are needed, means 16 hours. 4 hours are already
past, so 12 more hours are needed.
The correct answer is D
4. Let x equal the population immediately after that division
4x = x + 3,750 Or 3x = 3,750 Or x = 1,250 Ans. 80,000
(2 hours earlier) 1250, 2500, 5000 (now), 10000, 20000, 40000, 80000 (4 hours from now)
24. Of the 60 animals on a certain farm, 2/3 are either cows or pigs. How many of the animals are
cows?
(1) The farm has more than twice as many cows as pigs
(2) The farm has more than 12 pigs
25.
A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators
as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of
the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the
shipment, how many calculators were in the shipment?
A. 24
B. 25
C. 26
D. 28
E. 30
Concept # 5: Speed and Distance
Average Speed = Total Distance / Total Time
Time to overtake (same directions) = Gap distance / difference of speeds
Time to meet (opposite directions: towards each other) = Gap distance / sum of speeds
26. Lexy walks 5 miles from point A to point B in one hour, then bicycles back to point A along the
same route at 15 miles per hour. Ben makes the same round trip, but does so at half of Lexy’s
average speed. How many minutes does Ben spend on his round trip?
40 80 120 160 180
27. Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If
Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average
rate for the entire trip in miles per minute?
1/8 2/15 3/15 ¼ 3/8
28. What is the distance between Harry’s home and his office?
(1) Harry’s average speed on his commute to work this Monday was 30 miles per hour.
(2) If Harry’s average speed on his commute to work this Monday had been twice as fast, his trip
would have been 15 minutes shorter.
29. The ‘moving walkway’ is a 300-foot long conveyor belt that moves continuously at 3 feet per
second. When Bill steps on the walkway, a group of people that are also on the walkway stands
120 feet in front of him. He walks toward the group at a combined rate (including both walkway
and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary
until the walkway ends. What is Bill’s average rate of movement for his trip along the moving
walkway?
2 feet per second 2.5 feet per second 3 feet per second
4 feet per second 5 feet per second
36.
A certain truck traveling at 55 miles per hour gets 4.5 miles per gallon of diesel fuel consumed. Traveling
at 60 miles per hour, the truck gets only 3.5 miles per gallon. On a 500-mile trip, if the truck used a total
of 120 gallons of diesel fuel and traveled part of the trip at 55 miles per hour and the rest at 60 miles
per hour, how many miles did it travel at 55 miles per hour?
A. 140
B. 200
C. 250
D. 300
E. 360
37.
A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in
the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per
gallon did the car travel in the city?
(A) 14
(B) 16
(C) 21
(D) 22
(E) 27
38.
When a certain stretch of highway was rebuilt and straightened, the distance along the stretch was
decreased by 20 percent and the speed limit was increased by 25 percent. By what percent was the
driving time along this stretch reduced for a person who always drives at the speed limit?
16%
36%
37%
45%
56%
Concept # 6: Work:
If A alone takes X hours and B alone takes Y hours to do a piece of work, and if Z is the total time
taken when they work together, then we have: 1/X + 1/Y = 1/Z or Z = XY/(X + Y)
A’s contribution = Y/(X + Y) B’s contribution = X/(X + Y)
It is best to take the LCM of times taken to avoid using fractions in such questions.
39. Machine A and Machine B can produce 1 widget in 3 hours working together at their respective
constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in
2 hours working together at their respective rates. How many hours does it currently take
Machine A to produce 1 widget on its own?
½ 2 3 5 6
40. Tom, working alone, can paint a room in 6 hours. Peter and John, working independently, can
paint the same room in 3 hours and 2 hours, respectively. Tom starts painting the room and
works on his own for one hour. He is then joined by Peter and they work together for an hour.
Finally, John joins them and the three of them work together to finish the room, each one working
at his respective rate. What fraction of the whole job was done by Peter?
1/9 1/6 1/3 7/18 4/9
41. Machine A can fill an order of widgets in a hours. Machine B can fill the same order of widgets in
b hours. Machines A and B begin to fill an order of widgets at noon, working together at their
respective rates. If a and b are even integers, is Machine A's rate the same as that of Machine B?
(1) Machines A and B finish the order at exactly 4:48 p.m.
(2) (a + b)2 = 400
42. Six machines, each working at the same constant rate, together can complete a certain job in 12
days. How many additional machines, each working at the same constant rate, will be needed to
complete the job in 8 days?
A) 2 B) 3 C) 4 D) 6 E) 7
43. Pumps A, B and C operate at their respective constant rates. Pumps A and B, simultaneously, can
fill a certain tank in 6/5 hours. Pump A and C, operating simultaneously, can fill the tank in 3/2
hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many
hours does it take pumps A, B, and C, operating simultaneously, to fill the tank?
(A) 1/3 (B) 1/2 (C) 2/3 (D) 5/6 (E) 1
44. In the first 2 hours after Meadow's self-service laundry opens, m large washing machines and n
small washing machines are in continual use. Including the time for filling and emptying the
washing machines, each load of laundry takes 30 minutes in a large washing machine and 20
minutes in a small washing machine. What is the total number of loads of laundry done at
Meadow's self-service laundry during this 2-hour period?
1. n = 3m
2. 2m + 3n = 55
45.
𝑹 𝒏
Concept # 7: Compound Interest formula: 𝑨 = 𝑷 (𝟏 + 𝟏𝟎𝟎) CI = A – P.
𝑅 2𝑛
For half yearly calculation of the interest: 𝐴 = 𝑃 (1 + 200)
𝑅 4𝑛
For quarterly calculation of the interest: 𝐴 = 𝑃 (1 + 400)
In all these results: A = Final Amount, P = Principal (Initial Amount), R = Rate per annum, n = number
of years.
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
𝐹𝑖𝑛𝑎𝑙 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 [𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛]𝑇𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑜𝑛𝑒 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛
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Quant Session 3: Statistics + Numbers
Part 1: Statistics
Mean (Average)
1. Average or mean or AM = Sum of n quantities (or numbers) / number of them (n) OR Arithmetic
Mean (A.M.) is given by X = x / N .
2. Mean of the Combined Series If N1 and N2 are the sizes and M1 and M2 are the respective means of
M N + M2 N2
two series then the mean M of the combined series is given by M = 1 1 or we can write:
N1 + N 2
N 1 M 2 − M D2
= = ... This is the most important result in Mean.
N 2 M − M 1 D1
3. If a man (or train or boat or bus) covers some journey from A to B at X km/hr (or m/sec) and returns
to A at a uniform speed for Y km/hr, then the average speed during the whole journey is
[2XY / (X + Y)] km/hr. TIP: The average speed in such a case will be a bit less than the simple
average.
4. The sum of first “n” natural numbers is given by n (n + 1)/2.
5. For consecutive integers or for equally spaced numbers (AP), Mean = (First term + Last term) / 2.
6. If the average of a few consecutive integers is 0, then there will be an odd number of integers.
7. The average of an odd number of consecutive integers is an integer and the average of an even
number of consecutive integers is a non–integer.
8. If in a set of numbers, the average = the highest or the lowest number, all the numbers will have to
be equal.
N 1 M 2 − M D2
Problems: Use = = to solve the following 8 questions:
N 2 M − M1 D1
1. In a work force, the employees are either managers or directors. What is the percentage of directors?
(1) The average salary for managers is $5,000 less than the total average salary.
(2) The average salary for directors is $15,000 more than the total average salary.
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2. Committee X and Committee Y, which have no common members, will combine to form Committee
Z. Does Committee X have more members than Committee Y?
(1) The average (arithmetic mean) age of the members of Committee X is 25.7 years and the average
age of the members of Committee Y is 29.3 years.
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years
3. In a certain senior class, 72% of the male students and 80% of the female students have applied to
college. What fraction of the students in the senior class is male?
(1) There are 840 students in the senior class
(2) 75% of the students in the senior class have applied to college
4. At a certain company the average number years of experience are 9.8 years for males and 9.1 years
for females. What is the ratio of the number of the company's male employees to the number of the
company's female employees?
(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company's males and females combined is
9.3 years.
5. During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25. Did
the store sell more sweaters than shirts during the sale?
(1) The average (arithmetic mean) of the prices of all the shirts and sweaters that the store sold
during the sale was $21.
(2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420
6. A convenience store currently stocks 48 bottles of mineral water. The bottles have two sizes of
either 20 or 40 ounces each. The average volume per bottle the store currently has in stock is 35
ounces. How many 40–ounce bottles are in stock? ______________________
7. Every day, Walter burns 500 calories from cardio exercise. On some days, he also burns an
additional 600 calories from weight training. If, over a 240–day period, Walter burns an average of
850 calories per day from cardio exercise and weight training combined, then on how many days
did Walter engage in cardio exercise only?
40 60 80 100 140
8. A group of men and women gathered to compete in a marathon. The average weight of the entire
group was twice as far from the average weight of the women as it was from the average weight of
the men. What was the percentage of women in the group? _______________
Median:
• Median is the middle value or the average of two middle values when the values are arranged in
an order, either ascending or descending.
• If there are odd number of observations, median is directly the middle number.
• If there is an even number of observations, median is the average of the two middle numbers.
• For consecutive integers or for equally spaced numbers (AP),
Median = (First term + Last term) / 2. So, Median = Mean in this case.
• Median is the 50th percentile.
Median of a continuous series
Example:
Explanation:
6. If set Y consists of the consecutive integers p, q, r, s, and t such that p < q < r < s < t, then all of the
following are true EXCEPT
A. increasing only t will increase the range of set Y
B. increasing only t will increase the average (arithmetic mean) of set Y
C. decreasing only p will increase the range of set Y
D. decreasing only r will decrease the average (arithmetic mean) of set Y
E. increasing only t will increase the median of set Y
7. Set S contains exactly four distinct positive integers. When the range of S is added to the sum of all
the terms in S, the resulting sum is equal to the smallest term in S plus three times the largest term
in S. Then:
(A) Mean = Median
(B) Mean < Median
(C) Mean > Median
(D) No relationship between Mean and Median can be derived
8. Set A consists of all even integers between 2 and 100, inclusive. Set X is derived by reducing each
term in set A by 50, set Y is derived by multiplying each term in set A by 1.5, and set Z is derived by
dividing each term in set A by –4. Which of the following represents the ranking of the three sets in
descending order of standard deviation?
(A) X, Y, Z (B) X, Z, Y (C) Y, Z, X (D) Y, X, Z (E) Z, Y, X
9. If M is a negative integer and K is a positive integer, which of the following could be the standard
deviation of a set {–7, –5, –3, M, 0, 1, 3, K, 7}?
I. –1.5 II. –2 III. 0
(A) I only (B) II only (C) III only (D) I and III only (E) None
10. If y = ax + b, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?
11. Let Set T = {2, 4, 5, 7}. Which of the following values, if added to Set T, would most increase the
standard deviation of Set T? 1 3 6 8 14
12. What is the standard deviation of Q, a set of consecutive integers?

(1) Q has 21 members. (2) The median value of set Q is 20.
13. A certain list of 100 data has an average of 6 and a standard deviation of d, where d is positive. Which
of the following pairs of data, when added to the list, must result in a list of 102 data with standard
deviation less than d? A. –6 and 0 B. 0 and 0 C. 0 and 6 D. 0 and 12 E. 6 and 6
14. During an experiment some water was removed from each of the 6 tanks. If the standard deviation
of the volumes of the water at the beginning of the experiment was 10 gallons, what was the
standard deviation of the volumes of the water after the experiment?
(1) For each tank 30% of the volume of the water that was in the tank before the beginning of the
experiment was removed during the experiment
(2) The average (mean) volume of water in the tanks at the end of the experiment was 63 gallons
15. Set A consists of four distinct numbers; set B consists of five distinct numbers, including all four
numbers that are in set A. The average (arithmetic mean) of set A is equal to the average (arithmetic
mean) of set B. Which set has greater standard deviation?
A. A B. B C. Both equal D. Nothing can be said conclusively
Quant Topic: Numbers
For the purpose of the GMAT, all numbers are real.
REAL numbers are basically of two types:
1. Rational numbers: A rational number can always be represented by a fraction of the form p/q
where p and q are integers and q ≠ 0. Examples: finite decimal numbers, infinite repeating decimals,
whole numbers, integers, fractions i.e. 3/5, 16/9, 2, 0.666.... ∞ = 2/3 etc.
2. Irrational numbers: Any number which cannot be represented in the form p/q where p and q are
integers and q ≠ 0 is an irrational number. AN INFINITE NON–RECURRING DECIMAL IS AN
IRRATIONAL NUMBER. Examples – √2, , √5, √7.
INTEGERS: The set of Integers I = {0, ±1, ±2, ±3, … ∞}
EVEN NUMBERS: The numbers divisible by 2 are even numbers. E.g., 0, ±2, ±4, ±6, ±8, ±10....... Even
numbers are expressible in the form 2n where n is an integer. Thus –2, –6 etc. are also even
numbers. Remember that ‘0’ is an even number.
ODD NUMBERS: The numbers not divisible by 2 are odd numbers e.g. ±1, ±3, ±5, ±7, ±9.... Odd
numbers are expressible in the form (2n + 1) where n is an integer other than zero (not necessarily
prime). Thus, –1, –3, –9 etc. are all odd numbers.
You must remember:

Even ± Even = Even Even ± Odd = Odd Odd ± Odd = Even Odd ± Even = Odd
Even × Even = Even Even × Odd = Even Odd × Odd = Odd
POSITIVE INTEGERS: The numbers 1, 2, 3, 4, 5.... are known as positive integers.

• 0 is neither positive nor negative.
• 0 is an even number.
• 0 is not a factor of any integer.
• 0 is a multiple of all integers.
Prime numbers: A natural number which has no other factors besides itself and unity is a prime
number. Examples: 2, 3, 5, 7, 11, 13, 17, 19 ......
• If a number has no factor equal to or less than its square root, then the number is prime.
This is a test to judge whether a number is prime or not.
• The only even prime number is 2
• 1 is neither prime nor composite (by definition)
• The smallest composite number is 4.
Composite numbers: A composite number has other factors besides itself and unity, e.g., 8, 72, 39 etc.
Alternatively, we might say that a natural number greater than 1 that is not prime is a composite
number.
Problems:
1. If a and b are both positive integers, is ba+1 – bab odd?

(1) a + (a + 4) + (a – 8) + (a + 6) + (a – 10) is odd
(2) b3 + 3b2 + 5b + 7 is odd
2. If n is an integer between 10 and 99 is n < 80?

(1) The sum of the two digits of n is a prime number.
(2) Each of the two digits of n is a prime number.
3. For all positive integers m, [m] = 3m when m is odd and [m] = ½ m when m is even. Which of
the following is equivalent to [9]  [6]?
[81] [54] [36] [27] [18]
FACTORS / HCF (GCD / GCF) & LCM OF NUMBERS
Prime factors:
A composite number can be uniquely expressed as a product of prime factors.

E×. 12 = 2 × 6 = 2 × 2 × 3 = 22 × 31 20 = 4 × 5 = 2 × 2 × 5 = 22 × 51
124 = 2 × 62 = 2 × 2 × 31 = 22 × 31 etc.
If k and n are both integers greater than 1 and if k is a factor of n, k cannot be a factor of (n + 1).
NOTE:
The number of divisors (factors) of a given number N (including one and the number itself) where
N = am × bn × cp .... where a, b, c are prime numbers is given by (m + 1) (n + 1) (p + 1) ......
e.g. (1) 90 = 2 × 3 × 3 × 5 = 21 × 32 × 51
Hence here a = 2 b = 3 c = 5, m = 1 n = 2 p = 1
Number of divisors = (m + 1) (n + 1) (p + 1) .... = 2 × 3 × 2 = 12
Number of factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 = 12
HCF: It is the greatest factor common to two or more given numbers. It is also called GCF OR GCD
(greatest common factor or greatest common divisor); e.g. HCF of 10 & 15 = 5, HCF of 55 & 200 = 5, HCF
of 64 & 36 = 4
To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the
common term(s) from them and multiply them. This is the required HCF.
LCM: Lowest common multiple of two or more numbers is the smallest number which is exactly
divisible by all of them.
E.g. LCM of 5, 7, 10 = 70, LCM of 2, 4, 5 = 20, LCM of 11, 10, 3 = 330
To find the LCM resolve all the numbers into their prime factors and then pick all the quantities (prime
factors) but not more than once and multiply them. This is the LCM.
NOTE:
1. LCM × HCF = Product of two numbers (valid only for “two”)
2. HCF of fractions = HCF of numerators ÷ LCM of denominators
3. LCM of fractions = LCM of numerators ÷ HCF of denominators
Q. Find the LCM of 25 and 35 if their HCF is 5. LCM = 25 x 35/5 = 175
Calculating LCM: After expressing the numbers in terms of prime factors, the LCM is the product of
highest powers of all factors.
Q. Find the LCM of 40, 120, and 380.

