Mathematics (From www.examveda.

com)
Interest Compound interest
The interest of the previous years is added to the
principal for the calculation of the compound
Simple Interest (SI) interest. In such cases, interest for the first time
It is calculated on the basis of a basic amount interval is added to the principal and this
borrowed for the entire period at a particular amount becomes the principal for the second
rate of interest. The amount borrowed is the time interval, and so on. e.g. A sum of Rs. 100
principal for the entire period of borrowing at 10% per annum will have
Interest (I): Interest is the money paid for the Simple interest ===================
use of money borrowed. Compound interest
Principal (P): The sum borrowed is called the Rs. 100 ====> First year <====== Rs.100;
principal. Rs. 100 ====> 2nd year <======= Rs.110;
Amount (A): The sum of interest and principal Rs. 100 ====> 3rd year <======= Rs.121;
is called Amount. Compound Interest: The difference between the
A=I+P amount and the money borrowed is called the
Rate (r): The interest of 1 year for every Rs. 100 compound interest for given period of time.
is called the Interest rate or rate. If we say "the Formula:
rate of interest per annum is 10%". We meant Case 1: Let principal =P; time =n years; and rate
that Rs. 10 is the interest on a principal of Rs. = r% per annum and let A be the total amount at
100 for a year. the end of n years, then
A = P*[1+ (r/100)]n;
Time (t): The period for which money is
CI = {P*[1+ (r/100)]n -1};
deposited or borrowed is called time. Case 2: When compound interest reckoned half
yearly, then r% become r/2% and time n become
Relation Among Principal, Time, Rate per 2n;
annum and Total interest A= P*[1+ (r/2*100)]2n;
If P is the principal, R is rate; T is time and SI, Case 3: for quarterly,
i.e, the simple Interest. Then A= P*[1+ (r/4*100)]4n;
SI = (P*T*R)/100; P = (SI*100)/(R*T);
R = (SI*100)/ (P*T); Key facts
T = (SI*100)/ (P*R); The difference between compound interest and
Amount = Principal + Total interest; simple interest over two years is given by
Amount = Principal + (P*T*R)/100; Pr2/1002 or P(r/100)2;
Time = [(total interest)/ (interest on the principal The difference between compound interest and
for one year)] *years. simple interest over three years is given by
Note: = P(r/100)2*{(r/100)+3}
The rate of interest is normally specified in Example: If the difference between the simple
terms of annual rate of interest. In such case we interest and the compound interest on the same
take time t for the number of years. However, if sum at 5% per annum for two years is Rs. 25,
what is the sum?
the rate of interest is specified in terms of 6-
Solution:
monthly rate, we take time in terms of 6 months.
Given, difference, d= Rs. 25;
Also, the half-yearly rate of interest is half the R = 5%;
annual rate. That is if the interest is 10% per P =?
annum is to be charged six-monthly, we have to Difference= P(r/100)2;
add interest in every six month @ 5%. ==> 25 =P (5 /1002)
Or, P = (25*100*100)/(5*5);
Or, P = Rs. 10000.

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Depreciation of Value P1=100[1+ (20/100)]3;
The value of machine or any other article P1= 100*[6/5]3;
subject to wear and tear decreases with the time. P1= 100*6*6*6/5*5*5= 172.8
This decrease is called its depreciation. Net percentage change graphics is quite time
Thus if V0 is the value at a certain time and r% saving in exam situations for anything more
per annum is the rate of depreciation per year, than 3 years.
then the value V1 at the end of t years is, Example: we need to calculate the CI on Rs. 100
V1 = V0 *[1-(r/100)]t. at the rate of 10% per annum for a period of 10
Value of machine t years ago = V0 / [1-(r/100)]t. years
Solution:
Population If we go through formula of compound interest,
Concept of Depreciation and Population is the we get
just expansion of concept of compound interest. CI = [P0*[1+ (r/100)]n -1];
Let the population of a town be P now and CI = [100* (1+ (10/100)) 10 -1];
suppose it increases at the rate of R% per CI = [100*(11/10)10-1]; ............ (1)
annum, then: Particularly, this calculation is quite time
a) Population after n years = P*[1+ (r/100)]n; consuming.
b) Population n years ago = P/[1+ (r/100)]n; But, if we use net percentage change graphics
If the present population P decreases at the rate then we can save some time.
of R% per annum, then: 100==10%(increase)==>110==10%(increase)=
n
a) Population after n years = P*[1- (r/100)] ; =>121==10%(increase)==>133.1==10%(increa
b) Population n years ago = P/[1- (r/100)]n; se)==>146.4
If the present population P increase at the rate of ==10%(increase)==>161.04==10%(increase) =
R1% for the first year and decreases at the rate =>177.14==10%(increase)==> .........
with rate of R2 for second year and again 214.3==10%(increase)==>235.7==10%(increas
increases with the rate of R3% for the third year, e)==>259.2
then: After 10 years Rs. 100 become Rs. 259.2
a) Population after 3 years = P[1+(R1/100)]*[1- (approx.), that means CI would be 159.2
(R2/100)]*[1+(R3/100)]
b) If the same is reversed from now , then:
Population 3 years ago,
= P/ {[1+ (R1/100)]*[1-(R2/100)]*[1+
(R3/100)]}.

