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 = $\frac{10}{2}=5\mathrm{\%}$

$A=P{(1+\frac{r}{100})}^n$

= $A=10,000{(1+\frac{5}{100})}^3$

$A=10,000\times \frac{21}{20}\times \frac{21}{20}\times \frac{21}{20}=11576.25$

C.I = A – P = 11576.25 – 10000 = ₹1576.35

Simple Interest = $S.I=\frac{(P\times R\times T)}{100}$

= $\frac{(10000\times 10\times 1.5)}{100}$

= ₹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 + $\frac{P\times (t+2)\times 6}{100}$

1105 = P + $\frac{P\times (t+2)\times 6}{100}$

1105 – P = $\frac{P\times (t+2)\times 6}{100}$…….(1)

Amount obtained from simple interest at 10% = P + $\frac{P\times t\times 10}{100}$

1105 = P + $\frac{P\times t\times 10}{100}$

1105 – P = $\frac{P\times t\times 10}{100}$……..(2)

From (1) and (2),

$\frac{P\times (t+2)\times 6}{100}$ = $\frac{P\times t\times 10}{100}$

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 = $\frac{9500\times 2\times 20}{100}$

= ₹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 + $\frac{\text{P}\times 8\times \text{R}}{100}$

$\Rightarrow$  P =  $\frac{\text{P}\times 8\times \text{R}}{100}$

$\Rightarrow$  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 + $\frac{\text{P}\times \text{T}\times 12.5}{100}$

$\Rightarrow$  3P = $\frac{\text{P}\times \text{T}\times 12.5}{100}$

$\Rightarrow$  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 + $\frac{\text{P}\times 5\times \text{R}}{100}$ = $\frac{9}{8}$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${(1+\frac{10}{100})}^3-$ P = P${\left(\frac{110}{100}\right)}^3-$ P = P$\frac{1331}{1000}-$ P = $\frac{331}{1000}$P

Simple interest on the sum = $\frac{P\times 3\times 10}{100}$ = $\frac{3}{10}$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

$\Rightarrow$  $\frac{P\times 4\times 8}{100}+\frac{P\times 8\times 8}{100}=960$

$\Rightarrow$  $\frac{32P}{100}+\frac{64P}{100}=960$

$\Rightarrow$  $\frac{96P}{100}=960$

$\Rightarrow$   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 = $8000{(1+\frac{10}{100})}^1=8000\left(\frac{110}{100}\right)=8800$

When the interest is compounded yearly, time period = 2 half-years

Rate = $\frac{10}{2}$ = 5%

Compound interest when compounded half yearly = $8000{(1+\frac{5}{100})}^2=8000{\left(\frac{105}{100}\right)}^2=8820$

$\therefore$ 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 + $\frac{P\times 5\times R}{100}$ = $\frac{160}{100}\text{P}$

$\Rightarrow$  $\frac{P\times 5\times R}{100}$ = $\frac{160}{100}\text{P}$ – $\text{P}$

$\Rightarrow$  $\frac{P\times 5\times R}{100}$ = $\frac{60}{100}\text{P}$

$\Rightarrow$  R = 12%

Compound interest on ₹ 6,250 for two years at 12% = $6250{(1+\frac{12}{100})}^2-6250$

$=6250{\left(\frac{112}{100}\right)}^2-6250$

$=6250{\left(1.12\right)}^2-6250$

$=6250\left(1.2544\right)-6250$

$=6250\left(0.2544\right)$

$=$ ₹ 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 + $\frac{P\times 4\times 8}{100}=87750$

$\Rightarrow$  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

$=$>  $\frac{x\times 15\times 5}{100}+\frac{(10000-x)\times 10\times 5}{100}=6500$

$=$>  $\frac{3x}{4}+\frac{1}{2}(10000-x)=6500$

$=$>  $\frac{3x}{4}-\frac{x}{2}+5000=6500$

$=$>  $\frac{3x-2x}{4}=6500-5000$

$=$>  $\frac{x}{4}=1500$

$=$>  $x=6000$

$\therefore$Difference between the parts of amounts he lent = $x-(10000-x)=6000-(10000-6000)=$₹ 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) $\text{Math input error}$

b) $\text{Math input error}$

c) $\text{Math input error}$

d) $\text{Math input error}$

15) Answer (A)

