# Simple Compound and Interest Questions for CMAT

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**Question 1:Â **What is the difference (in â‚¹) between the compound interest, when interest is compounded 6-monthly, and the simple interest on a sum of â‚¹10,000 for $1\frac{1}{2}$ years at 10% p.a.?

a)Â 102.25

b)Â 87

c)Â 76.25

d)Â 91.5

**1)Â AnswerÂ (C)**

**Solution:**

If interest is half-yearly, Then

Time =Â $1.5\times\ 2=3$

Rate =Â $\frac{10}{2}=\ 5\%$

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

=Â $A=10,000\left(1+\frac{5}{100}\right)^3$

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

C.I = AÂ – P = 11576.25 – 10000 = â‚¹1576.35

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

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

=Â â‚¹1500

Required Difference = â‚¹1576.25 – â‚¹1500 =Â â‚¹76.25

Hence,Â **Option C** is correct.

**Question 2:Â **A sum of â‚¹25600 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 â‚¹5832. How much money(in â‚¹) was invested at 9% per annum?

a)Â 18000

b)Â 7600

c)Â 9600

d)Â 16000

**2)Â AnswerÂ (B)**

**Solution:**

Let the amount invested in 9% per annum = P

Amount invested in 7% per annum =Â 25600 – P

The total interest at the end of 3 years is â‚¹5832.

P x 3 x $\frac{9}{100}$ +Â (25600 – P) x 3 xÂ $\frac{7}{100}$ = 5832

$\frac{9}{100}$P + 1792 –Â $\frac{7}{100}$P = 1944

$\frac{2}{100}$P = 152

P = â‚¹7600

$\therefore$Â Amount invested at 9% per annum = P = â‚¹7600

Hence, the correct answer is Option B

**Question 3:Â **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 â‚¹1105. What was the sum (in â‚¹) lent in each scheme?

a)Â 900

b)Â 850

c)Â 936

d)Â 891

**3)Â AnswerÂ (B)**

**Solution:**

Let the sum lent in each scheme = P

Let the sum at 6% and 10%Â simple interest was recovered after ‘t+2’ years and ‘t’Â years respectively.

Amount obtained fromÂ simple interest at 6% = P +Â $\frac{P\times(t+2)\times6}{100}$

1105 =Â P +Â $\frac{P\times(t+2)\times6}{100}$

1105 – P =Â $\frac{P\times(t+2)\times6}{100}$…….(1)

Amount obtained from simple interest at 10% = P + $\frac{P\times t\times10}{100}$

1105 =Â P + $\frac{P\times t\times10}{100}$

1105 – P =Â $\frac{P\times t\times10}{100}$……..(2)

From (1) and (2),

$\frac{P\times(t+2)\times6}{100}$ =Â $\frac{P\times t\times10}{100}$

6t + 12 = 10t

t = 3

Substituting t = 3 in equation (2), we get P = 850

The sum lent in each scheme =Â â‚¹850

Hence, the correct answer is Option B

**Question 4:Â **A sum of â‚¹9500 amounts to â‚¹11495 in 2 years at a certain rate percent per annum, interest compounded yearly. What is the simple interest (in â‚¹) on the same sum for the same time and double the rate?

a)Â 3990

b)Â 3420

c)Â 4560

d)Â 3800

**4)Â AnswerÂ (D)**

**Solution:**

Let the rate of interest = R

11495 = 9500(1+$\frac{R}{100}$)$^2$

(1+$\frac{R}{100}$)$^2$ =Â $\frac{11495}{9500}$

(1+$\frac{R}{100}$)$^2$ = $\frac{121}{100}$

1+$\frac{R}{100}$ =Â $\frac{11}{10}$

$\frac{R}{100}$ =Â $\frac{1}{10}$

R = 10%

Simple interest =Â $\frac{9500\times2\times20}{100}$

=Â â‚¹3800

Hence, the correct answer is Option D

**Question 5:Â **A person borrowed a sum of â‚¹30800 at 10% p.a. for 3 years, interest compounded annually. At the end of two years, he paid a sum of â‚¹13268. At the end of 3rd year, he paid â‚¹ x to clear of the debt. What is the value of x ?

a)Â 26400

b)Â 26510

c)Â 26200

d)Â 26620

**5)Â AnswerÂ (A)**

**Solution:**

Amount to be paid after 2 years = 30800(1+$\frac{10}{100}$)$^2$

=Â 30800($\frac{11}{10}$)$^2$

=Â 30800($\frac{121}{100}$)

