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Simple Interest and Compound Interest P...

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Simple Interest and Compound Interest Practice Quiz

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Simple interest is calculated on the principal, or original, amount of a loan. Compound interest is calculated on the principal amount and also on the accumulated interest of previous periods, and can thus be regarded as “interest on interest”. The article Simple Interest and Compound Interest Practice Quiz useful to the candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

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1. Ramu wanted to borrow Rs. 1000 from a money lender for a period of one year. However, the money lender deducted Rs. 200 as interest charges and gave Rs. 800 t o Ramu. Ramu returned Rs. 1000 at the end of the year. The rate of interest charged by the money lender is

    A. 20
    B. 22.5
    C. 25
    D. 27.5


Answer – Option C

Explanation –
Effectively he borrowed Rs. 800 and returned Rs. 1000 after one year. So he paid Rs. 200 as interest on Rs. 800

i.e, Rate of interest = \(\frac {200} {800} * 100\) = 25%

2. Rajan borrowed Rs. 50000 from Rakesh at simple interest. After 3 years, Rakesh got Rs. 3000 more than what he had given to Rajan. What was the rate of interest per annum ?

    A. 2%
    B. 5%
    C. 8%
    D. 10%


Answer – Option A

Explanation –

Rate = \((\frac {100 * 300} {5000 * 3} )\)% = 2%

3. Rakesh took a loan for 7 years at the rate of 6 % p.a. S.I. If total interest paid was Rs. 2100, then principal was

    A. Rs. 4400
    B. Rs. 4800
    C. Rs. 5000
    D. Rs. 5200


Answer – Option C

Explanation –

Principal = Rs\((\frac {2100 * 100} {7 * 6} )\) = Rs. 500

4. How much should money lender lend at simple rate of interest of 15% in order to have Rs. 3234 at the end of \( 1 \frac {1} {2}\) years

    A. Rs. 1640
    B. Rs. 2620
    C. Rs. 2610
    D. Rs. 2640


Answer – Option D

Explanation –

Let required money be x

Then \(( x + \frac {x * 15} {100} * \frac {3}{2}\) = 3234

\(\frac {49 x} {40}\) = 3234

i.e, x = \(\frac {3234 * 40} {49}\) = 2640

5. An amount Rs. 8000 becomes Rs. 9200 in 3 years at simple interest. If rate of interest is increased by 3%, it would amount to

    A. Rs. 9920
    B. Rs. 10560
    C. Rs. 11120
    D. Rs. 11820


Answer – Option A

Explanation –

. Principal = Rs. 8000, S.I. = Rs. 1200,

Time = 3 years.

i.e, Rate = \((\frac {100 * 1200} {8000 * 3}) %\) = 5%

New rate = 8%, Principal = Rs. 8000,

Time = 3 years.

S.I = Rs. \((\frac {8000 * 8 * 3} {100})\) = Rs. 1920

i.e, New amount = Rs. (8000 + 1920)

= Rs. 9920

1. Find the simple interest on Rs. 4800 at the rate of 8 \(\frac {1} {2}\)% per annum for a period of 2 years 3 months.

    A. Rs. 796
    B. Rs. 816
    C. Rs. 918
    D. Rs. 990


Answer – Option C

Explanation –

SI = \(\frac {4800 * 8.S * 2.2S} {100}\) = Rs. 918

2. Simple Interest on Rs. 500 for 4 years at 6.25% per annum is equal to the Simple Interest on
Rs.400 at 5% per annum for a certain period of time. The period of time is


    A. 4 years
    B. 5 years
    C. 6 \(\frac {1} {4}\)
    D. 8 \(\frac {2} {3}\)


Answer – Option C

Explanation –

500 * 4 * 6.25% = 400 * 5 * t

t = 6.25 years

3. A sum becomes Rs. 2916 in 2 years at 8% per annum compound interest. The sum is

    A. Rs. 2750
    B. Rs. 2500
    C. Rs. 2625
    D. Rs. 2560


Answer – Option B

Explanation –

Let the required sum be Rs. x.

x * \(\frac {108} {100}\) * \(\frac {108} {100}\) = 2916

x = Rs. 2500

4. If `200 becomes `240 in 4 years, then the rate of simple interest per annum is

    A. \(\frac {25} {6}\)%
    B. \(\frac {25} {3}\)%
    C. \(\frac {25} {2}\)%
    D. 5%


Answer – Option D

Explanation –

Given that Rs. 200 becomes Rs. 240 in 4 years, thus it would have become Rs. 210 at the end of first year.

Hence, rate of simple interest = \(\frac {10} {200}\) * 100 = 5%

5. A sum of money doubles itself in 5 years when the interest is compounded annually. The number of years when it will become eight times is

    A. 10
    B. 12
    C. 15
    D. 20


Answer – Option C

Explanation –
The money gets doubled in 5 years which means it becomes twice of itself after every 5 years. Hence, it will be increased to 4 times in 10 years and 8 times in 15 years.