Relation between Net Percentage change

graphic and Compound Interest
Suppose, a company increases its sales by 20%
in the first year and then again increases its sales
by 20% in the second year and also third year.
It can be visualized by,
100===20%
(increase)===>120===20%(increase)===>144=
==20%(increase)===>172.8
This calculation is very similar to the calculation
of compound interest,
P1= P0*[1+ (r/100)]n;

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1. Find the simple interest on Rs. 5200 for 2 Solution: We can use formula of compound
years at 6% per annum. interest; A = P*[1+ (r/100)]n
A. Rs. 450 A = 2100*[1+(5/100)]2
B. Rs. 524 A = 2100*[105/100]2
C. Rs. 600 A = (2100*11025)/10000
D. Rs. 624 Hence, Amount, A = Rs. 2315.25.
Answer: Option D
Solution: I = PTR/100
I = 5200*2*6/100 5. Find the difference between the simple
I = 624. interest and the compound interest at 5% per
annum for 2 years on principal of Rs. 2000.
2. Rs. 1200 is lent out at 5% per annum simple A. 5
interest for 3 years. Find the amount after 3 B. 10.5
years. C. 4.5
A. Rs. 1380 D. 5.5
B. Rs. 1290 E. None of these
C. Rs. 1470
Answer: Option A
D. Rs.1200
Solution: The difference between compound
E. Rs. 1240
Answer: Option A interest and simple interest over two years is
Solution: A = P+I given by
A = 1200+(PTR/100) Pr2/1002 or P(r/100)2
A = [1200+(1200*5*3/100)] Here, Principal (P) = Rs. 2000
Amount, A = Rs. 1380. Rate (r) = 5%
Now difference, D = (2000*5*5)/(100*100)
3. Interest obtained on a sum of Rs. 5000 for 3 D = Rs. 5.
years is Rs. 1500. Find the rate percent.
A. 8% 6. Find the rate of interest if the amount after 2
B. 9% years of simple interest on a capital of Rs. 1200
C. 10% is Rs. 1440.
D. 11% A. 8%
E. 12% B. 9%
Answer: Option C C. 10%
Solution: Let rate is R%. D. 11%
We have, I = PTR/100 E. 12%
Here, 1500 = 5000*3*R Answer: Option C
Thus, R = 10%. Solution: Amount, A = Rs. 1440
Principal, P = Rs. 1200
4. Rs. 2100 is lent at compound interest of 5% Interest, I = Rs. (1440-1200) = 240
per annum for 2 years. Find the amount after R = (240*100)/(1200*2) = 10%.
two years.
A. Rs. 2300
Alternatively,
B. Rs. 2315.25
We can go through a thought process i.e.
C. Rs. 2310
D. Rs. 2320 1200----20% (240 in 2 years)--->1440.
E. None of these That means 10% rise in each year.
Answer: Option B

