IIFT Compound Intrest Questions PDF [Important]
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Question 1: What is the difference (in ₹) between the compound interest, when interest is compounded 6-monthly, and the simple interest on a sum of ₹10,000 for $1\frac{1}{2}$ years at 10% p.a.?
a) 102.25
b) 87
c) 76.25
d) 91.5
1) Answer (C)
Solution:
If interest is half-yearly, Then
Time = $1.5\times\ 2=3$
Rate = $\frac{10}{2}=\ 5\%$
$A=P\left(1+\frac{r}{100}\right)^n$
= $A=10,000\left(1+\frac{5}{100}\right)^3$
$A=10,000\ \times\ \frac{21}{20}\times\ \frac{21}{20}\times\ \frac{21}{20}=11576.25$
C.I = A – P = 11576.25 – 10000 = ₹1576.35
Simple Interest = $S.I\ =\frac{\left(P\times\ R\times\ T\right)}{100}$
= $\frac{\left(10000\times\ 10\times\ 1.5\right)}{100}$
= ₹1500
Required Difference = ₹1576.25 – ₹1500 = ₹76.25
Hence, Option C is correct.
Question 2: A sum of ₹25600 is invested on simple interest partly at 7% per annum and the remaining at 9% per annum. The total interest at the end of 3 years is ₹5832. How much money(in ₹) was invested at 9% per annum?
a) 18000
b) 7600
c) 9600
d) 16000
2) Answer (B)
Solution:
Let the amount invested in 9% per annum = P
Amount invested in 7% per annum = 25600 – P
The total interest at the end of 3 years is ₹5832.
P x 3 x $\frac{9}{100}$ + (25600 – P) x 3 x $\frac{7}{100}$ = 5832
$\frac{9}{100}$P + 1792 – $\frac{7}{100}$P = 1944
$\frac{2}{100}$P = 152
P = ₹7600
$\therefore$ Amount invested at 9% per annum = P = ₹7600
Hence, the correct answer is Option B
Question 3: Two equal sums were lent on simple interest at 6% and 10% per annum respectively. The first sum was recovered two years later than the second sum and the amount in each case was ₹1105. What was the sum (in ₹) lent in each scheme?
a) 900
b) 850
c) 936
d) 891
3) Answer (B)
Solution:
Let the sum lent in each scheme = P
Let the sum at 6% and 10% simple interest was recovered after ‘t+2’ years and ‘t’ years respectively.
Amount obtained from simple interest at 6% = P + $\frac{P\times(t+2)\times6}{100}$
1105 = P + $\frac{P\times(t+2)\times6}{100}$
1105 – P = $\frac{P\times(t+2)\times6}{100}$…….(1)
Amount obtained from simple interest at 10% = P + $\frac{P\times t\times10}{100}$
1105 = P + $\frac{P\times t\times10}{100}$
1105 – P = $\frac{P\times t\times10}{100}$……..(2)
From (1) and (2),
$\frac{P\times(t+2)\times6}{100}$ = $\frac{P\times t\times10}{100}$
6t + 12 = 10t
t = 3
Substituting t = 3 in equation (2), we get P = 850
The sum lent in each scheme = ₹850
Hence, the correct answer is Option B
Question 4: A sum of ₹9500 amounts to ₹11495 in 2 years at a certain rate percent per annum, interest compounded yearly. What is the simple interest (in ₹) on the same sum for the same time and double the rate?
a) 3990
b) 3420
c) 4560
d) 3800
4) Answer (D)
Solution:
Let the rate of interest = R
11495 = 9500(1+$\frac{R}{100}$)$^2$
(1+$\frac{R}{100}$)$^2$ = $\frac{11495}{9500}$
(1+$\frac{R}{100}$)$^2$ = $\frac{121}{100}$
1+$\frac{R}{100}$ = $\frac{11}{10}$
$\frac{R}{100}$ = $\frac{1}{10}$
R = 10%
Simple interest = $\frac{9500\times2\times20}{100}$
= ₹3800
Hence, the correct answer is Option D
Question 5: A person borrowed a sum of ₹30800 at 10% p.a. for 3 years, interest compounded annually. At the end of two years, he paid a sum of ₹13268. At the end of 3rd year, he paid ₹ x to clear of the debt. What is the value of x ?