40 = 4 × 10 = 2 × 2 × 2 × 5 = 23 × 51,
120 = 4 × 30 = 2 × 2 × 2 × 5 × 3 = 23 × 51 × 31
380 = 2 × 190 = 2 × 2 × 95 = 2 × 2 × 5 × 19 = 22 × 51 × 191
Required LCM = 23 × 51 × 31 × 191 = 2280.
Calculating HCF: After expressing the numbers in term of the prime factors, the HCF is product of
COMMON factors.
Ex. Find HCF of 88, 24, and 124

88 = 2 × 44 = 2 × 2 × 22 = 2 × 2 × 2 × 11 = 23 × 111
24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 = 23 × 31
124 = 2 × 62 = 2 × 21 × 311 = 22 × 311 HCF = 22
Problems:
1. If x is a prime number, what is the value of x?

(1) 2x + 2 is the cube of a positive integer.
(2) The average of any x consecutive integers is an integer.
2. If the integer n is greater than 1, is n equal to 2?

(1) n has exactly two positive factors
(2) The difference between any two distinct positive factors is odd.
3. The function f is defined for all positive integers n by the following rule: f(n) is the number of
positive integers each of which is less than n and also has no positive factor in common with n
other than 1. If p is a prime number then f(p) =?
p–1 p–2 (p + 1) / 2 (p – 1) / 2 2
4. For every positive even integer n, the function h(n) is defined to be the product of all the even
integers from 2 to n, inclusive. If p is the smallest prime factor of h (100) + 1, then p is
A. between 2 and 10 B. between 10 and 20 C. between 20 and 30
D. between 30 and 40 E. greater than 40
5. Is the integer n odd?

(1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n
6. How many different prime numbers are factors of the positive integer n?
(1) four different prime numbers are factors of 2n
(2) four different prime numbers are factors of n2.
7. Does the integer k have a factor p such that 1 4! (2) 13! + 2 ≤ k ≤ 13! + 13.
8. The positive integer k has exactly two positive prime factors, 3 and 7. If K has a total of 6
positive factors, including 1 and k, what is the value of K?
(1) 32 is a factor of k (2) 72 is NOT a factor of k
Divisibility / Remainders
TESTS FOR DIVISIBILITY:
1. A number is divisible by 2 if its unit’s digit is even or zero e.g. 128, 146, 34 etc.
2. A number is divisible by 3 if the sum of its digits is divisible by 3 e.g. 102, 192, 99 etc.
3. A number is divisible by 4 when the number formed by last two right hand digits is divisible by ‘4’
e.g. 576, 328, 144 etc.
4. A number is divisible by 5 when its unit’s digit is either five or zero: e.g. 1111535, 3970, 145 etc.
5. A number is divisible by 6 when it’s divisible by 2 and 3 both. e.g. 714, 509796, 1728 etc.
6. A number is divisible by 8 when the number formed by the last three right hand digits is divisible
by ‘8’. e.g. 512, 4096, 1304 etc.
7. A number is divisible by 9 when the sum of its digits is divisible by 9 e.g. 1287, 11583, 2304 etc.
8. A number is divisible by 10 when its unit’s digit is zero. e.g. 100, 170, 10590 etc.
9. A number is divisible by 11 when the difference between the sums of digits in the odd and even
places is either zero or a multiple of 11. e.g. 17259, 62468252, 12221 etc. For the number 17259:
Sum of digits in even places = 7 + 5 = 12, Sum of digits in the odd places = 1 + 2 + 9 = 12 Hence 12 –
12 = 0.
10. A number is divisible by 12 when it is divisible by 3 & 4 both. e.g. 672, 8064 etc.
11. A number is divisible by 25 when the number formed by the last two Right hand digits is divisible
by 25, e.g., 1025, 3475, 55550 etc.
Problems:
1. If t is a positive integer and r is the remainder when t2 + 5t + 6 is divided by 7, what is the value
of r?
(1) when t is divided by 7, the remainder is 6 (2) when t2 is divided by 7, the remainder is
1
2. If p is a positive odd integer, what is the remainder when p is divided by 4?

(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
3. If p, x, and y are positive integers, y is odd, and p = x2 + y2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5. (2) x – y = 3.
4. If n is a positive integer and r is the remainder when (n – 1) (n + 1) is divided by 24, what is the
value of r?
(1) n is not divisible by 2 (2) n is not divisible by 3
5. If N is a positive integer, is (N3 – N) divisible by 4?

(1) n = 2k + 1, where K is an integer. (2) n2 + n is divisible by 6
Power of a Prime Number in a Factorial: If we have to find the power of a prime number p in n!, it is
n  n   n  n
found using a general rule, which is   +  2  +  3  + ........ , where   denotes the greatest integer
 p  p   p   p
n
≤ to   etc.
 p
 100   100   100   100   100 
For example, power of 3 in 100! =   +  2  +  3  +  4  +  5  + .... = 33 + 11 + 3 + 1 + 0 = 48.
 3  3  3  3  3 
 200   200   200 
For example, power of 5 in 200! =   +  2  +  3  + ... = 40 + 8 + 1 + 0 = 49.
 5  5  5 
Number of Zeroes at the end of a Factorial: It is given by the power of 5 in the number.
Actually, the number of zeroes will be decided by the power of 10, but 10 is not a prime number, we
have 10 = 5 × 2, and hence we check power of 5.
For example, the number of zeroes at the end of 100! = = 20 + 4 = 24.
The number of zeroes at the end of 500! = = 100 + 20 + 4 = 124.
The number of zeroes at the end of 1000! = = 200 + 40 +8 + 1 = 249.
Unit’s digits in powers: Every digit has a cyclicity of 4. The fifth power of any single digit number has
the same right-hand digit as the number itself.
Example: What will be the unit’s digit in 12896?
In all such questions, divide the power by 4 and check the remainder.
If the remainder is 1, 2 or 3, then convert the question to LAST DIGIT RAISED TO REMAINDER.
If the remainder is 0, convert the question to LAST DIGIT RAISED TO FOUR.
In this question, 96/4 = 0, so the question converts to 84 = 82 x 82 = 64 x 64 = 4 x 4 = 16 = 6
Problems:
1. If d is a positive integer, f is the product of the first 30 positive integers, what is the value of d?
(1) 10d is a factor of f (2) d > 6
2. If n and m are positive integers, what is the remainder when 34n+2 + m is divided by 10?
(1) n = 2 (2) m = 1
DECIMALS and FRACTIONS
Recurring Decimals (Conversion to a Rational Number): If in a decimal fraction a figure or a set of

figures is repeated continually, then such a number is called a recurring decimal.
(i) 2/3 = 0.6666.... (ii) 22/7 = 3.142857142857 …
Rule: Write the recurring figures only one in the numerator and take as many nines in the denominator
as the number of repeating figures.
Ex. (1) 0.666666666 … = 6/9 = 2/3 (2) 0.234234234234 … = 234/999
Rounding Off
Number Nearest tenth Nearest hundredth Nearest thousandth
1.2346 1.2 1.23 1.235
31.6479 31.6 31.65 31.648
9.7462 9.7 9.75 9.746
Whether a fraction will result in a terminating decimal or not? To determine this, express the
fraction in the lowest form and then express the denominator in terms of Prime Factors. If the
denominator contains powers of only 2 and 5, it is terminating. If the denominator contains any power
of any other prime number, it is non-terminating.
Problems
1. If k is a positive integer and the ten’s digit of k + 5 is 4, what is the ten’s digit of k?
(1) k > 35 (2) The units digit of k is greater than 5.
2. Is the hundredth digit of decimal d greater than 5?

(1) The tenth digit of 10d is 7 (2) The thousandth digit of d/10 is 7
3. What is the tens digit of the positive integer r?

(1) The tens digit of r/10 is 3. (2) The hundreds digit of 10r is 6.
4. What is the result when x is rounded to the nearest hundredth?

(1) When x is rounded to the nearest thousandth the result is 0.455
(2) The thousandth digit is 5
Factor Theorem: If f(x) is completely divisible by (x – a), then f(a) = 0. So, (x – a) is a factor of f(x), then
f(a) = 0
Check whether (x + 1) is a factor of f(x) = 4x2 + 3x – 1. Putting x + 1 = 0, i.e., x = –1 in

the given expression we get f(–1) = 0. So, (x + 1) is a factor of f(x).
Remainder Theorem: If an expression f(x) is divided by (x – a), then the remainder is f(a).
Let f(x) = x3 + 3x2 – 5x + 4 be divided by (x – 1). Find the remainder.

Remainder = f(1) = 13 + 3 × 12 – 5 × 1 + 4 = 3.
Some properties of square numbers:
• A square number always has odd number of factors.
• A square number cannot end with 2, 3, 7, 8 or an odd number of zeroes.
• Every square number is a multiple of 3, or exceeds a multiple of 3 by unity.
• Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.
• If a square number ends in 9, the preceding digit is even.
Questions based on Number Line
1. If m and r are two numbers on a number line, what is the value of r?

(1) The distance between r and 0 is 3 times the distance between m and 0
(2) 12 is halfway between m and r
2. If n denotes a number to the left of 0 on the number line such that the square of n is less than
1/100, then the reciprocal of n must be
A. less than –10 B. between –1 and –1/10 C. between –1/10 and 0
D. between 0 and –1/10 E. greater than 10
3. On the number line, the segment from 0 to 1 has been divided into fifths, as indicated by the large
tick marks, and also into sevenths, as indicated by the small tick marks. What is the least possible
distance between any two of the tick marks? ______________________
4. If s and t are two different numbers on the number line, is s + t = 0?

(1) The distance between s and 0 is the same as the distance between t and 0
(2) 0 is between s and t
5.
(1)
N sh M 2 − M 25 − 21 2
= = =
N sw M − M1 21 − 15 3
So, sweaters > shirts.
(2) 15x + 25y = 420 or 3x + 5y = 84. (x, y) = (23, 3) and (3, 15) both satisfy so x > y or x < y. Not certain.
Not sufficient.
Ans. A
6.
M1 = 20, M2 = 40, M = 35, N1 + N2 = 48
N1 / N2 = (M2 – M) / (M – M1)
So N1/N2 = (40 – 35) / (35 – 20) = 5 / 15 = 1/3
48 divided in the ratio 1: 3 is 12 and 36.
Answer: 36
7.
M1 = 500, M2 = 1100, M = 850
N1/N2 = (1100 – 850) / (850 – 500) = 250/350 = 5/7
N1 + N2 = 240
240 in the ratio 5:7 is 100 and 140
8.
W _______________________________________ overall _________________________________ M

M – M1 = D1 M2 – M = D2
You are given: D1 = 2D2
Nw / NM = D2/D1 = 1/2
So % of women = 33.33%
Median:
1.
Let T, J, and S be the purchase prices for Tom’s, Jane’s, and Sue’s new houses. Given that the average
purchase price is 120,000, or T + J + S = (3)(120,000), determine the median purchase price.
(1) Given T = 110,000, the median could be 120,000 (if J = 120,000 and S = 130,000) or 125,000 (if J =
125,000 and S = 125,000); NOT sufficient.
(2) Given J = 120,000, the following two cases include every possibility consistent with T + J + S =
(3)(120,000), or T + S = (2)(120,000). (i) T = S = 120,000 (ii) One of T or S is less than 120,000 and the
other is greater than 120,000. In each case, the median is clearly 120,000; SUFFICIENT. The correct
answer is B; statement 2 alone is sufficient.
2
2.
In order to solve the question easier, we simplify the numbers such as 150, 000 to 15, 130,000 to 13,
and so on.
I. Median is 13, so, the greatest possible value of sum of eight prices that no more than median is
13*8=104. Therefore, the least value of sum of other seven homes that greater than median is (15*15–
104)/7=17.3>16.5. It's true.
II. According the analysis above, the price could be, 13, 13, 13, 13, 13,13,13,13, 17.3, 17.3, 17.3... So, II
is false.
III. Also, false.
Answer: only I must be true.
3.
To find the mean of the set {6, 7, 1, 5, x, y}, use the average formula: A = S/n where A = the
average, S = the sum of the terms, and n = the number of terms in the set. Using the
information given in statement (1) that x + y = 7, we can find the mean as 4.33: Regardless of
the values of x and y, the mean of the set is 4.33 because the sum of x and y does not change. To
find the median, list the possible values for x and y such that x + y = 7. For each case, we can
calculate the median.
x y DATA SET MEDIAN
1 6 1, 1, 5, 6, 6, 7 5.5
2 5 1, 2, 5, 5, 6, 7 5
3 4 1, 3, 4, 5, 6, 7 4.5
4 3 1, 3, 4, 5, 6, 7 4.5
5 2 1, 2, 5, 5, 6, 7 5
6 1 1, 1, 5, 6, 6, 7 5.5
Regardless of the values of x and y, the median (4.5, 5, or 5.5) is always greater than the mean (4.33).
Therefore, statement (1) alone is sufficient to answer the question. Now consider statement (2).
Because the sum of x and y is not fixed, the mean of the set will vary. Additionally, since there are
many possible values for x and y, there are numerous possible medians. The following table
illustrates that we can construct a data set for which x – y = 3 and the mean is greater than the
median. The table ALSO shows that we can construct a data set for which x – y = 3 and the median is
greater than the mean.
x Y DATA SET MEDIAN MEAN

22 19 1, 5, 6, 7, 19, 22 6.5 10
4 1 1, 1, 4, 5, 6, 7 4.5 4
Thus, statement (2) alone is not sufficient to determine whether the mean is greater than the
median. The correct answer is (A): Statement (1) alone is sufficient, but statement (2) alone is not
sufficient.
3
4.
Let 'X' stand for the sum of each of the sets.
(1) is clearly insufficient, as we know nothing whatsoever about set t.

