**Compound Interest Questions for SSC CHSL PDF:**

SSC CHSL Physics Previous Year Questions download PDF based on previous year question paper of SSC exams. 25 Very important Physics Previous Year questions for SSC CHSL Exam.

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**Question 1: **On a certain sum of money, the simple interest for 2 years is Rs. 350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned ?

a) Rs. 3.50

b) Rs. 7

c) Rs. 14

d) Rs. 35

**Question 2: **The compound interest on a sum of Rs. 5000 at 8% per annum for 9 months when interest is compound quarterly is:

a) Rs. 300

b) Rs. 300.12

c) Rs. 306.04

d) Rs. 308

**Question 3: **A certain amount grows at an annual interest rate of 12%, compounded monthly. Which of the following equations can be solved to find the number of years, y, that it would take for the investment to increase by a factor of 64 ?

a) $64=(1.01)^{12y}$

b) $\frac{1}{64}= (1.04)12y$

c) $64=(1.04)^{12y}$

d) $8=(1.01)^{6y}$

**Question 4: **What Is the compound interest (in Rs.) on a principal sum of Rs. 2800 for 2 years at the rate of 12% per annum?

a) 687.18

b) 634.46

c) 712.32

d) 568.68

**Question 5: **If interest being compound half yearly then what sum (in Rs.) will amount to Rs. 38416 in 2 years at the rate of 80% per annum at compound interest ?

a) 14000

b) 15000

c) 10000

d) 12000

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**Question 6: **A sum of Rs. 12,000, deposited at compoundd interest becomes double after 5 years. How much will it be after 20 years ?

a) Rs. 1,44,000

b) Rs. 1,20,000

c) Rs. 1, 50,000

d) Rs. 1,92,000

**Question 7: **The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 48. Then the sum is

a) Rs. 1,000

b) Rs. 1,200

c) Rs. 1,500

d) Rs. 2,000

**Question 8: **A sum of Rs. 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become

a) Rs. 48,000

b) Rs. 96,000

c) Rs. 1,90,000

d) Rs. 1,92,000

**Question 9: **The difference between the compound interest and simple interest for the amount Rs. 5,000 in 2 years is Rs.32. The rate of interest is

a) 5%

b) 8%

c) 10%

d) 12%

**Question 10: **There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs. 8000 after 2 years at the same rate of interest.

a) Rs. 2500

b) Rs. 2000

c) Rs. 2250

d) Rs. 2125

**Question 11: **The compound interest generated on a sum of Rs 5000, in two years at 5% per annum, the interest is being compounded half yearly is.

a) Rs 557

b) Rs 489

c) Rs 519

d) Rs 362

**Question 12: **The difference between simple interest and compound interest (at the same rate of interest) for two years on a sum of Rs. 25000 is Rs. 1000. What is the common rate of interest?

a) 10%

b) 20%

c) 30%

d) 40%

**Question 13: **Sohan deposited Rs 1 lac in a bank at a certain simple interest rate such that the amount got doubled after 5 years. If Sohan had deposited the same amount at the same interest rate compounded annually, the amount after 2 years would have been

a) Rs 1.44 lacs

b) Rs 1.21 lacs

c) Rs 1.25 lacs

d) Rs 1.69 lacs

**Question 14: **A certain sum of money will amount to 24200 in 2 years at 10% per annum compounded annually. What will this sum amount to if it is invested at 8% per annum simple interest for 3 years?

a) 25000

b) 24800

c) 24200

d) 24000

**Question 15: **What is the difference between simple interest and compound interest (at the same rate of interest) for two years on a sum of Rs. 20000? The common rate of interest is 10%.

a) Rs. 200

b) Rs. 400

c) Rs. 600

d) Rs. 800

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**Question 16: **If the difference between simple interest and compound interest for two years on a sum of Rs. 50000 is Rs. 12500 when the rate of interest is the same, what is the common rate of interest?

a) 40%

b) 50%

c) 45%

d) 35%

**Question 17: **Ganesh invests a sum of money in a compound interest which pays 20% annually. How many years will it take for Ganesh to get back double the amount he invested?

a) 3

b) 4

c) 5

d) Cannot be determined

**Question 18: **A bank offers 12% compound interest on a sum of Rs. 10,000 compounded every quarter. What will be the approximate interest earned at the end of 1 year?

