100 DI Mains Questions

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100
HIGH LEVEL
DATA
INTERPRETATION
QUESTIONS & SOLUTIONS
FOR IBPS & SBI MAIN EXAMS

100 Mains Level DI Questions Free PDF for PO & Clerk Mains
Directions [Set of 3 questions]: Line graph C) 13/25

given below shows the probability per cent of
D) 12/25
randomly drawing a different colour ball
(Black, Red, White, Blue and Yellow) from two E) None of these
bags A and B.
========================
Q3. If 10 balls are taken out from bag A and

put into bag B and now the probability of
selecting a blue ball from bag A is 3/20, then
what is the probability of selecting a blue ball
from bag B after transferring?
A) 1/10
B) 2/25
Note 1: Total number of blue colour balls in C) 3/50

bag A is 10 and total number of yellow
colour balls in bag B is 10. D) 4/25
Note 2: Probability of drawing a yellow colour E) None of these

ball from bag A is 8/25.
========================
Note 3: Probability of drawing a blue colour
Directions [Set of 5 questions]: The line graph
ball from bag B is 1/10.
shows different Discount% Schemes and the
Q1. What is the ratio of probability of bar graph shows the Cost Price of those Food
drawing two black colour balls from bag A at Items in Rs per kilogram. Food items are
random to the probability of drawing two classified according to their quality: Normal
blue colour balls from bag B at random? Quality, Good Quality and Great Quality.
A) 14: 33
B) 104: 35
C) 14: 135
D) 104: 135
E) None of these
========================
Q2. If one ball from each bag is drawn at

random, then what is the probability of the
event, that one red ball is from bag B and
one yellow ball is from bag A?
A) 2/5
B) 8/125
1
B) 8
C) 6
D) 12
E) 15
========================
Q6. What is the ratio of the difference

between the Selling price of Rice and Wheat if
Q4. A shopkeeper mixes 4 kgs of Normal all 3 qualities of Wheat are mixed equally and
Quality Wheat with 6 kgs of Good Quality Normal, Good and Great quality of Rice are
Wheat sells dishonestly as Great Quality mixed in the ratio 2:2:1 and mark-up for both
Wheat claiming to offer discount on Cost items is 50%, to the difference between the
price. If Discount% A and C are successively selling price of Apples and Oranges if
applied, the profit% earned by this Normal, Good and great quality of Oranges
Transactions is (x/23)%. In other case, for are mixed equally and Apples are mixed in
Rice, Great Quality, Good Quality and Normal the ratio 2:1:1 mark-up for both is 40%?
Quality are mixed in the ratio 1:1:2 but sold Given that Discount% A is applied for Wheat
mistakenly as Normal Quality Rice applying and Rice and Discount % C is applied for
successive discounts% A and B. For no Profit Apples and Oranges.
or loss, the rice is marked up by (y/9)%. What
is the value of (y+200)/x? A) 22:7
A) 8 B) 27:7
B) 10 C) 23:11
C) 12 D) 27:11
D) 15 E) 23:13
E) 20 ========================
======================== Q7. The ratio in which a shopkeeper must mix

Normal and Great quality rice such that he
Q5. 4 kgs of Normal quality Apples are mixed earns a profit of Rs 5 after a mark-up of Rs
with 'x' kgs of Good Quality Apples and the 45 and applying Discount% B and C, is x:1
mixture is marked up by 25%. Discount% B and the ratio in which the shopkeeper must
and C are applied successively such that the mix Good and Great Quality Apples such that
overall loss from selling the mixture is Rs 26. the selling price is Rs 441/kg after marking
Normal quality Oranges are mixed with a new up by 40% and using Discount% B, is 1:y.
type of Orange with price Rs 'y'/kg in the What is the value of 2x2 - 3y2?
ratio 3:2. The mixture is marked up by Rs
320/kg and Discount% A and B are applied, A) 12
and it is sold at a profit of 20%. What is the
value of (y - 10x)/13? B) 8
A) 30 C) 10
D) 20
2
E) 15 B) 6
======================== C) 4
Q8. If Normal and Good quality Apples are D) 2

mixed in the ratio 3:2 and sold dishonestly as
Great quality Apple applying all 3 discount E) Can't be determined
schemes successively claiming them to be
applied on Cost price, what is the profit ========================
earned by the seller? Q10. If the difference between the volumes of
A) 2% water and honey in solution 2 is 270 ml and
the difference between the volumes of milk
B) 1% and water in solution 2 is 600 ml, then what
is the volume of honey in solution 2?(it is
C) 2.66% given that the respective part of honey is
least in the solution 2)
D) 1.33%
A) 90 ml
E) 3.33%
B) 100 ml
========================
C) 300 ml
Directions : Following table shows the ratio
of volumes of Milk, Water and Honey in four D) 30 ml
solutions.
E) None of the above.
Ratio of volumes of Milk, Water and ========================
Solution
Honey
Q11. If the ratio of total volumes of solutions
1 45: W: 2 1 and 4 is 5:3 and the total volume of water
in solutions 1 and 4 together is 570 also the
total volume of honey in solutions 1 and 4
2 30: 10: H together is 60 ml, then what is the total
volume of water in solution A?
3 M: 5: 1
A) 460 ml
4 18: 7: 2 B) 420 ml
C) 430 ml
Q9. The total volume of solution 4 is 540 ml
and ratio of total volume of water in solutions D) 420 ml
3 and 4 together to the total volume of
Honey in solutions 3 and 4 together is 95: E) 427 ml
22. If the ratio of total volume of milk in
solutions 3 and 4 together to the total ========================
volume of water in solutions 3 and 4
Directions [Set of 5 questions]: The data
together is 42: 19, then what is the value of
given below is about 10 employees:
M?
employees A, B, C, D, E and F are working in
A) 10
3
same company X and employees P, Q, R, S, T and employee P if speed of employees B is

and U are working in same company Y. 18 km/h?
Persons A and P lives together, B and Q lives A) 10 km/h

together, C and R lives together, D and S
lives together, E and T lives together and F B) 6 km/h
and U lives together respectively in same
room. C) 0 km/h
Line graph given below shows the difference D) 12 km/h

between distances (km) travelled by them E) None of these
from their room to their office and bar graph
given below shows the difference between ========================
time taken (minutes) to go from their room to
their office. All of the room partners left their Q13. Employee R travels more distance than
room for office at the same time respectively. his room partner to reach his office Y and if
speed of employee C is 24 km/h which 4%
less than the speed of employee R. If
employee C takes less time to reach his office
than his room partner, then what is the
difference between the time taken by them to
reach their office when they switch their
speed with each other?
A) 17.6 minutes
B) 23.2 minutes
C) 28.4 minutes
D) 0 minutes
E) None of these
========================
Q14. Person A and C take more time than

their room partners while person A and R
travels more distance than their room
partners to reach their office. Speed of
person A and P is same while the speed of
Q12. Ratio of distances between the rooms
person C is 10 km/h less than the speed of
of employees A and Q from their respective
his room partner. Distance between company
offices is 12: 7 and speed of employee A is
X and rooms of A and C is 60 km and 32 km
32 km/h which is double of the speed of
respectively. If another person M travels a
employee Q. Employees A and B travel more
distance equal to the distance travelled by A
than from their respective room partners to
to reach his office from room with speed
reach their offices and employees P and Q
equal to the speed of P and person N travels
takes more time to reach their offices than
a distance equal to the distance travelled by
their room partners, then what is the
C to reach his office from room with speed
difference between the speed of employee B
equal to the speeds of R, then find the ratio
4
of time taken by M and N to cover their C) 20 km

journey.
D) 12 km
A) 25: 16
E) 18 km
B) 16: 9
========================
C) 75: 32
Directions : Read the given information in the
D) 75: 49 paragraph carefully and answer the following
questions:
E) None of these
There are two Cab services in Bangalore: Ola
======================== and Uber and there are two types of cabs of
each cab services: Mini and Micro.
Q15. Distance of office of S from his room is
68% of the distance of office of D from his Fare of Ola is calculated as the sum of Base
room and person D takes more time than S Fare, extra charges and taxes while the fare
to reach office from room. Total time taken by of Uber is calculated as the sum of Base fare
S to reach office is what per cent of total time and Taxes.
taken by D to reach office if speed of S is
25% less than the speed of person D? Below given formulas are for calculating the
Assume person D take more time to reach fare for Ola and Uber of both types (Mini and
office than S. Micro):
A) 90(2/3)% Ola:
B) 75(1/3)% Fare of Ola vehicles = Base Fare + 5% GST

on Base Fare + Extra charges + 20% GST on
C) 80(1/6)% extra charges
D) 60(2/3)% Base Fare for Ola Mini = Rs.5 * Total distance
travelled (km) + Rs.20 * Total time taken in
E) None of these (hours)
======================== Base Fare for Ola Micro = Rs.10 * Total
Q16. Speed of persons E, F, T and U is 24 distance travelled (km) + Rs.30 * Total time
km/h, 48 km/h, 16km/h and 40 km/h taken in (hours)
respectively. Persons E and F reach their Extra charges = 25% of Base Fare
office earlier than their respective room
partners and also, they travel less distance to Uber:
reach their office than their room partners.
What is the difference between the sum of Fare of Uber vehicles = Base Fare + 5% GST
distance of home of employees E and T from on Base Fare
their respective offices and the sum of
distance of home of employees F and U from Base Fare for Uber Mini = Rs.10 * Total
their respective offices? distance travelled (km) + Rs.20 * Total time
taken in (hours)
A) 24 km
Base Fare for Uber Micro = Rs.15 * Total
B) 16 km distance travelled (km) + Rs.20 * Total time
taken in (hours)
5
Five persons P, Q, R, S and T of a company what is the total amount of fare paid by
uses any one of these cabs for travelling from person U?
their home to Office and distances of their
home from office is different. Following A) Rs.3650
information is also known:
B) Rs.2950
Person P covers the distance between home
and office on a vehicle that travels with C) Rs.4250
average speed of 20 km/hr and takes total D) Rs.3045
1.5 hours. Ratio of time taken by person P to
that by person Q to reach office from home is E) None of these
5: 4 while the speed of vehicle in which
person Q is travelling is 5 km/hr less than ========================
that of vehicle on which person P travels.
Ratio of distance between home and office for Q19. If person T books a private cab while
persons Q to that for person R is 3: 4 and returning from office to home and charges of
time taken by R to reach office from home is private cab is mentioned below, then how
1 hour. Distance between home and office of much less amount he pays to private than the
person S is 21 km more than the average of amount that he pays when he goes to office
distance between home and office of persons from home by Uber Mini?* Fare for private
P, Q and R while the speed of person S is 15 cab = Base Fare + Taxes* Base Fare for
km/hr which is 5 km/hr less than the speed private cab = 8 * Total distance (km), and
of person T. Time taken by person T to reach Taxes = 25% of Base Fare.
office from home is 3 hours.
A) Rs.72
Q17. What is the difference between Fare
B) Rs.81
amount if person P goes to office from home
by Ola Mini and Fare amount when he returns C) Rs.93
from office to home by Uber Micro?
D) Rs.104
A) Rs.243
E) None of these
B) Rs.261
========================
C) Rs.282
Directions : Study the data carefully and
D) Rs.295 answer the following questions.
E) None of these In a company three kind of coffee such as
café Mocha, Americano and café Cubano like
========================
by the employees. Total number of employees
Q18. A person U starts from Delhi at 8:00 working in the company is 800 and 42.5%
AM and reaches Agra at 6:00 PM on the employees does not like coffee.
same day, From Delhi to point A he books
The number of employees who like café
Ola Micro that runs with speed 25 km/hr and
mocha is 10 less than the number of
from point A to Agra he books Uber Mini that
employees who like café Cubano and number
runs speed 20 km/hr. Ratio of distance
of employees who like all three kind of coffee
between Delhi to point A and distance
is 5% of total employees. The number of
between A to Agra is 5: 4 respectively, then
employees who like café mocha and
americano but not café Cubano is half of the
6
number of employees who like all three kind C) 100

of coffee and number of employees who like
café mocha and café Cubano but not D) 110
americano is 10 less than the number of
employees who like only americano. The E) 120
number of employees who like americano and ========================
café Cubano but not café mocha is 10 more
than the number of employees who like café Directions: Study the data carefully and
mocha and americano but not café Cubano. answer the following questions:
Number of employees who like only café
Mocha is 10 more than the twice of Six students M, N, O, P, Q and R give two
employees who like only americano. tests. In first test, total number of questions
are 100 and in second test, total number of
Q20. The number of employees who like only questions are 200. There are three sections
Americano is what percentage of the number A, B and C in first test and two sections X
of employees who like Café mocha and café and Y in second test. Section A, B and C
Cubano but not americano? consist of 35, 15 and 50 questions
respectively. Section X and Y consist of 150
A) 120% and 50 questions respectively.
B) 110% Below graph shows the number of questions
C) 125% attempted by six students M, N, O, P, Q and R
in first test and 18 times of the percentage of
D) 100% marks obtained by them in first test.
E) None of these
========================
Q21. What is the difference between the

number of employees who like Americano
and café mocha?
A) 50
B) 60
C) 70
D) 80
E) 90
========================
Q22. What is the number of employees who Below graph shows the number of questions
like exactly two kind of coffee? attempted by six students M, N, O, P, Q and R
in second test and five times of percentage of
A) 80 marks obtained by them in second test.
B) 90
7
than and equal to 12 wrong answers then -2

marks for each wrong answer and there are
more than 12 wrong answers then -4 marks
for each wrong answer. In section Y, if there
are upto 2 wrong answers then -2 marks for
each wrong answer, if there are more than 2
and less than equal to 6 wrong answers then
-4 marks for each wrong answer and there
are more than 6 wrong answers then-8 marks
for each wrong answer.
For example
In section X if student attempted 10 wrong

answers, the negative marks is calculated as
follows
Below table shows the number of questions
attempted by M, N, O, P, Q and R in section 4 * (- 1) + 6 * (- 2)
A, section B and section X.
In section Y if student attempted 10 wrong
First test Second test answers, the negative marks is calculated as
follows
Section a Section b Section x 2 * (- 2) + 4 * (- 4) + 4 * (- 8)
Marks percentage in first test = (total marks

M 28 11 65 obtained in section A, B and C) / 360 * 100
N 22 6 110 Marks percentage in second test = (total

marks obtained in section X and Y) / 1000 *
100
O 31 12 118
Q23. What is ratio of total marks obtained by
N to total marks obtained by R in both tests?
P 28 10 60
A) 301: 299
Q 21 7 140
B) 311: 290
R 25 8 110 C) 311: 299
D) 301: 290
Note- There is 4, 8 and 2 marks for each
correct answer and 1, 2 and 0.5 negative E) None of these
marking for every wrong answer in section A,
B and C respectively in first test. ========================
There is 4 and 8 marks for each correct Q24. What are the average marks obtained
answer in section X and Y respectively in by O, P and Q in section A and C together if
second test. In section X, if there are upto 4 number of wrong questions attempted in
wrong answers then -1 mark for each wrong section A and C by O is 18 and 8, by P is 16
answer, if there are more than 4 and less and 6 and by Q is 4 and 8 respectively?
8
A) 96 The following pie chart represents the

percentage-wise distribution of part of a work
B) 97 completed by five boys.
C) 98
D) 99
E) None of these
========================
Q25. Marks obtained by N in section X in

second test is what percentage more or less
than marks obtained by M in same section in
second test if M and N attempted 5 and 10
wrong questions in section Y in second test? The following bar graph represents the time
taken by them to complete their respective
A) 162.50% part of the work.
B) 164.25%
C) 166.66%
D) 168.75%
E) None of these
========================
Q26. Which of the following sequence is

correct for total marks obtained by all six
students in section A and C together in first Aman and Ankita together can complete the
test if M, N, O, P, Q and R attempted 6, 3, 4, work in 200/13 days, with the help of Binita,
2, 1 and 4 wrong questions respectively in they can complete the work in 100/9 days.
section B in first test? Ankita and Sushma together can complete
the work in 50/3 days. Tapan and Anjana
A) O>P>M>Q>R>N together can complete the work in 12 days.
B) M>O>P>Q>R>N Kamini and Madan together can complete the
work in 15 days.
C) P>O>M>Q>R>N
Q27. Aman started the work and Binita and
D) O>M>P>Q>R>N Anjana joined him every third day. In this way
they worked for 9 days and left. Find the time
E) M>Q>P>O>R>N taken by Bhuvan to complete the remaining
part of the work.
========================
A) 32 days
Directions : Study the following information
carefully and answer the questions given B) 36 days
below:
C) 44 days
9
D) 40 days reserved for emergency and can't be booked

by any means.
E) None of these
Note: Male passengers can sit only in the
======================== seats booked for male passengers and vice
versa.
Q28. Time taken by Suman and Kamini to
complete the work together is approximately · Fare for one adult male and one adult
what percent of the time taken by Tapan, female passenger is Rs.120 and Rs.100
Ankita and Sushma to complete the work respectively.
together.
· Fare of one male Kid and female kid is 75%
A) 139% of fare of one male adult and female adult
respectively.
B) 133%
· To calculate the maximum fare the
C) 123% emergency seats are also considered.
D) 143% Average number of male and female
E) 147% passengers in bus on Monday is 15 while
male passengers in the bus on Tuesday is 2
======================== more than the male passengers in the bus on
Monday. Total number of female passengers
Q29. Madan and Bhuvan started the work in bus on Tuesday and Wednesday is same
and left after 10 days. Ankita and Suman which is same as the total number of male
started the remaining part of the work but passengers in the bus on Friday. Total
Ankita left 1 day before the completion of the number of male passengers in bus on
work and Suman worked till the completion Thursday is 2 more than the total number of
of the work. Find the number of days for male passengers in the bus on Friday while
which Ankita worked. total number of female passengers in the bus
on Thursday is 4 less than male passengers
A) 159/17 days in the bus on that day. Total number of
passengers (male + female) in the bus on
B) 163/21 days
Thursday is same as the number of male
C) 165/22 days passengers in the bus on Tuesday. Total
number of female passengers in the bus on
D) 164/19 days Friday is 15 while the average number of
female passengers in the bus on all the five
E) None of these days is 11 and average number of male
passengers in the bus on all the five days is
======================== 14.
Directions : Read the data given below
carefully and answer the questions: Percent of
Average of male
female
Days kid and female kid
A bus runs on five different days from passengers that
passengers
Monday to Friday and it has certain number are kid
of passenger seats. Total seats in the bus is
45 out of which ratio of seats reserved for Monday 3 ----
male and female is 5: 4 respectively. 20% of
male seats and 25% of female seats are
10
A) 0
Tuesday ---- 30%
B) 3
Wednesday 2 ----
C) 5
Thursday 5 75% D) 4
E) can't be determined
Friday ---- ----
========================
Q30. Total amount of fare collected form the
Q33. If average number of female kid
bus on Monday is Rs.3195, then what is the
passengers in the bus on Wednesday and
ratio of male kids to female kids in the bus
Thursday is 3, then what is the average of
on Monday?
total fare collected from the bus on
A) 6: 5 Wednesday and Thursday together?
B) 5: 4 A) Rs.2150
C) 3: 4 B) Rs.2025
D) 1: 1 C) Rs.3275
E) 2: 3 D) Rs.2125
======================== E) Rs.1850
Q31. If out of total passengers in the bus on ========================