still, take the time to interpret it: it says that the middle number of set s is 0, which also means that
the sum of the elements in set s is 0 (by the fact above).
(2)
INCORRECT LOGIC: "the sum of these two sets (sets of consecutive integers) will be equal only when
the sum is zero.
Ex: Set S could be 5,6,7,8 & 9 while Set T could be 2,3,4,5,6,7 & 8 and these sets have equal sums.
There are umpteen other examples.
Using the fact above, we have that the average (whether mean or median – they're the same) of the
numbers in set s is X/5, and the average (again, mean or median) of the numbers in set t is X/7.
It's tempting to say 'sufficient' here, because at first glance X/5 and X/7 appear to be necessarily
different, but they aren't: in the singular case X = 0, the two will be identical. Therefore, insufficient.
Together
this tells us that X = 0, which means that the median of both sets is 0/7 = 0/5 = 0. Sufficient.
5.
This is an AP… common difference either positive or negative. There are 15 terms, so the 8th term will
be the median. 7 terms will be less than the median and 7 terms will be more than the median. If
median is 10, then we know that 7 terms are more than 10 and 7 terms are less than 10. Ans. B
a(8) is the middle term and hence the Median of this series. The number k can be positive or negative.
If k is negative, then the numbers a(1), a(2) etc. will be in descending order and if it is positive, then
the series will be in ascending order. No matter what, there will be 7 terms which will be greater than
10 (Please note that since k is non–zero, and hence all the terms in the series will be DISTINCT). It
could a(1) to a(7) if k is negative, and a(9) to a(15) if k is positive.
6.
Statement (1) tells us that 25 percent of the projects had 4 or more employees assigned. There is no
information given about the middle values of the number of employees per project. Hence statement
(1) is insufficient.
Statement (2) tells us that 35 percent of the projects have 2 or fewer employees but there is no
information about the middle values of the number of employees per project. Hence statement (2) is
insufficient.
Combining statement (1) and (2), we can gather than 100 – (25+35) = 40 percent of the projects have
exactly 3 employees. Therefore, when listing the number of employees per project in ascending order,
35 percent of the numbers are less than 2 and 36th to the 75th projects (overall 40 projects in the
middle) will have 3 employees each …
So (35 + 40) = 75 percent are 3 or less. Since the median lies in that middle 40 percent, the median is
3. Hence the correct answer is (C).
4
7.
The mean must be 50, and at least two terms in the original set must be 50. If the original set is 50, 50,
50, 50, 50, 50, the mean and median were 50 originally. The new term may be any value and the
median will not change.
If the original set is 0, 0, 0, 50, 50, 200, the mean was 50 and the median was 25 originally. If the new
term is greater than 25, the median increases. If the new term is less than 25, the median decreases.
Ans. D
8.
A ≤ B ≤ C ≤ D ≤ E. The problem indicates that: A + B + C + D + E = 250 (The set of five numbers has an
average of 50.) E = 5 + 3A (The largest element is five greater than three times the smallest element in
the set.) C = 50 (The median of the set equals the mean.) You're asked to maximize E. We can
maximize E by minimizing D. Therefore, make D = C = 50. Maximize E by minimizing B, so make B = A.
A + B + C + D + E = 250
A + (A) + 50 + 50 + (5 + 3A) = 250
105 + 5A = 250
5A = 145
A = 29
E = 5 + 3A = 5 + 3(29) = 5 + 87 = 92. The correct answer is (D).
9.
If the average of the cost of the three models is $900, then the sum is $2,700. Call the three models a, b,
and c, in order from least expensive to most.
c = 1.25b, so a + b + 1.25b = 2,700.
a + b + 1.25b = $2,700
a + 2.25b = $2,700
By definition, a < b, so plug LTb (less than b) into the equation for a: LTb + 2.25b = 2,700
LTb = 2,700/3.25 so, LTb = 828.
So, b > 828
So, 1.25 b > 1000
Ans. B
10.
Those last two terms have to make up for the –8 on the other side, so each x must be 4 over the
average, or 22.
Ans. D
5
11.
A group of 10 students could have held these scores: 46, 46, 46 | 50, 92, 92, 92, 92, 92, 92
Check: The 7 students who passed have an average score of [50 + 6(92)]/7 = 602/7 = 86.
Check: The average score for the entire group is [3(46) + 602]/10 = 74.
In this case, Median score for the entire group = 92 (average of the 5th and 6th students’ scores).
Range + SD
1.
Before analyzing the statements, let’s consider different scenarios for the range and the median of set
A. Since we have an even number of integers in the set, the median of the set will be equal to the
average of the two middle numbers. Further, note that integer 2 is the only even prime and it cannot
be one of the two middle numbers, since it is the smallest of all primes. Therefore, both of the middle
primes will be odd, their sum will be even, and their average (i.e. the median of the set) will be an
integer. However, while we know that the median will be an integer, it is unknown whether this
integer will be even or odd. For example, the average of 7 and 17 is 12 (even), while the average of 5
and 17 is 11 (odd). Next, let’s consider the possible scenarios with the range. Remember that the
range is the difference between the greatest and the smallest number in the set. Since we are dealing
with prime numbers, the greatest prime in the set will always be odd, while the smallest one can be
either odd or even (i.e. 2). If the smallest prime in the set is 2, then the range will be odd, otherwise,
the range will be even. Now, let’s consider these scenarios in light of each of the statements.
(1) SUFFICIENT: If the smallest prime in the set is 5, the range of the set, i.e. the difference between
two odd primes in this case, will be even. Since the median of the set will always be an integer, the
product of the median and the range will always be even.
(2) INSUFFICIENT: If the largest integer in the set is 101, the range of the set can be odd or even (for
example, 101 – 3 = 98 or 101 – 2 = 99). The median of the set can also be odd or even, as we discussed.
Therefore, the product of the median and the range can be either odd or even. The correct answer is A.
2.
In a set consisting of an odd number of terms, the median is the number in the middle when the terms
are arranged in ascending order. In a set consisting of an even number of terms, the median is the
average of the two middle numbers. If S has an odd number of terms, we know that the median must
be the middle number, and thus the median must be even (because it is a set of even integers). If S has
an even number of terms, we know that the median must be the average of the two middle numbers,
which are both even, and the average of two consecutive even integers must be odd, and so therefore
the median must be odd. The question can be rephrased: “Are there an even number of terms in the
set?”
(1) SUFFICIENT:
Consecutive even means AP … so Mean = Median.
So, the median must be even.
(2) INSUFFICIENT:
For 2, 4, 6, 8, median is odd.
For 2, 4, 6, 8, 10, 12, 14, the median is even. The correct answer is A.
6
3.
You must read the word ‘different’

Prior to median 25, there are 7 numbers.
To make the greatest number as greater as possible, these 7 numbers should cost the range as little as
possible. They will be, 24, 23, 22, 21, 20, 19, 18.
So, the greatest value that can fulfill the range is: 18+25=43
4.
The lowest number in Set S is −81. Use this result to jump to the consecutive integers portion of the
solution. In this case, −81 is the smallest or “First” element in the set:
Last – First + 1 = Count
Last –(–81) + 1 = 100
Last + 82 = 100
Last = 18
Therefore, the highest or “Last” number in Set S is 18.
Finally, calculate the average of the positive terms in the set using 1 as the smallest positive integer
and 18 as the largest positive integer: (1 + 18)/2 = 9.5 … The average of the positive numbers in the
set is 9.5.
5.
To deal with sets, the first step is to get the elements in ascending order:
X = {9, 10, 14, 14, 17, 20, 22}
If we increase both the highest and lowest selling prices by 2 dollars, we have the following new set
(we can call this set Y):
Y = {11, 10, 14, 14, 17, 20, 24}
Notice that adding 2 to the 9 has changed the order of our set, so we will need to re–order Set Y:
Y = {10, 11, 14, 14, 17, 20, 24}
Now we can compare the statistics for the two sets:
Range (the difference of the extreme values of the set):
Set X = 22 – 9 = 13
Set Y = 24 – 10 = 14
Median
Set X = 14
Set Y = 14
The mean would obviously increase.
The correct answers are I and III.
6.
The problem states, "...all of the following are true EXCEPT..." Therefore, we are looking for the
statement that is NOT TRUE.
(A) TRUE: The range is defined as the difference between the largest value and the smallest value in a
set, or in this case, t – p. Increasing the value of t will increase the value of t – p. Therefore, the range
of set Y will increase.
(B) TRUE: The mean, or average, is defined as the sum of the members in a set divided by the number
of members in the set, or in this case:
Mean = ( p + q + r + s + t)/5
Increasing t will increase the sum in the top of the fraction. Therefore, the mean of set Y will increase.
(C) TRUE: The range is defined as the difference between the largest value and the smallest value in
the set, or in this case, t – p. Decreasing the value of p will increase the value of t – p. Therefore, the
range of set Y will increase.
(D) TRUE: The mean, or average, is defined as the sum of the members in a set divided by the number
of members in the set, or in this case:
7
Mean = ( p + q + r + s + t)/5
Decreasing r will decrease the sum in the top of the fraction. Therefore, the mean of set Y will
decrease.
(E) NOT TRUE: The median is defined as the middle number in a set that is ordered from least to
greatest, or in this case, r. After increasing the value of t, r will still be the middle number in the set,
and the value of r will be unchanged. Therefore, the value of the median will be unchanged.
The correct answer is E.
7.
a < b < c < d. The mean of S is (a + b + c + d) / 4. The median of S is the average of the two middle
terms, (b + c) / 2. Express the question algebraically:
If mean = median, we have (a + b + c + d) / 4 = (b + c) / 2

Or
a+d=b+c
If the sum of the range of S and all the terms in S is equal to the smallest term in S plus three times the
largest term in S, then: (d − a) + (a + b + c + d) = a + 3d
b + c + 2d = a + 3d
b+c=a+d
Ans. A
8.
Let SD of A be .
SD of X will remain the same .
SD of Y will be multiplied by 1.5 so 1.5
SD of Z will be first divided by –ve sign … no change … then divided by 4 … so /4. As SD is always +ve,
1.5 >  > /4 … Ans. D
9.
SD can’t be negative. SD is zero only when all numbers are same (which is impossible in this case).
Ans. E
10.
To get the value of y, first we multiply x (so that it becomes ax) by a so SD also gets multiplied by a
(becomes a S).
Next, we add b to ax, so the SD doesn’t change.
Final answer |a| S … we put the mod sign so that the SD value is not negative.
11.
In this case, 14 creates the largest spread from the mean, and will therefore be the value that most
increases the standard deviation of Set T. The correct answer is E.
8
12.
Since, in any consecutive set with an odd number of terms, the middle value is the mean of the set, we
can represent the set as 10 terms on either side of the middle term x: [x – 10, x – 9, x – 8, x – 7, x – 6, x –
5, x – 4, x – 3, x – 2, x – 1, x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6, x + 7, x + 8, x + 9, x + 10] .
Standard Deviation will be the same for any 21 consecutive integers.

You don’t have to do this step but here is HOW (just for your understanding):
102 + 92 + 82 + 72 + 62 + 52 + 42 + 32 + 22 + 12 + 02 + (–1)2 + (–2)2(–3)2 + (–4)2 + (–5)2 + (–6)2(–

7)2 + (–8)2 + (–9)2 + (–10)2 = 770.
(2) NOT SUFFICIENT: Since the set is consecutive, we know that the median is equal to the mean.
Thus, we know that the mean is 20. However, we do not know how big the set is so we cannot identify
the difference between each term and the mean. Therefore, the correct answer is A.
13.
The denominator in SD formula (N) will become 102, so we have to find numbers that will change the
SD by the least amount: such numbers should be as close to the mean as possible. So, the answer is E.
14.
* if you ADD OR SUBTRACT A CONSTANT to/from all the values in a set, then the standard
deviation will remain exactly the same.
* if you INCREASE OR DECREASE ALL THE VALUES BY A FIXED FACTOR / PERCENTAGE, then the
standard deviation will increase or decrease by the same percentage.
Make sure you know that, when ALL numbers in a set are multiplied or divided by some number, the
mean and standard deviation are multiplied /divided by the same number.
This includes increasing or decreasing all the numbers in the set by some percentage (which
can be accomplished by multiplication: e.g., 30% increase = multiplication by 1.3).
Using this principle, statement (1) tells us that both the mean and the standard deviation of the set
will decrease by 30%. Therefore, the new standard deviation will decrease to 7 gallons. SUFFICIENT.
Statement (2) tells us nothing about standard deviation, which measures SPREAD of numbers. If we
achieved the 63 gallons by taking most of the water out of the tanks that were already lowest, then the
standard deviation will be huge (because you'll have some tanks almost full and some almost empty).
If we got there by taking most of the water out of the fullest tanks, then the standard deviation will be
a lot smaller. INSUFFICIENT.
9
15.
If including a fifth number does not change the mean of the set, then that fifth number must equal the
original mean of set A. Additional terms equal to the mean value will always reduce the standard
deviation of a set, provided the original standard deviation is not 0 (a fact that is guaranteed here, as
the sets contain numbers that are distinct from one another). Therefore, the standard deviation of the
new set B must be less than that of the original set A. The correct answer is A.
Numbers
Problems:
1.
We can first simplify the exponential expression in the question:

ba+1 – bab
b(ba) - b(ab)
b(ba - ab)
So we can rewrite this question then as is b(ba - ab) odd? Notice that if either b or ba - ab is even, the
answer to this question will be no.
(1) SUFFICIENT: If we simplify this expression we get 5a - 8, which we are told is odd. For the
difference of two numbers to be odd, one must be odd and one must be even. Therefore 5a must be
odd, which means that a itself must be odd. To determine whether or not this is enough to dictate the
even/oddness of the expression b(ba - ab), we must consider two scenarios, one with an odd b and one
with an even b:
a b b(ba - ab) odd/even
3 1 1(1 - 3 ) = -2
3 1 even
3 2 2(2 - 3 ) = -2
3 2 even
It turns out that for both scenarios, the expression b(ba - ab) is even.
(2) SUFFICIENT: It is probably easiest to test numbers in this expression to determine whether it
implies that b is odd or even.
b b3 + 3b2 + 5b + 7 odd/even
2 2 + 3(2 ) + 5(2)+ 7 = 37 odd
3 2
1 13 + 3(12) + 5(1) + 7 = 16 even

We can see from the two values that we plugged that only even values for b will produce odd values
for the expression b3 + 3b2 + 5b + 7, therefore b must be even. Knowing that b is even tells us that the
product in the question, b(ba - ab), is even so we have a definitive answer to the question.
The correct answer is D, EACH statement ALONE is sufficient to answer the question.
2. (1) Let's take, say, a sum of 13, which is prime, and somewhat big.
This could be 9+4 ––> 94 (more than 80), or 4+9 ––> 49 (less than 80). Not sufficient.
(2) The maximum can be 77 … sufficient.

Ans. B
3. The answer is (D) [27]
since 9 is odd, [9] =(3)(9) = 27

and 6 is even, [6] = (1/2)(6) = 3
27 x 3 = 81 So, the correct answer is (D) [27] = 3 x 27 = 81
10
Problems:
1.
(1) INSUFFICIENT: Start by listing the cubes of some positive integers: 1, 8, 27, 64, 125. If we set each
of these equal to 2x + 2, we see that we can find more than one value for x which is prime. For example
x = 3 yields 2x + 2 = 8 and x = 31 yields 2x + 2 = 64. With at least two possible values for x, the
statement is insufficient.
(2) INSUFFICIENT:
x must be odd. This is not sufficient to give us a specific value for the prime number x.
(1) AND (2) INSUFFICIENT: The two x values that we came up with for statement (1) also satisfy the
conditions of statement (2).
The correct answer is E.
2. Answer: B
(1)
this is a disguised way of saying 'n is prime' … n could be 2 or 3 or 5 or 7 … therefore,
insufficient. We can’t say for sure that n = 2.
(2) The most important word here is ANY.
Imagine n = 3 … factors are 1, 3. The difference will be even

Imagine n = 5 … factors are 1, 5. The difference will be even
Imagine n = 9 … factors are 1, 3, 9. The difference of 1 and 3, 1 and 9, and 3 and 9, will all be
even Notice the word ANY).
So for all odd numbers, (2) can’t be satisfied … as each number will have 1 and itself as a factor
… so the difference between the number (odd) and 1 will always be even.
Imagine n = 4 … factors = 1, 2, 4 … difference of 2 and 4 = even … OUT

Imagine n = 12 … factors = 1, 2, 3, 4, 6, 12 … differences (2, 4) (2, 6), (2, 12), (4, 6), (4, 12), (6,
12) etc… are all even … OUT.
So all even numbers > 2 cannot satisfy (2) … each number will have 2 and itself (even) as a
factor and the difference of 2 and an even number will always be even.
If n = 2, the factors are 1 and 2 and the difference is odd.
By the above analysis, 2 is the only such number. Ans. B
3. Pick a prime number for p. Let's say p=5.
The positive integers less than 5 are 4, 3, 2, and 1.
5 and 4 share only 1 as a factor

There are four positive integers, therefore, that are both less than 5 and share only 1 as a
factor.
Ans. p – 1.
11
4. The basic logic behind this question: If k and n are both integers greater than 1 and if k
is a factor of n, k cannot be a factor of (n + 1).
Let's first consider the prime factors of h (100). According to the given function,
h (100) = 2*4*6*8*...*100
By factoring a 2 from each term of our function, h(100) can be rewritten as
h (100) = 250 (1*2*3*...*50).
2, 3, 4 … 50 are factors of h (100) … so 2, 3, 4 … 50 can’t be factors of h (100) + 1. So the

smallest factor will be more than 50 (which is always more than 40). Ans. E
5. (1) n could be 6 or 9. NS
(2) 2n has twice as many factors as n
n = 1 … one factor; 2n = 2 … two factors
n = 2 … two factors; 2n = 4 … three factors
n = 3 … two factors; 2n = 6 … four factors
n = 4 … three factors; 2n = 8 … four factors
n = 9 … three factors; 2n = 18 … six factors
… we can see that for each of the odd numbers (n), the number of factors of 2n is exactly double
but for an even number, it is never the case. So n must be odd … Sufficient.
Ans. B
Let's say a number has "n" different factors.