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7. What is the difference between the simple 9. Find the compound interest on Rs. 1000 at the
interest on a principal of Rs. 500 being rate of 20% per annum for 18 month when
calculated at 5% per annum for 3 years and 4% interest is compounded half yearly.
per annum for 4 years? A. Rs. 331
A. Rs. 5 B. Rs. 10 B. Rs. 1331
C. Rs. 20 D. Rs. 40 C. Rs. 320
E. None of these D. Rs. 325
Answer: Option A E. None of these
Solution: I = PT R /100 Answer: Option A
I = (500*3*5)/100 = Rs. 75 Solution: Given, principal, P = Rs. 1000
I = PT R /100 Compound rate, R = 20% per annum = 20/2 =
I = (500*4*4)/100 = Rs. 60 10% half-yearly
Difference = Rs. 5. Time = 18 month = 3 half-years
Amount, A = {P*[1+ (R/100)]n }
Alternatively, = {1000*[1+(10/100)]3}
The interest is calculated simply and then it will = {(1000*11*11*11/10*10*10)}
have a rise of 15% in first case and 16% in 2nd A = Rs. 1331
case. Hence, compound interest = Rs. 331.
Difference = 1% 0n 500 = Rs. 5.
Otherwise, 500-----15% ----> 575 (1st case); 10. Find the principal if the interest
500-----16% ----> 580 (2nd case); compounded at the rate of 10% per annum for
We can see clear difference of Rs. 5. two years is Rs. 420.
A. Rs. 1000
8. What is the simple interest on a sum of Rs. B. Rs. 2200
700 if the rate of interest for the first 3 years is C. Rs. 2000
8% per annum and for the last 2 years is 7.5% D. Rs. 1100
per annum? E. Rs. 1200
A. Rs. 269.5 B. Rs. 283 Answer: Option C
C. Rs. 273 D. Rs. 280 Solution: Given,
E. None of these Compound rate, R = 10% per annum
Answer: Option C Time = 2 years
Solution: 1st case: CI = Rs. 420
I = 700*3*8/100 = Rs. 168 Let P be the required principal.
2nd case: A = (P+CI)
I = 700*2*7.5/100 = Rs. 105 Amount, A = {P*[1+ (R/100)]n }
2
Then total interest for five years = (I +I ) = Rs. (P+CI) = {P*[1+10/100] 2
}
273 (P+420) = P*[11/10]
P-1.21P = -420
Alternatively, 0.21P = 420
As interest is calculated as simple interest so we Hence, P = 420/0.21 = Rs. 2000.
can add up rates for all given 5 years and
calculate it easily i.e. Alternatively,
For the five years rate = (8*3+7.5*2) =39%. We can go through the method option checking.
Now, 700-----39% ---> 973. In option checking method, it is always better to
Interest= Rs. 273. choose the middle option first to proceed.
(The thought can go this way, we internally 2000(P)---10% --- 2200-----10% --
calculated 10% of 700 = (700/10) = 70. 2420(A).
Then, 39% 0f 700 = (40%- 1%) of 700 Then, required principal is Rs. 2000 as it is
= (280-7) = 273.) giving CI Rs. 420 for two years.