Tuesday, approximately 26.67% are kids,
Q34. Out of total passengers in the bus on
then total fare collected from the bus on
Friday, 40% are kids and ratio of male kids
Tuesday is what percent of maximum fare
to female kids is 1: 4. If 'x' more passengers
amount that can be collected from the bus?
boarded to the bus on that day out of which
A) 67.5% one is kid and remaining are adults. Total fare
collected from the bus on Friday is increased
B) 72.5% by Rs.570 from the original fare, then what is
the value of 'x'?
C) 63.5%
A) 5
D) 75%
B) 4
E) 48.5%
C) 1
========================
D) 3
Q32. If total fare collected from the bus on
Wednesday is 41.8% of maximum fare that E) 2
can be collected from the bus, then what is
the difference between male kid passengers ========================
and female kid passengers in the bus on
Wednesday?
carefully and answer the related questions.
11
There are total 8000 people in a locality who Following data is also given:
live in 5 different buildings A(10%), B(25%),
C(15%), D(35%) and remaining in E. All of Number of people in building A who like
them use different brand clothes among four white and blue color are in the ratio 3: 1
types i.e. Code, Nike, Adidas and Levis. Each respectively and 150 people in building A
person uses only one brand clothes. Each like white color which is 100 less than the
person like only one color among four number of people in building C who like
different colors i.e., red, grey, white and blue. same color. Number of people who like white
and blue color in building E are 300 and in
Following equation is used for calculating the building B are 1000. Total 4500 people are
number of persons who like two given colors there in locality who like white and blue color
among people live in all given buildings: and 550 people in building C like blue color.
For red: R = 2X - 2.5Y Q35. If 350 people from building A like grey
color and the ratio of people who like red
For grey: G = 2.5X - 3Y and grey color in building B are in the ratio
2: 3 respectively, then what will be the
Where, X and Y represents the number of difference between total number of Adidas
males and females respectively in each users from building A and B?
building and value of X and Y will be different
for each building. A) 410
Code: Number of users in building B and B) 420
building C are in the ratio 3: 2 respectively.
Total 1780 users are there out of which 250 C) 430
from building E. Users from building D are
twice of users from building C and users in D) 440
building A are 20 less than number of people
who like red color in building D. E) 450
Nike: Users in building B are 6% of total ========================

population of given locality and are 200 Q36. If there are 1200 females in building D,
more than the number of users in building E. then what will be the respective ratio of total
Total number of users are 30 less than total number of Code users in building A and D
number of users of Adidas. Number of users together to the total number of Levis users in
in A and C are in the ratio 5: 4 respectively. building C and E together?
Adidas: Number of users from building C and A) 19: 17
D are in the ratio 9:10 respectively. Number
of users in building B are 50 more than the B) 22: 19
number of people who like grey color from
same building. Users in building A are 80 less C) 28: 23
than the number of females in same building.
270 users are in building E. D) 33: 29
Levis: Total 2270 users are there out of E) 39: 35

which 300 live in building C. Users in
========================
building D are 4 times the number of people
in building C who like white color. Users in Q37. If the ratio of number of Levis users in
building B are 45 more than the number of building A and B are in the ratio 5: 14
people in building E who like red color. respectively, then total number of females in
12
building E are approximately what percent of D) 86

total population of given locality?
E) 92
A) 4%
========================
B) 6%
C) 8% carefully and answer the related questions.
D) 10% The following charts represent the database

of a health camp held in two states during
E) 12% the two different years.
========================
Q38. If males in building C are 200 more

than females in same building and number of
people who like white color in building B are
200 less than the number of people who like
blue color in same building, then number of
people in building C who like grey color are
approximately what percent of number of
people in building B who like white color? (Note: Values given in percentage for state P
for a particular year are out total population
A) 62.5% of state P in that year and values given in
percentage for state Q for a particular year
B) 54.5% are out of total population of state Q in that
year.)
C) 66.5%
Q40. if the number of sick people
D) 58.5% recommended for immediate treatment from
E) 60.5% state P in 2013 is 1344 less than the
previous year, then what is the value of 'C'?
========================
A) 6
Q39. In building E, 66%, 45% and 52%
Code, Nike and Levis users are females and B) 7
total number of male users of Code, Nike and C) 8
Levis in building C are 24 more than total
number of male users of same brands in D) 9
building E. If there are total 800 users of
Code, Nike and Levis in building C, then what E) 10
will be the difference between number of
male users of same brands in building E and ========================
number of female users of same brands in
Q41. If A - H - E = 61132 - 1200F and B +
building C?
G = 2(F - 15), then how many people visited
A) 84 health camp in state P is 2012?
B) 88 A) 81900
C) 90 B) 78750
13
C) 77700 Q44. If 20237 people from state Q in 2012

and 2013 taken together who found sick did
D) 79800 not recommend for immediate treatment,
then what is the difference between number
E) 80850 of sick people recommended for immediate
======================== treatment from state Q in 2012 and number
of people who were not sick according to test
Q42. In 2012 the ratio of number of people report in state P in 2013?
found sick according to test report in state Q
and number of people found sick according A) 40652
to test report but did not recommend for B) 39862
immediate treatment from state P is 1: 1. If
population of state Q is increased by 15% in C) 36572
2013 than previous year and number of
people who visited health camp but did not D) 38142
complete their health check-up test in state Q
in 2013 are 74676 less than the number of E) 41882
people visited health camp from state P in
same year, then what is the average of ========================
number of people from state P who did not Directions : There are five mixtures P, Q, R, S
visit camp in 2012 and 2013 together? and T of three liquids Oil, Petrol and
A) 22350 Kerosene. Pie chart given below shows the
distribution (degree) of total amount of Oil in
B) 23850 those five mixtures and line graph given
below shows the per cent more/less amount
C) 24750 of Petrol and Kerosene with respect to the
amount of Oil in those five mixtures.
D) 25650
E) 26250
========================
Q43. Average of number of people

completed their health check-ups tests from
state Q in 2012 and 2013 taken together is
63063. How many people from state Q
visited health camp but did not complete
their health check-up test in 2013?
A) 15682
B) 10226
C) 13524
D) 11426
E) 14388
========================
14
will be the part of Oil in total amount of

mixture B if '156' litres of mixture S is mixed
with 'X' litres of mixture T to from mixture B?
A) (45/119)
B) (38/117)
C) (37/117)
D) (23/121)
Q45. When mixture P and Q are mixed in the E) None of these

ratio 4: 5 to form a final mixture, then
different between amount of Oil and Petrol in ========================
the final mixture becomes 12 litres, then what
is the total amount of newly formed mixture? Directions : A manufacturing company of
bottles manufactures certain number of
A) 172 litres bottles and out of which some are sold, and
some remains unsold. Out of total sold
B) 151 litres bottles, some are sold at 50% profit and
remaining are sold at 20% profit.
C) 162 litres
Table given below shows the percent of
D) 144 litres unsold bottles out of total manufactured
bottles, Percent of bottles sold at 50% profit
E) None of these out of total bottles sold.
========================
Percent of bottles
Percent of unsold
Q46. In what ratio mixtures P and R must be sold at 50% profit
Years bottles out of total
mixed to form a final mixture so that in the out of total sold
manufactured bottles
final mixture, amount of Oil and Petrol is bottles
same while the amount of Kerosene is 15
litres less than that of Petrol and total 2011 20% 75%
amount of final mixture is 105 litres?
A) 5: 2 2012 16(2/3)% 60%
B) 7: 3
2013 12.5% 50%
C) 6: 5
2014 10% 40%
D) 8: 5
E) 4: 3 Q48. What will be the total number of unsold

bottles manufactured by the manufacturer in
======================== the years 2011 and 2014 together?I: Ratio
of total bottles sold at 50% profit in 2011 to
Q47. When 'X' litres of mixture S is mixed
the total bottles sold at 20% profit in 2014
with '156' litres of mixture T to form mixture
is 4: 3.II: Total unsold bottles in 2011 is 70
A, then ratio of Oil, Petrol and Kerosene in
more than the total unsold bottles in 2014.III:
the mixture A becomes 76: 75: 83, then what
If total unsold bottles in 2011 are assumed
15
to be sold and this increased the number of Total items sold at 50% in 2011 is 15 more
bottles sold at 50% in 2011 by 16(2/3)% than total items sold at 20% in 2013.
and number of bottles sold at 20% in 2011
becomes same as the number of bottles sold A) Quantity I > Quantity II > Quantity III
at 50% in 2014.
B) Quantity I = Quantity II < Quantity III
A) I and II together or II and III together are
sufficient. C) Quantity I = Quantity II = Quantity III
B) III alone is sufficient. D) Quantity I < Quantity II < Quantity III
C) I and II together or I and III together are E) Quantity I < Quantity II = Quantity III
sufficient. ========================
D) Any two of them together are sufficient. Directions : Read the data given below
E) I and III together or II and III together are carefully and answer the questions:
sufficient. Shrinkage can be defined as the time for
======================== which people are paid but not available to
handle calls or perform their task.
Q49. Quantity I: What will be the difference
between total items sold at 50% profit and Shrinkage percent = (1 - Total staffed
20% profit by the manufacturing company in hours/Total scheduled hours)
the year 2012 if items sold at 20% profit are Total Staffed hours = (Total answered calls *
80 more than the total unsold items.Quantity AHT) + Available time in minutes +
II: What will be the difference between sold Productive auxiliary time (AUX) in minutes
and unsold items manufactured in the year
2013 if unsold items in 2013 is 100. Total scheduled hours = Total Agents * Total
working hours in a day
A) Quantity I > Quantity II
Average Handle time (AHT) = Average Talk
B) Quantity I > Quantity II Time (ATT) + Average Call Work (ACW)
C) Quantity I < Quantity II Average-Talk-Time (ATT) is the average
D) Quantity I < Quantity II amount of time agents talk to customers.
E) Quantity I = Quantity II or relation can't be After-Call-Work (ACW) is the average amount

determined. of time an agent takes to wrap-up a call.
======================== Available time is defined as the amount of

time an agent is ready to take a call.
Q50. What is the total number of items sold
by the manufacturer in the years 2011 and There are total five BPO centres A, B, C, D
2013 together if ratio of unsold items in and E and table given below shows the Total
2011 to that in 2013 is 6: 5?Quantity I: Total Agents, Total working hours in a day,
number of sold items at 20% profit in 2013 Average Talk Time and Average Call Work in
is 115 more than that in 2011.Quantity II: those BPO centres:
Total number of unsold items in the years
2011 and 2013 together is 110.Quantity III:
16
that BPO is 8: 13. If value of Q is 65, then

Total Working ATT ACW
BPO what is the approximate shrinkage percent of
Agents hour (minutes) (minutes)
BPO B?
A 75 8 4 1.5 A) 7.5%
B) 8%
B 60 9 5 2
C) 5%
C 80 7.5 3 1 D) 12.5%
D 120 6 6 2 E) 10%
========================
E 90 8.5 4.5 1.5
Q53. If difference between available and AUX
time for BPO C is 50 minutes and its
Some other information is also known:
shrinkage percent is 35%, then what will be
Available time for BPO A is (P) minutes while the shrinkage percent of BPO B if available
AUX time of the same BPO is 30 minutes. and AUX time for BPO B is 90 minutes and
Total answered calls by BPO B is (Q) which is 46 minutes respectively?
5 less than total answered calls by BPO C.
A) 15%
Average of total answered calls by BPO D
and E is 70 while answered calls by BPO E is B) 10%
40 more than that by BPO D. Ratio of
Available time to AUX time of the BPO E is 5: C) 20%
1 while the same ratio of BPO C is 11: 6.
Available time and AUX time of BPO D is 175 D) 25%
minutes and 91 minutes respectively.
E) 5%
Q51. If average of total answered calls by
BPOs A, D and E together is 73.33 ========================
approximately and shrinkage percent of BPO Q54. If numerical value of shrinkage percent
A is 10%, then what is the value of 'P'? of BPO A and C is 10 and 17.5 respectively
A) 90 more than that of BPO D and total answered
calls by BPO A is 70 and AUX time of BPO C
B) 65 is 60 minutes, then what is the ratio of 'P' to
'Q'?
C) 75
A) 16: 15
D) 80
B) 14: 11
E) 70
C) 13: 12
========================
D) 16: 13
Q52. Available time for BPO B is 22 minutes
more than the AUX time for that BPO and E) 18: 13
ratio of numerical value of available time to
numerical value of total answered calls by ========================
17
Q55. If shrinkage percent of BPO E is 20%,

then numerical value of AUX time for that
BPO is approximately what percent of
number of total answered calls by the BPO?
A) 16.67%
B) 13.33%
C) 23.33%
D) 8.88%
E) 26.67%
========================
Q56. If profit earned by C and E after first '8
Directions : Study the following information + n' months partnership are in the ratio 768:
carefully and answer the related questions. 677 respectively and the investment of E
Following first pie-chart represents the becomes more than investment of C after first
percentage distribution of amount invested '8 + n' months, then what is the value of 'n'?
by five friends A, B, C, D and E respectively A) 4
and second pie-chart represents percentage
distribution of amount added or withdrew by B) 8
all five friends after first 8 months.
C) 6
Total initial investment = Rs.50000
D) 10
E) Cannot be determined.
========================
Q57. If only B and D added amount after first

8 months then after next four months B and
D left and A, C and E added Rs.2000, then
what is the ratio of profit earned by A, B, C, D
and E after 18 months partnership?
A) 340: 195: 240: 244: 214
B) 240: 214: 340: 244: 195
C) 340: 214: 240: 244: 195

Sum of positive values of amount
D) 240: 244: 340: 195: 244
added/withdrew by all given five friends =
Rs.15000 E) 340: 214: 240: 195: 244
========================
18
Q58. If only D and E added amount after first

B 3: 2
8 months then after first '8 + ___' months, D
invested Rs.4000 more and E invested Rs.___
more. Only after next ____ months both D C 4: 3
and E withdrew Rs.8000 and the profit
earned by them after next 2 months D 7: 5
partnership is in the ratio ___
respectively.Which of the following option is
suitable to fill the blanks in same order? E 5: 1
A) 6, 2000, 8, 201: 169 Q59. What is the average of length of all

B) 4, 4000, 10, 213: 277 given rectangles?
C) 12, 2000, 2, 109: 129 A) 71.6 cm
D) 8, 4000, 6, 211: 169 B) 73.6 cm
E) None of these C) 75.6 cm
======================== D) 77.6 cm
Directions : Study the following information E) 79.6 cm

carefully and answer the related questions. ========================
Following pie-chart represents the Q60. What is the respective ratio of length of
percentage break-ups of perimeter of five rectangles C and E together to the breadth of
rectangles and table represents the rectangles B and A together?
respective ratio of their length and breadth.
A) 23: 15
Total perimeter = 1200 cm
B) 18: 11
C) 38: 21
D) 56: 33
E) None of these
========================
Q61. What is the difference between the

perimeter of rectangle B to the sum of
breadths of all given rectangles?
A) 72 cm
Rectangle Length: Breadth B) 78 cm
C) 64 cm
A 5: 4
D) 88 cm
19
E) 84 cm ========================
======================== Q63. If speed of boat U in still water is 600%

of the speed of stream, then find that the
Directions : The data given below indicates speed of boat U in downstream is how much
the speeds of 6 boats and corresponding times the speed of boat U in upstream?
speeds of stream (some of the data is
deliberately left blank). A) 0.4
B) 1.4
C) 4
D) 1/4
E) None of these
========================
Q64. Find the difference between time taken

by boat P and boat R to cover a distance of
252 km in downstream.
Boats Speed of stream (in km/h)
A) 6 h
P 3
B) 7 h
Q ---- C) 8 h
D) 7.7 h
R 2
E) None of these
S ---- ========================
Q65. Boat T goes from point A to point B

T 6
and comes back to the original point A in 10
hours. What is distance between the points A
U ---- and B?
A) 126 km
Q62. If boat Q travels 2 km upstream in the
same time as it travels 3 km downstream, B) 138 km
than find the speed of Stream?
C) 108 km
A) 3 km/h
D) 144 km
B) 4 km/h
E) None of these
C) 5 km/h
========================
D) 6 km/h
Directions : First radar graph given below
E) None of these shows the total income (in thousand) of two
20
persons A and B in three different months Q67. Amount spent on Food by person B in
January, February and March respectively. February and March together is Rs.35000
which is 80% of his total expenditure on
Second radar graph shows the total savings Rent taking both the given months together,
as a per cent of income for A and B in those then what per cent of his total income in both
three different months. the months together he spent on shopping?
Total expenditure = Expenses on Food + A) 12.5%
Expenses on Rent + Expenses on Shopping.
B) 7.5%
C) 5.25%
D) 8.75%
E) None of these
========================
Directions : Study the data carefully and

answer the following questions:
Mohit started two businesses. He started

catering business with his brothers Sunil and
Punit and another Garment business with his
friends Rohan and Prakash.
In catering business, the initial investment by