11. Compound interest on rupees 8000 for 1 year at 10% per annum compounded half yearly is

    A. 800
    B. 1680
    C. 840
    D. 736.20


Answer – Option D

Explanation –

8000 * \(\frac {209} {200} * \frac {209} {200}\) = 8736.20

So, interest = 8736.20 − 8000 = 736.20

12. In how many years rupees 500 will amount to rupees 800 at simple interest of 10% per year

    A. 6
    B. 8
    C. 10
    D. 16


Answer – Option A

Explanation –

Simple interest = Amount – Principle

= 800 – 500 = 300

SI = = \(\frac {PTR} {100}\) = 300

500 * = \(\frac {10} {100}\) * T

T = 6 years

13. Compound interest Rs. 16000 for 1 year at 10% per annum compounded half yearly is

    A. 1600
    B. 1640
    C. 1680
    D. 3360


Answer – Option B

Explanation –

Compound interest for \(\frac {1} {2}\)year = \(\frac {PTR} {100}\)

= 16000 * \(\frac {1} {10} * \frac {1} {2}\) = 800

Compound interest for 2nd \(\frac {1} {2}\)year

= 800 + 800 * \(\frac {1} {10} * \frac {1} {2}\) = 840

i.e, Total I = 800 + 840 = 1640

14. In how many years `500 will amount to `700 at simple interest of 5% per annum?

    A. 4
    B. 5
    C. 6
    D. 8


Answer – Option D

Explanation –

P = Rs. 500

A = Rs. 700

Interest = 200 = \(\frac {500 * 5 * T} {100}\)

T = 8 years

15. In how many years Rs. 2000 will amount to Rs. 2100 at 10% per annum compounded half yearly

    A. 2
    B. 1.5
    C. 1
    D. 0.5


Answer – Option D

Explanation –

A = \(({1 + \frac {r} {100}})^{T}\)

2100 = 2000 \(({1 + \frac {\frac {10}{2}} {100}})^{T}\)

1.05 = \(({1 + \frac {5} {100}})^{T}\)

No. of years = 0.5

1. Find compound interest on Rs. 7,300 at the rate of 4% per annum for 2 year s, compounded annually?

    A. Rs. 612
    B. Rs. 300
    C. Rs. 600
    D. Rs. 630


Answer – Option A

Explanation –

A = 7500 * \({(1 +\frac {4} {100})}^{2}\) = 812

i.e, CI = 8112 – 7500 = 612

2. In how many years, a sum will be thrice of it at simple interest @10% per annum ?

    A. 15 years
    B. 20 years
    C. 30 years
    D. 40 years


Answer – Option B

Explanation –

According to the question sum becomes thrice.

If Rs. P is invested, it becomes 3P

i.e, Interest earned = 2P

2P = \(\frac {P * 10 * T} {100}\)

T = 20years

3. A sum of money amounts to Rs. 9680 in 2 years and Rs. 10648 in 3 years. The rate of interest per annum on compounded basis is

    A. 5%
    B. 10%
    C. 15%
    D. 20%


Answer – Option B

Explanation –

According to the question

9680 = P \(({1 + \frac {r} {100}})^{2}\)

10648 = P \(({1 + \frac {r} {100}})^{3}\) = P \(\frac {10648} {9680}\) = P \(1 + \frac {r} {100}\)

1.1 = 1 + \(\frac {r} {100}\)

r = 10% P.a

4. A man buys a TV by making cash down payment of Rs. 4945 and agrees to pay two more yearly installments of equivalent amounts at the end of first year and second year. If the rate of interest is 7 \(\frac {1} {2}\)% per annum, compounded annually, the cash value of the TV (in Rs.) is nearest to

    A. 12840
    B. 13804
    C. 13824
    D. 14835


Answer – Option C

Explanation –

Let c be the cost of T.V.

c = 4945 \(( 1 + \frac {1} {1.075}+ \frac {1} {{1.075}^{2}})\)

= 4945 + 4600 + 4279 = 13824

5. A sum of Rs.5000 amounts to Rs 8640 at compound interest in a ain times, then the same sum
amounts to what in one-third of the time?


    A. Rs 5886
    B. Rs 6000
    C. Rs 6214
    D. Rs 7000


Answer – Option B

Explanation –

8640 = 5000 \(({1 + \frac {r} {100}})^{T}\) …….(i)

k = 5000 \(({1 + \frac {r} {100}})^{\frac{T}{3}}\)

cubing both sides

\({k}^{3}\) = \({5000}^{3}\) \(({1 + \frac {r} {100}})^{T}\) …….(ii)

Divide (1) and (2)

\({k}^{3}\) = \({5000}^{2}\) * 8640

k = 6000

6. A loan of Rs.62496 is to be paid back in three equal annual installments. I f the inter est is compounded annual ly at 12 \(\frac {1} {2}\) %, t hen each installments will be of (in Rs.)