When you multiply this number by 2, you POTENTIALLY create "n" MORE factors - by doubling
each factor.
HOWEVER,
the only way that ALL of these factors can be NEW (i.e., not already listed in the
original n factors) is if they are ALL ODD.
If there are ANY even factors to start with, then those factors will be repeated in the original
list. Therefore, if the number is even, then the number of factors will be less than
doubled because of the repeat factors.
Thus if statement (2) is true, then the number must be odd.

Ans. B
12
6. The minimum value of a product of four different prime integers = 2 * 3 * 5 * 7 = 210
(1)
if 2n = 210 (4 prime factors), n = 105 (3 prime factors)

if 2n = 420 (4 prime factors), n = 210 (4 prime factors) … so we are not sure … n may have 3 or
4 prime factors. NS
(2)
If n is a positive integer, n and n2 will have an exact number of factors.
Imagine … n2 has four factors 2, 3, 5, 7 … then the minimum value of n2 will be 22 * 32 * 52 * 72 …

when we take the square root, we must have even powers of each of the factors.
So n will have the same number of prime factors as n2.
Ans. B
7. The question is whether k is non-prime?
HOW?
the question is asking whether k has a factor that is greater than 1, but less than itself.
if you're good at these number property rephrasing, then you can realize that this question is
equivalent to "is k non–prime?"
(1)
k > 24
if k = 25, it is not prime
if k = 29, it is prime
NS
(2)
The possible values of k are each of the positive integers from 13! + 2 to 13! + 13.
key realization:
every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.
13! + 2 can be written as 2k + 2 = multiple of 2 … not prime

.
.
.
So each of the values of k is a non prime number. Sufficient.

Ans. B
8. We know that the number of divisors (factors) of a given number N (including one and the
number itself) where N = am x bn x cp ..... where a, b, c are prime numbers is given by (1 + m)
(1 + n) (1 + p) ......
Here k = 3m * 7n so the number of factors = (1 + m) (1 + n) = 6

6 can be written as 1 × 6 or 2 × 3 or 3 × 2 or 6 × 1.
If 1 + m = 1, m = 0 … means we will not have any power of 3 … so this is unacceptable.
13
If 1 + n = 1, n = 0 … means we will not have any power of 7 … so this is unacceptable.
If 1 + m = 2 and 1 + n = 3, m = 1, n = 2 and the number will become 31 × 72

If 1 + m = 3 and 1 + n = 2, m = 2, n = 1 and the number will become 32 × 71
So the only possible values of k can be 31 × 72 or 32 × 71
(1) says 32 is a factor of k … so k = 32 × 71

(2) says 72 is a not factor of k … so k = 32 × 71
Each of the statements gives the value of k … Ans. D
Problems:
1. (1) t = 7k + 6
(t2 + 5t + 6) / 7 = [(7k + 6)2 + 5 (7k + 6) + 6] / 7 = (49k2 + 84k + 35k + 72) / 7 … the first three
terms will give 0 remainders as all of them are multiples of 7 … so the remainder = the
remainder when 72 is divided by 7 = 2. Sufficient
(2) t = 1 and t = 6 satisfy the conditions.

If t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.
If t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
Insufficient …
Ans. A
2. (1) p = 8k + 5 … 8k + 5 divided by 4 will give a remainder = 1 … 8k/4 will give 0 remainder and
5/4 will give a remainder of 1. Sufficient.
(2) p = x2 + y2 … p is odd … this means that one of the numbers is even and the other number is
odd.
Let x be even = 2a and y be odd = 2b + 1
x2 + y2 = 4a2 + 4b2 + 4b + 1 … which when divided by 4 will give a remainder of 1. Sufficient.
Ans. D
3. (1) p = 8k + 5 … x2 = p – y2 = p – (2b + 1)2 = 8k + 5 – (4b2 + 4b + 1) = 8k – 4b2 – 4b + 4 =

So x2 = 4 (2k – b2 – b + 1) = 4 [2k – (b2 + b) + 1] = 4 [even – even + 1] = 4 * Odd
b2 + b is always even whether b is even or odd. So x2 = 4 (odd number).
If x is a multiple of 4, x2 has to be a multiple of 16, which has to be an even multiple of 4.
So x cannot be a multiple of 4 … Sufficient
(2)
y = 1 , x = 4 … div by 4 … y = 3 , x = 6 … Not Div by 4. NS
Ans. A
14
4. (1)
if n = 3, then (n – 1)(n + 1) = 8, so the remainder is 8
insufficient
(2)
insufficient
(together)
the best approach, unless you're really good at number properties, is to try the first few
numbers that satisfy both statements, and watch what happens.
...you can see where this is headed.
here's the theory:

– if n is not divisible by 2, then n is odd, so both (n – 1) and (n + 1) are even. Moreover, since
every other even number is a multiple of 4, one of those two factors is a multiple of 4. So the
product (n – 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2
x 2 x 2 = three 2's in its prime factorization.
– if n is not divisible by 3, then exactly one of (n – 1) and (n + 1) is divisible by 3, because every
third integer is divisible by 3. Therefore, the product (n – 1)(n + 1) contains a 3 in its prime
factorization.
– thus, the overall prime factorization of (n – 1)(n + 1) contains three 2's and a 3.
– therefore, it is a multiple of 24.
– sufficient
answer = c
5. N3 – N = (N – 1) * N * (N + 1).
(1) N = Odd … so (N – 1) and (N + 1) both will be even and their product will be divisible by 4.
Sufficient
(2) n (n + 1) is divisible by 6 … N = 2 and N = 3 can be taken as 2 trial values.
If N = 2, N3 – N = 6 … not divisible by 4.
If N = 3, N3 – N = 24 … divisible by 4.
Not sufficient.
Ans. A
15
Problems:
1. f = 30! …
(1) Number of Zeroes at the end of a Factorial = Maximum Power of 10 in the factorial: It
is given by the maximum power of 5 in the number.
Maximum power of 5 in 30! = (30/5) + (30/52) + (30/53) + …

=6+1+0…=7
But 101 to 107 all the powers can be factors of 30!
So d can be any number from 1 to 7.
NS
(2)
NS
Combining: d = 7
Ans. C
Detailed background theory behind the above logic.

So how many 10's are in f?
write down the numbers that contain 2s and 5s (only those)
30*28*26*25*24*22*20*18*16*15*14*12*10*8*6*5*4*2
Now ask yourself Is my limiting factor going to be 5 or is it going to be 2?
It's going to be 5 because there are many more 2's up there. So circle the numbers that contain
5's:
30, 25, 20, 15, 10, 5
How many 5's do you have? Seven 5's (don't forget – 25 has two 5's!), so you can make seven
10's.
"limiting factor" means "which is least common or likely." Think of it this way: there are many
more multiples of 2 than there are multiples of 5. In probability terms, a number is more likely
to be even than to be a multiple of 5. In divisibility terms, take some large number that is
divisible by both 2 and 5, and it is likely to have more factors of 2 than 5.
For example: 400 = 4*10*10 = (2*2)(2*5)(2*5) = (2^4)(5^2).
I know, numbers with more factors of 5 than factors of 2 exist...this is just a bet we make to
ease the computation.
In general, the larger the factor, the less likely it is to divide evenly into a number. The larger
the factor, the more of a "limiting factor" it is.
here's all you have to do:
forget entirely about 10, 20, and 30, and ONLY THINK ABOUT PRIME FACTORIZATIONS.
(TAKEAWAY: this is the way to go in general – when you break something down into primes,
you should not think in hybrid terms like this. instead, just translate everything into the
language of primes.)
each PAIR OF A '5' AND A '2' in the prime factorization translates into a '10'.
there are seven 5's: one each from 5, 10, 15, 20, and 30, and two from 25.
there are waaaaaaayyyyy more than seven 2's.
therefore, 30! can accommodate as many as seven 10's before you run out of fives.
16
2. Unit’s digit of a number is the same as the remainder when the number is divided by 10.
REMAINDERS UPON DIVISION BY 10 are simply UNIT’S DIGITS.
For instance, when 352 is divided by 10, the remainder is 2.

So we need to know the unit’s digit of 34n + 2 + m.
Unit’s digit of 34n+2 can be divided by the rule of cyclicity.
In all such questions, divide the power by 4 and check the remainder.
If the remainder is 1, 2 or 3, then convert the question to LAST DIGIT RAISED TO REMAINDER.
If the remainder is 0, convert the question to LAST DIGIT RAISED TO FOUR.
So (4n + 2) / 4 gives a remainder of 2.

So the power will be reduced to 2.
So now the question becomes “What’s the unit’s digit of 32 + m?” or “What’s the unit’s digit of 9 +
m?” … the answer will depend only on the value of m … so Ans. B
Questions based on Number Line
1. If m and r are two numbers on a number line, what is the value of r?

(1) The distance between r and 0 is 3 times the distance between m and 0
(2) 12 is halfway between m and r
(1) |r| = 3|m| so r = 3m or r = -3m

NS
(2) (m + r) / 2 = 12 or m + r = 24
NS
Combining:
m + 3m = 24 or m = 6 so r = 3m = 18
OR
m – 3m = 24 or m = -12 so r = -3m = 36. Ans. E
2. If n denotes a number to the left of 0 on the number line such that the square of n is less than
1/100, then the reciprocal of n must be
A. less than –10 B. between –1 and –1/10 C. between –1/10 and 0
D. between 0 and –1/10 E. greater than 10
Given that n < 0 … A

Also, n2 < 1/100 … i.e. |n| < 1/10 … i.e. –1/10 < n < 1/10 … using A from above,
–1/10 < n < 0 … taking reciprocal, (applies since both are on same side of 0) …
–10 > 1/n or 1/n < –10. Ans. A
3. 1/5, 2/5, 3/5, 4/5=>7/35, 14/35, 21/35, 28/35

1/7, 2/7, 3/7, 4/7, 5/7, 6/7=>5/35, 10/35, 15/ 35, 20/35, 25/35, 30/35
It is easy to find that the least distance between any two of the marks is 1/35
17
4. If s and t are two different numbers on the number line, is s + t = 0?
(1) The distance between s and 0 is the same as the distance between t and 0
(2) 0 is between s and t
(1) |s| = |t|  s = t or s = -t.

But s ant t are different so s = t is not possible.
So s = -t or s + t = 0 … Sufficient.
(2) 0 is between s and t. But that means s can equal –5 and t can equal +3. In such a case, 0 is
still between s and t but that does not make them equidistant from 0. Or, s and t can be –4
and +4 respectively in which case they are equidistant from 0. Therefore, this statement
doesn't necessarily answer the question because it can have different results.
Ans. A
Miscellaneous questions
1. The correct answer is B

Your first task is to decode the given fact that the tens digit of (k+5) is 4. This means
40 < k+5 < 49, which means that 35 < k < 44.
(1)
All this tells you is that k isn't 35. That still leaves everything from 36 to 44, so the tens digit
could still be either 3 or 4: insufficient.
(2)
Of the aforementioned possibilities, only 35, 36, 37, 38, 39 fit this bill. therefore the tens digit
must be 3. Sufficient.
2. Let us say the number is 0.ABCD (the decimal d)
(1) d = 0.ABCD, hence 10d = A.BCD 10th digit of 10d = B = 7. which is 100th digit of d (=0.A7CD
and thus bigger than 5) Sufficient
(2) d = 0.ABCD, hence d/10 = 0.0ABCD 1000th digit of 10d = B = 7. which is 100th digit of d
(=0.A7CD and thus bigger than 5) Sufficient
Ans. D
3. (1)
Suppose r/10 = abc3d.e
So r = abc3de and ten’s digit = d, which is not known to us.
(2)
Let the number be
10r = abcd6ef
So r = abcd6e.f
so ten’s digit has to be 6 in original number.
Ans. B
18

Quant Session: Inequalities + Mods (Absolute Values)
Inequalities Basics:
1. a b
Examples:
• 3>2
• 3>0
• 3 > –3
• 0 > –3
4. a ≥ b
Examples:
• 3≥2
• 3≥3
• 3≥0
• –3 ≥ –3
• 0≥–3
5. So long as multiplication or division aren’t involved, we can cancel or shift quantities just as we
do in equations
• For example, x + y – 1 > x – y + 1 means either we can transfer all terms from the RHS to
the LHS and write x + y – 1 – x + y – 1 > 0 or 2y – 2 > 0 or 2y > 2 or y > 1 OR we can directly
cancel x from both sides and write y – 1 > –y + 1 or 2y > 2 or y >1.
6. If xy > 0, then x and y are of the same sign. Either both positive or both negative.
7. If x/y > 0, then x and y are of the same sign. Either both positive or both negative
• So, xy > 0 means x/y > 0
8. If xy < 0, then x and y are of the opposite sign. One positive and the other negative
9. If x/y < 0, then x and y are of the opposite sign. One positive and the other negative
• So, xy < 0 means x/y < 0
10. If a > b, then ax > bx, if x is positive … this means the sign of the inequality doesn’t change if we
multiply both sides by a positive quantity
• Similarly, if ax > bx, then a > b, if x is positive … this means the sign of the inequality
doesn’t change if we cancel a positive quantity from both sides
Note: the same rule applies for division: a > b, then a/x > b/x if x is positive and vice versa
11. If a > b, then ax < bx, if x is negative … this means the sign of the inequality changes if we multiply
both sides by a negative quantity
• Similarly, if ax > bx, then a < b, if x is negative … this means the sign of the inequality
changes if we cancel a negative quantity from both sides
12. If a/b > c/d, then we can’t just cross multiply to write ad > bc. Unless we know the sign of the
quantities, we can’t cross multiply.
• On the other hand, if all a, b, c, and d are positive, then we can surely cross multiply and
write ad > bc
13. The concept of number line is very useful in checking inequalities. The common values to check
are x = 0, 1, –1, >1 (preferred value = 2), between 0 and 1 (preferred values = 1/2 and 0.9),
between – 1 and 0 (preferred values = –1/2 and –0.9), and less than –1 (preferred value = –2).
So, in short, there are 9 points: –2, –1, –0.9, –1/2, 0, 1/2, 0.9, 1, 2.
14. If (x – a) (x – b) < 0, then x lies between a and b. OR a < x < b.
• Here a is less than b.
• (x – 3) (x – 5) < 0, then x lies between 3 and 5
• (x + 3) (x – 5) < 0 we can write this as [x – (–3)] (x – 5) < 0
So, x lies between –3 and 5
• (x + 5) (x + 3) < 0 we can write this as [x – (–5)] [x – (–3)] < 0
So, x lies between –5 and –3
15. If (x – a) (x – b) > 0, then x lies outside a and b. OR x < a, x > b
• Here a is less than b.
• (x – 3) (x – 5) > 0, then x doesn’t lie between 3 and 5. So either x is less than 3 or x is

greater than 5.
• (x + 3) (x – 5) > 0 We can write this as [x – (–3)] (x – 5) > 0
So, x doesn’t lie between –3 and 5. Either x is less than –3 or x is greater than 5.
• (x + 5) (x + 3) > 0 We can write this as [x – (–5)] [x – (–3)] > 0
So, x doesn’t lie between –5 and –3. Either x is less than –5 or x is greater than –3.
16. If x2 > x, then either x > 1 or x is negative (x < 0).
17. If x2 < x, then x lies between 0 and 1. (0 < x < 1).
18. If x2 = x, then x = 0 or x = 1.
19. If x3 > x, then either x > 1 or x is between –1 and 0 (either x > 1 or –1 < x < 0).
20. If x3 < x, then either x lies between 0 and 1 or x is less than –1.
(Either 0 < x < 1 or x < –1)
21. If x3 = x, then x = 0 or x = 1 or x = –1.
22.
• If 1/x > 0, then x > 0 Substitute x as –ve / 0 / +ve to verify.
• If 1/x < –x, then x must be negative Substitute x as –ve / 0 / +ve to verify.
23. If x2 > y2, then x > y and x < y both results are possible, and x and y can be of the same sign and
also of the opposite sign
• 52 > 32 and 5 > 3 (–5)2 > (–3)2 but –5 < –3 both same sign
• 52 > (–3)2 and 5 > –3 (–5)2 > (3)2 but –5 < 3 opposite signs
24. If x > y2, then x > y and x < y both results are possible
• 5 > 32 and 5 > 3 1/3 > (1/2)2 but 1/3 < ½
25. If x > y4, then x > y and x < y both results are possible
• 100 > 34 and 100 > 3 1/3 > (1/2)4 but 1/3 < ½
26. If x > y, it is necessarily true that x3 > y3 or etc. So, odd powers and roots don’t change sign.
27. Two inequalities with the same sign can be added just in the same way as two equations can be
added
So, if
a+b>c+d and
e+f>g+h
Then a + b + e + f > c + d + g + h
28. Two inequalities with different signs can be added after we change the sign of one of the
inequalities by multiplying it by a negative sign.
So, if
a+b>c+d and
e+f<g+h Check the less than sign
Then we can write
a+b>c+d and
–(e + f) > –(g + h)
Or
a+b>c+d and
–e – f > –g – h
So, a + b – e – f > c + d – g – h
29. If X is positive, then