Mohit, Sunil and Punit in the ratio of 4: 4: 3.
After 6 months, their cousin Shyamaly joined
the business with 2.5 times Mohit's initial
investment and at that time Mohit, Sunil and
Punit again invested into the business in the
Q66. Total amount of savings of person A in ratio of 4: 3: 3. Further 6 more months,
all the three months together is P% of his Mohit, Sunil, Punit and Shyamaly again
total income and total amount savings of invested into the business in the ratio of 6: 4:
person B in all the three months together is 5: 5. After 2 years from start of the business,
Q% of his total income, then what is the ratio Mohit and Punit withdraw 50% of mohit's
of 'P' to 'Q'? initial investment and 80% of Punit's
previous investment and Sunil invest 3.5
A) 217: 248 times of his initial investment and Shyamaly
invest 1.6 times of her initial investment.
B) 2312: 3457 Further 6 more months, Mohit, Sunil, Punit
and Shyamaly again invested in the ratio of 2:
C) 117: 448
4: 1: 4.
D) 2117: 2448
In Garment business, Mohit, Rohan and
E) None of these Prakash initially invested in the ratio of 8: 7:
5 and after 6 months Megha also joined the
======================== business with half of the initial investment of
Mohit. After 1 year start of the business,
21
Mohit, Rohan, Prakash and Megha invested in E) statement V

the ratio of 3: 2: 2: 3 and further 6 more
months, Mohit, Prakash and Mega again ========================
invested in the ratio of 5: 3: 2. After 2.5
years start of the business, Mohit withdraw Q69. In Garment business, initial investment
20% of his previous investment, Rohan of Megha is equal to the Rohan's investment
withdraw 25% of investment made by him after 1 years and Prakash's investment after
after 1 year start of the business, Megha 18 months is Rs. 18000 which is 50% more
withdraw 25% of her initial investment and than Mohit's investment after 1 year. What is
Prakash withdraw half of his investment made the weightage of capital invested by Megha
after 1 year start of the business. and Rohan in the garment business? (If the
weightage of capital investment means initial
They invested the whole amount for three investment +/- money added or withdrawal
years in both the businesses and the profit and so on till the end of the business)
earned in the business is proportional to the
investment and the period of investment. A) Rs. 48000
Q68. What is the Mohit's share of profit from B) Rs. 50000

both the business if the total profit in C) Rs. 40000
catering business is Rs. 854000 and in
garment business is Rs. 909000? D) Rs. 52000
I. After 1 year, Prakash investment is 50% of
Mohit's initial investment in garment business E) None of these
which is four times of Mohit's initial
investment in catering business. ========================
II. After 6 months, Sunil's investment is Rs.
Q70. In catering business, Sunil's investment
6000 which is 3 times of Mohit's investment
after 6 months is 40% less than Shyamaly's
after 2.5 years in catering business.
initial investment which is 50% of Punit's
III. After 1 year, Punit's investment in the
investment after 1 year. If Sunil investment
catering business is equal to Prakash's initial
after 2.5 years is Rs. 4000 which is equal to
investment in the garment business.
his initial investment and Mohit gets 5% of
IV. Megha's initial investment in garment
total profit for managing the catering
business is 33(1/3)% less than of her
business which Rs. 254400 less than the
investment after 18 months and Prakash's
share of Punit's profit, then what is the total
initial investment is Rs. 7000 more than the
profit at the end of three years?
initial investment of Punit.
V. Prakash initial investment is 33(1/3)% of A) Rs. 2114000
Mohit's investment after 18 months. Rohan's
initial investment in the garment business is B) Rs. 2024000
Rs. 10000 more than the Sunil's initial
investment in catering business. C) Rs. 368600
Which of the above statement/statements
is/are redundant to answer the question? D) Rs. 2284000
A) Either IV or V E) None of these
B) Either I or IV ========================
C) statement IV Q71. Initial investment by Mohit in catering

business is Rs. T which is equal to the
D) statement IV and statement V Shyamaly's investment after 2.5 years from
22
the start of business and Punit's investment The unit of Power is Megawatt(MW) and it is
after 6 months, is Rs. (T + 2000). After 1 an instantaneous quantity. The amount of
year, Sunil's investment is Rs. 4T and electricity produced = Power*Time
difference between Mohit's and Punit's initial
investment is Rs. 1000. Find the ratio of For example, if a unit has capacity 700 MW
profit of Mohit, Sunil, Punit and Shyamaly. and it works at 60% efficiency for 3 hours,
electricity produced = (60/100)*700*3 =
A) 158: 150: 97: 166 1260 MWh(Megawatt hour)
B) 150: 158: 97: 166 On a particular day, a number of units are

operated for some time at a particular level
C) 158: 150: 197: 166 and shut down after working through all
levels. The following is the information about
D) 158: 150: 97: 66 4 different levels of efficiency that a unit
E) None of these works at and the time period for which it can
work at that particular level before it has to
======================== be brought down one level or stopped
completely. So, from the table, if a unit is
Directions : In a power plant, there are 3 working at level 4 for 2 hr, it will be brought
stages. A stage consists of some combination down to level 3 and after working at level 3
of units that produce power and have some for 4 hours it will be brought down to level 2
maximum capacity. The efficiency of a unit(or and so on or it may be stopped at any
a stage) is the percentage of maximum position.
capacity it actually produces. This table shows
the number of units of each type of maximum Efficiency Range Working time
power capacity in a stage.
Level 1 40% - 50% 8

Number Number Number
Overall
of 200 of 250 of 500
capacity
MW units MW units MW units Level 2 55% - 65% 4
Stage I 3 1850 Level 3 60% - 80% 4
Stage Level 4 80% - 90% 2

5 2
II
At every level, there are 3 modes: Minimum

Stage mode, Normal mode and Maximum mode. For
2 2000
III example, if a unit is working at level 2 in max
mode, its efficiency = 65% and if a unit is
Overall 13 working at level 3 in normal mode, its
efficiency = (60 + 80)/2 = 70%. If a unit is
started in any mode, it will remain in the
The overall capacity of stage II is (60/137)
same mode throughout the levels.
part of the overall capacity of the power plant
and the ratio of the total capacity of 500 MW Q72. What is the difference between the total
units in all stages to the total capacity of 250 number of 250 MW units and total number of
MW units in Stage I is 8:3. 500 MW units in the power plant?
A) 5
23
B) 8 Q75. If, in stage III, all 500 MW units were

shut down and all 200 MW units were started
C) 3 in level 3 in minimum mode and after
completing level 2, some of them were shut
D) 10 down and all 250 MW units were started in
E) 2 level 3 in normal mode, the electricity
produced was 10280 MWh, how many 200
======================== MW units were shut down after level 2?
Q73. If in Stage I, all 200 MW units were A) 1

started in level 3 in max mode, all 250 MW
units were started in level 2 in normal mode B) 2
and all 500 MW units were started in level 4 C) 3
in max mode, what will be the electricity
produced by stage I if the units are not D) 4
stopped in between?
E) 5
A) 14900 MWh
========================
B) 12200 MWh
Directions : There are five boats P, Q, R, S
C) 12000 MWh and T and radar graph given below shows the
speed of stream for those five boats as a per
D) 12400 MWh cent of their speed in still water.
E) 16180 MWh
========================
Q74. If in Stage II, all 200 MW were started

in level 4 in min mode, all 250 MW units
were started in level 4 in normal mode but
stopped after completing level 3 and all 500
MW units were started in level 2 in normal
mode, the electricity produced will be what
percentage of the electricity produced had all
units in stage II be working at 100%
efficiency continuously for 10 hours?
A) 48(1/3)%
Line graph given below shows the upstream
B) 66(1/3)% distance travelled by those five boat and time
taken by them to cover its upstream
C) 58(1/3)% distances.
D) 62(1/3)%
E) 72(2/3)%
========================
24
========================
Q78. Boat Q goes 'x' km in upstream and

comes back to the original point in time 'a'
while boat S goes 'y' km in upstream and
comes back to the original point in time 'b'.
Ratio of x: y is 3: 4 and value of 'a' is 1.8
more than the value of 'b', then what is the
distance travelled by boat S in downstream in
(a + b) hours?
Q76. If boat P goes certain distance in A) 328 km

upstream and comes back to the same point B) 192 km
in total of 3.52 hours when travels with
original speed. When it goes same distance in C) 420 km
upstream and comes back to the same point
with its changed speed in still water, it takes D) 280 km
43.2 minutes less, then by what per cent
boat P changes its speed in still water and E) None of these
assume speed of stream remains same?
========================
A) 15%
Directions : There are five boats A, B, C, D
B) 30% and E that are travelling in different rivers.
Upstream and downstream distance travelled
C) 20% by all the five boat are same.
D) 25% Bar graph given below shows the difference

between upstream and downstream speed
E) None of these (km/h) and difference between time taken
(minutes) to go upstream and comes back to
======================== the same point by those five boats.
Q77. Boats R and T start from two different
points A and B respectively on a river and
speed of stream for boat T also becomes 5
km/h and both the boats meet after travelling
for 48 minutes towards each other. If both
the boats start from point A and start
travelling in upstream, then how far is the
boat T from point B when boat R reaches
there? Speed of the stream remains 5 km/h.
A) 11.25 km
B) 8.75 km
C) 12.5 km Line graph given below shows the one-way

distance (km) travelled by those five boats.
D) 15 km
E) None of these
25
Q81. In a different river, time taken by boat E

to go 45 km upstream is 1.8 hours and there
are two points P and Q on the same river and
the river is flowing from point Q to P. Two
boats D and E start simultaneously from
points P and Q respectively in upstream and
after travelling for 't' hours the distance
between both the boats becomes 30 km
before meeting, then after how much more
time the distance between them again
becomes 30 km if the distance between P
and Q is 80 km?
Q79. Total time taken by boat B to go 'D - A) 2.4 hours
20' km in upstream and come back 'D + 4'
km in downstream is 1.6 hours. If upstream B) 1.2 hours
and downstream speeds of the boat B are
decreased by 32% and 12.5% respectively, C) 2 hours
then what is the time taken by boat B to
cover total distance of '2D - 16' km in still D) 3 hours
water with changed speed? E) None of these
A) 2.4 hours ========================
B) 2 hours Q82. Total time taken by boat B to go 'X' km
C) 1.6 hours in upstream and comes back to 'X - 12' km in
downstream is same as the total time taken
D) 1.8 hours by boat D to go 'X - 15' km in upstream and
comes back to 'X' km in downstream and
E) None of these which is equal to 't', then total upstream
distance travelled by boat C in 't - 0.3' hours
======================== is what per cent of value of 'X'?
Q80. Ratio of total time taken by boat C to A) 60%
go 'D' km upstream and come back 'D - 15'
km in downstream to the total time taken to B) 50%
cover the same distance when the stream
reverses its direction is 23: 22, then what is C) 80%
the total time by boat C in its journey when
the stream is flowing in its original direction? D) 72%
A) 4 hours 15 minutes E) None of these
B) 2 hours 30 minutes ========================
C) 3 hours 25 minutes Directions : The line graph given below gives

the information about salary of five different
D) 3 hours 50 minutes persons.
E) None of these
========================
26
C) Rs. 2840
D) Rs. 1880
E) None of the above
========================
Q84. On food, what is the total expenditures

of all the five persons together?
A) Rs. 41248
The table given below gives the partial
information about their expenditures on food, B) Rs. 39568
travelling, petrol and others. (some values are
in percentage and some value are in Rupees) C) Rs. 36464
D) Rs. 38848
Food Travelling Petrol Others
E) None of the above
A 18% ----- --- Rs. 8640 ========================
Q85. The expenditures of B and C together

B Rs. 10240 ------- 16% 28%
on Food is how much less than the
expenditures of D and E together on
C ------ 24% Rs. 10368 32% Travelling?
A) Rs. 3248
D ----- Rs. 7200 ------- 22%
B) Rs. 3360
E 24% -------- Rs. 3072 ------
C) Rs. 3248
Each of them saves 20% of their salary D) Rs. 4229
For the person A, the ratio of the E) None of the above

expenditures on Travelling to that on Petrol
was 16: 13 ========================
Person D spends 50% more on food Directions : Bar graph given below shows the
compare to that on Petrol. profit per cent earned and marked up per
cent over cost price by shopkeeper P on five
For the person E, the expenditures on different items A, B, C, D and E respectively.
Travelling is Rs. 1024 more than that on
Others.
Q83. For the person A, the expenditures on

Travelling was how much more than that on
Petrol?
A) Rs. 2160
B) Rs. 4320
27
Q87. Shopkeeper sold two units of item D

and overall, he earns 24% profit, he sold first
unit at mentioned selling price and second
unit at new selling price. If on second unit,
ratio of numerical value of marked up per
cent to that of discount per cent is 3: 1, then
marked price of first unit is what per cent of
that of second unit where all the per cent
values are integer?
A) 81.25%
B) 76.25%
Pie chart given below shows the distribution
(degree) of selling price of the five different C) 88.25%
items and sum of the pie chart is Rs.12000.
D) 72.25%
E) None of these
========================
Q88. If the shopkeeper sold two items E and

F together and overall, he earns 22% profit.
If he marked up the price of item F by 40%
above its cost price and allows 20%
discount, then what is the ratio of total profit
amount earned after selling both the items to
the total marked up amount of both the items
together?
A) 64: 145
Q86. A customer went to shopkeeper P to B) 8: 15
purchase item C, while billing the shopkeeper
by mistake interchanges the marked up per C) 48: 125
cent and discount per cent, then new amount
paid by the customer to purchase that item is D) 88: 185
approximately what per cent of the actual
E) None of these
cost price of the item?
========================
A) 120%
Directions : The following bar graph shows
B) 75%
amount invested by five persons either on SI
C) 80% or on CI.
D) 65%
E) 135%
========================
28
B) 2:3
C) 3:1
D) 2:1
E) None of these
========================
Q91. Prakash invested 2/3rd of his amount

on Compound Interest for two years and rest
The line graph represents the rate percent at on SI for 5 years. Find the total interest
which they invested the amount. earned by him.
A) Rs.4498
B) Rs.6858
C) Rs.8424
D) Rs.10944
E) None of these
========================
Q89. If Ramesh invested his amount on Directions : The following graph shows
simple interest for ten years and Suresh number of hours required for 6 different
invested his amount on compound interest persons individually to complete a task.
for three years, what will be the difference
between total amount collected by Ramesh
and Suresh after respective time period?
A) Rs.24768.08
B) Rs.28645.02
C) Rs.29223.04
D) Rs.26442.06
E) None of these
Q92. Abhishek and Satish started the task
======================== and worked for certain number of hours
together after which Vinay joined them.
Q90. Find the respective ratio of the interest Abhishek, Satish and Vinay worked together
earned by Dinesh and Naresh, if Dinesh for 3 hours less than the time taken by
invested his amount on simple interest for 8 Abhishek and Satish together. Now, Satish
years and Naresh invested his amount on left the task midway and the remaining task
simple interest for 5 years. was completed by Abhishek and Vinay in 2
hours 10 minutes. What is the total number
A) 3:2 of hours taken to complete the task?
29
A) 8(1/3) hrs Train A of length 250 meters can cross

another train B running in opposite direction
B) 9(1/6) hrs. with speed 30 km/h in 19.2 seconds. Speed
of train A is 'A' km/h and train B can cross a
C) 7(2/3) hrs. platform of similar length in 36 seconds.
D) 4(3/5) hrs. Ramesh is running inside train A in same
direction as that of train, can cross that
E) Other than above platform in 10.8 seconds. In a 400 meters
race Ramesh gives Mahesh a start of 50
======================== meters who is travelling with speed 4 km/h
and still beat him by 'B' meters. Ramesh got
Q93. Neha started the task and after 1 hour Rs.25000 prize amount for winning the race
Shruti joined Neha and after another hour and 'C'% of it he used to purchase a cycle
Nikita joined them. If after one hour and 30 whose price depreciates every year 10% and
minutes Vinay and Satish joined them and all remaining he invested in a scheme that offers
five worked for another 2 hours, What 20% simple rate of interest. After 2 years
percent of task did they complete? from present, his present worth(cycle +
investment) will increase by Rs.4100. From
A) 30%
the interest amount received after 2 years
B) 45% started a business along with Mahesh whose
initial capital is Rs.1200 more than that of
C) 40% Ramesh and after 5 months Ramesh and
Mahesh withdraw Rs.1000 and Rs.1200
D) 60% respectively from their capital. Total profit
from the business after a year is Rs.1320 out
E) Other than the above of which profit amount of Mahesh is 'D'.
======================== Mahesh went to a village fair where he has to
hit a target and he will be given three
Q94. If Vinay solved some number questions chances to do this and as soon as he hit the
for 2 hours, Nikita solved some number of target, he will be declared winner. Probability
questions for 7 hours and Neha solved some that he hit the target in an attempt is (1/5)
number of questions for 16.5 hours and the and probability that he won that game is 'E'.
total questions solved by them were 253, Now, after finishing his game Mahesh started
then how many questions can Abhishek, running towards his home with his usual
Shruti and Satish solve in 216 minutes? speed and takes 45 minutes to reach home
which is 'F' km far from the village fair.
A) 253.5
Q95. Value of A = ?
B) 245.2
A) 30
C) 225.1
B) 50
D) 232.3
C) 60
E) 240.6
D) 45
========================
E) None of these
Directions : Read the data given in the
paragraph carefully and answer the questions ========================
given below:
30
Q96. Value of B = ? Q100. Value of F = ?
A) 20 A) 2
B) 30 B) 6
C) 50 C) 4
D) 40 D) 5
E) None of these E) None of these
======================== ========================
Q97. Value of C = ? Explanation/Solution:-
A) 75 Probability of drawing a blue colour ball from

bag A = 1 - (16 + 24 + 8 + 32)/100 = 1/5
B) 60
Total number of blue colour balls in bag A =
C) 50 10
D) 40 Total number of balls of all colour in bag A =
10 * (5/1) = 50
E) None of these
Probability of drawing a yellow colour balls
======================== from bag B = 1 - (15 + 20 + 30 + 10)/100
Q98. Value of D = ? = 1/4
A) 720 Total number of yellow colour balls in bag B