a) 26400
b) 26510
c) 26200
d) 26620
5) Answer (A)
Solution:
Amount to be paid after 2 years = 30800(1+$\frac{10}{100}$)$^2$
= 30800($\frac{11}{10}$)$^2$
= 30800($\frac{121}{100}$)
= ₹37268
Amount paid by the person = ₹13268
Remaining amount = ₹37268 – ₹13268 = ₹24000
Amount to be paid at the end of 3rd year to clear debt(i.e, compound interest on ₹24000 for next 1 year) = 24000(1+$\frac{10}{100}$)$^1$
$\Rightarrow$ x = 24000($\frac{11}{10}$)
$\Rightarrow$ x = ₹26400
Hence, the correct answer is Option A
Question 6: At what rate percent per annum will ₹7200 amountto ₹7938 in one year, if interest is compounded half yearly?
a) 5
b) 8
c) 12
d) 10
6) Answer (D)
Solution:
Let the rate of interest per annum = R
According to the problem,
7938 = 7200(1+$\frac{\frac{R}{2}}{100}$)$^2$
3969 = 3600(1+$\frac{\frac{R}{2}}{100}$)$^2$
441 = 400(1+$\frac{\frac{R}{2}}{100}$)$^2$
(1+$\frac{\frac{R}{2}}{100}$)$^2$ = $\frac{441}{400}$
1+$\frac{\frac{R}{2}}{100}$ = $\frac{21}{20}$
$\frac{R}{200}$ = $\frac{1}{20}$
R = 10%
Hence, the correct answer is Option D
Question 7: A sum at simple interest becomes two times in 8 years at a certain rate of interest p.a. The time in which the same sum will be 4 times at the same rate of interest at simple interest is:
a) 25 years
b) 20 years
c) 30 years
d) 24 years
7) Answer (D)
Solution:
Let the principal amount = P
Rate of interest = R
Time = 8 years
Amount = 2P
$\Rightarrow$ 2P = Principal amount + Simple Interest
$\Rightarrow$ 2P = P + $\frac{\text{P}\times8\times \text{R}}{100}$
$\Rightarrow$ P = $\frac{\text{P}\times8\times \text{R}}{100}$
$\Rightarrow$ R = 12.5%
Let the time in which sum will be 4 times at the same rate of interest = T
i.e, Amount = 4P
$\Rightarrow$ 4P = P + $\frac{\text{P}\times \text{T}\times12.5}{100}$
$\Rightarrow$ 3P = $\frac{\text{P}\times \text{T}\times12.5}{100}$
$\Rightarrow$ 3 = $\frac{\text{T}}{8}$
$\Rightarrow$ T = 24 years
$\therefore\ $The time in which the same sum will be 4 times at the same rate of interest = 24 years
Hence, the correct answer is Option D
Question 8: Suresh lent out a sum of money to Rakesh for 5 years at simple interest. At the end of 5 years, Rakesh paid 9/8 of the sum to Suresh to clear out the amount. Find the rate of simple interest per annum.
a) 3.5% p.a.
b) 2.5% p.a.
c) 3% p.a.
d) 2% p.a.