(1) (a + X) / (b + X) > a/b if a b
30. On the GMAT, any square number is always greater than or equal to 0. So x2 ≥ 0. On the GMAT, a
square number can’t be negative.
21. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x4 – x3 (2) –12y2 – y2x + x2y2 > 0
22. If r + s > 2t, is r > t? (1) t > s (2) r > s
23. If p < q and p < r, is (p)(q)(r) < p? (1) pq < 0 (2) pr < 0
24. Is 5n < 0.04? (1) (1/5)n > 25 (2) n3 < n2
25. Is p2q > pq2? (1) pq < 0 (2) p < 0
26. Is m > n ? (1) n – m + 2 > 0 (2) n – m – 2 > 0
27. Is 3p > 2q ? (1) q = 2p (2) q > 0
28. Is mp greater than m? (1) m > p > 0 (2) p is less than 1
29. Is 2X – 3Y < X2? (1) 2X – 3Y = –2 (2) X > 2 and Y > 0

𝑥+1 𝑥
30. Is 𝑦+1 > 𝑦 ?
(1) 0 < x < y
(2) xy > 0
Absolute Values (Mods) – Concepts
1. |𝑥| is defined as the non–negative value of x and hence is never negative.
• So |x| ≥ 0, always (by definition) AND |x| < 0 is impossible (by definition)
2. |5| = 5, |–5| = 5
• |x| = x, if x is positive … If x = 5, then |5| = 5
• |x| = –x, if x is negative … If x is –5, then |–5| = –(–5) = 5. Here x is negative and –x is positive.
o So, when |x| = –x, x is a negative number and –x is a positive number
• |x| = –x (means x is negative)
o This still means that |x| is positive because in this case –x is a positive number
3. |𝑥| is defined as the distance of point x from 0 on the number line. The point x can be anywhere on
the line (positive or negative)
4. |𝑥 − 𝑎| is defined as the distance of point x from a on the number line. The point x and a can be
anywhere on the line (positive or negative).
5. We define √𝑥 2 = |𝑥| as both √𝑥 2 and |𝑥| can’t be negative.
• √𝑥 2 = |𝑥| … squaring both sides, we get 𝑥 2 = |𝑥| × |𝑥|
Q. If a2 < a, then is |a| > a? Yes / No? Ans. NO
If a2 < a, then 0 < a < 1, or a is positive. When a is positive, |a| = a
6. As square roots can’t be negative, then on the GMAT (by definition)
• √36 = 6 and not − 6
• BUT if 𝑥 2 = 36, we have √𝑥 2 = √36 OR |x| = 6, which gives x = 6 or –6.
Remember, we didn’t take √36 to be both 6 or –6. √36 is 6 only.
But because √𝑥 2 is |x|, we wrote |x| = 6, which gave us x = 6 or –6
This is the most misunderstood concept on the GMAT.
• So √𝑥 2 = 𝑥 or − 𝑥 both are possible.
• If x is positive, then √𝑥 2 = 𝑥
o Here x is positive and hence the square root is positive

• If x is negative, then √𝑥 2 = −𝑥
o Here x is negative, so –x is POSITIVE and hence the square root is positive
• The confusion arises because we assume x is positive and –x is negative. BUT x doesn’t have
a sign of its own, unless given. Please don’t assume anything.
Let’s see one real–GMAT question to understand the concept further:
4
If z is negative, then √(4𝑧 − 5)4 + √(2𝑧 − 3)2 + √−𝑧|𝑧| = ?
A. 5z – 8
B. 7z – 8
C. –8
D. 8 – 7z
E. 4z – 8
Sol. We can write this as

|4z – 5| + |2z – 3| + |z|
|4z – 5| = 4z – 5 OR –(4z – 5) = 5 – 4z, whichever is positive
Because z is negative, 4z – 5 will be negative and 5 – 4z will be positive
|2z – 3| = 2z – 3 OR –(2z – 3) = 3 – 2z, whichever is positive
Because z is negative, 2z – 3 will be negative and 3 – 2z will be positive
|z| = z or –z, whichever is positive
Because z is negative, –z will be positive
So, the answer will be: 5 – 4z + 3 – 2z + (–z) which gives 8 – 7z. Ans. D
7. |𝑥| = 𝑥 ⇒ 𝑥 ≥ 0 (substitute x as – / 0 / + and verify)
8. |𝑥| = −𝑥 ⇒ 𝑥 < 0 (substitute x as – / 0 / + and verify)
9. |𝑥| > 𝑥 ⇒ 𝑥 < 0 (substitute x as – / 0 / + and verify)
10. −𝑥|𝑥| > 𝑥 ⇒ 𝑥 < 0 (substitute x as – / 0 / + and verify)
11. −𝑥|𝑥| > 0 ⇒ 𝑥 < 0 (substitute x as – / 0 / + and verify)
12. |𝑥 − 𝑎| > 0 ⇒ 𝑥 ≠ 𝑎 Imagine |x – 3| > 0 … this expression is true for all values of x except x = 3.
Try to substitute x = –10, –5, 0, 1, 2, 4, 10, 100 … all of these will satisfy |x – 3| > 0. So, |x – 3| > 0
means 𝑥 ≠ 3
𝑥
13. |𝑥|
= 1 if x is positive Substitute any positive value of x and verify
𝑥
14. |𝑥|
= −1 if x is negative. Substitute any negative value of x and verify
Q. If x = y / |y|, what is |x|? Sol. x = 1 or –1, so |x| = 1

15. |𝑎| = |𝑏| ⇒ 𝑎 = 𝑏 𝑂𝑅 𝑎 = −𝑏 When we remove the mods, we substitute ±
So, we will have ±a = ±b which gives +a = +b, –a = –b, +a = –b, and –a = +b. So we get a = b or a = –b
16. If |x| = a, then x = a or x = –a.
• If |x| < a, then x < a or x > –a so –a < x < a.
• If |x – a| < b, then –b < x – a a, then x > a or x < –a.
• If |x – a| > b, then x – a > b or x – a < –b.
Q. If |7 – 3j| ≤ 8, what is the range for j?
–8 ≤ 7 – 3j ≤ 8
Subtract 7
–15 ≤ – 3j ≤ 1
Divide by 3
–5 ≤ –j ≤ 1/3
Multiply by a negative sign
–1/3 ≤ j ≤ 5 Ans.
Q. If |x|/|3| > 1, which of the following must be true?

A. x > 3
B. x < 3
C. x = 3
D. x ≠ 3
E. x < –3
Cross multiply: |x| > 3, which means either x > 3 or x < –3. In either case, x ≠ 3. Ans. D
Quant Session 3: Geometry
Concept # 1: 30-60-90 Right angled triangle:
Side opposite the angle 30° is the smallest (say a), so the side opposite the angle 60° will be a√3 and the
side the angle 90° will be 2a. The ratio is 1 : √3 : 2.
Example:
1. In the given figure, points P and Q lie on the circle with center O. What is the value of s?
A. ½
B. 1
C. √2
D. √3
E. 1/√2
Train with Sandeep Gupta (Official GMAT Score 800, Harvard Final Admit) … Contact: +91-97395-61394.
Explanation:
Drop two perpendiculars PA and QB from P and Q onto X axis. Now for the triangle on the left,
the sides are 1 and √3. So, the angles are 30 (opposite 1, the vertical side) and 60 degrees
(opposite √3, the horizontal side). The middle angle is 90 degrees, so the angle on the right has
to be 60 degrees. So, the angle at Q must be 30 degrees. Opposite 30 degrees, the side must be 1,
so s = 1.
Answer B
Example: If PQ = 1, what is the length of RS?
A. 1/12
B. (√3)/12
C. 1/6
D. 2/(3√3)
E. 2/(√12)
Explanation: Ans. B
If PQ = 1, QT = 1/2. If QT = 1/2, RT = 1/4. If RT is the long leg in triangle RST,

RT = x√3 = 1/4 and the short leg RS = x.
Concept # 2:
45-45-90 Right angled triangle (isosceles right triangle):
Ratio of sides = 1 : 1 : √2
Example:
The perimeter of a certain isosceles right triangle is 16 + 16√2, what is the length of the
hypotenuse of the triangle?
A. 8
B. 8√2
C. 16
D. 4
E. 4√2
Explanation: Let the hypotenuse be x, the other two sides will be x/√2 and x/√2
So, x + x/√2 + x/√2 = 16 + 16√2 Or x + 2*( x/√2) = 16 + 16√2 because 2/√2 = √2
Or x + x√2 = 16 + 16√2 Or x (1 + √2) = 16 (1 + √2)
Cancel (1 + √2) from both sides We get x = 16. Ans. C
Example:
Explanation:
Example:
Explanation:
• Concept: In an isosceles right-angled triangle (45-45-90 triangle), the foot of the perpendicular
from the right-angled vertex to the hypotenuse and the mid-point of the hypotenuse are the same
point.
Example:
If angle BAD is a right angle, what is the length of side BD? (figure not drawn to scale)
(1) AC is perpendicular to BD (2) BC = CD
Explanation:
(1) INSUFFICIENT: This tells us that AC is the height of triangle BAD to base BD. This does not
help us find the length of BD.
(2) INSUFFICIENT: This tells us that C is the midpoint of segment BD. This does not help us find
the length of BD.
(1) AND (2) SUFFICIENT: Using statements 1 and 2, we know that AC is the perpendicular
bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a
length of 5 (the same length as side AD). We also know that angle BAD is a right angle, so side BD
is the hypotenuse of right isosceles triangle BAD. If each leg of the triangle is 5, the hypotenuse
(using the Pythagorean theorem) must be 5√2. The correct answer is C.
Concept 3:
If the side of a cube is a, its face diagonal is a√2, body diagonal is a√3, its surface area is 6a2 and volume
a 3.
Example:
A sphere is inscribed in a cube with an edge of x centimeters. In terms of x what is the shortest
possible distance from one of the vertices of the cube to the surface of the sphere?
A. x(√3−1)
B. x/2
C. x(√2−1)
D. (x/2)( √3−1)
E. (x/2)( √2−1)
Explanation:
Say x=10 centimeters.
Then, since a sphere is inscribed in cube then the edge of the cube equals to the diameter of a
sphere, thus Diameter=10.
Next, body diagonal of the cube equals 10√3
The gap between the vertex of a cube and the surface of the sphere
= ½(Body Diagonal – Diameter)
Thus gap=(Diagonal−Diameter)/2=(10√3−10)/2=5(√3−1)
Since x=10 then 5(√3−1)=x/2(√3−1) Ans. D
Example:
In the cube shown, the length of line segment AB is 8. What is the surface area of the cube?
Explanation:
45-45-90 triangle
Side of the cube = 8/√2 = 4√2 … so the area of one face = (4√2)2 = 32. So the area of all faces = 32 * 6 =
192.
Concept 4: If ABC is a right-angled triangle with right angle at B, and BD is the perpendicular to the
hypotenuse AC, then:
• p = ac / b because ½ac = ½pb (both are areas of the triangle)

• p2 = h1 × h2
• AC² = AB² + BC² (Pythagoras theorem)
• All the three right-angled triangles are similar to each other (means all have same angles and
proportional sides)
• You should remember some of the Pythagorean triplets (e.g. 3, 4, 5 because 5² = 3² + 4²). Some
others are (5, 12, 13), (7, 24, 25), (9, 40, 41), (11, 60, 61), (13, 84, 85), (15, 112, 113), (20, 99,
101), (20, 21, 29), (8, 15,17), (12, 35, 37), (28, 45, 53) etc.
• If (3, 4, 5) is a triplet, so is (6, 8, 10) … (3, 4, 5) multiplied by 2
• The only basic right-angled triangle in which sides are in an AP is 3, 4, 5 (or its multiples)
• In a 45-45-90 triangle, the altitude from the right-angled vertex to the hypotenuse is half the
hypotenuse. So, if the diagram above is a 45-45-90 triangle, then p = b/2
How to generate triplets (you may skip this part):
Every basic triplet must contain
• one prime number
• two odd numbers
• one even number
3, 4, 5 is a basic triplet because it can’t be factorized further, but 6, 8, 10 isn’t a basic triplet because we
can cancel 2 from both terms.
BASIC TRIPLETS starting with an odd number: Square the smallest side (will be odd) and write it as
the sum of two consecutive numbers (always odd). Thus, we obtain the triplet. Like if the first number
is 3, then 32 = 9 = x + (x + 1) or x = 4 and x + 1 = 5, so the triplet is 3, 4, 5. Similarly, (5, 12, 13), (7, 24,
25), (9, 40, 41), (11, 60, 61), (13, 84, 85), (15, 112, 113) etc.
Note: (9, 12, 15), (15, 20, 25) are also triplets but not basic triplets. They are just multiples of (3, 4, 5).
So, the basic triplet starting with an odd number will always be unique.
BASIC TRIPLETS Starting with an even number: Square the smallest side (a) and keep on writing it
as:
a2 / 2n = x +(x + 2n), where n is an integer … till we do not get a fractional number.
The values of x and (x + 2n) will be the two other sides.
Let’s assume the smallest side is 20, then
202 / 2 = x + (x + 2) Or x = 99, so the triplet is (20, 99, 101)
202 / 4 = x + (x + 4) Or x = 48, so the triplet is (20, 48, 52) which is basically (5, 12, 13)
202 / 8 = x + (x + 8) Or x = 21, so the triplet is (20, 21, 29)
202 / 16 = x + (x + 16) … now x is fractional – so we stop here.
***
Example:
In the diagram, triangle PQR has a right angle at Q and has a perimeter of 60. Line segment QS is
perpendicular to PR and has a length of 12. PQ > QR. All the sides of all the three triangles are integers.
What is the ratio of the area of triangle PQS to the area of triangle RQS?
A. 9/4
B. 25/4
C. 25/16
D. 16/9
E. 25/9
Explanation:
Imagine right angled triangle with sides 12. As the smallest right triangle is 3, 4, 5… and 12 is a
multiple of 3 and 4 both, we may make
3, 4, 5 MULTIPLIED by 3 = 9, 12, 15… and