Solution:

Given,

Principal amount (P) = ₹ 8,000

Rate of interest (R) = 6.5% per annum = $\frac{6.5}{12}\mathrm{\%}$ per month

Time (n) = 18 months

$\therefore$Amount = $P{(1+\frac{R}{100})}^n$

$=8000{(1+\frac{6.5}{12\times 100})}^{1}8$

$=8000{(1+0.0054167)}^{18}$

$=8000{\left(1.0054167\right)}^{18}$

$=$₹ 8816.97

Hence, the correct answer is Option A

Question 16: Latha deposited an amount of $\text{Math input error}$ in a bank with simple interest 11% per annum. How much interest will she earn after one year?

a) $\text{Math input error}$

b) $\text{Math input error}$

c) $\text{Math input error}$

d) $\text{Math input error}$

16) Answer (D)

Solution:

Given,

Principal amount (P) = $\text{Math input error}$

Rate of Simple interest (R)% = 11%

Time (T) = 1 year

$\therefore$Simple interest earned by Latha = $\text{Math input error}$

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 $\text{Math input error}$ If the rate of interest is 10 % p.a, then what is the sum of money?

a) $\text{Math input error}$

b) $\text{Math input error}$

c) $\text{Math input error}$

d) $\text{Math input error}$

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 $\text{Math input error}$

$=$>  $\frac{P\times 10\times 10}{100}-\frac{P\times 8\times 10}{100}=200$

$=$>  $\frac{100P}{100}-\frac{80P}{100}=200$

$=$>  $\frac{20P}{100}=200$

$=$>  $\text{Math input error}$

$\therefore$Sum of money = $\text{Math input error}$

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

$=$>  $\frac{\text{P}\times 12\times \text{R}}{100}$ = 2P – P

$=$>  $\frac{\text{P}\times 12\times \text{R}}{100}$ = 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) $\text{Math input error}$

b) $\text{Math input error}$

c) $\text{Math input error}$

d) $\text{Math input error}$

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 = $\frac{x\times 3\times 1}{100}=\frac{3x}{100}$

Interest on amount $y$ at 4% after 1 year = $\frac{y\times 4\times 1}{100}=\frac{4y}{100}$

Interest on amount $z$ at 6% after 1 year = $\frac{z\times 6\times 1}{100}=\frac{6z}{100}$

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)

$=$>  $x+\frac{3}{4}x+\frac{3}{6}x=9000$

$=$>  $x+\frac{3}{4}x+\frac{1}{2}x=9000$

$=$>  $\frac{4x+3x+2x}{4}=9000$

$=$>  $\frac{9x}{4}=9000$

$=$>  $x=4000$

$\therefore$ $y=\frac{3}{4}x=\frac{3}{4}\times 4000=3000$ and

$z=\frac{3}{6}x=\frac{3}{6}\times 4000=2000$

$\therefore$The amount invested at 6% Simple interest = $\text{Math input error}$

Hence, the correct answer is Option C

Question 20: In how may years will a sum of $\text{Math input error}$ amount to $\text{Math input error}$ 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) = $\text{Math input error}$

Rate (R)% = 12.5%

Amount (A) = $\text{Math input error}$

Let the required number of years = n

$=$>  $\text{P}{(1+\frac{\text{R}}{100})}^n=405$

$=$>  $320{(1+\frac{12.5}{100})}^n=405$

$=$>  $320{\left(\frac{112.5}{100}\right)}^n=405$

$=$>  ${\left(\frac{1125}{1000}\right)}^n=\frac{405}{320}$

$=$>  ${\left(\frac{9}{8}\right)}^n=\frac{81}{64}$

$=$>  ${\left(\frac{9}{8}\right)}^n=\frac{9^2}{8^2}$

$=$>  ${\left(\frac{9}{8}\right)}^n={\left(\frac{9}{8}\right)}^2$

$=$>  $n=2$

Hence, the correct answer is Option A

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