8) Answer (B)
Solution:
Let the sum of money lent out by Suresh to Rakesh = P
Amount paid by Rakesh to Suresh after 5 years = $\frac{9}{8}$P
Let the rate of simple interest = R%
$\Rightarrow$ P + $\frac{\text{P}\times5\times \text{R}}{100}$ = $\frac{9}{8}$P
$\Rightarrow$ 1 + $\frac{R}{20}$ = $\frac{9}{8}$
$\Rightarrow$ $\frac{R}{20}$ = $\frac{9}{8}-1$
$\Rightarrow$ $\frac{R}{20}$ = $\frac{1}{8}$
$\Rightarrow$ R = 2.5%
$\therefore\ $Rate of simple interest per annum = 2.5%
Hence, the correct answer is Option B
Question 9: If the difference between the compound interest and simple interest on a certain sum of money for three years at 10% p.a. is ₹ 558, then the sum is:
a) ₹ 18,500
b) ₹ 15,000
c) ₹ 16,000
d) ₹ 18,000
9) Answer (D)
Solution:
Let the principal sum = P
Rate = 10%
Time = 3 years
Compound interest on the sum = P$\left(1+\frac{10}{100}\right)^3-$ P = P$\left(\frac{110}{100}\right)^3-$ P = P$\frac{1331}{1000}-$ P = $\frac{331}{1000}$P
Simple interest on the sum = $\frac{P\times3\times10}{100}$ = $\frac{3}{10}$P
According to the problem,
$\frac{331}{1000}$P $-$ $\frac{3}{10}$P = 558
$\Rightarrow$ $\frac{331P-300P}{1000}$ = 558
$\Rightarrow$ $\frac{31P}{1000}$ = 558
$\Rightarrow$ P = 18000
$\therefore\ $The principal sum = ₹ 18,000
Hence, the correct answer is Option D
Question 10: The sum of simple interest on a sum at 8% p.a. for 4 years and 8 years is ₹960. The sum is:
a) ₹800
b) ₹1100
c) ₹1000
d) ₹900
10) Answer (C)
Solution:
Let the Principal amount = P
Rate = 8%
Given, the sum of simple interest on P for 4 years and 8 years is ₹960
$\Rightarrow$ $\frac{P\times4\times8}{100}+\frac{P\times8\times8}{100}=960$
$\Rightarrow$ $\frac{32P}{100}+\frac{64P}{100}=960$
$\Rightarrow$ $\frac{96P}{100}=960$
$\Rightarrow$ P = 1000
$\therefore\ $The required sum = ₹1000
Hence, the correct answer is Option C
Question 11: The difference between the compound interest on a sum of ₹ 8,000 for 1 year at the rate of 10% per annum, interest compounded yearly and half yearly is:
a) ₹40
b) ₹10
c) ₹30
d) ₹20
11) Answer (D)
Solution:
Given, Principal amount = ₹ 8,000
Rate = 10%
When the interest is compounded yearly, time period = 1 year
Compound interest when compounded yearly = $8000\left(1+\frac{10}{100}\right)^1=8000\left(\frac{110}{100}\right)=8800$
When the interest is compounded yearly, time period = 2 half-years
Rate = $\frac{10}{2}$ = 5%
Compound interest when compounded half yearly = $8000\left(1+\frac{5}{100}\right)^2=8000\left(\frac{105}{100}\right)^2=8820$
$\therefore$ Required difference = 8820 – 8800 = ₹20
Hence, the correct answer is Option D
Question 12: There is a 60% increase in an amount in 5 years at simple interest. What will be the compound interest on ₹ 6,250 for two years at the same rate of interest, when the interest is compounded yearly?
a) ₹ 1,480
b) ₹ 1,560
c) ₹ 1,500
d) ₹ 1,590
12) Answer (D)
Solution:
Let the rate of interest = R%
Principal amount = P
Time = 5 years
$\Rightarrow$ Amount = $\frac{160}{100}P$
$\Rightarrow$ P + $\frac{P\times5\times R}{100}$ = $\frac{160}{100}\text{P}$
$\Rightarrow$ $\frac{P\times5\times R}{100}$ = $\frac{160}{100}\text{P}$ – $\text{P}$
$\Rightarrow$ $\frac{P\times5\times R}{100}$ = $\frac{60}{100}\text{P}$
$\Rightarrow$ R = 12%
Compound interest on ₹ 6,250 for two years at 12% = $6250\left(1+\frac{12}{100}\right)^2-6250$
$=6250\left(\frac{112}{100}\right)^2-6250$
$=6250\left(1.12\right)^2-6250$
$=6250\left(1.2544\right)-6250$
$=6250\left(0.2544\right)$
$=$ ₹ 1590
Hence, the correct answer is Option D
Question 13: If the present amount is ₹ 87,750 with 8% rate of interest in four years, then what was the principal amount?