a) Rs 1055

b) Rs 1155

c) Rs 1255

d) Rs 1355

**Question 19: **The compound interest on a certain sum of money for 2 years at 5% per annum is 410. The simple interest on the same sum at the same rate and for the same time is

a) 400

b) 300

c) 350

d) 405

**Question 20: **The compound interest on =1,800 at 10% per annum for a certain period of time is 378. Find the time in years.

a) 2.0 years

b) 2.8 years

c) 3.0 years

d) 2.5 year

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**Question 21: **If the compound interest on a certain sum for two years at 12% per annum is 2,544, the simple interest on it at the same rate for 2 years will be

a) 2,400

b) 2,500

c) 2,480

d) 2,440

**Question 22: **A man borrows Rs. 21000 at 10% compound interest. How much he has to pay equally at the end of each year, to settle his loan in two years ?

a) Rs. 12000

b) Rs. 12100

c) Rs. 12200

d) Rs. 12300

**Question 23: **The time in which 80,000 ambunts to 92,610 at 10% p.a. at compound interest, interest being compounded semi annually is :

a) 1.5 years

b) 2 years

c) 2.5 years

d) 3 years

**Question 24: **A sum of money at compound interest will amount to 650 at the end of the first year and 676 at the end of the second year. The amount of money is

a) 1,300

b) 650

c) 1,250

d) 625

**Question 25: **A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?

a) 12 years

b) 13 years

c) 8 years

d) 16 years

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**Answers & Solutions:**

**1) Answer (B)**

Rate of interest = 4% and time period = 2 years

Let principal sum = Rs. $x$

Simple interest = $\frac{P\times R\times T}{100}$

=> $\frac{x\times4\times2}{100}=350$

=> $x=\frac{35000}{8}=4375$

Now, interest earned under compound interest = $P[(1+\frac{R}{100})^T-1]$

= $4375[(1+\frac{4}{100})^2-1]$

= $4375[(\frac{26}{25})^2-1]$

= $4375\times\frac{676-625}{625}$

= $7\times51=Rs.$ $357$

$\therefore$ Difference in interest = $357-350=Rs.$ $7$

=> Ans – (B)

**2) Answer (C)**

Principal sum = Rs. 5000

Rate of interest = 8% and time period = $\frac{9}{12}=\frac{3}{4}$ years

Compound interest when interest is compound quarterly = $P[(1+\frac{R}{400})^{4T}-1]$

= $5000[(1+\frac{8}{400})^{\frac{3}{4}\times4}-1]$

= $5000[(1+\frac{1}{50})^3-1]$

= $5000[(\frac{51}{50})^3-1]$

= $5000\times(\frac{132651-125000}{125000})$

= $\frac{7651}{25}=Rs.$ $306.04$

=> Ans – (C)

**3) Answer (A)**

Rate of interest = 12% p.a. = 1% per month

Time = $12y$ months

Let principal = Re 1 and thus amount = Rs. 64

$\therefore$ $A=P(1+\frac{R}{100})^T$

=> $64=1(1+\frac{1}{100})^{12y}$

=> $64=(1.01)^{12y}$

=> Ans – (A)

**4) Answer (C)**

Principal sum = Rs. 2800

Rate of interest = 12% and time = 2years

Compound interest = $P[(1+\frac{R}{100})^T-1]$

= $2800[(1+\frac{12}{100})^2-1]$

= $2800[(\frac{28}{25})^2-1]$

= $2800\times(\frac{784-625}{625})$

= $2800\times\frac{159}{625}=Rs.$ $712.32$

=> Ans – (C)

**5) Answer (C)**

Let principal sum = Rs. $P$ and amount = Rs. 38,416

Rate of interest = 80% and time = 2 years

Amount if interest being compound half yearly = $P(1+\frac{R}{200})^{2T}$

=> $P(1+\frac{80}{200})^{2\times2}=38,416$

=> $P\times(\frac{7}{5})^4=38,416$

=> $P=38,416\times\frac{625}{343\times7}$

=> $P=16\times625=Rs.$ $10,000$

=> Ans – (C)

**6) Answer (D)**

For compound interest A = $p(1+\frac{r}{100})^{n}$ where p is principal amount, r is rate and t is time

after 5 years it gets doubled

hence putting the values we will get $(1+\frac{r}{100})^{5}$ = 2

now after 20 years total amount will be $p((1+\frac{r}{100})^{5})^{4}$ = 16p = $16 \times 12000$ = 192000

**7) Answer (B)**

Difference between simple interest and compound interest for two years will be

48 = $\frac{2pr}{100} – ( \frac{2pr}{100} + \frac{pr^2}{10^4})$ (where p is principal amount and r is rate per annum)

Putting r=20% and solving above equation for p, we will get p = 1200 rs.