= 10
B) 1080
Total number of balls of all colour in bag B =
C) 900 10 * (4/1) = 40
D) 800 Total number of black colour balls in bag A =

16% of 50 = 8
E) None of these
Total number of black colour balls in bag B =
======================== 15% of 40 = 6
Q99. Value of E = ? Similarly, we can calculate other values ........
A) 12/35
Colour/Bags Bag A Bag B
B) 56/125
Black 16% of 50 = 8 15% of 40 = 6
C) 61/125
D) 8/25 Red 24% of 50 = 12 20% of 40 = 8
E) None of these
White 8% of 50 = 4 30% of 40 = 12
========================
31
Total blue balls in bag B after transferring =

Blue 10 40 * (1/10) = 4
4+4=8
Yellow 50 * (8/25) =16 10 Total balls in bag B = 40 + 10 = 50
Probability of selecting a blue ball from bag B

Total 50 40 = 8/50 = 4/25
Q1 - B Explanation/Solution:-
Explanation/Solution:- Q4 - B
Probability of drawing two black colour balls Explanation/Solution:-

from bag A = 8C2/50C2 C.P of 1 kg of Mixed Wheat
= (8 * 7)/(50 * 49) = 4/175 = (4*40 + 6*50)/(4 + 6) = Rs 46
Probability of drawing two blue colour balls S.P of 1 kg of Mixed Wheat
from bag B = 4C2/40C2
= 60(1 - 10/100)(1 - 10/100)
= (4 * 3)/(40 * 39) = 1/130
= Rs 48.60
Required ratio = (4/175): (1/130) = 104: 35
Profit% = ((48.60 - 46)/46)*100 = 130/23%
Q2 - B
∴ x/23 = 130/23
Explanation/Solution:-
=> x = 130
Probability of drawing one red ball from bag
B = 8/40 = 1/5 C.P of 1 kg of Mixed Rice = (1*100 + 1*60 +
2*40)/(1 + 1 + 2) = Rs 60
Probability of drawing one yellow ball from
bag A = 16/50 = 8/25 Let it be Marked-up by z%
Required probability = (1/5) * (8/25) = Given, z = y/9
8/125
M.P of 1 kg of Mixed Rice = 40(1 + z/100)
Q3 - D
S.P of 1 kg of Mixed Rice = 40(1 + z/100)(1
Explanation/Solution:- - 10/100)(1 - 25/100)
Remaining balls in bag A = 50 - 10 = 40 = 27(1 + z/100)
Probability of selecting a blue ball from bag A For no profit or loss,
= 3/20
S.P = C.P
So, after transferring number of blue balls in
bag A = 40 * (3/20) = 6 => 27(1 + z/100) = 60
Total blue balls transferred from bag A to bag => 1 + z/100 = 60/27
B = (10 - 6) = 4
=> z/100 = 11/9
=> z = 1100/9
32
∴ y/9 = 1100/9 S.P of Mixture = (b + 320)*(1 - 20/100)*(1 -

30/100) = (56/100)*(b + 320)
=> y = 1100
Profit% = (((56/100)(b + 320) - b)/b)*100 =
∴ (y+200)/x = 1300/130 = 10 20
Q5 - A => ((56/100)*(b + 320))/b - 1 = 1/5
Explanation/Solution:- => ((56/100)*(b + 320))/b = 6/5
Let the C.P of the mixture be Rs 'a' kg. => 6b = 5*(56/100)*(b + 320)
M.P of Mixture = a(1 + 25/100) => 6b = 14b/5 + 896
S.P of Mixture = (125a/100)*(1 - 20/100)*(1 => 30b = 14b + 4480
- 10/100)
=> 16b = 4480
= (125a/100)*(80/100)*(90/100)
=> b = 280
= 9a/10
By Alligation Method:
Given, a - 9a/10 = 26
=> a/10 = 26
=> a = Rs 260
Given, (y - 280)/80 = 3/2
=> 2y - 560 = 240
=> y = 400
=> (y - 10x)/13 = 390/13 = 30

The ratio in which Normal and Good quality
Apples are mixed = 40:10 = 4:1 Q6 - B
∴x=1 Explanation/Solution:-
Now Normal quality Oranges are mixed with a C.P of Mixed Wheat = (40 + 50 + 60)/3 = Rs
new type of Orange with price Rs 'y'/kg. 50/kg
Let the C.P of the mixture be Rs 'b'/kg. C.P of Mixed Rice = (40*2 + 60*2 + 100)/(2
+ 2 + 1) = Rs 60/kg
M.P of the mixture = (b + 320)
33
M.P of Mixed Wheat = 50(1 + 50/100) = Rs

75/kg
M.P of Mixed Rice = 60(1 + 50/100) = Rs

90/kg
S.P of Mixed Wheat = 75(1 - 10/100) = Rs

67.5/kg
S.P of Mixed Rice = 90(1 - 10/100) = Rs

81/kg
Required difference = 81 - 67.5 = Rs 13.5

Ratio = 45:15 = 3:1
C.P of Mixed Apples = (250*2 + 300 +
400)/(2 + 1 + 1) = Rs 300/kg ∴x=3
C.P of Mixed Oranges = (200 + 250 + Let the C.P of Mixed Apples be Rs 'b'/kg
300)/3 = Rs 250/kg
M.P of Mixed Apples = b(1 + 40/100) =
M.P of Mixed Apples = 300(1 + 40/100) = 7b/5
Rs 420/kg
S.P of Mixed Apples = (7b/5)(1 - 10/100) =
M.P of Mixed Oranges = 250(1 + 40/100) = 63b/50
Rs 350/kg
Given, 63b/50 = 441
S.P of Mixed Apples = 420(1 - 20/100) = Rs
336/kg => b/50 = 7
S.P of Mixed Oranges = 350(1 - 5/100) = Rs => b = 350

332.5/kg
Required difference = 336 - 332.5 = Rs 3.5
Required Ratio = 13.5:3.5 = 27:7
Q7 - E
Let the C.P of the Mixed Rice be Rs 'a'/kg
M.P of mixed rice = (a + 45)
S.P of Mixed Rice = (a + 45)(1 - 25/100)(1 -

20/100) = (3/5)*(a + 45)
Ratio = 50:50 = 1:1
Given, Profit = (3/5)*(a + 45) - a = 5
∴y=1
=> a = 55
∴ 2x2 - 3y2 = 2(3)2 - 3(1)2 = 15
Q8 - D
34
Explanation/Solution:- (48 M +360) /(380) =42/19
C.P of Mixed Apples 48M +360 =840
= (250*3 + 300*2)/(3 + 2) 48M =480
= Rs 270/kg M=10
S.P of Mixed Apples Q10 - D
= 400(1 - 5/100)(1 - 10/100)(1 - 20/100) Explanation/Solution:-
= Rs 273.60/kg Let the volumes of milk, water and honey in

solution 2 be 30a ml,10a ml and Ha ml
Profit% = ((273.60 - 270)/270)*100 = respectively.
1.33%
Total volume of solution 2 = (40 + H) a ml
(10a) - (Ha) =270
Q9 - A
(10 -H) a =270------------------------(1)
and,
Let the total volumes of Milk, Water and
Honey in solution 4 be 18V ml,7V ml and 2V 30a -10a =600
ml respectively.
20a=600
27V =540
a = 30
V=20 ml
Substituting in equation 1 we get
Volumes of Milk, Water and Honey in solution
4 are 360 ml,140 ml and 40 ml respectively. 10-H =9
Let the total volumes of Milk, Water and H=1

Honey in solution 3 be Ma ml, 5a ml and 1a
ml respectively. Volume of honey =30 ml
But we have given (5a +140) / (1a +40) = Q11 - C

95/22 Explanation/Solution:-
110a + 3080 = 95a + 3800 Let the total volume of solution 1 be 5V and
15a =720 the total volume of solution 4 be 3V
a = 720/15 = 48 ml Total volume of Water in solutions 1 and 4

together = ((W/ (47 +W)) x 5V) + ((7/ (27)
Volumes of Milk, Water and Honey in solution x3V)
3 are 48M ml, 240 ml and 48 ml
respectively. = ((5W/ (47+ W)) +(7/9)) x V
We have given that (48M + 360)/ (240 = (52W +329) x (V/(9x (47 + W)))
+140) =42/19
35
(52W +329) x (V/ (9x (47 + W))) = 570 ------- [(12x - 12)/P] - [12x/32] = (30/60) .......... (1)
(1)
[7x/16] - [(7x + 2)/18] = (5/60)
Total volume of Honey in solutions 1 and 4
together = ((2/ (47 +W)) x 5V) + ((2/ (27) [7x/8] - [(7x + 2)/9] = (1/6)
x3V)
(63x - 56x - 16) = 12
= ((10/ (47+ W)) +(2/9)) x V
7x = 28
= ((2W +184) x (V/ (9x (47 + W)))
x=4
((2W +184) x (V/ (9x (47 + W))) = 60
....................(2) From equation (1)-
Dividing (1) by 2 we get [(48 - 12)/P] - [48/32] = (30/60)
(52W + 329)/ (2W + 184 ) =19/2 36/P = 0.5 + 1.5 = 2
104W +658 =38W + 3496 P = 18 km/h
66W = 2838 Required difference = 18 - 18 = 0
W = 43 Q13 - B
Substituting in equation 2, Explanation/Solution:-
(86 +184) x(V/90) =60 x9 Let distance travelled by C = D km
3V =60 x 9 Distance travelled by R = (D + 8) km
V =180 Speed of C = 24 km/h
Required volume = (43/90) x900 =430 ml Speed of R = 24 * (100/96) = 25 km/h
Explanation/Solution:- According to question-
Q12 - C [(D + 8)/25] - [D/24] = (16/60)
Explanation/Solution:- (24D + 192 - 25D) = 160
Let distance between the homes of D = 32 km/h

employees A and Q from their respective Time taken by C when he switches his speed
offices is 12x and 7x respectively. with R = D/25 = 32/25 = 1.28 hours
Distance between the home of employee P Time taken by R when he switches his speed
from his office = (12x - 12) with C = (D + 8)/24 = 40/24 = (5/3) hours
Distance between the home of employee B Required difference = [(5/3) - 1.28]*60 =
from his office = (7x + 2) 23.2 minutes
Let speed of employee P is 'P'. Q14 - C
According to question- Explanation/Solution:-
36
Let speed of A and P is 'A' km/h Let speed of D = 100D km/h
Let speed of C & R is 'C' and 'C + 10' km/h Speed of S = 75% of D = 75D
respectively.
According to question-
Distance between office of P from his room =
60 - 12 = 48 km [25/100D] - [17/75D] = (7/60)
Distance between office of R from his room = (75 - 68)/300D = 7/60

32 + 8 = 40 km
7/300D = 7/60
D = 0.2
[60/A] - [48/A] = (30/60)
Speed of D = 100D = 20 km/h
(12/A) = 0.5
Speed of S = 75D = 15 km/h
A = 24
Time taken by D to reach office = (25/20) =
Speeds of A and P is 24 km/h each. 1.25 hours
[32/C] - [40/(C + 10)] = (16/60) Time taken by S to reach office = (17/15)

hours
C2 + 40C - 1200 = 0
Required per cent = [(17/15)/1.25] * 100 =
(C - 20)(C + 60) = 0 272/3% = 90(2/3)%
C = 20 Q16 - E
Speed of C = C = 20 km/h Explanation/Solution:-
Speed of R = C + 10 = 30 km/h Let distance of E's office from his room = E

km
Required ratio = (60/24): (32/30) = 75: 32
Distance of T's office from his room = (E + 2)
Q15 - A km
Explanation/Solution:- Let distance of F's office from his room = F
km
Let distance of office of D from room = 100x
Distance of U's office from his room = (F + 8)
Distance of office of S from room = 68% of km
100x = 68x
100x - 68x = 8
[(E + 2)/16] - [E/24] = (30/60)
x = 0.25
(3E + 6 - 2E)/48 = 0.5
Distance of office of D from room = 100x =
25 km E + 6 = 24
Distance of office of S from room = 68x = 17 E = 18
km
37
Distance of E's office from his room = E = 18 Distance between office and home of S =
km [(30 + 18 + 24)/3] + 21 = 45 km
Distance of T's office from his room = (E + 2) Time taken by S to reach office from home =
= 20 km 45/15 = 3 hours
According to question- Time taken by T to reach office from home =

3 hours
[(F + 8)/40] - [F/48] = (18/60)
Distance between office and home of T = 3 *
(6E + 48 - 5E)/240 = 0.3 (15 + 5) = 60 km
F + 48 = 72
Distance between Time taken to reach
F = 24 Persons home and office office from home
(km) (hours)
Distance of F's office from his room = F = 24
km P 30 1.5
Distance of U's office from his room = (F + 8)
= 32 km Q 18 1.2
Sum of distance of room of employees E and

T from their respective offices = 18 + 20 = R 24 1
38 km
S 45 3
Sum of distance of room of employees F and
U from their respective offices = 24 + 32 =
56 km T 60 3
Required difference = 56 - 38 = 18 km
Q17 - B
Explanation/Solution:- Explanation/Solution:-
Time taken by P to reach office from home = Fare amount when person P goes to office
1.5 hours from home by Ola Mini = Base Fare + 5%
Distance between office and home of P = 1.5 GST on Base Fare + Extra charges + 20%
* 20 = 30 km GST on extra charges
Time taken by Q to reach office from home = Base Fare for Ola mini = 5 * 30 + 20 * 1.5 =
1.5 * (4/5) = 1.2 hours Rs.180
Distance between office and home of Q = 1.2 Extra charges = 25% of 180 = Rs.45
* (20 - 5) = 18 km Fare amount = 180 + 5% of 180 + 45 +
Distance between office and home of R = 18 20% of 45
* (4/3) = 24 km = 189 + 54
Time taken by R to reach office from home = = Rs.243
1 hours
38
Fare amount when person P return from Q19 - C

office to home by Uber Micro = Base Fare +
5% GST on Base Fare Explanation/Solution:-
Base Fare for Uber Micro = 15 * 30 + 20 * Base Fare for Uber Mini = 10 * 60 + 20 * 3
1.5 = Rs.480 = Rs.660
Fare amount = 480 + 5% of 480 = Fare of Uber Mini for person T when goes to
office = 660 + 5% of 660 = Rs.693
= 480 + 24
Base Fare for private cab = 8 * 60 = Rs.480
= Rs.504
Taxes = 25% of 480 = Rs.120
Required difference = 504 - 243 = Rs.261
Fare for private cab when person T returns
Q18 - D from office = 480 + 120 = Rs.600
Explanation/Solution:- Required difference = 693 - 600 = Rs.93
Let distance between Delhi to point A and Explanation/Solution:-

distance between point A to Agra is '5x' and
'4x' respectively. Let areas in Venn-diagram be a, b, c, d, e, f, g
and h.
According to the question:
(5x/25) + (4x/20) = 6:00 PM - 8:00 AM =

10 hours
(2x/5) = 10
x = 25
Distance between Delhi to point A = 5x =

125 km
Distance between point A to Agra = 4x =

100 km
Base Fare for Ola Micro = 10 * 125 + 30 *

(125/25) = Rs.1400
Extra charges = 25% of 1400 = Rs.350 Total number of employees working in the
company is 800
Fare of Ola Micro = 1400 + 5% of 1400 +
350 + 20% of 350 = Rs.1890 Number of employees who does not like
coffee = h = 800 * 42.5/100 = 340
Base Fare for Uber Mini = 10 * 100 + 20 *
(100/20) = Rs.1100 Number of employees who like coffee = 800
- 340 = 460
Fare of Uber Mini = 1100 + 5% of 1100 =
Rs.1155 So, a + b + c + d + e + f + g = 460 ---(1)
Total Fare = 1890 + 1155 = Rs.3045
39
The number of employees who like all three => 2a + b + d = 370

kind of coffee is 5% of total employees
From equation (4), we get
g = 5/100 * 800 = 40
=> 2(10 + 2b) + b + d = 370
mocha is 10 less than the number of => 5b + d = 350 ---(5)
employees who like café Cubano
Adding equation (3) and (5), we get
So, (a + d + e + g) = (c + d + g + f) - 10
=> b = 60 and d = 50
=> a + e = c + f - 10 ---(2)
So, a = c = 130
mocha and americano but not café Cubano is
half of the number of employees who like all
three kind of coffee
So, e = 1/2 * g = 1/2 * 40 = 20

mocha and café Cubano but not americano is
10 less than the number of employees who
like only americano
So, d = b - 10
=> b - d = 10 ---(3)
The number of employees who like americano

and café Cubano but not café mocha is 10 Q20 - A
more than the number of employees who like
café mocha and americano but not café Explanation/Solution:-
Cubano
Required % = 60/50 * 100 = 120%
So, f = 10 + e = 30
Q21 - E
From equation (2), we get
=> a + 20 = c + 30 - 10
Required difference = (130 + 20 + 40 + 50)
=> a = c - (60 + 30 + 40 + 20) = 90
And, Number of employees who like only café Q22 - C
Mocha is 10 more than the twice of
employees who like only americano. Explanation/Solution:-
So, a = 10 + 2b ---(4) Required total = 20 + 50 + 30 = 100
From equation (1), we get Explanation/Solution:-
a + b + c + d + 20 + 30 + 40 = 460 Tabulating data-
40
First test (216/5)

M 110 432 65 45
%
Numb Numb Numb

er of er of er of (266/5)
Total N 150 532 110 40
questi questi questi %
questi
ons ons ons
on Tot
Percent attemp attemp attemp
attemp al (264/5)
age ted in ted in ted in O 160 528 118 42
ted mar %
marks sectio sectio sectio
(maxim ks
nA nB nC
um =
(maxim (maxim (maxim
100) (178/5)
um = um = um = P 90 356 60 30
35) 15) 50) %
(735/1 14 (208/5)
M 75 28 11 36 Q 175 416 140 35
8)% 7 %
(450/1 (236/5)
N 60 90 22 6 32 R 140 472 110 30
8)% %
Q23 - B
(820/1 16
O 90 31 12 47
8)% 4 Explanation/Solution:-
(805/1 16 Marks percentage in first test of N = 450/18