a) ₹ 69,345.6
b) ₹ 78,456.34
c) ₹ 56,896.98
d) ₹ 66,477.2
13) Answer (D)
Solution:
Given, Amount = ₹ 87,750
Rate = 8%
Time = 4 years
Let Principal amount = P
$\Rightarrow$ P + $\frac{P\times4\times8}{100}=87750$
$\Rightarrow$ P + $\frac{32P}{100}=87750$
$\Rightarrow$ $\frac{132P}{100}=87750$
$\Rightarrow$ $\frac{P}{100}=664.7727$
$\Rightarrow$ P = 66477.27
$\therefore\ $Principal amount = ₹ 66477.27
Hence, the correct answer is Option D
Question 14: A man has ₹10,000. He lent a part of it at 15% simple interest and the remaining at 10% simple interest. The total interest he received after 5 years amounted to ₹6,500. The difference between the parts of the amounts he lent is:
a) ₹1,750
b) ₹2,500
c) ₹2,000
d) ₹1,500
14) Answer (C)
Solution:
Given,
Total Amount = ₹10,000
Let the amount lent at 15% = $x$
$=$> Amount lent at 10% = $10000-x$
Total interest he received after 5 years amounted to ₹6,500
$=$> $\frac{x\times15\times5}{100}+\frac{\left(10000-x\right)\times10\times5}{100}=6500$
$=$> $\frac{3x}{4}+\frac{1}{2}\left(10000-x\right)=6500$
$=$> $\frac{3x}{4}-\frac{x}{2}+5000=6500$
$=$> $\frac{3x-2x}{4}=6500-5000$
$=$> $\frac{x}{4}=1500$
$=$> $x=6000$
$\therefore\ $Difference between the parts of amounts he lent = $x-\left(10000-x\right)=6000-\left(10000-6000\right)=$₹ 2000
Hence, the correct answer is Option C
Question 15: Ram deposited an amount of ₹ 8,000 in a bank’s savings account with interest 6.5% compounded monthly. What amount will he get at the end of 18 months?
a) $₹ 8816.97$
b) $₹ 8788.98$
c) $₹ 8907.56$
d) $₹ 8790.54$
15) Answer (A)
Solution:
Given,
Principal amount (P) = ₹ 8,000
Rate of interest (R) = 6.5% per annum = $\frac{6.5}{12}\%$ per month
Time (n) = 18 months
$\therefore\ $Amount = $P\left(1+\frac{R}{100}\right)^n$
$=8000\left(1+\frac{6.5}{12\times100}\right)^18$
$=8000\left(1+0.0054167\right)^{18}$
$=8000\left(1.0054167\right)^{18}$
$=$₹ 8816.97
Hence, the correct answer is Option A
Question 16: Latha deposited an amount of $₹ 35,000$ in a bank with simple interest 11% per annum. How much interest will she earn after one year?
a) $₹ 3,370$
b) $₹ 3,220$
c) $₹ 3,500$
d) $₹ 3,850$
16) Answer (D)
Solution:
Given,
Principal amount (P) = $₹ 35,000$
Rate of Simple interest (R)% = 11%
Time (T) = 1 year
$\therefore\ $Simple interest earned by Latha = $\frac{\text{PTR}}{100}=\frac{35000\times1\times11}{100}=₹ 3,850$
Hence, the correct answer is Option D
Question 17: The difference of simple interest on a sum of money for 8 years and 10 years is $₹ 200.$ If the rate of interest is 10 % p.a, then what is the sum of money?
a) $₹ 1,600$
b) $₹ 1,000$
c) $₹ 1,200$
d) $₹ 1,400$
17) Answer (B)
Solution:
Let the Principal amount = P
Given, rate of interest = 10%
The difference of simple interest on the sum of money for 8 years and 10 years is $₹ 200$
$=$> $\frac{P\times10\times10\ }{100}-\frac{P\times8\times10}{100}=200$
$=$> $\frac{100P\ }{100}-\frac{80P}{100}=200$
$=$> $\frac{20P\ }{100}=200$
$=$> $P=₹ 1000$
$\therefore\ $Sum of money = $₹ 1000$
Hence, the correct answer is Option B
Question 18: At which rate of simple interest does an amount become double in 12 years?