**8) Answer (D)**

As we know $P(1+ \frac{r}{100})^t$ is amount of compound interest where r is rate, P is principal amount and t is time.

So $12000(1+ \frac{r}{100})^5 = 2 \times 12000$

or $(1+ \frac{r}{100}) = 2^(\frac{1}{5})$ eq(1)

Now after 20 years compound interest will be = $12000(1+ \frac{r}{100})^20$

or $12000 (2^{(\frac{1}{5})})^{20}$ (from eq.(1))

or $ 12000 \times 16 = 192000$

**9) Answer (B)**

Difference between compound interest and simple interest for 2 years will be

= $(P((1+\frac{r}{100})^2) – P) – 2P \frac{r}{100} = 32$ (where P is principal amount 5000 and r is rate )

after solving above equation we will get r = 8%

**10) Answer (D)**

I = PTR/100

there is 100% increase in amount means, interest = principle.

given, T = 8yrs.

I = P

P = P*8*R/100

R = 12.5%

compound interest of 8000/- at 12.5% for 2 years is

CI = total amount – 8000/-

= P$(1+R/100)^{n}$ – 8000/-

= 8000$(1+12.5/100)^{2}$ – 8000/-

= 10125 – 8000

= 2125/-

so the answer is option D.

**11) Answer (C)**

The formula to calculate compound interest is,

$A=P(1+\frac{r}{n})^{nt}$ where,

A = Amount

P = Principal

r = rate of interest

n = number of times compounded in a year

t = time period in years

Applying the above formula we get,

Compound interest generated after two years is

= Amount – Principal

= $5000 (1 + \frac{0.05}{2})^{(2*2)}$ – $5000$= $1.0509*5000 – 5000$ = $5519 – 5000$ = Rs $519$

Hence Option C.

**12) Answer (B)**

Simple interest for 2 years on Rs. 25000 at rate ‘r’ = $\frac{25000*2*r}{100}=500r$

Simple interest for each year = $250r$

Compound interest for 2 years on Rs. 25000 at rate ‘r’ =

Simple interest for 1st year + Simple interest for 2nd year

Simple interest for 1st year = $250r$

Simple interest for 2nd year = $\frac{(25000+250r)*r}{100}$ = $250r + 2.5r^2$

The difference between interest is given as Rs. 1000

So we get,

$250 + 2.5r^2 + 250r – 500r = 1000$

$2.5r^2 = 1000$

$r^2 = 400$

$ r = 20$%

Thus, Option B.

**13) Answer (A)**

If the amount gets doubled after 5 years, the simple interest accrued will be equal to P (principal amount).

=> $P = \frac{P \times r \times 5}{100}$ => r = 20%

So, the amount after 2 years under compound interest will be

$1(1+\frac{20}{100})^2 = 1(1.2)^2 = 1.44$ lacs.

Thus, A is the correct answer.

**14) Answer (B)**

We know that the formula for compound interest is given by

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

Here A = 24200, n = 2 and r = 10, so on putting these values in the equation, we get

24200 = $P( 1 + \frac{10}{100})^2$

=> P = 24200/1.21 = 20,000

If it is invested at 8% per annum simple interest for 3 years, the interest earned would be

I = (P*R*T)/100 = 20000*3*8/100 = 4800

Hence, the amount will be 20,000 + 4800 = 24800

**15) Answer (A)**

Simple interest for 2 years on Rs. 20000 at rate 10% = $\frac{20000*2*10}{100}=4000$ rupees

Compound interest for 2 years on Rs. 20000 at rate 10% =

Simple interest for 1st year + Simple interest for 2nd year

Simple interest for 1st year = $2000$ rupees

Principal for the second year will be 20000+2000 = Rs. 22000

Simple interest for 2nd year = $\frac{(22000)*10}{100}=2200$ rupees

Total simple interest = Rs. 4000

Total compound interest = Rs. 2000 + Rs. 2200 = Rs. 4200

The difference is Rs. 200

Thus, Option A.