=Â â‚¹37268

Amount paid by the person =Â â‚¹13268

Remaining amount =Â â‚¹37268 –Â â‚¹13268 =Â â‚¹24000

Amount to be paid at the end of 3rd year to clear debt(i.e, compound interest onÂ â‚¹24000 for next 1 year) = 24000(1+$\frac{10}{100}$)$^1$

$\Rightarrow$Â Â x = 24000($\frac{11}{10}$)

$\Rightarrow$Â Â x =Â â‚¹26400

Hence, the correct answer is Option A

**Question 6:Â **At what rate percent per annum will â‚¹7200 amountto â‚¹7938 in one year, if interest is compounded half yearly?

a)Â 5

b)Â 8

c)Â 12

d)Â 10

**6)Â AnswerÂ (D)**

**Solution:**

Let the rate of interest per annum = R

According to the problem,

7938 = 7200(1+$\frac{\frac{R}{2}}{100}$)$^2$

3969 =Â 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+$\frac{\frac{R}{2}}{100}$ =Â $\frac{21}{20}$

$\frac{R}{200}$ =Â $\frac{1}{20}$

R = 10%

Hence, the correct answer is Option D

**Question 7:Â **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)Â 25 years

b)Â 20 years

c)Â 30 years

d)Â 24 years

**7)Â AnswerÂ (D)**

**Solution:**

Let the principal amount = P

Rate of interest = R

Time = 8 years

Amount = 2P

$\Rightarrow$Â 2P = Principal amount + Simple Interest

$\Rightarrow$Â 2P = P + $\frac{\text{P}\times8\times \text{R}}{100}$

$\Rightarrow$Â P = Â $\frac{\text{P}\times8\times \text{R}}{100}$

$\Rightarrow$ Â R = 12.5%

Let the time in which sum will be 4 times at the same rate of interest = T

i.e, Amount = 4P

$\Rightarrow$Â 4P = P + $\frac{\text{P}\times \text{T}\times12.5}{100}$

$\Rightarrow$Â 3P =Â $\frac{\text{P}\times \text{T}\times12.5}{100}$

$\Rightarrow$Â 3 =Â $\frac{\text{T}}{8}$

$\Rightarrow$Â 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:Â **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)Â 3.5% p.a.

b)Â 2.5% p.a.

c)Â 3% p.a.

d)Â 2% p.a.

**8)Â AnswerÂ (B)**

**Solution:**

Let the sum of money lent out by Suresh to Rakesh = P

Amount paid by Rakesh to Suresh after 5 years =Â $\frac{9}{8}$P

Let the rate of simple interest = R%

$\Rightarrow$Â P + $\frac{\text{P}\times5\times \text{R}}{100}$ = $\frac{9}{8}$P

$\Rightarrow$Â 1 + $\frac{R}{20}$ =Â $\frac{9}{8}$

$\Rightarrow$ Â $\frac{R}{20}$ =Â $\frac{9}{8}-1$

$\Rightarrow$ Â $\frac{R}{20}$ =Â $\frac{1}{8}$

$\Rightarrow$Â R = 2.5%

$\therefore\ $Rate of simple interest per annum = 2.5%

Hence, the correct answer is Option B

**Question 9:Â **If the difference between the compound interest and simple interest on a certain sum of money for three years at 10% p.a. is â‚¹ 558, then the sum is:

a)Â â‚¹ 18,500

b)Â â‚¹ 15,000

c)Â â‚¹ 16,000

d)Â â‚¹ 18,000

**9)Â AnswerÂ (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 =Â $\frac{331}{1000}$P

Simple interest on the sum = $\frac{P\times3\times10}{100}$ = $\frac{3}{10}$P

According to the problem,

$\frac{331}{1000}$P $-$Â $\frac{3}{10}$P = 558

$\Rightarrow$ Â $\frac{331P-300P}{1000}$ = 558

$\Rightarrow$ Â $\frac{31P}{1000}$ = 558

$\Rightarrow$ Â P = 18000

$\therefore\ $The principal sum =Â â‚¹ 18,000

Hence, the correct answer is Option D

**Question 10:Â **The sum of simple interest on a sum at 8% p.a. for 4 years and 8 years is â‚¹960. The sum is:

a)Â â‚¹800

b)Â â‚¹1100

c)Â â‚¹1000

d)Â â‚¹900

**10)Â AnswerÂ (C)**

**Solution:**

Let the Principal amount = P

Rate = 8%

Given, the sum of simple interest on P for 4 years and 8 years is â‚¹960

$\Rightarrow$ Â $\frac{P\times4\times8}{100}+\frac{P\times8\times8}{100}=960$

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

$\Rightarrow$ Â $\frac{96P}{100}=960$

$\Rightarrow$ Â P = 1000

$\therefore\ $The required sum =Â â‚¹1000

Hence, the correct answer is Option C

**Question 11:Â **The difference between the compound interest on a sum of â‚¹ 8,000 for 1 year at the rate of 10% per annum, interest compounded yearly and half yearly is:

a)Â â‚¹40

b)Â â‚¹10

c)Â â‚¹30

d)Â â‚¹20

**11)Â AnswerÂ (D)**

**Solution:**

Given, Principal amount =Â â‚¹ 8,000

Rate = 10%

When the interest is compounded yearly, time period = 1 year

Compound interest when compounded yearly =Â $8000\left(1+\frac{10}{100}\right)^1=8000\left(\frac{110}{100}\right)=8800$

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

Rate =Â $\frac{10}{2}$ = 5%

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

$\therefore$ Required difference = 8820 – 8800 = â‚¹20

Hence, the correct answer is Option D

**Question 12:Â **There is a 60% increase in an amount in 5 years at simple interest. What will be the compound interest on â‚¹ 6,250 for two years at the same rate of interest, when the interest is compounded yearly?

a)Â â‚¹ 1,480

b)Â â‚¹ 1,560

c)Â â‚¹ 1,500

d)Â â‚¹ 1,590

**12)Â AnswerÂ (D)**

**Solution:**

Let the rate of interest = R%

Principal amount = P

Time = 5 years

$\Rightarrow$Â Amount = $\frac{160}{100}P$

$\Rightarrow$Â P +Â $\frac{P\times5\times R}{100}$ =Â $\frac{160}{100}\text{P}$

$\Rightarrow$ Â $\frac{P\times5\times R}{100}$ =Â $\frac{160}{100}\text{P}$ – $\text{P}$

$\Rightarrow$ Â $\frac{P\times5\times R}{100}$ =Â $\frac{60}{100}\text{P}$

$\Rightarrow$ Â R = 12%

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

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

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

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

$=6250\left(0.2544\right)$

$=$ â‚¹ 1590

Hence, the correct answer is Option D

**Question 13:Â **If the present amount is â‚¹ 87,750 with 8% rate of interest in four years, then what was the principal amount?

a)Â â‚¹ 69,345.6

b)Â â‚¹ 78,456.34

c)Â â‚¹ 56,896.98

d)Â â‚¹ 66,477.2

**13)Â AnswerÂ (D)**

**Solution:**

Given, Amount =Â â‚¹ 87,750

Rate = 8%

Time = 4 years

Let Principal amount = P

$\Rightarrow$Â P + $\frac{P\times4\times8}{100}=87750$

$\Rightarrow$Â P + $\frac{32P}{100}=87750$

$\Rightarrow$ Â $\frac{132P}{100}=87750$

$\Rightarrow$ Â $\frac{P}{100}=664.7727$

$\Rightarrow$Â P = 66477.27

$\therefore\ $Principal amount = â‚¹ 66477.27

Hence, the correct answer is Option D

**Question 14:Â **A man has â‚¹10,000. 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 â‚¹6,500. The difference between the parts of the amounts he lent is:

a)Â â‚¹1,750

b)Â â‚¹2,500

c)Â â‚¹2,000

d)Â â‚¹1,500

**14)Â AnswerÂ (C)**

**Solution:**

Given,

Total Amount = â‚¹10,000

Let the amount lent at 15% = $x$

$=$>Â Amount lent at 10% = $10000-x$

Total interest he received after 5 years amounted to â‚¹6,500

$=$> Â $\frac{x\times15\times5}{100}+\frac{\left(10000-x\right)\times10\times5}{100}=6500$

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

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

$=$> Â $\frac{3x-2x}{4}=6500-5000$

$=$> Â $\frac{x}{4}=1500$

$=$> Â $x=6000$

$\therefore\ $Difference between the parts of amounts he lent =Â $x-\left(10000-x\right)=6000-\left(10000-6000\right)=$â‚¹ 2000

Hence, the correct answer is Option C

**Question 15:Â **Ram deposited an amount of â‚¹ 8,000 in a bankâ€™s savings account with interest 6.5% compounded monthly. What amount will he get at the end of 18 months?