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11. In what time will Rs. 3300 becomes Rs. Answer: Option B
3399 at 6% per annum interest compounded Solution: Interest for the last 5 years = PTR/100;
half-yearly? = 360*5*6/100 = Rs. 108.
A. 6 months B. 1 year Interest for year = 540-360 = 180;
C. 1(1/2) years D. 3 months So, interest for first four years
E. None of these = 180-108 = Rs. 72;
Answer: Option A Now, rate for first four years
Solution: P = Rs. 3300 = (72*100)/360*4 = 5%.
A = Rs. 3399
R = 6% per annum 14. What will be the simple interest on Rs. 700
Let the time be n years. at 9% per annum for the period from February
Compound interest is taken half-yearly. 5, 1994 to April 18, 1994?
A = P*[1+ (R/2*100)]2n A. Rs. 12.60 B. Rs. 11.30
3399 = 3300(1+3/100)2n C. Rs. 15 D. Rs. 13
(1.03)2n = 3399/3300 E. None of these
(1.03)2n = (1.03)1 Answer: Option A
Thus, 2n = 1 year Solution: Here, time interval is given as
n = 1/2 year = 6 months. February 5, 1994 to April 18, 1994
= 72 days = 72/365 = 0.197 years;
Alternatively, Now, interest = PTR/100
3300-----3% (1st time interval, 99)--- 3399. = (700*9*0.197)/100 = Rs. 12.41.
Here, time interval is given as half-yearly i.e. 6
months. 15. Asif borrows Rs. 1500 from two
moneylenders. He pays interest at the rate of
12. Rahul purchased a Maruti van for Rs. 1, 12% per annum for one loan and at the rate of
96,000 and the rate of depreciation is 14(2/7) % 14% per annum for the other. The total interest
per annum. Find the value of the van after two he pays for the entire year is Rs. 186. How
years. much does he borrow at the rate of 12%
A. Rs. 1,40,000 B. Rs.1,44,000 A. Rs. 1200
C. Rs. 1,50,000 D. Rs. 1,60,000 B. Rs.1300
E. None of these C. Rs. 1400
Answer: Option B D. Rs. 300
Solution: Value of Maruti Van, V0= Rs. 196000 E. None of these
Rate of depreciation, r = 14(2/7)% = 100/7%; Answer: Option A
Time, t = 2 years Solution: Let Asif lent Rs. X at 14% per year.
Let V1 is the value after depreciation. Hence, Money lent at 12% = (1500-x);
V1 = V0*[1-(r/100)]t Given, total interest = Rs. 186.
V1 = 196000*[1-((100/7)/100)]2 {(X*14*1)/100}
V1 = 196000*(6/7)2 + {[(1500-x)*12*1/100]} = 186;
V1 = (196000*36)/49 = Rs. 144000. 14x/100 + (18000 -12x)/100 = 186;
14x+18000-12x = 186*100;
13. What is the rate of simple interest for the 2x = 18600-18000;
first 4 years if the sum of Rs. 360 becomes Rs. X = 600/2 = Rs. 300.
540 in 9 years and the rate of interest for the last Hence, money lent at 12% = 1500-300
5 years is 6%? = Rs. 1200.
A. 4% B. 5%
C. 3% D. 6%
E. 3.5%

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16. A sum was invested at simple interest at a 1750*9*T/100 = (2500*10.5*4)/100;
certain interest for 2 years. It would have Or, T = (2500*10.5*4)/1750*9;
fetched Rs. 60 more had it been invested at 2% Or, T = 6.66 = 6 years and 8 months.
higher rate. What was the sum?
A. Rs. 1500 B. Rs. 1300 19. Raju lent Rs. 400 to Ajay for 2 years and Rs.
C. Rs. 2500 D. Rs. 1000 100 to Manoj for 4 years and received together
Answer: Option A from both Rs. 60 as interest. Find the rate of
Solution: Let the rate be R at which Principal P interest, simple interest being calculated.
is invested for 2 years. A. 5% B. 6%
C. 8% D. 9%
According to question, Answer: Option A
(Interest at Rate (R+2)) % - (interest at rate R Solution:
%) = Rs. 60; Let rate is R%.
(P*2*(R+2))/100 - (P*2*R)/100 = 60; According to the question,
(2PR+4P-2PR)/100 = 60; [400*2*R/100] + [100*4*R/100] = 60;
4P = 60*100; 8R+4R = 60;
Or, P = 60*100/4; Hence, R = 5%.
Hence, P = Rs. 1500.
20. A sum becomes 4 times at simple interest in
17. The difference between simple and 10 years. What is the rate of interest?
compound interest on a sum of money at 20% A. 10% B. 20%
per annum for 3 years is Rs. 48. What is the C. 30% D. 40%
sum? Answer: Option C
A. Rs. 550 B. Rs. 500 Solution: 1st Method:
C. Rs. 375 D. Rs. 400 Let rate is R%.
Answer: Option C Now, P = 100; A = 400; I = 400-100 = 300;
Solution: Let sum is P. Time, T = 10 years;
The difference between compound interest and I = PTR/100;
simple interest over three years is given by Or, R = (100*I)/PT;
= P(r/100)2*{(r/100)+3}; Or, R = (100*300)/(100*10);
48 = P*(20/100)2*{(20/100)+3}; Hence, R = 30%;
48 = P* 4/100* 16/5;
48 = P*64/500; 2nd Method:
Or, 64P = 48*500; Here, the sum becomes 4 times that means 100,
Hence, P = Rs. 375. become 400;
Rate of such question is given by;
18. In what time will the simple interest on Rs. R= interest/time = 300/10 = 30%;
1750 at 9% per annum be the same as that on
Rs. 2500 at 10.5% per annum in 4 years? 3rd Method:
A. 6 years and 8 months Here, 300% of rise in the sum,so
B. 7 years and 3 months 100------300% ----400;
C. 6 years R=(total percentage rise/given time) = 300% /10
D. 7 years and 6 months = 30%.
Answer: Option A
Solution: Let time is T years.
According to questions,

