# Simple Interest and Compound Interest Questions for IBPS Clerk PDF

0
3043

## Simple Interest and Compound Interest Questions for IBPS Clerk PDF

Download Top-15 Banking Exams Simple Interest and Compound Interest Questions PDF. Banking Exams Simple Interest and Compound Interest questions based on asked questions in previous exam papers very important for the Banking  exams.

Question 1: Ms suchi deposits an amount of 24000 to obtain in a simple interest at the rate of S.I 14 p.c.p.a for 8 years .what total amont will ms suchi gets at the end of 8 years

a) 52080

b) 28000

c) 50880

d) 26880

e) none of these

Question 2: A sum of money was invested for 14 years was in Scheme A which offers simple interest at a rate of 8% p.a. The amount received from Scheme A after 14 years was then invested for two years in Scheme B which offers compound interest (compounded annually) at a rate of 10% p.a. If the interest received from Scheme B was Rs. 6,678, what was the sum invested in Scheme A?

a) Rs. 15,500

b) Rs. 14,500

c) Rs. 16,500

d) Rs. 12,500

e) Rs. 15,000

Question 3: The difference between the compound interest and the simple interest for a period of 2 years at the rate  of 10% per annum is Rs. 50. Find the principal.

a) Rs. 4000

b) Rs. 5000

c) Rs. 5500

d) Rs. 4500

e) None of these

Question 4: The simple interest accrued on an amount of Rs. 16,500 at the end of three years is Rs. 5,940. What would be the compound interest accrued on the same amount at the same rate in the same period? (rounded off to two digits after decimal)

a) Rs. 6681.31

b) Rs. 6218.27

c) Rs. 6754.82

d) Rs. 6537.47

e) None of these

Question 5: The interest received on a sum of money when invested in scheme A is equal to the interest received on the same sum of money when invested for 2 years in scheme B. Scheme A offers simple interest (p.c.p.a.) and scheme B offers compound interest (compounded annually). Both the schemes offer the same rate of interest. If the numerical value of the number of years for which the sum is invested in scheme A is same as the numerical value of the rate of interest offered by the same scheme, what is the rate of interest (p.c.p.a) offered by scheme A?

a) 3

b) $2{1 \over {77}}$

c) $3{4 \over {99}}$

d) $2{2 \over {99}}$

e) 2

Question 6: Sudhanshu invested Rs. 15,000 at interest 10 p.c.p.a. for one year. If the interest is compounded every six months what amount wit Sudhanshu get at the end of the year ?

a) Rs. 16,537.50

b) Rs. 16, 500

c) Rs. 16, 525.50

d) Rs. 18,150

e) None of these

Question 7: Ms. Anisha deposits an amount of Rs. 35000 to obtain a simple interest at the rateof 15 p.c.p.a. for 4 years. What total amount will Ms. Anisha get at the end of 4 years ?

a) Rs. 56000

b) Rs. 60500

c) Rs. 52000

d) Rs. 48500

e) None of these

Question 8: The simple interest accrued on an amount of Rs. 19,800 at the end of three years is Rs. 7,128. What would be the compound interest accrued on the same amount at the same rate in the same period ?

a) Rs. 8934.6784

b) Rs. 8017.5744

c) Rs. 7861.8754

d) Cannot be determined

e) None of these

Question 9: Mr. Vijay deposits an amount of Rs. 45,500 to obtain a simple interest at the rate of 11 p.c.p.a for 3 years. What total amount will Mr. Vijay get at the end of 3 years?

a) Rs. 60,515

b) Rs. 60,015

c) Rs. 65,515

d) Rs. 62,015

e) None of these

Question 10: The simple interest accrued on an amount of Rs. 10,530 at the end of 5 years is Rs. 6,318. What is the rate of interest p.c.p.a. ?

a) 8

b) 14

c) 10

d) 12

e) None of these

Question 11: Mr. Madhur deposits an amount of Rs. 58,750/- to obtain a simple interest at the rate of 12 p.c:p.a:for years. What total amount will Mr. Madhur get at the end of 4 years?