a)Â $â‚¹ 8816.97$

b)Â $â‚¹ 8788.98$

c)Â $â‚¹ 8907.56$

d)Â $â‚¹ 8790.54$

**15)Â AnswerÂ (A)**

**Solution:**

Given,

Principal amount (P) = â‚¹ 8,000

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

Time (n) = 18 months

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

$=8000\left(1+\frac{6.5}{12\times100}\right)^18$

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

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

$=$â‚¹ 8816.97

Hence, the correct answer is Option A

**Question 16:Â **Latha deposited an amount of $â‚¹ 35,000$ in a bank with simple interest 11% per annum. How much interest will she earn after one year?

a)Â $â‚¹ 3,370$

b)Â $â‚¹ 3,220$

c)Â $â‚¹ 3,500$

d)Â $â‚¹ 3,850$

**16)Â AnswerÂ (D)**

**Solution:**

Given,

Principal amount (P) =Â $â‚¹ 35,000$

Rate of Simple interest (R)% = 11%

Time (T) = 1 year

$\therefore\ $Simple interest earned by Latha =Â $\frac{\text{PTR}}{100}=\frac{35000\times1\times11}{100}=â‚¹ 3,850$

Hence, the correct answer is Option D

**Question 17:Â **The difference of simple interest on a sum of money for 8 years and 10 years is $â‚¹ 200.$ If the rate of interest is 10 % p.a, then what is the sum of money?

a)Â $â‚¹ 1,600$

b)Â $â‚¹ 1,000$

c)Â $â‚¹ 1,200$

d)Â $â‚¹ 1,400$

**17)Â AnswerÂ (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 $â‚¹ 200$

$=$> Â $\frac{P\times10\times10\ }{100}-\frac{P\times8\times10}{100}=200$

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

$=$> Â $\frac{20P\ }{100}=200$

$=$> Â $P=â‚¹ 1000$

$\therefore\ $Sum of money =Â $â‚¹ 1000$

Hence, the correct answer is Option B

**Question 18:Â **At which rate of simple interest does an amount become double in 12 years?

a)Â $7 \frac{4}{5}%$

b)Â $7 \frac{1}{2}%$

c)Â $8%$

d)Â $8 \frac{1}{3}%$

**18)Â AnswerÂ (D)**

**Solution:**

Given,

Time (T) = 12 years

Let the Principal amount = P

Rate of Simple Interest = R

According to the Problem,

Amount = 2P

$=$>Â Principal + Simple Interest = 2P

$=$>Â P +Â $\frac{\text{P}\times\text{T}\times \text{R}}{100}$ = 2P

$=$>Â Â $\frac{\text{P}\times12\times \text{R}}{100}$ = 2P – P

$=$> Â $\frac{\text{P}\times12\times \text{R}}{100}$ = P

$=$> Â $\text{R}=\frac{100}{12}$

$=$> Â $\text{R}=8\frac{1}{3}$%

Hence, the correct answer is Option D

**Question 19:Â **A person invested a total of â‚¹9,000 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)Â $â‚¹3,000$

b)Â $â‚¹5,000$

c)Â $â‚¹2,000$

d)Â $â‚¹4,000$

**19)Â AnswerÂ (C)**

**Solution:**

Let the amount invested by the person at 3%, 4% and 6% are $x$, $y$, $z$ respectively

$=$> Â $x+y+z=9000$ ………….(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 =Â $\frac{y\times4\times1}{100}=\frac{4y}{100}$

Interest on amount $z$ at 6% after 1 year =Â $\frac{z\times6\times1}{100}=\frac{6z}{100}$

Given, interest received after 1 year on $x$, $y$, $z$ amounts are equal

$=$> Â $\frac{3x}{100}=\frac{4y}{100}=\frac{6z}{100}$

$=$> Â $3x=4y=6z$ ……………….(2)

Sustituting values from (2) in (1)

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

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

$=$> Â $\frac{4x+3x+2x}{4}=9000$

$=$> Â $\frac{9x}{4}=9000$

$=$> Â $x=4000$

$\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 =Â $â‚¹2,000$

Hence, the correct answer is Option C

**Question 20:Â **In how may years will a sum of $â‚¹ 320$ amount to $â‚¹ 405$ if interest is compounded at 12.5% per annum?

a)Â 2 years

b)Â 1 year

c)Â $1 \frac{1}{2}$ years

d)Â $2 \frac{1}{2}$ years

**20)Â AnswerÂ (A)**

**Solution:**

Given,

Principal (P) =Â $â‚¹ 320$

Rate (R)% = 12.5%

Amount (A) =Â $â‚¹ 405$

Let the required number of years = n

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

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

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

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

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

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

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

$=$> Â $n=2$

Hence, the correct answer is Option A