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21. A sum of money placed at compound 23. A sum of money becomes 7/4 of itself in 6
interest doubles itself in 4 years. In how many years at a certain rate of simple interest. Find the
years will it amount to 8 times? rate of interest.
A. 9 years B. 8 years A. 12%
C. 27 years D. 12 years B. 12(1/2)%
Answer: Option D C. 8%
Solution: Let, D. 14%
Principal = Rs. 100. Answer: Option B
Amount = Rs. 200. Solution: Let sum is Rs. 100 ,then it become 7/4
Rate = r% times i.e. Rs. 700/4 in 6 years.
Time = 4 years. Interest = (700/4)-100 = Rs. 300/4;
Now, Hence, Rate = Total interest/ given time =
A = P*[1+ (r/100)]n; 300/4*6 = 12(1/2)%.
200 = 100*[1+(r/100)]4;
2 = [1+(r/100)]4; ........... (i) 24. If a certain sum of money becomes doubles
If sum become 8 times in the time n years, at simple interest in 12 years, what would be the
then, 8 = (1+(r/100))n; rate of interest per annum?
23 = (1+(r/100))n; ........ (ii) A. 8(1/3)
Using eqn (i) in (ii), we get; B. 10
([1+(r/100)]4)3 = (1+(r/100))n; C. 12
[1+(r/100)]12 = (1+(r/100))n; D. 14
Thus, n = 12 years. Answer: Option A
Solution: Let,
22. Divide Rs. 6000 into two parts so that Principal, P = Rs. 100;
simple interest on the first part for 2 years at 6% Amount, A = Rs. 200;
p.a. may be equal to the simple interest on the Time = 12 years;
second part for 3 years at 8% p.a. Interest = Rs. 100;
A. Rs. 4000, Rs. 2000 Rate of interest = total interest / given time =
B. Rs. 5000, Rs. 1000 100/12 = 8(1/3)%.
C. Rs. 3000, Rs. 3000
D. None of these 25. If a sum of Rs. 13040 is to be repaid in two
Answer: Option A equal installments at 3(3/4)% per annum, what
st nd
Solution: Let 1 part is x and 2 part is (6000- is the amount of each installment?
x). A. 7045 B. 8000
C. 65067 D. 6889
According to question, Answer: Option D
(X*2*6/100) = ((6000-x)*3*8)/100; Solution: Let each installment be P.
12x = 144000- 24x; Hence,
Or, 36x = 144000; [x/{(100/100+r) + (100/100+r)2}];
Or, x = 144000/36 = Rs. 4000. Or, x/(1+(15/400)) + x/(1+(15/400)2) = Rs.
1st part = Rs. 4000; 13040;
2nd part = Rs. 2000. On solving, it gives, x = Rs. 6889.

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26. What is the amount of equal installment, if a 28. The population of vultures in a particular
sum of Rs. 1428 due 2 years hence has to be locality is decreases by certain rate of interest
completely repaid in 2 equal annual installments (compounded annually). If the current
starting next year?. population of vultures be 29160 and the ratio of
A. 700 decrease in population for second year and 3rd
B. 800 year be 10:9. What was the population of
C. 650 vultures 3 years ago?
D. cannot be determined A. 30000 B. 35000
Answer: Option D C. 40000 D. 50000
Solution: As short-cut of installment for 2 E. Cannot determined
installments is given by, Answer: Option C
Installment = [P/{(100/100+R) + Solution: Decrease in second / decrease in third
(100/100+R)2}] year,
There is the need of rate (R) which is = 100/(100-r) = 10/9
unavailable in the question so, we cannot r = 10%
determine the answer. Let the population of vultures 3 years ago be P,
then
27. A milkman sells cow milk at the rate of Rs. P*[1-(10/100)]3 = 29160.
55 litre including a profit 12 per cent. He also P = 40000.
sells buffalo milk at the rate of Rs. 36 per litre
including a profit of 20%. How much profit will Alternatively,
he earn in five days if he sells 8 litres of cow Once, we get the rate of decrement of
milk and 10 litres of buffalo milk per day? population, we can use graphical method to
A. Rs. 632 calculate.
B. Rs. 624 100 ----10% --->90 ----10% --->81----10% ---
C. Rs. 646 >72.9
D. Rs. 642 Now comparing,
E. None of these 72.9% = 29160.
Answer: Option E So, 100% = 40000.
Solution: Total cow milk sold in five days = 5*8
= 40 litres 29. A man had 1000 hens at the beginning of
Total buffalo milk sold in five days = 5*10 = 50 year 2001 and the number of hens each year
litres increases by 10% by giving birth. At the end of
hence, each year we double the no. of hens by
SP of 40 litres cow milk = 40*55 = Rs. 2200. purchasing the same no. of hens as there is the
Hence, no. of hens with us at the time. What is the no.
Profit on cow milk = (2200*12)/112 = Rs. of hens at the beginning of 2004?
235.71 A. 10600 B. 10648
Profit on buffalo milk = (50*36*20)/120 = Rs. C. 8848 D. 8226
300 E. None of these
Thus, Answer: Option B
Total Profit = 235.71+300 = Rs. 535.71. Solution: 1000 ---10% --->1100----Double---
>2200 ---10% ---> 2420----Double--->4840 ---
10% --->5324----Double--->10648