P 80 28 10 42 = 25%
8)% 1
Q 50
(670/1 13
21 7 22 obtained in section A, B and C)/360 * 100
8)% 4
=> 25 = (total marks obtained in section A, B
and C)/360 * 100
(540/1 10
R 65 25 8 32
8)% 8
=> total marks obtained in section A, B and
C = 25 * 360/100
Second test
=> total marks obtained in section A, B and
C = 90
Number
Number
of
of Marks percentage in second test of N =
Total question 266/5 = 53.2%
question
question s
s
attempt Total attempt Marks percentage in second test = (total
Percenta attempt
ed mark ed in marks obtained in section X and Y)/1000 *
ge marks ed in
(maximu s section
section 100
m= X
Y
200) (maximu
(maximu => total marks obtained in section X and Y =
m=
m = 50) 53.2 * 1000/100 = 532
150)
Total marks obtained by N in both test = 90
+ 532 = 622
41
Marks percentage in first test of R = 540/18 Q25 - D

= 30%
obtained in section A, B and C)/360 * 100 Marks obtained by M:
=> total marks obtained in section A, B and Total marks obtained in second test = 432
C = 30 * 360/100 = 108
In section Y = 40 * 8 - 2 * 2 - 4 * 3 = 304
Marks percentage in second test of R =
236/5 = 47.2% So, in section X = 432 - 304 = 128
Marks percentage in second test = (total Marks obtained by N:

marks obtained in section X and Y)/1000 * Total marks obtained in second test = 532
100
In section Y = 30 * 8 - 2 * 2 - 4 * 4 - 8 * 4 =
=> total marks obtained in section X and Y) 188
= 472
So, in section X = 532 - 188 = 344
Total marks obtained by R in both test = 108
+ 472 = 580 So, required % = (344 - 128)/128 * 100 =
168.75%
So, required ratio = 622: 580 = 311: 290
Q26 - B
Q24 - D
Marks obtained in section A and C together
Marks obtained by O: in first test:
In section A = (31 - 18) * 4 - 18 * 1 = 34 By M = 147 - (5 * 8 - 6 * 2) = 147 - 28 =
In section C = (47 - 8) * 2 - 8 * 0.5 = 74 119
Total = 34 + 74 = 108 By N = 90 - (3 * 8 - 3 * 2) = 90 - 18 = 72
Marks obtained by P: BY O = 164 - (8 * 8 - 4 * 2) = 164 - 56 =

108
In section A = (28 - 16) * 4 - 16 * 1 = 32
By P = 161 - (8 * 8 - 2 * 2) = 161 - 60 =
In section C = (42 - 6) * 2 - 6 * 0.5 = 69 101
Total = 32 + 69 = 101 By Q = 134 - (6 * 8 - 1 * 2) = 134 - 46 = 88
Marks obtained by Q: By R = 108 - (4 * 8 - 4 * 2) = 108 - 24 = 84
In section A = (21 - 4) * 4 - 4 * 1 = 64 So, correct sequence is M (119) > O (108) >

P (101) > Q (88) > R (84) > N (72)
In section C = (22 - 8) * 2 - 8 * 0.5 = 24
Total = 64 + 24 = 88
Time taken by Aman to complete 20% of the
Required average = (108 + 101 + 88)/3 = work = 8 days
99
42
=> Time taken by Aman to complete 1% of => 1/Ankita = 1/25

the work = 8/20 days
1/Aman + 1/Ankita + 1/Binita = 9/100
=> Time taken by Aman to complete 100%
of the work = 8/20 x 100 = 40 days => 13/200 + 1/Binita = 9/100
Time taken by Bhuvan to complete 15% of => 1/Binita = 9/100 - 13/200

the work = 9 days
=> 1/Binita = (18 - 13)/200
=> Time taken by Bhuvan to complete 1% of
the work = 9/15 days => 1/Binita = 5/200
=> Time taken by Bhuvan to complete 100% => 1/Binita = 1/40

of the work = 9/15 x 100 = 60 days 1/Ankita + 1/Sushma = 3/50
Time taken by Suman to complete 10% of => 1/25 + 1/Sushma = 3/50
the work = 3 days
=> 1/Sushma = 3/50 - 1/25
=> Time taken by Suman to complete 1% of
the work = 3/10 days => 1/Sushma = (3 - 2)/50
=> Time taken by Suman to complete 100% => 1/Sushma = 1/50
of the work = 3/10 x 100 = 30 days
1/Tapan + 1/Anjana = 1/12
Time taken by Tapan to complete 25% of the
work = 5 days => 1/20 + 1/Anjana = 1/12
=> Time taken by Tapan to complete 1% of => 1/Anjana = 1/12 - 1/20

the work = 5/25
=> 1/Anjana = (5 - 3)/60
=> Time taken by Tapan to complete 100%
of the work = 5/25 x 100 = 20 days => 1/Anjana = 2/60
Time taken by Madan to complete 30% of => 1/Anjana = 1/30

the work = 12 days
1/Kamini + 1/Madan = 1/15
=> Time taken by Madan to complete 1% of
=> 1/Kamini + 1/40 = 1/15
the work = 12/30 days
=> 1/Kamini = 1/15 - 1/40
=> Time taken by Madan to complete 100%
of the work = 12/30 x 100 = 40 days => 1/Kamini = (8 - 3)/120
Now, => 1/Kamini = 5/120
1/Aman + 1/Ankita = 13/200 => 1/Kamini = 1/24
=> 1/40 + 1/Ankita = 13/200 Q27 - B
=> 1/Ankita = 13/200 - 1/40 Explanation/Solution:-
=> 1/Ankita = (13 - 5)/200 Let, required time = n days
=> 1/Ankita = 8/200 9/40 + 3/40 + 3/30 + n/60 = 1
43
=> 12/40 + 1/10 + n/60 = 1 Let, Ankita worked for n days.
=> 3/10 + 1/10 + n/60 = 1 10/40 + 10/60 + n/25 + (n + 1)/30 = 1
=> 4/10 + n/60 = 1 => ¼ + 1/6 + (6n + 5n + 5)/150 = 1
=> 2/5 + n/60 = 1 => (11n + 5)/150 = 1 - ¼ - 1/6
=> n/60 = 1 - 2/5 => (11n + 5)/150 = (12 - 3 - 2)/12
=> n/60 = (5 - 2)/5 => 11n + 5 = 150 x 7/12
=> n/60 = 3/5 => 11n = 175/2 - 5
=> n = 60 x 3/5 => n = 1/11 x (175 - 10)/2
=> n = 36 days => n = 165/22
Q28 - E Explanation/Solution:-
Explanation/Solution:- Total seats in the bus = 45
Let, time taken by Suman and Kamini to Male seats in the bus = 45 * (5/9) = 25
complete the work = m days
Female seats in the bus = 45 * (4/9) = 20
And time taken by Tapan, Ankita and Sushma
to complete the work = n days Male seat reserved for emergency = 20% of
25 = 5
m x (1/30 + 1/24) = 1
Female seat reserved for emergency = 25%
=> m x (4 + 5)/120 = 1 of 20 = 5
=> m = 120/9 Fare for one male adult = Rs.120
=> m = 40/3 days Fare for one female adult = Rs.100
And Fare for one male kid = 75% of 120 = Rs.90
n x (1/20 + 1/25 + 1/50) = 1 Fare for one female kid = 75% of 100 =
Rs.75
=> n x (5 + 4 + 2)/100 = 1
Maximum amount of fare that can be
=> n = 100/11 days collected from the bus on any day = (25 *
120) + (20 * 100) = Rs.5000
Required percentage = (40/3)/(100/11) x
100 Sum of male and female passengers in the
bus on Monday = 15 * 2 = 30
= 146.67%
Let male and female passengers in the bus on
= 147% approx. Monday is 'x' and '30 - x' respectively.
Q29 - C Male passengers in the bus on Tuesday = (x
Explanation/Solution:- + 2)
44
Let female passengers in the bus on Tuesday

Monday (2y - 2) = 18 (32 - 2y) = 12
and Wednesday is 'y' each.
Male passengers in the bus on Friday = y Tuesday 2y = 20 y = 10
Male passengers in the bus on Thursday = (y

+ 2) Wednesday z = 10 y = 10
Female passengers in the bus on Thursday =

Thursday (y + 2) = 12 (y - 2) = 8
(y + 2) - 4 = (y - 2)
According to the question: Friday y = 10 15

(y + 2) + (y - 2) = (x + 2)
Q30 - D
2y = x + 2
x = (2y - 2)
Sum of male and female kids in the bus on
Male passengers in the bus on Monday = x = Monday = 3 * 2 = 6
(2y - 2)
Let male and female kids in the bus on
Female passengers in the bus on Monday = Monday = 'x' and '6 - x' respectively.
(30 - x) = 30 - (2y - 2) = (32 - 2y)
Male adults in the bus on Monday = (18 - x)
Male passengers in the bus on Tuesday = (x
+ 2) = 2y Female adults in the bus on Monday = 12 -
(6 - x) = (6 + x)
Female passengers in the bus on Friday = 15
Total fare collected form the bus on Monday
According to the question: = [(18 - x) * 120] + [x * 90] + [(6 + x) * 100]
+ [(6 - x) * 75] = 3195
[(32 - 2y) + y + y + (y - 2) + 15]/5 = 11
2160 - 120x + 90x + 600 + 100x + 450 -
30 + y + 15 = 55 75x = 3195
y = 10 5x = 15
let male passengers in the bus on Wednesday x=3
=z
Required ratio = x: (6 - x) = 3: 3 = 1: 1
And,
Q31 - C
[(2y - 2) + 2y + z + (y + 2) + y]/5 = 14
6y + z = 70
Total passengers in the bus on Tuesday = 20
z = 70 - 60 + 10
z = 10 Total kid passengers in the bus on Tuesday =
26.67% of 30 = 8
Total Male Total Female
Days Total male kid passengers in the bus on
passengers passengers
Tuesday = 8 - 30% of 10 = 5
45
Total male adult passengers in the bus on Explanation/Solution:-

Tuesday = 20 - 5 = 15
Female kid passenger in the bus on Thursday
Total female adult passengers in the bus on = 75% of 8 = 6
Tuesday = 10 - 3 = 7
Female kid passenger in the bus on
Total fare collected from the bus on Tuesday Wednesday = 6 - (2 * 3) = 0
= [15 * 120] + [5 * 90] + [7 * 100] + [3 *
75] = 1800 + 450 + 700 + 225 = Rs.3175 Total kid passengers in the bus on
Wednesday = 2 * 2 = 4
Maximum fare amount that can be collected
from the bus = Rs.5000 Male kid passenger in the bus on Wednesday
=4-0=4
Required percent = (3175/5000) * 100 =
63.5% Male kid passenger in the bus on Thursday =
(2 * 5) - 6 = 4
Q32 - A
Total fare collected from the bus on
Explanation/Solution:- Wednesday = [6 * 120] + [4 * 90] + [10 *
100] + [0 * 75] = 720 + 360 + 1000 + 0 =
Total kid passengers in the bus on Rs.2080
Wednesday = 2 * 2 = 4
Total fare collected from the bus on Thursday
Let male kid and female kid passengers in the = [8 * 120] + [4 * 90] + [2 * 100] + [6 * 75]
bus on Wednesday is 'x' and '4 - x' = 960 + 360 + 200 + 450 = Rs.1970
respectively.
Required average = (2080 + 1970)/2 =
Male adult passengers in the bus on Rs.2025
Wednesday = (10 - x)
Q34 - A
Female adult passengers in the bus on
Wednesday = 10 - (4 - x) = (6 + x) Explanation/Solution:-
Total fare collected from the bus on Since, only 15 female passengers can sit in
Wednesday = [(10 - x) * 120] + [x * 90] + [(6 the bus and there are already 15 female
+ x) * 100] + [(4 - x) * 75] = 41.8% of 5000 passengers in the bus on Friday. So, 'x'
boarded passengers should be male.
1200 - 120x + 90x + 600 + 100x + 300 -
75x = 2090 Male adults boarded in the bus on Friday = (x
- 1)
5x = 10
Total kids in the bus on Friday before
x=2 boarding of 'x' passengers = 40% of (10 +
15) = 10
Male kids in the bus on Wednesday = x = 2
Male kid passengers in the bus = 10 * (1/5)
Female kids in the bus on Wednesday = (4 - +1=3
x) = 2
Female kid passengers in the bus = 10 *
Required difference = 2 - 2 = 0 (4/5) = 8
Q33 - B Male adult passengers in the bus = (10 - 2) +
(x - 1) = (7 + x)
46
Female adult passengers in the bus = 15 - 8 For building D:

=7
Total population = 35% of 8000 = 2800
Total original fare collected the bus before
boarding of 'x' passengers = [8 * 120] + [2 * White + blue = 4500 - 200 - 1000 - 800 -
90] + [7 * 100] + [8 * 75] = 960 + 180 + 300 = 2200
700 + 600 = Rs.2440
Then, 2800 = 2X - 2.5Y + 2.5X - 3Y + 2200
Total original fare collected the bus after
boarding of 'x' passengers = [(7 + x) * 120] + 9X - 11Y = 1200
[3 * 90] + [7 * 100] + [8 * 75] = 2440 + For building E:
570
840 + 120x + 270 + 700 + 600 = 3010
White + blue = 300
120x = 600
Then, 1200 = 2X - 2.5Y + 2.5X - 3Y + 300
x=5
9X - 11Y = 1800
Let total users in Nike = n
Total population = 8000
Then, total users in Adidas = n + 30
For building A:
So, 8000 = 1780 + n + n + 30 + 2270
n = 1960
White = 150, blue = 150 * 1/3 = 50
Code:
Then, 800 = 2X - 2.5Y + 2.5X - 3Y + 150 +
50 Total users = 1780
9X - 11Y = 1200 E = 250
For building B: B = 3C/2
Total population = 25% of 8000 = 2000 D = 2C
White + blue = 1000 A = (2X - 2.5Y)for building D - 20
Then, 2000 = 2X - 2.5Y + 2.5X - 3Y + 1000 Nike:
9X - 11Y = 2000 Total users = 1960
For building C: B = 6% of 8000 = 480
Total population = 15% of 8000 = 1200 E = 480 - 200 = 280
White + blue = 150 + 100 + 550 = 800 A = 5C/4
Then, 1200 = 2X - 2.5Y + 2.5X - 3Y + 800 Adidas:
9X - 11Y = 800 Total users = 1960 + 30 = 1990
47
C = 9D/10 Explanation/Solution:-
B = (2.5X - 3Y)for building B + 50 For building D:
A = females in building A - 80 Y = 1200
E = 270 Then, 9X - 11 * 1200 = 1200
Levis: X = 1600
Total users = 2270 Now, for Code:
C = 300 Number of Code users in building A = (2 *

1600 - 2.5 * 1200) - 20 = 180
D = 4 * (150 + 100) = 1000
Now, 1780 = 180 + (3C/2) + C + 2C + 250
Then, E = 1200 - 250 - 280 - 270 = 400
C = 300
B = (2X - 2.5Y)for building E + 45
Then, D = 2 * 300 = 600
Q35 - C
Now, number of Code users in building A and
Explanation/Solution:- D together = 180 + 600 = 780
For building A: For Levis:
350 = 2.5X - 3Y Number of Levis users in building C and E
together = 300 + 400 = 700
And, 9X - 11Y = 1200
Therefore, ratio = 780: 700 = 39: 35
Then, X = 500, Y = 300
Q37 - B
For building B:
9X - 11Y = 2000
Levis users:
And, 2X - 2.5Y = 1000 * 2/5
A = 5B/14
Then, X = 1200, Y = 800
Then, 2270 = (5B/14) + B + 300 + 1000 +
Now, Adidas uses from building A = females 400
in building A - 80
B = 420
= 300 - 80 = 220
Now, for building E:
Adidas users from building B = (2.5X - 3Y)for
building B + 50 420 = (2X - 2.5Y) + 45
= 1000 * 3/5 + 50 2X - 2.5Y = 375
= 650 And, 9X - 11Y = 1800
Therefore, difference = 650 - 220 = 430 So, X = 750, Y = 450
Q36 - E
48
Therefore, percentage = (450/8000) * 100 = The number of sick people recommended for
5.625% = 6% (approx.) immediate treatment from state P in 2013
are 1344 less than the previous year. Then,
Q38 - A
1344 = C% of A - 6% of 117600
A * C = 840000...(i)
For building C:
From second question:
X = 200 + Y
A - H - E = 61132 - 1200F...(ii)
Then, 9(200 + Y) - 11Y = 800
B + G = 2(F - 15)...(iii)
Y = 500, X = 700
From third question:
Then, number of people in who like grey
color = 2.5 * 700 - 3 * 500 = 250 It is given that population of state Q is
increased by 15% in 2013 than previous
For building B: year. Then,
Number of people in who like white color = E = 115% of 98000 = 112700
(1000 - 200)/2 = 400
E = 112700
Therefore, percentage = (250/400) * 100
= 62.5% In 2012 the ratio of number of people found
sick according to test report in state Q and
Q39 - D number of people found sick according to
test report but did not recommend for
Explanation/Solution:- immediate treatment from state P is 1: 1.
Number of female users of Code, Nike and Then,
Levis in building E = 66% of 250 + 45% of 15% of 98000 = (22 - C)% of A
280 + 52% of 400
1470000 = 22A - AC
= 499
Then, from (i), we have
Number of male users of Code, Nike and
Levis in building E = 250 + 280 + 400 - 1470000 = 22A - 840000
499 = 431
A = 105000
Number of female users of Code, Nike and
Levis in building C = 800 - 431 - 24 = 345 Then, C = 840000/105000 = 8
Therefore, difference = 431 - 345 = 86 C=8
Explanation/Solution:- Number of people who visited health camp

but did not complete their health check-up
Data given in all individual questions for the test in state Q in 2013 are 74676 less than
given graph is interrelated, so, we will the number of people visited health camp
determine the value of all unknown variables from state P in same year. Then,
by using data given in respective questions.
74676 = F% of 117600 - (78890 - G% of
From first question: E)
49
74676 = 1176F - 78890 + G% of 112700 Now, from equation (ii), we get
1176F = 153566 - 1127G...(iv) A - H - E = 61132 - 1200F
From fourth question: 105000 - H - 112700 = 61132 - 1200 * 75
Average of number of people completed their H = 21168

health check-ups tests from state Q in 2012
and 2013 taken together is 63063. Then, Q40 - C
2 * 63063 = B% of 98000 + G% of E Explanation/Solution:-
126126 = 980B + G% of 112700 Value of C = 8
980B + 1127G = 126126...(v) Q41 - A
From fifth question: Explanation/Solution:-
20237 people from state Q in 2012 and Number of people visited health camp in
2013 taken together who found sick did not state P is 2012
recommend for immediate treatment. Then,
= 78% of A
20237 = 15% of 98000 - D + (20 - 9)% of
E = 78% of 105000
20237 = 14700 - D + 11% of 112700 = 81900
D = 6860 Q42 - E
From equation (iii), (iv) and (v), we get Explanation/Solution:-
B + G = 2(F - 15) Average of number of people from state P

who did not visit camp in 2012 and 2013
1176F = 153566 - 1127G together
F = (153566 - 1127G)/1176 = ((100 - 78)% of A + (100 - F)% of