a) Rs. 91,230/-

b) Rs. 86,950/-

c) Rs. 74,760/-

d) Rs. 69,540/-

e) None of These

Question 12: How much will a sum of Rs. 7,250 amount to in a span of 2 years, at 6 p.c.p.a. rate of compound interest (Rounded off to the nearest integer) ?

a) Rs. 8,176

b) Rs.8,146

c) Rs. 8,216

d) Rs.8, 170

e) Rs. 8,190

Question 13: Mr. Khanna took a loan of Rs.10,000/- on simple interest for two years at the rate 3 p.c:p.a:The total amount that he will be paying as interest in 2 years is 3% of his monthly salary. What is his monthly salary?

a) Rs.30,000/ –

b) Rs. 16,000/-

c) Rs.20,000/-

d) Rs. 12,000/-

e) None of these

Question 14: Ankur invested a sum of Rs. 16800 for four years in a scheme A. The rate of interest in scheme A is 8% per annum compounded yearly for the first two years and 10% for the third and fourth years compounded yearly. What will be the compound interest at the end of 4 years ?

a) Rs. 6810

b) Rs. 6910

c) Rs. 6540

d) Rs. 6210

e) Rs. 6740

Question 15: Mary paid 15% of her monthly salary towards an EMI. From the remaining salary. she paid 10% as Internet bill and 20% as rent. If after the mentioned expenses she was left with Rs. 24,990 /-, what was Mary’s monthly salary ?

a) Rs.45,000/-

b) Rs.48,000/-

c) Rs.42,000-

d) Rs.36,000/-

e) Rs.40,000 /-

Amount deposited = Rs. 24,000

Rate = 14 % and time = 8 years under simple interest

=> $S.I. = \frac{P \times R \times T}{100}$

= $\frac{24000 \times 14 \times 8}{100}$

= $240 \times 112 = Rs. 26,880$

$\therefore$Total amount will ms Suchi gets at the end of 8 years

= $24000 + 26880 = Rs. 50,880$

Let the sum invested in scheme A = $Rs. 100x$

Time = 14 years and rate = 8% under simple interest

=> $S.I. = \frac{P \times R \times T}{100}$

= $\frac{100x \times 8 \times 14}{100} = 112x$

=> Amount invested in Scheme B = $100x + 112x = Rs. 212x$

Time = 2 years and rate = 10% under compound interest.

$C.I. = P [(1 + \frac{R}{100})^T – 1]$

=> $6678 = 212x [(1 + \frac{10}{100})^2 – 1]$

=> $6678 = 212x [(\frac{11}{10})^2 – 1] = 212x (\frac{21}{100})$

=> $0.21x = \frac{6678}{212} = 31.5$

=> $x = \frac{31.5}{0.21} = 150$

$\therefore$ Sum invested in scheme A = $100 \times 150 = Rs. 15,000$

Let the principal amount = $Rs. 100x$

Time = 2 years and rate = 10%

=> $C.I. = P [(1 + \frac{R}{100})^T – 1]$

= $100x [(1 + \frac{10}{100})^2 – 1]$

= $100x [(\frac{11}{10})^2 – 1] = 100x (\frac{121 – 100}{100})$

= $100x \times \frac{21}{100} = 21x$

=> $S.I. = \frac{P \times R \times T}{100}$

= $\frac{100x \times 10 \times 2}{100} = 20x$

It is given that difference between compound and simple interest = $21x – 20x = 50$

=> $x = 50$

$\therefore$ Principal = $100 \times 50 = Rs. 5,000$

Principal amount = Rs. 16,500

Simple interest earned = Rs. 5,940 and time = 3 years

Let rate of interest = $R \%$

=> $S.I. = \frac{P \times R \times T}{100}$

=> $5940 = \frac{16500 \times R \times 3}{100}$

=> $R = \frac{5940}{165 \times 3}$

=> $R = 12 \%$

Now, compound interest accrued on the same amount at the same rate in the same period = $P [(1 + \frac{R}{100})^T – 1]$

= $16,500 [(1 + \frac{12}{100})^3 – 1]$

= $16,500 [(\frac{28}{25})^3 – 1] = 16,500 (\frac{21952 – 15625}{15625})$

= $16,500 \times \frac{6327}{15625} = Rs. 6681.31$

Let sum invested in both schemes = $Rs. P$

Let rate of interest in both schemes = $R \%$

Time period in scheme A = $R$ years

=> Simple interest under Scheme A = $\frac{P \times R \times R}{100}$

= $Rs. \frac{P R^2}{100}$

Also, interest received from both schemes is also same, and time period under scheme B = 2 years