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30. The difference between simple and 72.9=Rs.5832
compound interest for the fourth year is Rs. 100=Rs.x
7280 at 20% p.a. What is the principal sum? by cross multiplication
A. 10000 B. 50000
x=(5832*100)/72.9
C. 70000 D. 40000
Answer: Option B we get,
Solution: Difference between Ci and SI for x=8000=worth of article three years ago.
nth year,
= (Pr /100) *[(1 +(r /100))n-1 -1] 33. A sum of money at simple interest amounts
7280 = (20P/100) *[(1.2)3-1] to Rs. 815 in 3 years and to Rs. 854 in 4 years.
P = 50000. The sum is:
A. Rs. 650 B. Rs. 690
31. A sum of Rs. 210 was taken as a loan. This C. Rs. 698 D. Rs. 700
is to be paid back in two equal installments. If E. None of these
the rate of the interest be 10% compounded Answer: Option C
annually, then the value of each installment is: Solution:
A. Rs. 121 B. Rs. 127 S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
C. Rs. 210 D. Rs. 225 S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Answer: Option A Therefore Principal = Rs. (815 - 117) = Rs. 698.
Solution: Let X = equal installment at the end of
one year( rate% annually) . 34. Mr. Thomas invested an amount of Rs.
Now 1st year, 13,900 divided in two different schemes A and
P =210, B at the simple interest rate of 14% p.a. and
Interest = PTR/100 = 210 *0.1 = 21. 11% p.a. respectively. If the total amount of
Let X is to be paid as an equal installment. simple interest earned in 2 years be Rs. 3508,
Now, at the beginning of 2nd year, what was the amount invested in Scheme B?
P =210+21- X, A. Rs. 6400 B. Rs. 6500
Interest at the end of 2nd year, C. Rs. 7200 D. Rs. 7500
= (231-X)*0.1 = 23.1-0.1X. E. None of these
Hence,total installment, Answer: Option A
2X = 210+21+23.1- 0.1X, Solution: Let the sum invested in Scheme A be
X = 254.1/2.1 = 121. Rs. x and that in Scheme B be Rs. (13900 - x).
Then,
32. There is a decrease of 10% yearly on an
article. If this article was bought 3 years ago and
present cost is Rs. 5,832 then what was the cost
of article at buying time? ⇒ 28x - 22x = 350800 - (13900 x 22)
A. Rs. 7,200 B. Rs. 7,862 ⇒ 6x = 45000
C. Rs. 8,000 D. Rs. 8,500 ⇒ x = 7500.
Answer: Option C So, sum invested in Scheme B = Rs. (13900 -
Solution: use effective percent formula for 10% 7500) = Rs. 6400.
for three years we got 27.1% decreased worth of
article
100-27.1=72.9=present worth of article

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35. A sum fetched a total simple interest of Rs. 38. A sum of Rs. 12,500 amounts to Rs. 15,500
4016.25 at the rate of 9 p.c.p.a. in 5 years. What in 4 years at the rate of simple interest. What is
is the sum? the rate of interest?
A. Rs. 4462.50 B. Rs. 8032.50 A. 3%
C. Rs. 8900 D. Rs. 8925 B. 4%
E. None of these C. 5%
Answer: Option D D. 6%
Solution: Principal E. None of these
Answer: Option D
Solution: S.I. = Rs. (15500 - 12500)
= Rs. 3000.