117600)/2
980B + 1127G = 126126
= (22% of 105000 + (100 - 75)% of
B = (126126 - 1127G)/980 117600)/2
Then, ((126126 - 1127G)/980) + G = 2 * = 26250

(((153566 - 1127G)/1176) - 15)
Q43 - C
G = 58
B = (126126 - 1127 * 58)/980 = 62
Number of people from state Q who visited
B = 62 health camp but did not complete their health
check-up test in 2013 = 78890 - G% of E
F = (153566 - 1127 * 58)/1176 = 75
= 78890 - 58% of 112700
F = 75
= 13524
50
Q44 - A
150% 75% of
T 30x of 30x 30x = 4: 6: 3
= 45x 22.5x
Number of sick people recommended for
immediate treatment from state Q in 2012 Q45 - C
= D = 6860 Explanation/Solution:-
Number of people who were not sick Part of Oil in mixture P = 4/12
according to test report in state P in 2013
Part of Petrol in mixture P = 5/12
= 68680 - H
Part of Oil in mixture Q = 4/15
= 68680 - 21168
Part of Petrol in mixture Q = 5/15
= 47512
By the rule of alligation:
Therefore, difference = 47512 - 6860
= 40652
Let total amount of Oil in all the five mixtures

together = 360x
[(4/15) - P]: [P - (4/12)] = 4: 5
Ratio (Oil: (4/3) - 5P = 4P - (4/3)
Mixtures Oil Petrol Kerosene Petrol:
Kerosene) 9P = (8/3)
P = 8/27
125%
75% of
of 90x By the rule of alligation:
P 90x 90x = 4: 5: 3
=
67.5x
112.5x
125% 150% of
Q 60x of 60x 60x = 4: 5: 6
= 75x 90x
[(5/15) - Q]: [Q - (5/12)] = 4: 5
75% of 50% of
R 72x 72x = 72x = 4: 3: 2 (5/3) - 5Q = 4Q - (5/3)
54x 36x
9Q = (10/3)
75% of 125% of Q = 10/27

S 108x 108x = 108x = 4: 3: 5
81x 135x Ratio Oil, Petrol and Kerosene in the final
mixture = 8: 10: (27 - 8 - 10) = 8: 10: 9
Total amount of newly formed mixture = 12 *

(27/2) = 162 litres
51
Q46 - E 48 - (152/3) = (76X/234) - (4X/12)
Explanation/Solution:- (8/3) = 2X/234
Total amount of final mixture = 105 litres X = 312
Let amount of Oil, Petrol and Kerosene in Let part of Oil in mixture B = 'B'
final mixture is 'x', 'x' and 'x - 15' litres
respectively. By the rule of alligation:
x + x + (x - 15) = 105
x = 40
Ratio of Oil, Petrol and Kerosene in final

mixture = x: x: (x - 15) = 40: 40: (40 - 15) =
8: 8: 5
[(4/13) - B]: [B - (4/12)] = 1: 2
Part of Oil in mixture P = 4/12
(8/13) - 2B = B - (1/3)
Part of Oil in mixture R = 4/9
3B = 37/39
Part of Oil in final mixture = 8/21
B = 37/117
2011:
Let total bottles manufactured = 100a

Ratio in which mixtures P and R must be
mixed = [(4/9) - (8/21)]: [(8/21) - (4/12)] = Total unsold bottles = 20% of 100a = 20a
(4/63): (4/84) = 84: 63 = 4: 3
Total sold bottles = 100a - 20a = 80a
Q47 - C
Total bottles sold at 50% profit = 75% of
Explanation/Solution:- 80a = 60a
Part of Oil in mixture S = 4/12 Total bottles sold at 20% profit = 80a - 60a
= 20a
Part of Oil in mixture T = 4/13
2012:
Part of Oil in mixture A = 76/234
Let total bottles manufactured = 300b
Total unsold bottles = 16(2/3)% of 300b =
50b
Total sold bottles = 300b - 50b = 250b
Total bottles sold at 50% profit = 60% of

[(4/13) - (76/234)]: [(76/234) - (4/12)] = X: 250b = 150b
156
52
Total bottles sold at 20% profit = 250b - Explanation/Solution:-

150b = 100b
From III:
2013:
Total bottles sold at 50% profit in 2011 if
Let total bottles manufactured = 100c unsold bottles are assumed to be sold =
116(2/3)% of 60a = 70a
Total unsold bottles = 12.5% of 100c =
12.5c Total bottles sold at 20% profit in 2011 if
unsold bottles are assumed to be sold = 20a
Total sold bottles = 100c - 12.5c = 87.5c + [20a - (70a - 60a)] = 30a
Total bottles sold at 50% profit = 50% of According to the question:
87.5c = 43.75c
30a = 36d
Total bottles sold at 20% profit = 87.5c -
43.75c = 43.75c a: d = 6: 5
2014: a = 1.2d ........ (1)
Let total bottles manufactured = 100d We can't calculate the required number.
Total unsold bottles = 10% of 100d = 10d Statement III alone is not sufficient.
Total sold bottles = 100d - 10d = 90d From I and II:
Total bottles sold at 50% profit = 40% of According to the question:

90d = 36d
60a: 54d = 4: 3
Total bottles sold at 20% profit = 90d - 36d
= 54d a: d = 6: 5
a = 1.2d ........ (1)

Sold Sold
Total
Year Total at at Unsol And,
manufacture
s Sold 50% 20% d
d
profit profit 20a - 10d = 70
From equation (1):

201
100a 80a 60a 20a 20a
1
24d - 10d = 70
201 250 d = 5 and a = 6

300b 150b 100b 50b
2 b
Total number of unsold bottles by the
manufactured in the years 2011 and 2014
201 87.5 43.75 43.75 together = 20a + 10d = 120 + 50 = 170
100c 12.5c
3 c c c
Statement I and II together are sufficient.
201 From I and III:
100d 90d 36d 54d 10d
4
Q48 - A
53
60a: 54d = 4: 3 Required difference = 150b - 100b = 50b =

80
a: d = 6: 5
Quantity II:
a = 1.2d ........ (1)
Unsold items in 2013 = 12.5c = 100
From statement III we get:
c=8
a = 1.2d
Sold items in 2013 = 87.5c = 700
Since, both the equations are same. Hence,
we can't determine the required number. Required difference = 700 - 100 = 600
Statements I and III together are not Hence, Quantity I < Quantity II
sufficient.
Q50 - B
From II and III:
According the question:
20a - 10d = 70 ...... (2)
20a: 12.5c = 6: 5
From statement III we get:
a = 0.75c ..... (1)
a = 1.2d .......... (1)
Quantity I:
From (1) and (2):
24d - 10d = 70
43.75c - 20a = 115 ...... (2)
d = 5 and a = 6
From (1) and (2):
Total number of unsold bottles by the
manufactured in the years 2011 and 2014 43.75c - 15c = 115
together = 20a + 10d = 120 + 50 = 170
28.75c = 115
Statement II and III together are sufficient.
c = 4 and a = 3
Q49 - C
Total number of sold items by the
Explanation/Solution:- manufacturer in the years 2011 and 2013
together = 80a + 87.5c = 240 + 350 = 590
Quantity I:
Quantity II:
Items sold at 20% profit in 2012 = 100b
Unsold items in 2012 = 50b
20a + 12.5c = 110 ...... (3)
From (1) and (3):
100b - 50b = 80
15c + 12.5c = 110
b = 1.6
27.5c = 110
54
c = 4 and a = 3
(140 -
6 + 2 = 120 * 6
D 175 91 40)/2 =
Total number of sold items by the 8 = 720
50
manufacturer in the years 2011 and 2013
together = 80a + 87.5c = 240 + 350 = 590
(140 +
4.5 + 90 * 8.5
Quantity III: E 5e e 40)/2 =
1.5 = 6 = 765
90
60a - 43.75c = 15 ....... (4) Q51 - E
From (1) and (4): Explanation/Solution:-
45c - 43.75c = 15 Total answered calls by A = (3 * 73.33) - (50

+ 90) = 80
1.25c = 15
Total Staffed hours = (Total answered calls *
c = 12 and a = 9 AHT) + Available time in minutes +
Productive auxiliary time (AUX) in minutes
Total number of sold items by the
manufacturer in the years 2011 and 2013 = 80 * 5.5 + P + 30
together = 80a + 87.5c = 720 + 1050 =
1770 = (470 + P)
Hence, Quantity I = Quantity II < Quantity III Shrinkage percent = [1 - {(470 + P)/600}] *
100
10/100 = (130 - P)/600
AHT for BPO A = 4 + 1.5 = 5.5 minutes
130 - P = 60
Total scheduled hours for BPO A = 75 * 8 =
600 hours P = 70
Similarly, we can calculate AHT and total Q52 - C

scheduled hours for other BPOs as well.
AUX Total answered calls by BPO B = Q = 65

AHT Total Availabl Total
BP time
(minute Schedule e Time answere
O (min Available time = 65 * (8/13) = 40 minutes
s) d hours (min) d calls
)
AUX time = 40 - 22 = 18 minutes
A 5.5 600 P 30 ---- Total Staffed hours = (Total answered calls *
AHT) + Available time in minutes +
5 + 2 = 60 * 9 = Productive auxiliary time (AUX) in minutes
B ---- ---- Q
7 540
= (65 * 7) + 40 + 18
3 + 1 = 80 * 7.5 = 513
C 11c 6c Q+5
4 = 600
Shrinkage percent = [1 - (513/540)] * 100
= 5%
55
Q53 - B Shrinkage percent of BPO D = [1 -

(666/720)] * 100 = 7.5%
Total Staffed hours of BPO A = (Total
According to the question: answered calls * AHT) + Available time in
minutes + Productive auxiliary time (AUX) in
11c - 6c = 50 minutes
c = 10 = (70 * 5.5) + P + 30 = (P + 415)
Total Staffed hours of BPO C = (Total Shrinkage percent of BPO A = [1 - {(P +
answered calls * AHT) + Available time in 415)/600}] * 100
minutes (7.5 + 10)/100 = (185 - P)/600
= [(Q + 5) * 4] + 11c + 6c 105 + 185 - P
= (4Q + 190) P = 80
Shrinkage percent of BPO C = [1 - {(4Q + Total Staffed hours of BPO C = (Total
190)/600}] * 100 answered calls * AHT) + Available time in
35/100 = (410 - 4Q)/600 minutes
410 - 4Q = 210 = [(Q + 5) * 4] + 110 + 60 = (4Q + 190)
4Q = 200 Shrinkage percent of BPO C = [1 - {(4Q +
Q = 50 190)/600}] * 100
Total Staffed hours of BPO B = (Total (7.5 + 17.5)/100 = (410 - 4Q)/600

answered calls * AHT) + Available time in 150 = 410 - 4Q
minutes 4Q = 260
= (50 * 7) + 90 + 46 Q = 65
= 486 Required ratio = P: Q = 80: 65 = 16: 13
Shrinkage percent of BPO B = [1 - Q55 - B
(486/540)] * 100
= 10%
Total Staffed hours = (Total answered calls *
Q54 - D AHT) + Available time in minutes +
Productive auxiliary time (AUX) in minutes
= (90 * 6) + 5e + e = (6e + 540)
Total Staffed hours of BPO D = (Total
answered calls * AHT) + Available time in Shrinkage percent = [1 - {(6e + 540)/765}] *
minutes + Productive auxiliary time (AUX) in 100
minutes
20/100 = (225 - 6e)/765
= (50 * 8) + 175 + 91 = 666
56
153 = 225 - 6e = 96000 + 9600n
6e = 72 Product of investment and period of

investment for E
e = 12
= 8 * 8000 + n * (8000 + 3900)
Required percent = (e/90) * 100
= 64000 + 11900n
= (12/90) * 100
Then, profit ratio, C: E = 768: 677 = (96000
= 13.33% + 9600n): (64000 + 11900n)
Explanation/Solution:- n=6
Amount Q57 - B
Initial
Partners added/withdrew after
investment Explanation/Solution:-
first 8 months
Product of investment and period of
18% of investment for A
A 50000 = 20% of 15000 = 3000
9000 = 9000 * 8 + (9000 - 3000) * 4 + (9000 -
3000 + 2000) * 6
20% of = 144000
B 50000 = 14% of 15000 = 2100
10000 Product of investment and period of
investment for B
24% of
C 50000 = 16% of 15000 = 2400 = 10000 * 8 + (10000 + 2100) * 4
12000
= 128400
22% of Product of investment and period of

D 50000 = 24% of 15000 = 3600 investment for C
11000
= 12000 * 8 + (12000 - 2400) * 4 +
(12000 - 2400 + 2000) * 6
16% of
E 50000 = 26% of 15000 = 3900
= 204000
8000
Q56 - C investment for D
Explanation/Solution:- = 11000 * 8 + (11000 + 3600) * 4
Since, investment of E becomes more than = 146400
investment of C after first '8 + n' months.
Then, Product of investment and period of
investment for E
investment for C = 8000 * 8 + (8000 - 3900) * 4 + (8000 -
3900 + 2000) * 6
= 8 * 12000 + n * (12000 - 2400)
57
= 117000 Profit ratio, D: E = (109200 + 14600 * 4 +

18600 * 10): (71800 + 11900 * 4 + 11900
Therefore, profit ratio, A: B: C: D: E = * 10 + 4000 * 10 + 2 * 4000)
144000: 128400: 204000: 146400:
117000 = 221: 179 ≠ 213: 277
= 240: 214: 340: 244: 195 Hence, this option is not possible.
Q58 - D From option 3: m = 12, n = 2, a = 2000
Explanation/Solution:- Profit ratio, D: E = (109200 + 14600 * 12 +
18600 * 2): (71800 + 11900 * 12 + 11900
Let after '8 + m' months, D invested Rs.4000 * 2 + 2000 * 2 + 2 * 2000)
and E invested Rs. a and also, only after next
'n' months both D and E withdrew Rs.8000. = 201: 154 ≠ 109: 129
Then,
Hence, this option is not possible.
investment for D From option 4: m = 8, n = 6, a = 4000
= 8 * 11000 + m * (11000 + 3600) + n * Profit ratio, D: E = (109200 + 14600 * 8 +

(11000 + 3600 + 4000) + 2 * (11000 + 18600 * 6): (71800 + 11900 * 8 + 11900 *
3600 + 4000 - 8000) 6 + 4000 * 6 + 2 * 4000)
= 109200 + 14600m + 18600n = 211: 169
Product of investment and period of Hence, this option is possible.

investment for E
= 8 * 8000 + m * (8000 + 3900) + n *
(8000 + 3900 + a) + 2 * (8000 + 3900 + a Perimeter of rectangle A = 15% of 1200 =
- 8000) 180 cm = 2 * (length of A + breadth of A)
= 71800 + 11900m + 11900n + an + 2a Length of rectangle A = 90 * 5/9 = 50 cm
Then, profit ratio, D: E = (109200 + 14600m Breadth of rectangle A = 90 - 50 = 40 cm

+ 18600n): (71800 + 11900m + 11900n +
Perimeter of rectangle B = 25% of 1200 =
an + 2a)
300 cm = 2 * (length of B + breadth of B)
From option 1: m = 6, n = 8, a = 2000
Length of rectangle B = 150 * 3/5 = 90 cm
Profit ratio, D: E = (109200 + 14600 * 6 +
Breadth of rectangle B = 150 - 90 = 60 cm
18600 * 8): (71800 + 11900 * 6 + 11900 *
8 + 2000 * 8 + 2 * 2000) Perimeter of rectangle C = 14% of 1200 =
168 cm = 2 * (length of C + breadth of C)
= 432: 323 ≠ 201: 169
Length of rectangle C = 84 * 4/7 = 48 cm
Hence, this option is not possible.
Breadth of rectangle C = 84 - 48 = 36 cm
From option 2: m = 4, n = 10, a = 4000
Perimeter of rectangle D = 20% of 1200 =
240 cm = 2 * (length of D + breadth of D)
58
Length of rectangle D = 120 * 7/12 = 70 cm Speed of boat in downstream = 20 *3/2 =

