299+ Compound Interest Questions and Answers

Q:

The compound interest on rs.30000 at 7% per annum is Rs.4347. The period is

A) 2 years	B) 2.5 years
C) 3 years	D) 4 years

Answer & Explanation Answer: A) 2 years

Explanation:

Amount = Rs.(30000+4347) = Rs.34347

let the time be n years

Then,30000(1+7/100)^n = 34347

(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2

$\therefore$ n = 2years

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379 95008

Q:

Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually

A) 2109	B) 3109
C) 4109	D) 6109

Answer & Explanation Answer: B) 3109

Explanation:

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

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341 58104

Q:

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is

A) 10000	B) 20000
C) 40000	D) 50000

Answer & Explanation Answer: B) 20000

Explanation:

Let sum=Rs.x

C.I. when compounded half yearly = $[x {(1 + \frac{10}{100})}^{4} - x] = \frac{4641}{10000}$

C.I. when compounded annually = $[x {(\frac{20}{100})}^{2} - x] = \frac{11}{25}$

$\frac{4641}{10000} x - \frac{11}{25} x = 482$

=> x=20000

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274 56342

Q:

A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum.

A) 4360	B) 4460
C) 4560	D) 4660

Answer & Explanation Answer: B) 4460

Explanation:

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

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252 71247

Q:

Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the interest being compounded half-yearly.

A) 524.32	B) 624.32
C) 724.32	D) 824.32

Answer & Explanation Answer: D) 824.32

Explanation:

Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
Amount == Rs. 10824.32.

$R s . [10000 * {(1 + \frac{2}{100})}^{4}] = R s . [10000 * \frac{51}{50} * \frac{51}{50} * \frac{51}{50} * \frac{51}{50}]$

$\therefore$ C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

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Q:

The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. if the interest were compounded half yearly , the difference in two interests would be nearly

A) Rs.24.64	B) Rs.21.85
C) Rs.16	D) Rs.16.80

Answer & Explanation Answer: A) Rs.24.64

Explanation:

For 1st year S.I =C.I.

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200

i.e S.I on the principal for 1 year is Rs.200

Principle = $R s . \frac{100 * 200}{8 * 1}$ = Rs.2500

Amount for 2 years, compounded half-yearly

$R s . [2500 * {(1 + \frac{4}{100})}^{4}] = R s . 2924.4$

C.I = Rs.424.64

Also, $S . I = R s . (\frac{2500 * 8 * 2}{100}) = R s . 400$

Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64

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121 42932

Q:

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A) 3.6	B) 6
C) 18	D) Cannot be determined

Answer & Explanation Answer: B) 6

Explanation:

Explanation:

Let rate = R% and time = R years.

Then, $(\frac{1200 * R * R}{100}) = 432$

$12 R^{2} = 432$

=> $R^{2} = 36 \Leftrightarrow R = 6$

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Q:

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.

A) 1261	B) 1271
C) 1281	D) 1291

Answer & Explanation Answer: A) 1261

Explanation:

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261.
.. C.I. = Rs. (9261 - 8000) = Rs. 1261.

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