3, 4, 5 MULTIPLIED by 4 = 12, 16, 20.
Now let’s check the perimeter. 16 + 9 + 20 + 15 = 60, so these are the correct combinations.
Area of the two triangles will be ½ * base * height …

½ * 16 * 12 and ½ * 9 * 12 … ratio = 16/9
Ans. D
If you notice, the triangles are:

9, 12, 15 3, 4, 5 MULTIPLIED by 3
12, 16, 20 3, 4, 5 MULTIPLIED by 3
15, 20, 25 3, 4, 5 MULTIPLIED by 3
All the three triangles in such a situation are always similar (the ratio of corresponding sides is
always the same)
Example:
In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?
A. 3
B. 15/4
C. 5
D. 16/3
E. 20/3
Explanation:
p2 = h1 × h2 so 42 = 3 × CD or CD = 16/3 Ans. D
• Concept: If the perimeter of a triangle is fixed, the largest area will be for the equilateral triangle
√3 2
(most symmetrical shape). The area of an equilateral triangle of side a is A = 𝑎 and the height
4
√3
of the triangle is h = 𝑎
2
Example:
Explanation:
Example:
If points A and B are on the y–axis in the figure, what is the area of equilateral triangle ABC?
(1) Coordinates of point B are (0, 5√3)

(2) Coordinates of point C are (6, 3√3).
Explanation:
To find the area of equilateral triangle ABC, we need to find the length of one side. The area of an
equilateral triangle can be found with just one side since there is a known ratio between the side
and the height (using the 30: 60: 90 relationship). Alternatively, we can find the area of an
equilateral triangle just knowing the length of its height.
3
Or we know that height of an equilateral triangle is given by h = a and area of an equilateral
2
3 2
triangle is given by A = a , where ‘a’ is the side of the triangle.
4
(1) INSUFFICIENT: This does not give us the length of a side or the height of the equilateral
triangle since we don't have the coordinates of point A.
(2) SUFFICIENT: Since C has an x-coordinate of 6, the height of the equilateral triangle must be
6. We can determine the side and hence the area.
The correct answer is B.
• Concept: If two sides of a triangle are fixed and the included angle is a variable, then the area of
the triangle will be maximum when the included angle is 90°.
Proof: if a and b are two sides of a triangle and the included angle is θ, then the area of the
triangle is given by A = ½ ab sin θ … and sin θ will be maximum when θ = 90°.
You don’t have to know any trigonometry (or proofs) for the GMAT. Just know the result.
Example: what is the greatest possible area of a triangular region with one vertex at the center
of a circle of radius 1 and the other two vertices on the circle?
A. 1
B. ½
C. √3/2
D. √3/4
E. 2
Explanation:
The two sides are the radii, each equal to 1. The area will be maximum when the triangle is a
right-angled triangle.
So, the answer is ½ * 1 * 1 = ½ Ans. B
• Concept: For any figure, for a given (fixed or finite) perimeter, the more the symmetry, the more
will be the area. By this logic, a right-angled triangle will have the maximum area when the
triangle is isosceles right-angled triangle (45-45-90 triangle)
Example:
Is the area of the right-angled triangle ABC > 25?
(1) AC = 6 (2) AB = 10
Explanation:
(1) Imagine a right triangle with its legs as 6 and 1 … area = ½*6*1 = 3
Imagine a right triangle with its legs as 6 and 100 … area = ½*6*100 = 300 NS
(2) Now each side will be less than 10.

Here the perimeter is finite because the other two sides will have to be less than 10.
In such a case, the more the symmetry, the more the area. And the most symmetrical right-angled
triangle is going to be the isosceles right-angled triangle. Imagine the two legs are x each.
So x2 + x2 = 102 so x = 5√2
So, the maximum area of the triangle = ½*5√2*5√2 = 25
Our original question: Is the area of triangle ABE greater than 25? NO, it is not greater than 25,
because the maximum area is 25. Confirmed NO! Since we can answer the question using
Statement (2) alone, the correct answer is B.
More on symmetry and area:
• If with a given perimeter, different figures are formed like equilateral triangle, square, regular
hexagon, regular octagon … and eventually a circle (a regular polygon of infinite sides), then the
triangle will have the minimum area and circle will have the maximum area.
• Given a fixed perimeter, among all triangles, an equilateral triangle has the maximum area.
• If different triangles are inscribed in a circle, then the equilateral triangle will have the maximum
area.
• Concept: If two sides of a triangle are given and the area of the triangle is given, then the third
side of the triangle can have 2 different values.
Example:
In a triangle ABC, AB = 5, AC = 13, what is the length of segment BC?

(1) Angle ABC is 90 degrees.
(2) The area of the triangle is 30.
Explanation.
(1) SUFFICIENT: If we know that ABC is a right angle, then triangle ABC is a right triangle and we
can find the length of BC using the Pythagorean theorem. In this case, we can recognize the
common triple 5, 12, 13 - so BC must have a length of 12.
(2) If 2 sides of a triangle are given and the area of the triangle is given, the third side of the
triangle can have 2 possible values. NS
Visual Proof:
½ * 5 * 12 = ½ * 13 * h so h = 60/13 which is less than 5.
If the area of triangle ABC is 30, the height from point C to line AB must be 12 (We know that the
base is 5 and area of a triangle = 0.5 × base × height). There are only two possibilities for such a
triangle. Either angle CBA is a right triangle, and CB is 12, or angle BAC is an obtuse angle and the
height from point C to length AB would lie outside of the triangle. In this latter possibility, the
length of segment BC would be greater than 12. The correct answer is A.
• Concept: For two similar figures, the ratio of areas = ratio of squares of corresponding sides OR
A1 / A2 = (L1 / L2)2
Example:
In the given figure, if the area of the triangle on the right is twice the area of the triangle on the
left, then in terms of s, S=?
A. s/√2
B. s√3/2
C. s√2
D. s√3
E. 2s
Explanation.
In this problem, you have a2 : b2 = 2 : 1. If you know the result(s) above, then it follows at once
that a : b (the ratio of lengths, which is what you're looking for) is √2 : 1. Ans. C.
Example:
In the figure, AC = 3, CE = x, and BC is parallel to DE. If the area of triangle ABC is 1/12 of the area
of triangle ADE, then find the value of x.
A. 3
B. 6
C. 9
D. 3√3
E. 6√3 – 3
Explanation:
Use A1 / A2 = (L1 / L2)2 … So, we have 1/12 = [3/(3+x)]2 or 1/√12 = 3/(3+x) or 1/2√3 = 3/(3+x)
So, x = 6√3 – 3. Ans. E
• Concept: For two similar solids, the ratio of volumes = ratio of cubes of corresponding sides …
OR V1 / V2 = (L1 / L2)3
AND … the ratio of surface areas = ratio of squares of corresponding sides OR S1 / S2 = (L1 / L2)2
Example:
A cone has been divided into four slices parallel to the base. The heights of the various slices
from top to bottom are in the ratio 1 : 2 : 3 : 4. If the volume of the bottommost slice is 784 cm 3,
what is difference in volumes of the third slice from the top and the second slice from the top (in
cm3)?
A. 576
B. 216
C. 729
D. 125
E. 163
Explanation:
The heights of the various slices from top to bottom are in the ratio 1:2:3:4.
The heights of the cones will be 1:(1 + 2):(1 + 2 + 3):(1 + 2 + 3 + 4) = 1:3:6:10
So, the volumes will be in the ratio: 13:33:63:103 = 1:27:216:1000
Let’s say the volumes are: V:27V:216V:1000V
So, the volume of the 4th slice from the top = 1000V – 216V = 784V
The volume of the 3rd slice from the top = 216V – 27V = 189V
The volume of the 2nd slice from the top = 27V –V = 26V
The volume of the topmost slice = V
We are given 784V = 784, so V = 1
We are asked the value of 189V – 26V = 163V = 163.
Ans. E
• Concept: Angle in a semicircle is a right angle.
Example:
Triangle ABC is inscribed in a semicircle centered at D. What is the area of triangle ABC?
(A) 12/√3
(B) 6√3
(C) 12
(D) 12√3
(E) 18√3
Explanation:
• Concept: The sum of any two sides of a triangle is greater than the third side and the difference
of any two sides of a triangle is less than the third side
Example:
Is the perimeter of a rectangle greater than 8 inches?

(1) The diagonal of the rectangle is twice as long as its shorter side
(2) The diagonal of the rectangle is 4 inches longer than its shorter side
Explanation.
The question asks whether Perimeter=2(a+b)>8, or whether a+b>4, (where a and b are the
length of the sides of the rectangle).
(1) The diagonal of the rectangle is twice as long as its shorter side. Clearly insufficient, we know
the shape of the rectangle but not its size.
(2) The diagonal of the rectangle is 4 inches longer than its shorter side. This statement basically
says that the length of the diagonal is greater than 4 inches: d>4. Now, consider the triangle made
by the diagonal and the two sides of the rectangle: since the length of any side of a triangle
must be smaller than the sum of the other two sides, then we have that a+b>d, so a+b>d>4.
Sufficient. Ans. B
Similar Triangles:
Proportionality Theorem:
Intercepts made by two transversal lines (cutting lines) on three or more parallel lines are
proportional. In the figure, lines X1Y1 & X2Y2 are transversals cutting the three parallel lines AB,
CD, EF. Then AC, CE, BD, DF are intercepts Also,
AC/BD = CE/DF
Midpoint Theorem:
A triangle, the line joining the mid points of two sides is parallel to the third side and half of it.
Example:
In the figure, if point C is the center of the circle and DB = 7, what is the length of DE?
(1) x = 60° (2) DE || CA
Explanation:
(1) INSUFFICIENT: Just knowing that x = 60° tells us nothing about triangle EDB. To illustrate,
note that the exact location of point E is still unknown. Point E could be very close to the circle,
making DE relatively short in length. However, point E could be quite far away from the circle,
making DE relatively long in length. We cannot determine the length of DE with certainty.
(2) SUFFICIENT: If DE is parallel to CA, then (angle EDB) = (angle ACB) = x. Triangles EBD and
ABC also share the angle ABC, which of course has the same measurement in each triangle. Thus,
triangles EBD and ABC have two angles with identical measurements. Once you find that
triangles have 2 equal angles, you know that the third angle in the two triangles must also be
equal, since the sum of the angles in a triangle is 180°. So, triangles EBD and ABC are similar. This
means that their corresponding sides must be in proportion:
CB/DB = AC/DE
radius/diameter = radius/DE
3.5/7 = 3.5/DE
Therefore, DE = diameter = 7. The correct answer is B.
Basic Proportionality Theorem:
A line parallel to any one side of a triangle divides the other two sides proportionally. If DE is
parallel to BC, then
(a) AD/BD = AE/EC
(b) AB/AD = AC/AE
(c) AD/DE = AB/BC and so on.
Example: If BE || CD, and BC = AB = 3, AE = 4 and CD = 10, what is the area of trapezoid BEDC?
A. 12
B. 18
C. 24
D. 30
E. 48
Explanation:
Since BE||CD, triangle ABE is similar to triangle ACD (parallel lines imply two sets of equal
angles). We can use this relationship to set up a ratio of the respective sides of the two triangles:
AB/AC = AE/AD, or 3/6 = 4/AD or AD = 8
We can find the area of the trapezoid by finding the area of triangle CAD and subtracting the area
of triangle ABE. Triangle CAD is a right triangle since it has side lengths of 6, 8 and 10, which
means that triangle BAE is also a right triangle (they share the same right angle). Area of
trapezoid = area of triangle CAD – area of triangle BAE = 0.5(6)(8) – 0.5(3)(4) = 24 – 6 = 18
The correct answer is B
Concept # 7: Lines and Angles:
• Two angles whose sum is 90° are complementary. Each one is the complement of the other.
• Two angles whose sum is 180º are supplementary. Each one is the supplement of the other.
• Sum of the three interior angles of a triangle is 180°
• In a triangle, an exterior angle = Sum of the other two interior angles not adjacent to it
• In a triangle, sum of any two sides is greater than the third side and the difference of any two
sides is less than the third side.
• In a triangle, the side opposite to the greatest angle will be the greatest and vice versa.
• If a, b, c denote the sides of a triangle then
o if c² < a² + b², Triangle is acute angled
o if c² = a² + b², Triangle is right angled
o if c² > a² + b², Triangle is obtuse angled
• Sum of the four interior angles of a quadrilateral = 360°
• If a quadrilateral can be inscribed in a circle, it is called a cyclic quadrilateral. Here opposite
angles are supplementary.
• Parallelogram: Opposite sides equal and parallel. P = 2 (a + b)
• Rectangle: A quadrilateral whose opposite sides are equal and each internal angle equal to
90°, is called a rectangle.
l = length, b = breadth, A = l x b, P = 2 (l + b), Diagonal = √(l² + b²)
• Diagonals are equal and bisect each other.

• Of all rectangles of given perimeter, a square has max. area
• When inscribed in a circle, a rectangle will have maximum area when it’s a square.
• Rhombus: Area of a rhombus (quadrilateral with all four sides equal but diagonals unequal)
= ½ * D1 * D2, where D1 and D2 are its diagonals.
If the side of the rhombus is a, the perimeter is 4a and D12 + D22 = 4a2. The two diagonals are
always perpendicular.
• Square All four sides equal, diagonals equal
If the side of the square is a, then

Area = a2
Diagonal = a√2
Area = Diagonal2 / 2
Diagonals are perpendicular.
The figure formed by joining the mid-points of the sides of a square is also a square.
• Trapezium: in this, the two opposite sides are parallel.
Ares of a trapezium = ½ (sum of parallel sides) * height = ½ (b1 + b2) * h
Isosceles trapezium (oblique sides equal)
• Polygons: In any polygon, the sum of exterior angles = 360°. In any polygon, the sum of Interior
angles = (2n – 4) 90° or (n – 2) 180°.
For a regular hexagon of side a, perimeter = 6a and area = (3√3a2)/2