**16) Answer (B)**

Simple interest for 2 years on Rs. 50000 at rate ‘r’ = $\frac{50000*2*r}{100}=1000r$

Simple interest for each year = $500r$

Compound interest for 2 years on Rs. 50000 at rate ‘r’ = Simple interest for 1st year + Simple interest for 2nd year

Simple interest for 2nd year = Simple interest for 1st year +simple interest on interest of first year

=$(500r)+(500r)+500r(\frac{r}{100})$ =$1000r+5r^{2}$

Difference between the simple and the compound interest = $1000r+5r^{2} – 1000r$ = $5r^{2}$

$5r^{2} = 12500$

$r = 50$%

Hence, Option B.

**17) Answer (B)**

Le the amount he invested be $P$.

After 3 years the amount will be,

$A=P(1+r)^t$

$=>A=P(1+0.2)^3$

$=>A=P(1.2)^3$

$=>A=1.728P$

After 4 years the amount will be,

$A=P(1.2)^4$

$=>A=2.0736P$

Thus the amount will get doubled after 4 years.

Hence Option B.

**18) Answer (C)**

If the rate of interest is 12% per annum, the corresponding rate for 3 months in 12%/4 = 3% per quarter.

So, the amount at the end of the year = $10,000 * (1+\frac{3}{100})^4$ = Rs 11255.09

Hence the interest earned = Amount – Principal = 11255.09 – 10000 = Rs 1255.09

**19) Answer (A)**

we know that :

1. For first year , compound interest and simple interest is same if the principal amount and rate of interest is same in both cases.

2. From 2nd year onwards ,the compound interest is normal interest plus the interest on accumulated amount due to interest untill last cycle.

3. Every year simple interest remains same if Rate of Interest and principal amount remains same .

Let the compound interest for 1st year be be Rs y

For two years ,CI = Rs 410

y + y + $\frac{5}{100}$y = 410

$\frac{41y}{20}$ = 410

y = 200

So for two years , Simple Interest = 200 + 200 = Rs 400

**20) Answer (A)**

Principal Amount (P) = Rs 1800

Rate of Interest = 10%

Compound Interest = Rs 378

Let the time period be T

So

C.I = P$(1+ \frac{R}{100})^T$ – P

378 = 1800$(1+ \frac{10}{100})^T$ – 1800

T = 2 years

**21) Answer (A)**

Let the Rate of Interest be R and Principal Amount be Rs P

So, R = 12%

Time Period (T) = 2 years

Compound Interest = P$(1 + \frac{R}{100})^T$ – P

2544 = P$(1 + \frac{12}{100})^2$ – P

P = Rs 10000

Now SI with same rate of interest = $\frac{P \times R \times T}{100}$ = $\frac{10000 \times 12 \times 2}{100}$ = Rs 2400

**22) Answer (B)**

We now that if Rs z is the amount to be paid by a person after n years then the present value of that Rs z is given as = $\frac{z}{(1 + \frac{R}{100})^n}$

where R is the Rate of Interest (Compounded Interest Rate)

So let assume that the amount of equal installment be Rs y

and hence we can say that if R = 10% per annum

and Principal Amount = Rs 21000

then present value of 1st installment + present value of 2nd installment = Rs 21000

$\frac{z}{(1 + \frac{10}{100})^1}$ +$\frac{z}{(1 + \frac{10}{100})^2}$ = 21000

$\frac{2.1z}{(1.21)}$ = 21000

z = Rs 12100

**23) Answer (A)**

Given that Rs 80000 becomes Rs 92610 at 10% per annum .

semi-annual Rate become = $\frac{10}{2}$ = 5%

Let number of semi-annual cycles required = T

So Using ,

Compounded Amount = P$(1+\frac{R}{100})^T$

92610 = 80000$(1+\frac{5}{100})^T$

T = 3 semi annual cycles = 1.5 years

**24) Answer (D)**

let the sum of money be Rs P

and Rate of Interest be R% per annum

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

n – number of time periods

Compounded amount after 1 year = P$(1 + \frac{R}{100})^1$ = 650………….(1)

Compounded amount after 2 year = P$(1 + \frac{R}{100})^2$ = 676………….(2)

Dividing both equation 1 and equation 2

$(1 + \frac{R}{100})$ = 1.04

and from here it can be concluded that R = 4% . Now using equation 1 and value of R

P$(1 + \frac{4}{100})^1$ = 650

P = Rs 625

**25) Answer (C)**

it is given that A sum of money placed at compound interest doubles itself in 4 years

here we need to make the money 4 times

imagine that we invested Rs P

and hence ,

P becomes 2P in 4 years and so this 2P will become 4P in another 4 years and hence total 8 years are required to make Rs P –> Rs 4P