=> Compound interest under scheme B = $P [(1 + \frac{R}{100})^T – 1]$

=> $\frac{P R^2}{100} = P [(1 + \frac{R}{100})^2 – 1]$

=> $\frac{R^2}{100} = [1 + (\frac{R}{100})^2 + \frac{2 R}{100}] – 1$

=> $\frac{R}{100} = \frac{R}{10000} + \frac{1}{50}$

=> $\frac{R}{100} (1 – \frac{1}{100}) = \frac{1}{50}$

=> $\frac{R}{100} \times \frac{99}{100} = \frac{1}{50}$

=> $R = \frac{100 \times 100}{99 \times 50}$

=> $R = \frac{200}{99} = 2 \frac{2}{99} \%$

Interest is compounded half yearly

=> rate of interest = 10/2 = 5%

and time period = 1 * 2 = 2

=> Amount = $15000 (1 + \frac{5}{100})^2$

= $15000 * \frac{21}{20} * \frac{21}{20}$ = Rs. 16,537.50

Principal amount = Rs. 35,000

Rate of interest = 15% and time period = 4 years

Simple interest = $\frac{35000 \times 15 \times 4}{100}$

= Rs. $21,000$

Total amount Ms. Anisha will get = 35000 + 21000

= Rs. 56,000

Let rate of interest be $r$ %

=> Simple interest = $\frac{19800 \times r \times 3}{100} = 7128$

=> $r = 12$%

Now, C.I. = $19800 [(1 + \frac{12}{100})^3 – 1]$

= $19800 (1.12^3 – 1)$

= $19800 * 0.405 = 8019$

= Rs. 8017.5744

Principal = Rs. 45,500

Rate of interest = 11% and time period = 3 years

S.I. = $\frac{45500 \times 11 \times 3}{100}$

= Rs. 15,015

$\therefore$ Required amount = 45,500 + 15,015 = Rs. 60,515

Principal amount = Rs. 10,530

Let rate of interest = $r$ %

=> Simple interest = $\frac{10530 \times r \times 5}{100} = 6318$

=> $r = \frac{6318 \times 100}{10530 \times 5}$

=> $r = 12$ %

Principal amount = Rs. 58,750

Rate of interest = 12% and time period = 4 years

Simple interest = $\frac{58750 \times 12 \times 4}{100}$

= Rs. $28,200$

Total amount Mr. Madhur will get = 58750 + 28200

= Rs. 86,950

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

= $7250 (1 + \frac{6}{100})^2$

= $7250 \times (1.06)^2$ = Rs. 8,146

$S.I. = \frac{Principal \times Rate \times time}{100}$

= $\frac{10,000 \times 3 \times 2}{100}$

= Rs. $600$

$\therefore$ Mr. Khanna’s monthly salary

= $600 \times \frac{100}{3}$

= Rs. $20,000$

We know that, in compound interest, amount A :

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

=> $A = 16800 (1 + \frac{8}{100})^2 (1 + \frac{10}{100})^2$

=> $A = 16800 \times (\frac{27}{25})^2 \times (\frac{11}{10})^2$

=> $A = 23710$

$\therefore$ C.I. = 23710 – 16800

= Rs. 6,910

Let Mary’s monthly salary = Rs. $100x$

Money spent on EMI = $\frac{15}{100} \times 100x = 15x$

Amount left = $100x – 15x = 85x$

% amount left after paying internet bill and rent = $100 – (10 + 20) = 70 \%$

=> Amount left = $\frac{70}{100} \times 85x = 24,990$

=> $\frac{7 \times 17x}{2} = 24,990$

=> $x = \frac{24,990 \times 2}{7 \times 17}$

=> $x = 210 \times 2 = 420$

$\therefore$ Mary’s monthly salary = $100 \times 42$ = Rs. $42,000$

We hope this Simple Interest and Compound Interest Questions for IBPS Clerk  will be highly useful for your preparation.