a) $7 \frac{4}{5}%$
b) $7 \frac{1}{2}%$
c) $8%$
d) $8 \frac{1}{3}%$
18) Answer (D)
Solution:
Given,
Time (T) = 12 years
Let the Principal amount = P
Rate of Simple Interest = R
According to the Problem,
Amount = 2P
$=$> Principal + Simple Interest = 2P
$=$> P + $\frac{\text{P}\times\text{T}\times \text{R}}{100}$ = 2P
$=$> $\frac{\text{P}\times12\times \text{R}}{100}$ = 2P – P
$=$> $\frac{\text{P}\times12\times \text{R}}{100}$ = P
$=$> $\text{R}=\frac{100}{12}$
$=$> $\text{R}=8\frac{1}{3}$%
Hence, the correct answer is Option D
Question 19: A person invested a total of ₹9,000 in three parts at 3%, 4% and 6% per annum on simple interest. At the end of a year, he received equal interest in all the three cases. The amount invested at 6% is:
a) $₹3,000$
b) $₹5,000$
c) $₹2,000$
d) $₹4,000$
19) Answer (C)
Solution:
Let the amount invested by the person at 3%, 4% and 6% are $x$, $y$, $z$ respectively
$=$> $x+y+z=9000$ ………….(1)
Interest on amount $x$ at 3% after 1 year = $\frac{x\times3\times1}{100}=\frac{3x}{100}$
Interest on amount $y$ at 4% after 1 year = $\frac{y\times4\times1}{100}=\frac{4y}{100}$
Interest on amount $z$ at 6% after 1 year = $\frac{z\times6\times1}{100}=\frac{6z}{100}$
Given, interest received after 1 year on $x$, $y$, $z$ amounts are equal
$=$> $\frac{3x}{100}=\frac{4y}{100}=\frac{6z}{100}$
$=$> $3x=4y=6z$ ……………….(2)
Sustituting values from (2) in (1)
$=$> $x+\frac{3}{4}x+\frac{3}{6}x=9000$
$=$> $x+\frac{3}{4}x+\frac{1}{2}x=9000$
$=$> $\frac{4x+3x+2x}{4}=9000$
$=$> $\frac{9x}{4}=9000$
$=$> $x=4000$
$\therefore\ $ $y=\frac{3}{4}x=\frac{3}{4}\times4000=3000$ and
$z=\frac{3}{6}x=\frac{3}{6}\times4000=2000\ $
$\therefore\ $The amount invested at 6% Simple interest = $₹2,000$
Hence, the correct answer is Option C
Question 20: In how may years will a sum of $₹ 320$ amount to $₹ 405$ if interest is compounded at 12.5% per annum?
a) 2 years
b) 1 year
c) $1 \frac{1}{2}$ years
d) $2 \frac{1}{2}$ years
20) Answer (A)
Solution:
Given,
Principal (P) = $₹ 320$
Rate (R)% = 12.5%
Amount (A) = $₹ 405$
Let the required number of years = n
$=$> $\text{P}\left(1+\frac{\text{R}}{100}\right)^n=405$
$=$> $320\left(1+\frac{12.5}{100}\right)^n=405$
$=$> $320\left(\frac{112.5}{100}\right)^n=405$
$=$> $\left(\frac{1125}{1000}\right)^n=\frac{405}{320}$
$=$> $\left(\frac{9}{8}\right)^n=\frac{81}{64}$
$=$> $\left(\frac{9}{8}\right)^n=\frac{9^2}{8^2}$
$=$> $\left(\frac{9}{8}\right)^n=\left(\frac{9}{8}\right)^2$
$=$> $n=2$
Hence, the correct answer is Option A