30 km/h
Breadth of rectangle D = 120 - 70 = 50 cm
=> x - y = 20 and x + y = 30
Perimeter of rectangle E = 26% of 1200 =
312 cm = 2 * (length of E + breadth of E) After solving:
Length of rectangle E = 156 * 5/6 = 130 cm x = 25 and y = 5
Breadth of rectangle E = 156 - 130 = 26 cm Hence speed of stream = 5 km/h
Q59 - D Q63 - B
Average of length of all given rectangles Let speed of stream be y. Then, speed of boat
U in still water = 600% of y = 6y
= (50 + 90 + 48 + 70 + 130)/5
Speed of boat U in upstream = 6y - y = 15
= 77.6 cm (Given in chart)
Q60 - E => y = 3 km/h
Explanation/Solution:- Speed of boat U in downstream = 6y + y =
7y = 7 * 3 = 21 km/h
Length of rectangles C and E together = 48
+ 130 = 178 cm Required answer = 21/15 = 1.4 times
Breadth of rectangles B and A together = 40 Q64 - B
+ 60 = 100 cm
Therefore, ratio = 178: 100 = 89: 50
Speed of boat P in downstream = 12 + 3 +
Q61 - D 3 = 18 km/h
Explanation/Solution:- Speed of boat R in downstream = 8 + 2 + 2
= 12 km/h
Perimeter of rectangle B = 300 cm
Required difference in time = 252/12 -
Sum of breadths of all given rectangles = 40 252/18 = 21 - 14 = 7 h
+ 60 + 36 + 50 + 26 = 212 cm
Q65 - D
Therefore, difference = 300 - 212 = 88 cm
Let distance between A and B = x km.
Q62 - C
Upstream speed of boat T = 24 km/h
Downstream speed of boat T = 24 + 6 + 6 =
Let speed of boat in still water is 'x' km/h and 36 km/h
speed of stream is 'y' km/h.
Speed of boat in upstream = 20 km/h
59
=> x/24 + x/36 = 10 Total income of person B in all the three

months together = 45000 + 40000 +
=> x = 144 km 60000 = Rs.145000
Explanation/Solution:- Q% = (34000/145000) * 100
A B Q = (680/29)
Persons/ Month P: Q = (365/18): (680/29)
s Incom Saving Incom Saving
e s e s = 2117: 2448
Q67 - C
20% 40%
of of Explanation/Solution:-
3500 4500
January 35000 45000
0 0
= = Total savings of person B on February and
7000 18000 March together = 4000 + 12000 =
Rs.16000
15% 10%
of of Total amount spent on Food by person B on
2500 4000 February and March together = Rs.35000
February 25000 40000
0 0
= =
3750 4000 Total amount spent on Rent by person B on
February and March together = 35000 *
(100/80) = Rs.43750
25% 20%
of of Total amount spent on Shopping by person B
3000 6000
March 30000 60000 on February and March together = (40000 +
0 0
= =
7500 12000
60000) - (16000 + 35000 + 43750) =
Rs.5250
Q66 - D Required per cent = (5250/100000) * 100
Explanation/Solution:- = 5.25%
Total amount of savings of person A in all the Explanation/Solution:-
three months together = 7000 + 3750 +
7500 = Rs.18250 In Catering business,
Total income of person A in all the three Initial investment by Mohit, Sunil and Punit
months together = 35000 + 25000 + are Rs. 4a, Rs. 4a and Rs. 3a respectively
30000 = Rs.90000
After 6 months,
P% = (18250/90000) * 100
Shyamaly joins with initial investment Rs.
P = (365/18) 10a.
Total amount of savings of person B in all the Mohit, Sunil and Punit again invested Rs. 4b,
three months together = 18000 + 4000 + 3b and 3b respectively.
12000 = Rs.34000
After 1 year,
60
Investment of Mohit, Sunil, Punit and Investment of Mohit, Rohan, Prakash and
Shyamaly Rs. 6c, Rs. 4c, Rs. 5c and Rs. 5c Megha are Rs. 3n, Rs. 2n, Rs. 2n and Rs. 3n
respectively. respectively.
After 2 years, After 1.5 years,
Mohit withdraw = Rs. 2a Investment of Mohit, Prakash and Mega are

5k, 3k and 2k respectively.
Punit withdraw = Rs. 4c
After 2.5 years,
Sunil invest = Rs. 3.5 * 4a = 14a
Mohit withdraw = Rs. K
Shyamaly invest = 16a
Rohan withdraw = 25/100 * 2n = Rs. n/2
After 2.5 years,
Megha withdraw = 25/100 * 4m = Rs. m
Mohit, Sunil, Punit and Shyamaly investment
are Rs. 2d, Rs. 4d, Rs. d, Rs. 4d respectively. Prakash withdraw = 1/2 * 2n = Rs. n
Ratio of profit of Mohit, Sunil, Punit and Ratio of profit of Mohit, Rohan, Prakash and
Shyamaly Megha
= [4a * 6 + (4a + 4b) * 6 + (4a + 4b + 6c) * = [8m * 12 + (8m + 3n) * 6 + (8m + 3n +

12 + (4a + 4b + 6c - 2a) * 6 + (4a + 4b + 5k) * 12 + (8m + 3n + 5k - k) * 6]
6c - 2a + 2d) * 6]
: [7m * 12 + (7m + 2n) * 18 + (7m + 2n -
: [4a * 6 + (4a + 3b) * 6 + (4a + 3b + 4c) * n/2) * 6]
12 + (4a + 3b + 4c + 14a) * 6 + (4a + 3b +
4c + 14a + 4d) * 6] : [5m * 12 + (5m + 2n) * 6 + (5m + 2n + 3k)
* 12 + (5m + 2n + 3k - n) * 6]
: [3a * 6 + (3a + 3b) * 6 + (3a + 3b + 5c) *
12 + (3a + 3b + 5c - 4c) * 6 + (3a + 3b + : [4m * 6 + (4m + 3n) * 6 + (4m + 3n + 2k) *
5c - 4c + d) * 6] 12 + (4m + 3n + 2k - m) * 6]
: [10a * 6 + (10a + 5c) * 12 + (10a + 5c + = (48m + 12n + 14k) : (42m + 15n/2) :

16a) * 6 + (10a + 5c + 16a + 4d) * 6] (30m + 7n + 9k) : (19m + 12n + 6k)
= (20a + 20b + 24c + 2d) : (52a + 15b + Q68 - A

16c + 4d) : (18a + 15b + 12c + d) : (82a +
20c + 4d) Explanation/Solution:-
In Garment business, From I:
Initial investment of Mohit, Rohan and 2n = 50/100 * 8m

Prakash are Rs. 8m, 7m and 5m respectively. => n = 2m ----(1)
After 6 months, Megha joins the business 8m = 4 * 4a
with initial investment of Rs. 4m
=> m = 2a ----(2)
After 1 year,
From II:
3b = Rs. 6000
61
=> b = 2000 ----(3) Hence, Mohit's share of profit in catering

business = 110/427 * 854000 = Rs.
And, 3 * 2d = 6000 220000
=> d = 1000 ----(4) In Garment business,
From III: Ratio of profit of Mohit, Rohan, Prakash and
Megha
5c = 5m
= (48m + 12n + 14k) : (42m + 15n/2) :
=> c = m ----(5) (30m + 7n + 9k) : (19m + 12n + 6k)
From IV: = (96000 + 48000 + 84000) : (84000 +
4m = 2/3 * 2k 30000) : (60000 + 28000 + 54000) :
(38000 + 48000 + 36000)
=> m = k/3 ----(6)
= 228000: 114000: 142000: 122000
And, 5m = 7000 + 3a ---(7)
= 228: 114: 142: 122
From V:
= 114: 57: 71: 61
5m = 1/3 * 5k
Hence, Mohit share of profit in Garment
=> m = k/3 ---(8) business = 114/303 * 909000 = Rs.
342000
And, 7m = 10000 + 4a ---(9)
Thus, statement I, II, III, IV are sufficient to
From statement I, II, III and IV: answer
By solving equations 1, 2, 3, 4, 5, 6, and 7, From Statement I, II, III and V:
we get
Similarly, from equations (1), (2), (3), (4), (5),
a = 1000, m = 2000, c = 2000, n = 4000, k (8) and (9), we get
= 6000, b = 2000 and d = 1000
a = 1000, m = 2000, c = 2000, n = 4000, k
In catering business, = 6000, b = 2000 and d = 1000
Ratio of profit of Mohit, Sunil, Punit and Thus, from these values we can find the
Shyamaly answer.
= (20a + 20b + 24c + 2d) : (52a + 15b + Thus, statement I, II, III, and V together are
16c + 4d) : (18a + 15b + 12c + d) : (82a + sufficient to answer
20c + 4d)
Therefore, either statement IV or V is
= (20000 + 40000 + 48000 + 2000) : redundant to answer the question.
(52000 + 30000 + 32000 + 4000) :
(18000 + 30000 + 24000 + 1000) : Q69 - B
(82000 + 40000 + 4000)
= 110000: 118000: 73000: 126000
4m = 2n
= 110: 118: 73: 126
=> n = 2m -----(1)
62
And, 3k = 18000 c = 4000
=> k = 6000 Ratio of profit of Mohit, Sunil, Punit and

Shyamaly
18000 = 3/2 * 3n
= (20a + 20b + 24c + 2d) : (52a + 15b +
=> 18000 * 2/9 = n 16c + 4d) : (18a + 15b + 12c + d) : (82a +
20c + 4d)
=> n = 4000
= (20000 + 40000 + 96000 + 2000) :
From equation (1), we get (52000 + 30000 + 64000 + 4000) :
m = 2000 (18000 + 30000 + 48000 + 1000) :
(82000 + 80000 + 4000)
Total investment by Megha = 4m + 3n + 2k -
m = 3m + 3n + 2k = 6000 + 12000 + = 158000: 150000: 97000: 166000
12000 = Rs. 30000 = 158: 150: 97: 166
Total investment by Rohan = 7m + 2n - n/2 Let total profit x.
= 7m + 3n/2 = 14000 + 6000 = 20000
For managing business Mohit got = 5x/100
Therefore, total investment = 30000 +
20000 = Rs. 50000 Punit's profit = 95x/100 * 97/571
Q70 - D According to the question,
Explanation/Solution:- 5x/100 = 95x/100 * 97/571 - 254400
In catering business, => 95x/100 * 97/571 - 5x/100 = 254400 *
571
3b = 60% of 10a
=> 9215x - 2855x = 25440000 * 571
=> 3b = 3/5 * 10a
=> 6360x = 25440000 * 571
=> b = 2a ---(1)
=> x = 25440000/6360 * 571
And, 10a = 50% of 5c
=> x = 4000 * 571
=> 10a = 1/2 * 5c
=> x = Rs. 2284000
=> c = 4a ---(2)
Q71 - A
And, 4d = 4000
=> d = 1000 ---(3)
4a = Rs. T ---(1)
And, 4a = 4000
And, T = 4d ---(2)
=> a = 1000 ---(4)
3b = T + 2000 ---(3)
From (1) and (4) we get
4c = 4T
b = 2000
=> c = T ---(4)
And, from (2) and (4), we get
63
And, 4a - 3a = 1000 Overall capacity of Stage II = 6850 - 1850 -

2000 = 3000 MW
=> a = 1000
Total capacity of 250 MW units in Stage II =
From (1), we have 3000 - 5*200 - 2*500 = 1000 MW
T = 4000 Number of 250 MW units in Stage II =
1000/250 = 4
Thus, d = 1000
Let the number of 250 MW units in Stage I
And, from (3) we get be 'x'
3b = 4000 + 2000 Total capacity of 500 MW units in Stage I =
=> b = 2000 1850 - 3*200 - x*250 = (1250 - 250x) MW
Hence, c = 4000 So, Total capacity of 500 MW units in all

stages = (8/3)*(250x) = 2000x/3 MW
Ratio of profit of Mohit, Sunil, Punit and
Shyamaly So, 1250 - 250x + 3*500 = 2000x/3
= (20a + 20b + 24c + 2d) : (52a + 15b + 2750x/3 = 2750

16c + 4d) : (18a + 15b + 12c + d) : (82a + x=3
20c + 4d)
Number of 500 MW units in stage I = (1250
= (20000 + 40000 + 96000 + 2000) : - 250*3)/500 = 1
(52000 + 30000 + 64000 + 4000) :
(18000 + 30000 + 48000 + 1000) : Completing the table:
(82000 + 80000 + 4000)
= 158000: 150000: 97000: 166000 Number Number Number

Overall
of 200 of 250 of 500
capacity
= 158: 150: 97: 166 MW units MW units MW units
Explanation/Solution:- Stage I 3 3 1 1850

Number of 200 MW units in Stage III = 13 - 3
-5=5 Stage
5 4 2 3000
II
Total capacity of 500 units in Stage III =
2000 - 5*200 - 2*250 = 500 MW
Stage
5 2 1 2000
Number of 500 MW units in Stage III = 1 III
The overall capacity of stage II is (60/137) Overall 13 9 4 6850

part of the overall capacity of the power
plant.
Q72 - A
So, The overall capacity of stage I and stage
III combined is (1 - 60/137) or (77/137) part Explanation/Solution:-
of the overall capacity of the power plant
Required difference = 9 - 4 = 5
Overall capacity of the power plant =
Q73 - E
(137/77)*(1850 + 2000) = 6850 MW
64
Explanation/Solution:- 105/(8p - p) = 180/60 = 3
For Stage I: p=5
Electricity produced Ratio of speed of boat Q in still water to

speed of stream = 100: 20 = 5: 1
= 3*200*0.8*4 + 3*200*0.65*4 +
3*200*0.5*8 + 3*250*0.6*4 + 60/(5q - q) = 300/60 = 5
3*250*0.45*8 + 500*0.9*2 + 500*0.8*4 +
500*0.65*4 + 500*0.5*8 = 16180 MWh q=3
Q74 - B Ratio of speed of boat R in still water to

speed of stream = 100: 20 = 5: 1
80/(5r - r) = 240/60 = 4
For Stage II:
r=5
electricity produced = 5*200*0.8*2 +
5*200*0.6*4 + 5*200*0.55*4 + Ratio of speed of boat S in still water to
5*200*0.4*8 + 4*250*0.85*2 + speed of stream = 100: 25 = 4: 1
4*250*0.7*4 + 2*500*0.6*4 +
2*500*0.45*8 48/(4s - s) = 120/60 = 2
= 19900 MWh s=8
If all units were working at 100% efficiency Ratio of speed of boat T in still water to
for 10 hours, electricity produced speed of stream = 100: 20 = 5: 1
= 3000*10 = 30000 MWh 112/(5t - t) = 420/60 = 7
Required percentage = t=4

(19900/30000)*100% = 66(1/3)%
Speed Speed
Upstream
Q75 - C Boats
in still of
speed
Downstream
water stream speed (km/h)
(km/h)
Explanation/Solution:- (km/h) (km/h)
Let 'y' 200 MW units were shut down.

8p =
P p=5 35 45
So, 5*200*0.6*4 + 5*200*0.55*4 + (5 - 40
x)*200*0.4*8 + 2*250*0.7*4 + 2*250*0.6*4
+ 2*250*0.45*8 = 10280 5q =
Q q=3 12 18
15
640(5 - x) = 1280
5-x=2 R
5r =
r=5 20 30
25
x=3
Explanation/Solution:- 4s =
S s=8 24 40
32
Ratio of speed of boat P in still water to
speed of stream = 100: 12.5 = 8: 1
65
In 1.8 hours, distance travelled by boat T in

5t =
T t=4 16 24 upstream = 1.8 * (20 - 5) = 27 km
20
Required distance = 36 - 27 = 9 km
Q76 - D
Q78 - A
Let distance = 'D' km
(x/12) + (x/18) = a
(D/35) + (D/45) = 3.52
5x = 36a
(9D + 7D) = 3.52 * 315
3.75y = 36(b + 1.8) ...... (1) [x: y = 3: 4 and
D = 69.3 km (a - b) = 1.8]
Let changed speed of boat P in still water = (y/24) + (y/40) = b
'x' km/h
y = 15b ..... (2)
[69.3/(x + 5)] + [69.3(x - 5)] = 3.52 -
(43.2/60) = 2.8 From (1) and (2):
990(x - 5 + x + 5) = 40(x + 5)(x - 5) 3.75 * 15b = 36(b + 1.8)
1980x = 40x2 - 1000 56.25b = 36b + 64.8
40x2 - 1980x - 1000 = 0 20.25b = 64.8
x = 50 b = 3.2 and a = 5
Required per cent change = [(50 - 40)/40] * Distance travelled by boat S in downstream in
100 = 25% (a + b) hours = 40 * (a + b) = 328 km
Q77 - E Explanation/Solution:-
Explanation/Solution:- After tabulating the information given in the

graphs:
Since, speed of stream for both the boats are
same and both the boats are travelling in One-way
opposite direction. So, speed of stream will Difference in Difference in
Boats distance
not affect their effective speed. speeds (km/h) time (minutes)
(km)
Effective speed of both the boat when

travelling towards each other = 25 + 20 = A 45 20 36
45 km/h
B 80 30 36
Distance between A and B = 45 * (48/60) =
36 km
C 75 15 50
Time taken by boat R to reach point B when
travelling in upstream = 36/(25 - 5) = 1.8
hours
66
b = 65
D 90 30 48
Boat C:
E 60 10 30
[75/(c - 7.5)] - [75/(c + 7.5)] = (50/60) =
(5/6)
Difference between upstream and
downstream distance shows the double of (c + 7.5 - c + 7.5)/(c - 7.5)(c + 7.5) = 1/90
the speed of stream. Since upstream and
distance covered by any boat is same. So, 90 * 15 = c2 - 56.25
time taken in upstream will be more then the
time taken by downstream. c2 = 1406.25
Speed of stream for boat A = 20/2 = 10 c = 37.5

km/h
Boat D:
Speed of stream for boat B = 30/2 = 15
[90/(d - 15)] - [90/(d + 15)] = (48/60) = 0.8
km/h
(d + 15 - d + 15)/(d - 15)(d + 15) = 1/112.5
Speed of stream for boat C = 15/2 = 7.5
km/h 112.5 * 30 = d2 - 225
Speed of stream for boat D = 30/2 = 15 d2 = 3600
km/h
d = 60
Speed of stream for boat E = 10/2 = 5 km/h
Boat E:
Let speed of boats A, B, C, D and E in still
water is 'a', 'b', 'c', 'd' and 'e' respectively. [60/(e - 5)] - [60/(e + 5)] = (30/60) = 0.5
Now, (e + 5 - e + 5)/(e - 5)(e + 5) = 1/120
Boat A: 120 * 10 = e2 - 25
[45/(a - 10)] - [45/(a + 10)] = (36/60) = 0.6 e2 = 1225
(a + 10 - a + 10)/(a - 10)(a + 10) = 1/75 e = 35
75 * 20 = a2 - 100
Speed in Upstream
Downstream
Boats still water Speed
a2 = 1600 Speed (km/h)
(km/h) (km/h)
a = 40
40 - 10 =
Boat B: A a = 40 40 + 10 = 50
30
[80/(b - 15)] - [80/(b + 15)] = (36/60) = 0.6
65 - 15 =
(b + 15 - b + 15)/(b - 15)(b + 15) = 1/100 B b = 65 65 + 15 = 80
50
133(1/3) * 30 = b2 - 225
37.5 - 7.5 = 37.5 + 7.5 =
C c = 37.5
b2 = 3225 30 45
67
Total time by boat C in its journey when the