    A. 26736
    B. 26244
    C. 25736
    D. 24244


Answer – Option B

Explanation –

Let equal instruments be of Ps. x

x(\({1.125})^{2} + 1.125 + 1\) = 62496 \(({1.125})^{3}\)

i.e, x = 26244


7. Two equal sum are lent out at 6 % and 5% simple interest per annum respectively at the same time. The first is recovered 24 years earlier than the second one and the amount received in each case was Rs. 28800. Each sum (in Rs.) was

    A. 18000
    B. 20000
    C. 20500
    D. 2200


Answer – Option A

Explanation –

Let each sum be Rs. x

According to the question

\(\frac {x * 6* (t – 24)} {100}\) = \(\frac {2 * 5 * t} {100}\)

Also, x + \(\frac {x * 5 * f} {100}\) = 28800

x = 18,000

8. A computer is available for Rs. 22750 cash payment or for Rs. 6200 cash down payment and three equal annual installments of Rs. x. If the interest charged is 10% per annum. Compounded annually, the value of x is

    A. 5517
    B. 5578
    C. 6565
    D. 6655


Answer – Option D

Explanation –

(22750 – 6200) \( {(1.2)}^{3}\) = x \( {(1.1)}^{2}\) + (1.1) + 1

x = 6655

9. A sum of money at simple interest amounts to Rs.13800 in 3 years . If rate of interest is increased by 30%,the same sum amounts to Rs.14340 in the same time. The rate of interest per annum is

    A. 3%
    B. 4%
    C. 5%
    D. 8%


Answer – Option C

Explanation –

\(\frac {P * (1.3r) * 3} {100}\) – \(\frac {P * r * 3} {100}\) = 540

Pr = 60000

Also, P + \(\frac {P * r * 3} {100}\) = 13,800

r = 5%

10. A person borrowed some money on compound interest and returned it in three years in equal annual installments. If the rate of interest in 15% per annum and the annual installment is Rs.48668, then the sum borrowed was (in Rs )

    A. 101020
    B. 111050
    C. 111120
    D. 146004


Answer – Option C

Explanation –

Let money borrowed be Rs. x

According to the question

48668 (\({1.5})^{2} + 1.15 + 1\) = x \(({1.15})^{3}\)

x = 111120

11. A sum of Rs. x at simple interest amounts to Rs. 14160 i n 3 year s. If the rate of interest is increased by 25 % the same sum amounts to Rs.14700 in the same time. The value of x is

    A. 120000
    B. 12400
    C. 13000
    D. 13400


Answer – Option A

Explanation –

According to the question

\(x + \frac {x * r* 3} {100}\) = 14160 ….(1)

\(x + \frac {x *(r + \frac{r}{4}) * x3} {100}\) = 14700 ….(2)

solving (1) & (2)

i.e, x =12000.

12. A certain sum of many is borrowed at compound interest for 3 year s at 5% per annum. The interest for the third year is greater than that of second year by Rs.642.60. t he sum (in Rs.) borrowed is

    A. 24480
    B. 185400
    C. 244800
    D. 368400


Answer – Option C

Explanation –

I = Interest

\( {n}_{1st year}\) = \(\frac {P} {20}\)

\( {n}_{2nd year}\) = \(\frac {P} {20} + \frac {P} {400} +\frac {21P} {400}\)

\( {n}_{3nd year}\) = \(\frac {P} {20} + \frac {P} {400} + \frac {21P} {20 * 400}\)

\(\frac {4P} {8000}\) = 642.6

i.e, P = 244800

13. A sum of Rs. 129780 is be paid back in three equal half yearly installments. If the interest is compounded half yearly at the rate of 3 \(\frac {1} {3}\)% per annum, then each installment is of Rs.

    A. 44690
    B. 45960
    C. 46080
    D. 49152


Answer – Option D

Explanation –

Let each installment be Rs. x

\((129780) {(1.0665)}^{3}\)

x (\( {1.0665}^{2} + (1.0665) + 1\))

x = 49152

14. A loan of Rs. 26480 is to be paid back in three equal yearly installment s. If the interest is compounded yearly at 10% per annum, then each installment is of Rs.

    A. 11548
    B. 10864
    C. 10648
    D. 8827


Answer – Option C

Explanation –

(26480) \( {1.1}^{3}\)

(\( {1.1}^{2} + (1.1) + 1\))x

x = 10648

15. A sum of Rs.78060 is divided between A and B, so that the amount of A after 3 \(\frac {1} {2}\) years is equal to the amount of B after 4 \(\frac {1} {2}\) year, the interest is compounded half yearly at 8% per annum. The share of B in the given sum is (in Rs.)

    A. 40560
    B. 38560
    C. 37800
    D. 37500


Answer – Option D

Explanation –

Let sum with B = x

i.e, sum with A = (78060 – x)

According to the question

\((1 + \frac {4} {100})^{9}\) = (78060 – X) \((1 + \frac {4} {100})^{7}\)

x\((1 + \frac {4} {100})^{2}\) = 78060 – x

x = 37500