39. An automobile financier claims to be

36. How much time will it take for an amount of lending money at simple interest, but he
Rs. 450 to yield Rs. 81 as interest at 4.5% per includes the interest every six months for
annum of simple interest? calculating the principal. If he is charging an
A. 3.5 years B. 4 years interest of 10%, the effective rate of interest
C. 4.5 years D. 5 years becomes:
E. None of these A. 10%
Answer: Option B B. 10.25%
Solution: Time = C. 10.5%
D. None of these
Answer: Option B
Solution: Let the sum be Rs. 100.
Then,
37. Reena took a loan of Rs. 1200 with simple S.I. for first 6 months =
interest for as many years as the rate of interest.
If she paid Rs. 432 as interest at the end of the
loan period, what was the rate of interest?
A. 3.6 B. 6
C. 18 D. Cannot be determined S.I. for last 6 months =
E. None of these
Answer: Option B
Solution: Let rate = R% and time = R years.

So, amount at the end of 1 year

= Rs. (100 + 5 + 5.25)
= Rs. 110.25
∴ Effective rate
= (110.25 - 100)
= 10.25%

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 Mathematics (From www.examveda.com)
40. A lent Rs. 5000 to B for 2 years and Rs. 42. A man took loan from a bank at the rate of
3000 to C for 4 years on simple interest at the 12% p.a. simple interest. After 3 years he had to
same rate of interest and received Rs. 2200 in all pay Rs. 5400 interest only for the period. The
from both of them as interest. The rate of principal amount borrowed by him was:
interest per annum is: A. Rs. 2000 B. Rs. 10,000
A. 5% B. 7% C. Rs. 15,000 D. Rs. 20,000
C. 9% D. 10% Answer: Option C
E. None of these Solution: Principal =
Answer: Option D
Solution: Let the rate be R% p.a.
Then,

43. A sum of money amounts to Rs. 9800 after 5

years and Rs. 12005 after 8 years at the same
rate of simple interest. The rate of interest per
annum is:
A. 5% B. 8% C. 12% D. 15%
E. None of these
Answer: Option C
Solution: S.I. for 3 years = Rs. (12005 - 9800) =
41. A sum of Rs. 725 is lent in the beginning of
Rs. 2205.
a year at a certain rate of interest. After 8
S.I. for 5 years =
months, a sum of Rs. 362.50 more is lent but at
the rate twice the former. At the end of the year,
Rs. 33.50 is earned as interest from both the
loans. What was the original rate of interest? ∴ Principal = Rs. (9800 - 3675) = Rs. 6125.
A. 3.6% B. 4.5% Hence, rate =
C. 5% D. 6%
E. None of these
Answer: Option E
Solution: Let the original rate be R%. Then, new
44.
rate = (2R)%.
What will be the ratio of simple interest earned
Note:
by certain amount at the same rate of interest for
Here, original rate is for 1 year(s); the new rate
6 years and that for 9 years?
is for only 4 months i.e. 1/3 year(s).
A. 1 : 3 B. 1 : 4
C. 2 : 3 D. Data inadequate
E. None of these
Answer: Option C
Solution: Let the principal be P and rate of
interest be R%.
Therefore Required ratio =

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 Mathematics (From www.examveda.com)
45. A certain amount earns simple interest of
Rs. 1750 after 7 years. Had the interest been 2%
more, how much more interest would it have
earned?
A. Rs. 35
B. Rs. 245
C. Rs. 350
D. Cannot be determined
E. None of these
Answer: Option D
Solution: We need to know the S.I., principal
and time to find the rate.
Since the principal is not given, so data is
inadequate.

46. A person borrows Rs. 5000 for 2 years at

4% p.a. simple interest. He immediately lends it
to another person at 6(1/4) p.a for 2 years. Find
his gain in the transaction per year.
A. Rs. 112.50
B. Rs. 125
C. Rs. 150
D. Rs. 167.50
Answer: Option A

Solution: Gain in 2 years

∴ Gain in 1 year

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