60 - 15 =
D d = 60 60 + 15 = 75 stream is flowing in its original direction = (D
45
- 6)/18 = 3 hours 50 minutes
E e = 35 35 - 5 = 30 35 + 5 = 40 Q81 - A
Q79 - B
Let speed of stream = 'x' km/h
45/(35 - x) = 1.8
35 - x = 25
[(D - 20)/50] + [(D + 4)/80] = 1.6
x = 10
8D - 160 + 5D + 20 = 1.6 * 400 = 640
Upstream speed of D = 60 - 10 = 50 km/h
13D = 780
Upstream speed of E = 35 - 10 = 25 km/h
D = 60
Effective speed of D and E when travelling in
Decreased upstream speed = 68% of 50 = upstream = 50 - 25 = 25 km/h
34 km/h
Decreased downstream speed = 87.5% of
80 = 70 km/h t = (80 - 30)/25 = 2 hours
Changed speed of boat B in still water = (34 Now,
+ 70)/2 = 52 km/h
Time at which distance between then will be
Required time taken = (2D - 16)/52 = 2 again 30 km = (80 + 30)/25 = 4.4 hours
hours
Required time = 4.4 - 2 = 2.4 hours
Q80 - D
Q82 - E
Total time taken to go 'D' km upstream and
come back 'D - 15' km in downstream by According to the question:
boat C = [D/30] + [(D - 15)/45] = (D - 6)/18
[X/50] + [(X - 12)/80] = [(X - 15)/45] +
Total time taken to go 'D' km upstream and [X/75]
come back 'D - 15' km in downstream by
boat C when stream reverses its direction = 72X + 45X - 540 = 80X - 1200 + 48X
[D/45] + [(D - 15)/30] = (D - 9)/18
11X = 660
X = 60
(D - 6)/18: (D - 9)/18 = 23: 22
Now,
22D - 132 = 23D - 207
[X/50] + [(X - 12)/80] = t
D = 75
(60/50) + (48/80) = t
68
t = 1.2 + 0.6 Person D spends 50% more on food

compare to that on Petrol
t = 1.8 hours
Therefore, the ratio of the expenditures on
Upstream distance travelled by boat C in 't - Food to that on Petrol = 3: 2
0.3' hours = (1.8 - 0.3) * 30 = 45 km
7200 = 18% of 4000
Required per cent = (45/60) * 100 = 75%
The expenditures on Food and petrol
Explanation/Solution:- together = 100 - 18 - 22 = 60%
Total expenditures of A = 80% of 45000 = Expenditures on Food = 3*60/5 = 36%
36000
Expenditures on Petrol = 2*60/5 = 24%
Total expenditures of B = 80% of 40000 =
32000 For the person E,
Total expenditures of C = 80% of 36000 = Expenditures on Food = 24% of 25600 =

28800 6144
Total expenditures of D = 80% of 50000 = The expenditures on Travelling and others

40000 together = 25600 - 6144 - 3072 = 16384 -
--- (i)
Total expenditures of E = 80% of 32000 =
25600 The difference between the expenditures on
Travelling and others = 1024 ---- (ii)
For the person A,
By solving, expenditures on Travelling =
8640 = 24% of 36000 (16384 + 1024)/2 = 8704
The expenditures on Travelling and Petrol Expenditures on others = 8704 - 1024 =
together = 100 - 24 - 18 = 58% 7680
The expenditures on travelling = 58*16/29 =
32% Food Travelling Petrol Others
The expenditures on Petrol = 38*13/29 = 18% of 32% of 26% of

26% A 36000 = 36000 = 36000 = Rs. 8640
Rs. 6480 Rs. 11520 Rs. 9360
For person B,
10240 = 32% of 32000 24% of 16% of 28% of

Rs.
B 32000 = 32000 = 32000 =
The expenditures on travelling = 100 - 32 - 10240
Rs. 7680 Rs. 5120 Rs. 8960
16 - 28 = 24%
For person C, 8% of 24% of 32% of

Rs.
C 28800 = 28800 = 28800 =
10368
10368 = 36% of 28800 Rs. 2304 Rs. 6912 Rs. 9216
Expenditures on Food = 100 - 24 - 36 - 32

= 8%
For person D,
69
Marked price of item B = 150% of 1280 =

36% of
24% of 22% of Rs.1920
40000 =
D Rs. 7200 40000 = 40000 =
Rs.
Rs. 9600 Rs. 8800 Selling price of item C = 12000 * (96/360) =
14400
Rs.3200
24% of Cost price of item C = 3200 * (100/128) =

E 25600 = Rs. 8704 Rs. 3072 Rs. 7680 Rs.2500
Rs. 6144
Marked price of item C = 140% of 2500 =
Q83 - A Rs.3000
Explanation/Solution:- Selling price of item D = 12000 * (90/360) =

Rs.3000
The required difference = 11520 - 9360 =
Rs. 2160 Cost price of item D = 3000 * (100/120) =
Rs.2500
Q84 - B
Marked price of item D = 130% of 2500 =
Explanation/Solution:- Rs.3250
The required sum = 6480 + 10240 + 2304 Selling price of item E = 12000 * (54/360) =
+ 14400 + 6144 = Rs. 39568 Rs.1800
Q85 - B Cost price of item E = 1800 * (100/144) =

Rs.1250
Marked price of item E = 160% of 1250 =
The expenditures of B and C together on Rs.2000
Food = 10240 + 2304 = Rs. 12544
Items Selling price Cost price Marked price
The expenditures of D and E together on
Travelling = 7200 + 8704 = Rs. 15904
A 2400 2000 2700
The required difference = 15904 - 12544 =
Rs. 3360
B 1600 1280 1920
Selling price of item A = 12000 * (72/360) = C 3200 2500 3500

Rs.2400
D 3000 2500 3250
Cost price of item A = 2400 * (100/120) =
Rs.2000
E 1800 1250 2000
Marked price of item A = 135% of 2000 =
Rs.2700 Q86 - D
Selling price of item B = 12000 * (48/360) = Explanation/Solution:-
Rs.1600
Item C:
Cost price of item B = 1600 * (100/125) =
Rs.1280 Actual marked up per cent = 40% [Given]
70
Actual discount per cent = (300/3500) * 100 Total selling price of both the item together =
= (60/7) = 8(4/7)% 122% of (1250 + x) = (1.22x + 1525)
New, marked price = 108(4/7)% of 2500 = Selling price of item F = (1.22x + 1525) -
Rs.(19000/7) 1800 = (1.22x - 275)
New, selling price = 60% of (19000/7) = Marked price of item F = 140% of x = 1.4x
Rs.(11400/7)
Selling price of item F = 80% of 1.4x =
Required per cent = [(11400/7)/2500] * 100 1.12x = 1.22x - 275
= 65%
0.1x = 275
Q87 - A
x = 2750
Cost price of item F = x = Rs.2750
Item D:
Marked price of item F = 1.4x = Rs.3850
Total selling price of both the items together
= 124% of 5000 = Rs.6200 Selling price of item F = 1.12x = Rs.3080
Selling price of second unit = 6200 - 3000 = Total profit amount earned after selling both
Rs.3200 the items = (1800 + 3080) - (1250 + 2750)
= Rs.880
Let marked up and discount per cent is '3x'%
and 'x'% respectively. Total marked up amount of both the items
together = (2000 + 3850) - (1250 + 2750)
Marked price of second unit = (100 + 3x)% = Rs.1850
of 2500 = 25(100 + 3x)
Required ratio = 880: 1850 = 88: 185
Selling price of second unit = (100 - x)% of
25(100 + 3x) = 3200 Explanation/Solution:-
[(100 - x)(100 + 3x)]/4 = 3200 Q89 - C
10000 + 200x - 3x2 = 12800 Explanation/Solution:-
3x2 - 200x + 2800 = 0 Amount of Ramesh after 10 years = (P x r x

t)/100 + P
x = 20
= (65000 x 10 x 10)/100 + 65000
Marked price of second unit = 25(100 + 3x)
= Rs.4000 = 65000 + 65000
Required per cent = (3250/4000) * 100 = = Rs.130000

81.25%
Amount of Suresh after 3 years = P x (1 +
Q88 - D r/100)t
Explanation/Solution:- = 80000 x (1 + 8/100)3
Let cost price of item F = 'x' = 80000 x 108/100 x 108/100 x 108/100
= Rs.100776.96
71
Required difference = 130000 - 100776.96 Per hour work done by Vinay =1/40
= Rs.29223.04
Per hour work done by Abhishek and Satish
Q90 - D together = (1/16) + (1/20) = (5 +4)/ (80)
=9/80
Let Satish and Abhishek worked for 'n' hours
SI for Dinesh after 8 years = (P x r x t)/100
Work done by Satish and Abhishek =(9n)/80
= (75000 x 12 x 8)/100
Per hour work done by Abhishek, Vinay and
= Rs.72000 Satish together = (1/16) + (1/20) + (1/40)
SI for Naresh after 5 years = (P x r x t)/100 = (5+ 4 + 2)/ (80) = (11/80)
= (90000 x 8 x 5)/100 Satish, Vinay and Abhishek worked for (n-3)
hours
= Rs.36000
Work done by Satish, Vinay and Abhishek =
Required ratio = 72000 : 36000 = 2:1 (11 x(n-3))/ (80)
Q91 - D Work done by Vinay and Abhishek = (1/20)
Explanation/Solution:- +(1/40) = (6/80)
Interest earned by Prakash on CI = P x [(1 + Abhishek and Vinay worked for =2(1/6)
r/100)t - 1] =13/6 hours.
= 2/3 x 60000 x [(1 + 6/100)2 - 1] Work done by Vinay and Abhishek = (13/6) x
(6/80) =13/80
= 40000 x 106/100 x 106/100 - 40000
(9n/80) +((11(n-3))/80) + (13/80) =1
= 44944 - 40000
9n + 11 x (n-3) + 13 =80
= Rs.4944
9n +11n -33 +13 =80
Interest earned by Prakash on SI = (P x r x
t)/100 20n =100
= (1/3 x 60000 x 6 x 5)/100 n = 5 hours
= Rs.6000 Total number of hours =5 +2 +(13 /6) = 7 +

(13/6) = 9(1/6) hrs.
Total interest earned by him = 4944 + 6000
= Rs.10944 Q93 - E
Q92 - B Percentage of work completed by Satish in 2

hours = 2 x (1/16) x 100% =12.5%
Percentage of work completed by Vinay in 2
Per work done by Satish =1/16 hours = 2 x (1/40) x 100% = 5%
Per hour work done by Abhishek = 1/20

72
Percentage of work completed by Nikita in (250 + x) = (A + 30) * 19.2 * (5/18)

3.5 hours = 3.5 x (1/35) x 100% = 10%
(250 + x) = 16(A + 30)/3
Percentage of work completed by Shruti in
4.5 hours = 4.5 x (1/36) x 100% = 12.5% (16A/3) - x = 90 ...... (1)
Percentage of work completed by Neha in 5.5 Now,

hours = 5.5 x (1/55) x 100% = 10%
(x + x) = 30 * 36 * (5/18)
Required percentage = (12.5% + 5% + 10%
+ 12.5% +10%) = 50%. x = 150 meters
Q94 - D From equation (1):
Explanation/Solution:- (16A/3) - 150 = 90
Work done by Vinay in 2 hours = 2 x (1/40) A = 45 km/h

=1/20 Let the running speed of Ramesh = 'y' km/h
Work done by Nikita in 7 hours = 7 x (1/35) Effective speed of Ramesh to cross the
=1/5 platform = (45 + y)
Work done by Neha in 16.5 hours = 16.5 x Now,
(1/55) =3/10
150 = (45 + y) * 10.8 * (5/18)
Work done by Vinay, Nikita and Neha =
(1/20) + (1/5) + (3/10) 50 = 45 + y
= (11/20) y = 5 km/h
Work done by Abhishek, Shruti and Satish in Time taken by Ramesh to finish the race =
one hour = (1/20) + (1/36) + (1/16) (400/5) * (1/1000) = 0.08 hours
= (36 + 20 + 45)/ (720) Distance covered by Mahesh in 0.08 hours =
(0.08 * 4) = 320 meters
= 101/720
Distance by which Ramesh beat Mahesh in
Work done in 216 minutes =3.6 x(101/720) the race = B = 400 - 320 - 50
=101/200
B = 30 meters
We have (11/20) = 253
Cost price of cycle = C% of 25000 = 250C
Then (101/200) =?
Amount invested in scheme = (25000 -
=> ? = ((253) x (101/200))/ (11/20) 250C)
=> ? = 232.3 Total worth of cycle after 2 years = (90% of
Explanation/Solution:- 90% of 250C) = 202.5C
Let the length of train B is 'x' meters. Total amount from the scheme after 2 years =
(25000 - 250C) + [{(25000 - 250C) * 20 *
According to the question: 2}/100] = (25000 - 250C) + (10000 -
100C) = (35000 - 350C)
73
According to the question: Q95 - D
202.5C + (35000 - 350C) = 25000 + 4100 Explanation/Solution:-
5900 = 147.5C A = 45 km/h
C = 40 Q96 - B
Amount invested in scheme = (25000 - Explanation/Solution:-

250C) = Rs.15000
B = 30 meters
Amount of interest after 2 years = (15000 *
20 * 2)/100 = Rs.6000 Q97 - D
Investment of Ramesh = Rs.6000 Explanation/Solution:-
Investment of Mahesh = 6000 + 1200 = C = 40%

Rs.7200
Q98 - A
Total profit weightage of Ramesh = (6000 *
5) + (5000 * 7) = 65000 Explanation/Solution:-
Total profit weightage of Mahesh = (7200 * D = Rs.720

5) + (6000 * 7) = 78000 Q99 - C
Ratio of their profit = 65000: 78000 = 5: 6 Explanation/Solution:-
profit amount of Mahesh = D = 1320 * E = 61/125
(6/11)
Q100 - E
D = Rs.720
Probability that he hit the target in an
attempt = (1/5) F = 3 km
Probability that he will win the game =
Probability of hitting the target in first
attempt + Probability of hitting the target in
second attempt + Probability of hitting the
target in third attempt
E = (1/5) + [(4/5) * (1/5)] + [(4/5) * (4/5) *

(1/5)]
E = (1/5) + (4/25) + (16/125)
E = 61/125
Running speed of Mahesh = 4 km/h
Distance between his home from the village

fair = F = 4 * (45/60)
F = 3 km
74
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Directions [Set of 3 questions]: Line graph C) 13/25

Q3. If 10 balls are taken out from bag A and

Note 1: Total number of blue colour balls in C) 3/50

Note 2: Probability of drawing a yellow colour E) None of these

Q2. If one ball from each bag is drawn at

Q6. What is the ratio of the difference

======================== Q7. The ratio in which a shopkeeper must mix

Q8. If Normal and Good quality Apples are D) 2

same company X and employees P, Q, R, S, T and employee P if speed of employees B is

Persons A and P lives together, B and Q lives A) 10 km/h

Line graph given below shows the difference D) 12 km/h

Q14. Person A and C take more time than

of time taken by M and N to cover their C) 20 km

B) 75(1/3)% Fare of Ola vehicles = Base Fare + 5% GST

number of employees who like all three kind C) 100

Q21. What is the difference between the

than and equal to 12 wrong answers then -2

In section X if student attempted 10 wrong

Marks percentage in first test = (total marks

N 22 6 110 Marks percentage in second test = (total

R 25 8 110 C) 311: 299

A) 96 The following pie chart represents the

Q25. Marks obtained by N in section X in

Q26. Which of the following sequence is

D) 40 days reserved for emergency and can't be booked

Q31. If out of total passengers in the bus on ========================

Nike: Users in building B are 6% of total ========================

Levis: Total 2270 users are there out of E) 39: 35

building E are approximately what percent of D) 86

D) 10% The following charts represent the database

Q38. If males in building C are 200 more

C) 77700 Q44. If 20237 people from state Q in 2012

Q43. Average of number of people

will be the part of Oil in total amount of

Q45. When mixture P and Q are mixed in the E) None of these

A) 5: 2 2012 16(2/3)% 60%

E) 4: 3 Q48. What will be the total number of unsold

B) III alone is sufficient. D) Quantity I < Quantity II < Quantity III

E) Quantity I = Quantity II or relation can't be After-Call-Work (ACW) is the average amount

======================== Available time is defined as the amount of

that BPO is 8: 13. If value of Q is 65, then

Q55. If shrinkage percent of BPO E is 20%,

Q57. If only B and D added amount after first

A) 340: 195: 240: 244: 214

B) 240: 214: 340: 244: 195

C) 340: 214: 240: 244: 195

Q58. If only D and E added amount after first

A) 6, 2000, 8, 201: 169 Q59. What is the average of length of all

C) 12, 2000, 2, 109: 129 A) 71.6 cm

D) 8, 4000, 6, 211: 169 B) 73.6 cm

E) None of these C) 75.6 cm

Directions : Study the following information E) 79.6 cm

Q61. What is the difference between the

Rectangle Length: Breadth B) 78 cm

======================== Q63. If speed of boat U in still water is 600%

Q64. Find the difference between time taken

∴ y/9 = 1100/9 S.P of Mixture = (b + 320)(1 - 20/100)(1 -

= (2503 + 3002)/(3 + 2) 48M =480