Sum of interior angles = 720°, each interior angle = 120°, each exterior angle = 60°.
Example: In the figure, what is x?
Explanation:
The polygon in the center of the figure is a pentagon.
The sum of the interior angles of an n-sided polygon is given by (n – 2) × 180°.
For a pentagon, (5 – 2) × 180° = 540°
x + (180 – z) + (180 – y) + x + 110 = 540, or 2x = 70 + y + z.
From the sum of the angles in the triangle at the upper right of the figure: 50 + y + z = 180
So, y + z = 130. Thus, 2x = 70 + 130 = 200, so x = 100.
• Circle: Circumference = 2R = D (R = Radius, D = Diameter = 2R)
Area = A = R² = D²/4  = 22/7 or 3.14
Length of arc of a circle is given by l = (θ °/360°) x 2R
Area of Sector ABC = (θ /360) x R²
Distance travelled by a wheel in n revolutions = n x circumference
If two chords intersect inside a circle, then the product of the lengths of the segments of one
chord is equal to the product of the lengths of the segments of the other chord.
In the figure, EA ⋅ EB = EC ⋅ ED
In a circle, angle at the center made by an arc = twice the angle made by the arc at any point on
the remaining part of the circumference.
We have APB = ½ AOB = 30° = AQB
Example:
In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the
diameter. What is the perimeter of the shaded region?
A. (5/3) + 5√3
B. (5/3)  + 10√3
C. (10/3)  + 5√3
D. (10/3)  + 10√3
E. (10/3)  + 20√3
Explanation:
Given that line CD is parallel to the diameter, we know that angle DCB and angle CBA are equal.
Thus x = 30°. First, let's calculate the length of arc CAE. Since arc CAE corresponds to an inscribed
angle of 60° (2x = 2×30° = 60°), it must correspond to a central angle of 120° which is 1/3 of the
360° of the circle. Thus we can take 1/3 of the circumference to give us the arc length CAE. The
circumference is given as 2πr, where r is the radius. Thus the circumference equals 10π and arc
length CAE equals (10/3)π.
Now we need to calculate the length of CB and BE. Since they have the same angle measure, these
lengths are equal so we can just find one length and double it. Let us find the length of CB. If we
draw a line from A to C we have a right triangle because any inscribed triangle that includes the
diameter is a right triangle. Also, we know that x = 30° so we have a 30–60–90 triangle. The
proportions of the length of the sides of a 30–60–90 triangle are 1–√3–2 for the side opposite
each respective angle. We know the hypotenuse is the diameter which is 2*r = 10. So length AC
must equal 5 and length CB must equal 5√3. Putting this all together gives us (10/3)π + 2 × 5√3
= (10/3)π + 10√3 The correct answer is D.
Example:
If PQ is a diameter of the circle shown, and the lengths of AX, BX, PX, and QX are integers, what is
the area of the circle?
(1) AX × BX = 16 (2) QX > AB
Explanation:
Use the product of chord segments property: AX × BX = PX × QX
The question asks for the area of the circle, so it is necessary to find the radius.
(1)
(AX)(BX) = (PX)(QX) = 16
PX = 2 and QX = 8
Diameter = 10
OR
PX = 1 and QX = 16
Diameter = 17 Not Sufficient
(2) NOT SUFFICIENT: This statement gives no indication of the scale of the diagram—i.e., there
is no indication of the size of anything whatsoever—so it is insufficient to determine the area of
the circle.
(1) AND (2) SUFFICIENT:
(AX)(BX) = 16 = 1 × 16 = 2 × 8 = 4 × 4 (in any order). So AB = 17 or 10 or 8.
(PX)(QX) = 16 = 1 × 16 = 2 × 8 = 4 × 4 (in any order).
But AB ≥ 8
QX > AB is only possible if QX = 16
Therefore, QX = 16, PX = 1, and so the diameter of the circle is 17; this is enough information to
calculate the area of the circle.
The correct answer is C.
Example:
The circle has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of arc
BXC, what is the length of arc BXC?
(A) 2π
(B) 8π/3
(C) 3π
(D) 4π
(E) 16π/3
Explanation:
Because AD is parallel to BC, the measure of angle ACB is also 45°. Angles CAD and ACB are both
inscribed angles of the circle. The measures of the corresponding central angles are twice 45°,
or 90° each. Therefore, taken together, minor arcs AB and CD make up 180° of the entire circle,
leaving 180° for arcs BXC and AYD. Because arc AYD is twice the length of arc BXC, arc BXC must
correspond to a 60° central angle and arc AYD to a 120° central angle. Therefore, arc BXC is
60/360 = 1/6 of the entire circumference of the circle, which equals

2πr = 16π. The length of arc BXC is thus 16π/6 = 8π/3.
Solids
Quant: Co–ordinate Geometry + Functions and Graphs
www.top–one–percent.com | info@top–one–percent.com | +91–97395–61394

Co–ordinates:
• If 2 points (a, b) and (c, d) lie in the same quadrant, then a and c should have the same sign; and
b and d should have the same sign.
• Distance between 2 points (x1, y1) and (x2, y2) is given by 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
• Distance of the point (a, b) from the origin (0, 0) is given by √𝑎2 + 𝑏 2
• Any point on the X axis can be taken as (a, 0)
• Any point on the Y axis can be taken as (0, b)
• In order to find the X–intercept of a line, put Y = 0 in the equation of the line and find X
• In order to find the Y intercept of a line, put X = 0 and find Y.
Straight line (slope):
o Equation of x axis: y = 0
o Equation of y axis: x = 0
o Equation of line parallel to x axis: y = k (k is a constant)
o Equation of line parallel to y axis: x = k (k is a constant)
o The equation of a line which makes an angle of θ (measured anticlockwise) with the X axis and
has the y–intercept c is given by y = mx + c, where m = tan θ. Here m is the slope of the line and
c is the y intercept of the line. For this line, slope = m, x intercept = –c/m and y intercept = c. So,
to find the slope of any line, write it in Y = mX + c form first.
o For example, the line 2x – y + 10 = 0 can be written as y = 2x + 10, so the slope will be 2. The line
4x = y + 5 can be written as y = 4x – 5, so the slope will be 4.
o Slope = RISE / RUN
o Slope = (y2 – y1) / (x2 – x1)
𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
o Slope = − 𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
o For the line ax + by + c = 0, slope = –a/b, x intercept = –c/a, y intercept = –c/b
o If we write it in y = mx + c form, the line can be written as y = –(a/b)x –c/b

o So, the slope = –a/b
o To find the x intercept, substitute y = 0, and find x → x intercept = –c/a
o To find the y intercept, substitute x = 0, and find y → y intercept = –c/b
o If two lines are parallel then their slopes are equal (m1 = m2). If two lines do not intersect, they
are parallel.
o If two lines are perpendicular to each other, the product of their slopes is –1. (m1 m2 = –1).
o The point of intersection of two lines (X, Y) is obtained by simultaneously solving both the
equations.
o To plot a line, first put y = 0, find the point on x axis; then put x = 0, fins the point on y axis. Join
the two points to get the desired graph.
o Positive and Negative Slopes: Whether a line has a positive or negative slope is easy to tell just
by looking at a graph of the line. If the line slopes uphill as you trace it from left to right, the slope
is positive. If a line slopes downhill as you trace it from left to right, the slope is negative. Uphill
= positive. Downhill = negative.
o You can get a sense of the magnitude of the slope of a line by looking at the line’s steepness. The
steeper the line, the greater the slope; the flatter the line, the smaller the slope. Note that an
extremely positive slope is larger than a moderately positive slope, while an extremely negative
slope is smaller than a moderately negative slope.
o Check out the lines below and try to determine whether the slope of each line is negative or
positive and which has the greatest slope:
Lines a and b have positive slopes, and lines c and d have negative slopes. In terms of slope
magnitude, line a > b > c > d.
Q.
Does the line L whose equation is y = mx + c cut the x–axis in the positive direction of x–axis?
(1) The intercepts of Line K, perpendicular to L, are of the opposite signs.
(2) Line L passes through the 4th quadrant.
Sol. E
𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
Slope = − 𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
From (1), if the intercepts of K are of opposite signs, by the above formula, slope of K is positive.
L and K are perpendicular, so the product of their slopes = –1. Since the slope of K is positive, the slope
of L must be negative for the product of slopes to be –1. In the two figures below, L has a negative slope
and passes through the 4th quadrant, but in one case, the x intercept is positive and in the other case,
the x–intercept is negative. So, E
Q.
The y intercept of a line L is 4. If the slope of L is negative, which of the following could be the x
intercept of L?
I. –1
II. 0
III. 6
Sol.
X intercept of y = mx + c is Xintercept = –c/m = –4/m = –4/ a –ve number = positive.
Only III is possible. OR
Let's draw a bunch of lines with a negative slope AND have a y–intercept of 4, means passes through
the point (0,4).
As we can see, the x–intercept can have ANY POSITIVE value. So, the x–intercept COULD be 6, but it
could NOT be –1 or 0
Q. Does line Ax + By + C = 0 (A is not 0) intersect the x–axis on the negative side?
(1) BA < 0.
(2) AC > 0.
Sol.
The question is asking whether the x intercept of this line is negative. Or whether –C/A is negative or
not or whether C/A is positive or not or whether A and C are of same sign or not.
(2) gives us that A and C are of the same sign. Ans. B
Q.
An (x, y) coordinate pair is to be chosen at random from the xy–plane. What is the probability that y≥|x|?
A. 1/10
B. 1/8
C. 1/6
D. 1/5
E. 1/4
Sol. The graph of y = |x| is
All points which satisfy y≥|x| condition lie above that graph. You can see that portion of the plane which
is above the graph is 1/4. Answer: E.
Q. In the figure below, which line has greater slope, AB or AC?
Sol. While there are no numbers on the graph, both lines have positive slopes (the lines rise upward
when reading from left to right) and segment AB is steeper than segment AC. Thus, segment AB has a
greater slope.
Q. Which line has greater slope, L1 or L2?
Sol.
Ans. L1 has a greater slope.
Parabola Basics
If we plot y = ax2 + bx + c, we get the equation of a parabola.
For x-axis intersection, y = 0, so ax2 + bx + c = 0
• If b2 > 4ac, the parabola will cut the x axis in 2 distinct points (distinct real roots)
• If b2 = 4ac, the parabola will touch the x axis at one point (equal roots)
• If b2 < 4ac, the parabola will not cut the x axis (imaginary roots)
If the x term is absent (b = 0), then the parabola will have its axis of symmetry
on y axis.
If the x term is present, the axis of symmetry will be parallel to y axis, but not y-
axis itself.
If a and c are positive, then we have four possibilities for y = ax2 + c
1. y = ax2 + c Opens upwards, vertex above x axis.

2. y = ax2 – c Opens upwards, vertex below x axis.
3. y = –ax2 + c Opens downwards, vertex above x axis.
4. y = –ax2 – c Opens downwards, vertex below x axis.
Q. Graph the curve: y = –3x2 + 3
Graph of y = a(x−h)2 + k
Here (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this
form is also known as the vertex form of a quadratic function.
y=−3(x+2)2 + 4. y=x2−3
Q.
Does the curve y=b(x−2)2 + c lie completely above the x–axis?
(1) b>0, c<0
(2) b>2, c<2
Sol.
(1)
b is positive and c is negative.
b is positive. (x − 2)2 cannot be negative. It will be 0 (when x = 2) or positive. So, at x = 2, the first term
will be 0 but c will be negative. This means y will be negative. So, the curve will not lie completely above
x axis.
(2)
b>2, c<2 c could be positive or negative.
b is positive so first term would be 0 or positive. Depending on the value of c, the graph could lie
completely above x axis (say if c = 1) or below too (if c = –1)
Not sufficient. Answer (A)
Q.
Does y = ax2 + bx + c intersect the x axis?
(1) a < 0
(2) c > 0
Sol.
b2 is either zero or positive (squares can’t be negative). ac will be negative, 4ac will also be negative, so
b2 will always be greater than 4ac. Ans. C
Q.
Does the curve h(x)=y=ax2 + c intersect the x–axis?
(1) h(4) > 0
(2) a>0
Sol.
Is b2 > 4ac?
b = 0, so the question becomes: is 0 > 4ac? Or is 4ac < 0 or is ac < 0? Or are a and c of opposite signs?
Given (1) 16a + c > 0 (2) a > 0
Combine, (a, c) could be (1, 1) or (a, c) could be (1, –1) … in either case, the conditions from both the
statements are satisfied by the values above, so ac could be positive or negative. Ans. E
Q.
Does the graph of y = mx2 + h intersect the x–axis?
(1) m < 0
(2) h > 0
Sol.
Is b2 > 4ac? Here, b = 0, a = m, c = h, so the question becomes: is 0 > 4mh? Or is 4mh < 0 or is mh < 0? Or
are m and h of opposite signs? Ans. C
Q.
In the xy–plane, at what two points does the graph of y = (x + a)(x + b) intersect the x–axis?
(1) a + b = –1
(2) The graph intersects the y–axis at (0, –6)
Sol.
x axis means y = 0, so (x + a) (x + b) = 0 or x2 + (a + b)x + ab = 0
(1) gives a + b, but not ab. NS
(2) gives ab = –6. NS
Combining:
The equation becomes x2 – x – 6 = 0
So, (x – 3) (x + 2) = 0 or x = 3 or –2
So, the two points are (3, 0) and (–2, 0). Ans. C
Quant: Permutations and Combinations | Probability
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Definitions:
• Each of the different orders of arrangements, obtained by taking some, or all, of a number of things,
is called a Permutation.
• Each of the different groups, or collections, that can be formed by taking some, or all, of a number of
things, irrespective of the order in which the things appear in the group, is called a Combination.
Example:
Suppose, there are four quantities A, B, C, and D. The different orders of arrangements of these four
quantities by taking three at a time, are:
ABC, ACB, BAC, BCA, CAB, CBA, ... (1) ABD, ADB, BAD, BDA, DAB, DBA, ... (2)
ACD, ADC, CAD, CDA, DAC, DCA, ... (3) BCD, BDC, CDB, CBD, DBC, DCB. ... (4)
Thus, each of the 24 arrangements, of the four quantities A, B, C, and D by taking three at a time, is called
a permutation. Hence, it is clear that the number of permutations of four things taken three at a time is
24.
Again, it may be easily seen, from the above that out of these 24 permutations, the six, given in (1), are
all formed of the same three quantities A, B, C in different orders; hence, they all belong to the same
group. Similarly, the permutations, given in (2), all belong to a second group; those given in (3), belong
to a third and those in (4), belong to a fourth. Hence, we see that there are only four different groups
that can be formed of four quantities A, B, C, and D by taking three at a time. Thus, the number of
combinations of four things taken three at a time is only four.
If there are m ways of doing a thing and n ways of doing a second thing and p ways of doing a third
thing, then the total number of “distinct” ways of doing all these together is m × n × p.
Ex 1.
Suppose, there are five routes for going from a place A to another place B and six routes for going from the
place B to a third place C. Find the numbers of different ways through which a person can go from A to C
via B.
Sol.
Since there are five different routes from A to B, person can go from A to B in five different ways. After
reaching B, he has six different ways of finishing the second part of his journey (i.e. going from B to C).
Thus, for one way of going from A to B there are six different ways of completing the journey from A to
C via B. Hence, the total number of different ways of finishing both parts of the journey (i.e. A to B and
then from B to C) = 5 times six different ways = 5 × 6 = no. of ways from the first part to the second point
⨯ number of ways from the second part to the third point
Ex 2.
Find the number of different ways in which four persons can be accommodated in three different chairs.
Sol.
Let’s assume that the four persons are P, Q, R, and S. Since all the three different chairs are vacant, any
one of the four persons can occupy the 1st chair. Thus, there are four ways of filling up the 1st chair.
When the 1st chair has been filled up by any one of the four people, say P, the 2nd chair can be filled up
by any one of the remaining three persons Q, R and S. Thus, for each way of filling up the 1st chair, the
2nd chair can be filled up in three different ways. Hence, total no. of ways in which the first two chairs
can be filled up is equal to 4⨯3 = 12 ways. Again, when the 1st and 2nd chairs are filled up in any one
way (i.e. the 1st by P and the 2nd by Q), the 3rd chair can be filled up by any one of the two remaining
persons, R and S. Thus, for each way of filling up the first two chairs, there are 4⨯3⨯2 i.e. 24 ways of
filling up the third chair along with the first two chairs. Hence, the total no. of ways in which four persons
can be accommodated in the three given chairs is equal to 4⨯3⨯2 = 24. We therefore conclude that, the
total number of different orders of arrangements of 4 different things, taken 3 at a time, is the same as
the total number of different ways in which 3 places can be filled up by 4 different things.
PERMUTATIONS
• Permutations of n different things taken ‘r’ at a time is denoted by nPr and is given by
nPr = n! / (n – r)!
• The total number of arrangements of n things taken r at a time, in which a particular thing always
occurs = r ⨯ n – 1Pr – 1
• The total number of permutations of n different things taken r at a time in which a particular thing
never occurs = n – 1Pr
• The total number of permutations of n dissimilar things taken r at a time with repetitions = nr
• No. of circular permutations of n things taken all at a time = (n – 1)!
• No. of circular permutations of n different things taken r at a time = nPr/r
• The number of permutations when things are not all different: If there be n things, p of them of one
kind, q of another kind, r of still another kind and so on, then the total number of permutations is
given by n! / (p! q! r!)
COMBINATIONS
• Number of combinations of n dissimilar things taken ‘r’ at a time is denoted by nCr and is given by
nCr = n! / [ (n – r)! r!]
• Number of combinations of n different things taken r at a time in which p particular things will
always occur is n – pCr – p
• No. of combinations of n dissimilar things taken ‘r’ at a time in which ‘p’ particular things will never
occur is n – pCr
• nCr = nCn – r
PROBABILITY
Probability of an event occurring = Number of favorable outcomes

Number of all possible outcomes
Note:
• If an event E is sure to occur, we say that the probability of the event E is equal to 1 and we write
P (E) = 1.
• If an event E is sure not to occur, we say that the probability of the event E is equal to 0 and we write
P (E) = 0.
• Therefore, for any event E, 0 ≤ P (E) ≤ 1
• The probability of E not occurring, denoted by P (not E), is given by P (not E) = 1 – P(E)
• Odds in favor = No. of favorable cases / No. of unfavorable cases
• Odds against = No. of unfavorable cases / No. of favorable cases
Mutually Exclusive Events:
Two events are mutually exclusive if one happens, the other can’t happen and vice versa. In other words,
the events have no common outcomes. For example:
In rolling a die
E: – The event that the no. is odd
F: – The event that the no. is even
G: – The event that the no. is a multiple of three
In drawing a card from a deck of 52 cards

E: – The event that it is a spade
F: – The event that it is a club
G: – The event that it is a king
In the above 2 cases events E and F are mutually exclusive but the events E and G are not mutually
exclusive or disjoint since they may have common outcomes.
ADDITION LAW OF PROBABILITY:
If E and F are two mutually exclusive events, then the probability that either event E or event F will
occur in a single trial is given by:
P (E or F) = P(E) + P(F)
If the events are not mutually exclusive, then
P (E or F) = P(E) + P(F) – P (E and F together)
P (neither E nor F) = 1 – P (E or F).
Independent Events and Multiplication Law:
Two events are independent if the happening of one has no effect on the happening of the other.
For ex:
On rolling a die and tossing a coin together
E: – The event that no. 6 turns up.
F: – The event that head turns up.
In shooting a target
E: – Event that the first trial is missed.
F: – Event that the second trial is missed.
In both these cases events E and F are independent.
BUT, in drawing a card from a well shuffled pack
E: – Event that first card is drawn
F: – Event that second card is drawn without replacing the first
G: – Event that second card is drawn after replacing the first
In this case E and F are not Independent but E and G are independent.
MULTIPLICATION LAW OF PROBABILITY:
If the events E and F are independent then P (E and F) = P (E) ⨯ P (F)
and
P (not E and F) = 1 – P (E and F together).

SUMMARY of concepts
Arrangements - keywords – seating, sitting, sequence, order, alphabets, schedule, ranking, itinerary,
codes
Order important – gives unique arrangements
For e.g. A and B sitting on chair can be AB or BA so these are two distinct arrangements
It is basically selection followed by arrangement. So n Pr = n Cr  r !
n!
n
Pr =
(n − r )!
Selection - keywords – team, committee, balls, handshakes, matches, picking
Order not important – For example choosing A and B from a group of 3 or four alphabets. The order
does not matter. India playing a match against Australia is the same as Australia playing against India.
n!
n
Cr =
(n − r )!  r !
Different formulae
n!
1.
n
Pr =
(n − r )!
When to use? When n distinct items present and r have to be selected and then arranged.
E.g. – how many ways can you arrange 4 people in 5 chairs = 5 P 4
2. nr
All n distinct selection of r but repetition is allowed.
In how many ways can you wear three different rings on four fingers?
43
n!
3.
p! q! r !
Arranging n things in which p are of one type, q of a second type and r of third type:
Ex: In how many ways can you arrange the letters of word Banana?
6!
Ans.
3! 2 !
4. Special Cases
5 people A, B, C, D, E to be arranges in which A and B are together.

4! × 2!
5 people A, B, C, D, E to be arranged in which A and B are not together.

5! – 4! × 2!
5. Block diagrams - Some problems cannot be done with any formula but with a block diagram
Combinations
10
1. Select 5 people out of 10 Ans. C5
Particular Cases – Select 5 out of 10 people such that A and B are always selected. This means only
3 of the remaining 8 are to be selected 8C3
Select 5 out of 10 such that A and B are never selected. This means that out of remaining 8, 5 have
to be selected so it is 8C5
2. Select 5 out of 10 so that A and B are never together.

= Total – Together = 10C5 – 8C3
AND denotes Multiplication
OR denotes Addition
Circular Permutations: (n – 1)!
Multiple trials of a single event: If multiple independent trials of a single event are performed, then
the probability of r successes out of a total of n trials can be determined by n C r  p r  q n− r
Where
n = number of times the event is performed
r = number of successes
p = probability of success in one trial
q = probability of failure in one trial = 1 – p.
Solved examples for building key concepts:
1. Find the number of ways in which the letters of the word “machine” can be arranged such that the
vowels may occupy only odd positions?
2. Sixteen jobs are vacant; how many different batches of men can be chosen out of twenty candidates?
How often may any particular candidate be selected?
3. How many numbers greater than a million can be formed with the digits 2, 3, 0, 3, 4, 2, 3?
4. In how many ways can 3 letters be posted in four letter boxes in a village? If all the three letters are
not posted in the same letter box, find the corresponding number of ways of posting.
5. In rolling two dice, find the probability that (1) there is at least one ‘6’ (2) the sum is 5.
6. A single card is selected from a deck of 52 bridge cards. What is the probability that (1) it is not a
heart, (2) it is an ace or a spade?
7. A box contains 2 red, 3 yellow and 4 blue balls. Three balls are drawn in succession with
replacement. Find the probability that (1) all are yellow, (2) the first is red, the second is yellow, the
third is blue, (3) none are yellow, (4) all three are of the same color.
8. With the data in Example 7, answer those questions when the balls are drawn in succession without
replacement.
9. There are 7 Physics and 1 Chemistry book in shelf A. There are 5 Physics books in shelf B. One book
is moved from shelf A to shelf B. A student picks up a book from shelf B. Find the probability that the
Chemistry book: (1) is still in shelf A, (2) is in shelf B, (3) is taken by the student.
10. The ratios of number of boys and girls in X-A and X-B are 3: 1 and 2: 5 respectively. A student is
selected to be the chairman of the students’ association. The chance that the student is selected from
X-A is 2/3. Find the probability that the chairman will be a boy.
11. The probability that a man will be alive in 25 years is 3/5 and the probability that his wife will be
alive in 25 years is 2/3. Find the probability that: (1) both will be alive, (2) only the man will be alive,
(3) only the wife will be alive, (4) at least one will be alive.
SOLUTIONS
1.
“Machine” consists of seven letters: four of them are consonants and three vowels. Let us mark out the
position to be filled up as follows:
1 2 3 4 5 6 7
(a) () (i) () (e) () ()
Since the vowels can be placed only in three out of the four positions marked 1,3,5,7, the total number
of ways in which they can be made to occupy odd positions =
4P3 = 4.3.2 = 24. ... (1)
Suppose one arrangement of the vowels is as shown in the diagram; then for this particular
arrangement of the vowel, the number of ways in which the 4 consonants can be made to occupy the
remaining positions (marked 2,4,6,7) =
4P4 = 4.3.2.1 = 24.
Hence, for each way of placing the vowels in odd positions there are 24 arrangements of the whole set.
Consequently, the total number of arrangements of the given letters under the given condition = 24⨯24
= 576 Ans.
2
We have only to find out the number of different groups of 16 men that can be formed out of 20 without
any reference to the appointment to be given to each.
Hence, the required number of ways = 20C16 = 20C4
= 20⨯19⨯18⨯17 / (1⨯2⨯3⨯4) = 5⨯19⨯3⨯17 = 4845.
Let us now find out how many times a particular candidate may be chosen.
Every time that a particular candidate is selected the other 15 candidates will have to be chosen from
the remaining 19 candidates.
Hence a particular man may be selected as many times as we can select a group of 15 men out of the
remaining 19. Hence, the required number of times = 19C15 = 19C4
= 19⨯18⨯17⨯16 / (1⨯2⨯3⨯4) = 19⨯3⨯17⨯4 = 3876
3.
Since, each number is to consist of not less than 7 digits, we shall have to use all the digits in forming
the numbers. Now, among these 7 digits there are 2 two’s and 3 three’s; hence the total number of ways
of arranging the digits = 7! / (2! 3!) = 420. But out of these arrangements we have to reject those that
begin with zero, for they are six–digit numbers. Now, evidently there are as many such arrangements
as there are ways of arranging the remaining 6 digits among themselves
Their no. = 6! /2! 3! = 60 Hence, the required number = 420 – 60 = 360.

4.
We can post the first letter in 4 ways. Similarly, the second and third can be posted in 4 ways each. So,
the total number of ways = 4⨯4⨯4 = 64. Now all the three letters together can be posted in any letterbox.
In this case there will be four ways and when all the letters are not posted together, the number of ways
= 64 – 4 = 60.
5.
The total possible outcomes are 36 as shown below.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6); (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6); (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6); (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
The outcomes with at least one ‘6’ are
(1,6), (2,6), .... (6,6). There are 11 such pairs.
(1)
P (at least one ‘6’) = 11/36 Ans.
(2)
The pairs with a sum of 5 are (1,4), (2,3), (3,2), (4,1).
P (the sum is 5) = 4/36 = 1/9 Ans.
6.
A deck of bridge cards has 4 suits – spade, heart, diamond and club. Each suit has 13 cards.
Ace, two, three, ...., ten, jack, Queen, King.
(1) P (not a heart) = 1 – P (a heart) = 1 – 13/52 = 39/52 = 3/4 Ans.
(2) There are 4 aces and 12 spades besides the ace of spades …
P (an ace or a spade) = 16/52 = 4/13 Ans.
7.
(1)
In a draw, P (red) = 2/9, P (yellow) = 3/9, P (blue) = 4/9.
In 3 draws, Prob of all are yellow = (3/9). (3/9). (3/9) = 1/27 Ans.
(2)
Required probability = P (1st red). P (2nd yellow). P (3rd blue)
= 2/9. 3/9. 4/9 = 8/243 Ans.

(3)
Probability that none are yellow = P (1st not yellow). P (2nd not yellow). P (3rd not yellow)
= (1 – 3/9) ⨯ (1 – 3/9) ⨯ (1 – 3/9) = 8/27 Ans.
(4)
Probability that all three are of the same color
= P (all red) + P (all yellow) + P (all blue) {mutually exclusive}
= (2/9)3 + (3/9)3 + (4/9)3 = 11/81 Ans.
8.
(1)
Prob of all yellow = P (1st yellow). P (2nd yellow). P (3rd yellow)
= 3/9. 2/8. 1/7 = 1/84 Ans.
Since when the first yellow ball has been drawn, there are 8 balls remaining in the bag of which 2 are
yellow.
(2)
Required probability = P (1st red). P (2nd yellow). P (3rd blue)
= 2/9. 3/8. 4/7 = 1/21 Ans.
(3)
Probability that none are yellow
= P (1st not yellow). P (2nd not yellow). P (3rd not yellow)
= (1 – 3/9) (1 – 3/8) (1 – 3/7) = 6/9. 5/8. 4/7 = 5/21 Ans.
(4)
Probability that all three are of the same color
= P (all red) + (all yellow) + P (all blue)
= 2/9. 1/8. 0/7 + 3/9. 2/8. 1/7 + 4/9. 3/8. 2/7 = 5/84 Ans.
9.
(1)
The probability that it is in shelf A = 7/8 Ans. (this means that Physics book was picked up)
(2)
The probability that it is in shelf B =
P (it is moved from A to B). P (it is not taken by the student) = 1/8. 5/6 = 5/48 Ans.
(3)
The probability that is it taken by the student
= P (it is moved from A to B). P (it is taken by the student) = 1/8. 1/6 = 1/48 Ans.
10.
Probability that the boy comes from X-A = 2/3. 3/4 = 1/2
Probability that the boy comes from X-B = 1/3. 2/7 = 2/21
The required probability = 1/2 + 2/21 = 25/42
11.
(1)
P (both alive) = P (man alive) ⨯P (wife alive) = 3/5⨯2/3 = 2/5
(2)
P (only man alive) = P (man alive) ⨯P (wife dead) = 3/5⨯1/3 = 1/5
(3)
P (only wife alive) = P (man dead) ⨯P (wife alive) = 2/5⨯2/3 = 4/15 Ans.
(4)
P (at least one will be alive) = 1 – P (both dead) = 1 – (2/5⨯1/3) = 13/15 Ans.

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By Sandeep Gupta | GMAT 800/800, Harvard Final Admit

To express a% as a fraction divide it by 100

To express a fraction as a percent multiply it by 100

a/b = [(a/b) ⨯ 100] %

10% can be expressed as 0.1.

6.5% = 0.065 etc.

Increase % = [ Increase / Original value] ⨯ 100%

Decrease % = [ Decrease / Original value] ⨯ 100%

Change % = [ Change / Original value] ⨯ 100%

In increase %, the denominator is smaller, whereas in decrease %, the denominator is larger.

X increased by 10% would become X + 0.1 X = 1.1 X

Also, let us remember that:

2 = 200% (or 100% increase)

3/2 = 1 + 1/2 = 100 + 50 = 150%

5/2 = 2 + 1/2 = 200 + 50 = 250% etc.

2/3 = 1 – 1/3 = 100 – 33.33 = 66.66%

4/3 = 1 + 1/3 = 100 + 33.33 = 133.33%,

5/3 = 1 + 2/3 = 100 + 66.66 % = 166.66%

7/3 = 2 + 1/3 = 200 + 33.33 = 233.33%

8/3 = 2 + 2/3 = 200 + 66.66 = 3 – 1/3 = 300 – 33.33 = 266.66%

5/4 = (1 + 1/4) = 125% (= 25% increase)

7/4 = (1 + 3/4 = 2 – 1/4) = 175% (= 75% increase)

9/4 = (2 + 1/4) = 225% (= 125% increase)

11/4 = 275% = (175% increase)

6/5 = (1 + 1/5) = 120%

7/5 = (1 + 2/5) = 140% etc.

7/6 (1 + 1/6) = 116.66%

9/8 = (1 + 1/8) = 112.5%

11/8 = (1 + 3/8) = 137.5%

10/9 = (1 + 1/9) = 111.11%

11/9 = (1 + 2/9) = 122.22% etc.

(i) population after n years = p [1 + (X/100)]n

(ii) population n years ago = p / [1 + (X/100)]n

(i) population (or value of machine) after n years = p [1 – (R/100)]n

(ii) population (or value of machine) n years ago = p / [1 –(R/100)]n

Profit % = (Profit /CP) x 100%

Loss % = (Loss/CP) x 100 %

In problems on DISCOUNT, remember the following:

Discount is calculated on Marked price and NOT on Cost price.

Let the pocket money be X

Expenditure = Consumption ⨯ Price

So, the expenditure increases by 12%

By formula, we have 1200 ⨯ (1 + R/100)² = 1323 so R = 5%

So, profit percent = 20%

We have 1440 = X ⨯ Y .................(1)

X = no. of apples, Y = price of one apple.

Now 1440 = (X + 120) ⨯ 0.8Y ......(2)

Equate the two (as both are 1440)

Substitute X = 480 in (1) Y = Rs 3

wages per hour ⨯ number of hours = earnings

Now: 12⨯12.5 = 150 From 100 to 150 → 50% increase

Let initially X = number of hours and Y = wages/hour

Now Daily earnings = 1.25X ⨯ 1.2Y = 1.5 XY Hence 50% increase.

Let CP = 100, Profit = 19, SP = 100 + 19 = 119

So, 85% of M.P. = 119 or MP = 119 ⨯ 100/85 = Rs 140

Answer = 40% above the C.P.

Sol. 160/340 = 8/17 = 47.05%.