Compound interest Questions and answers

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Aptitude

11). The difference between compound and simple rates of interest on RS. 10000 for 3 yr at 5% per annum is

A). RS.76.25

B). RS.76.75

C). RS.76.50

D). RS.76

View Answer

12). What is the difference between the compound interest and simple interest calculated on an amount of RS.16200 at the end of 3 yr at 25% pa? (Rounded off to two digits after decimal)

A). RS.3213.44

B). RS.3302.42

C). RS.3495.28

D). RS.3290.63

View Answer

Correct Answer: RS.3290.63

Let required difference be RS. D

By formula,

D =\( \Large P\times \left(\frac{R}{100}\right)^{2}\times \left(3+\frac{R}{100}\right) \)

= \( \Large 16200\times \left(\frac{25}{100}\right)^{2}\times \left(3+\frac{25}{100}\right) \)

= \( \Large 16200\times \frac{625}{10000}\times {13}{4} \)

= \( \Large \frac{162\times 625\times 13}{4\times 100} \) =RS.3290.63

13). If the difference between the compound interest and the simple interest on a certain sum for 2 yr at 8% per annum is RS. 32, then the sum is

A). RS.5000

B). RS.5500

C). RS.6000

D). RS.5250

View Answer

Correct Answer: RS.5000

Given, R = 8%, D = RS. 32 and P = ?

By formula, D = \( \Large P \left(\frac{R}{100}\right)^{2} \)

=> 32 = \( \Large P \left(\frac{8}{100}\right)^{2} \)

P = \( \Large \frac{32\times 100\times 100}{8\times 8} \) =RS.5000

14). The simple interest for certain sum in 2 yr at 4% pa is RS. 80. What will be the compound interest for the same sum, if conditions of rate and time period are same?

A). RS.91.60

B). RS.81.60

C). RS.71.60

D). RS.80

View Answer

Correct Answer: RS.81.60

Given, n = 2 yr, R = 4% and SI = 80

According to the formula,

CI = \( \Large SI \left(1+ \frac{R}{100}\right) \)

= \( \Large 80 \left(1+ \frac{4}{200}\right) \)=\( \Large \frac{80\times 51}{50} \)=RS.81.60

15). The compound interest is RS. 6.40 more than the simple interest, if a sum is lent for 2 yr at 8% compound interest. Find the sum.

A). RS.1800

B). RS.10000

C). RS.800

D). RS.1000

View Answer

Correct Answer: RS.1000

Given, CI - SI = 6.40

T=2yr and R=8%

=> \( \Large P \left[(1+ \frac{R}{100}\right)^{T}-1] \)-\( \Large \frac{P\times R\times T}{100} \)=6.40

=> \( \Large P \left[(1+ \frac{8}{100}\right)^{2}-1] \)-\( \Large \frac{P\times 8\times 2}{100} \)=6.40

=> \( \Large P \left[(\frac{27}{25}\right)^{2}-1] \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large P \left[\frac{27^{2}-25^{2}}{625}\right] \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large \frac{104 P}{625} \)-\( \Large \frac{16P}{100} \)=6.40

=> \( \Large \frac{416P-400P}{2500} \)=6.40

=> 16P = \( \Large 6.40\times 2500 \)

P = \( \Large \frac{6.40\times 2500}{16} \)= RS.1000

16). A certain sum at compound interest amounts to RS. 12960 in 2 yr and RS. 13176 in 3 yr. Find the rate per cent per annum.

A). \( \Large 1\frac{1}{3}% \)

B). \( \Large 2\frac{1}{3}% \)

C). \( \Large 1\frac{2}{3}% \)

D). \( \Large 2\frac{2}{3}% \)

View Answer

Correct Answer: \( \Large 1\frac{2}{3}% \)

SI for on RS.12960 for 1 yr

= 13176-12960=RS.216

By formula,SI=\( \Large \frac{P\times R\times T}{100} \)

=> 216=\( \Large \frac{12960\times R\times 1}{100} \)

R=\( \Large \frac{5}{3} \)=\( \Large 1\frac{2}{3}% \)

17). A sum amounts to RS. 2916 in 2 yr and RS. 3149.28 in 3 yr at compound interest. The sum is

A). RS.1500

B). RS.2500

C). RS.2000

D). RS.3000

View Answer

Correct Answer: RS.2500

Let required amount be RS. P

According to the question,

2916 =\( \Large P \left(1+ \frac{R}{100}\right)^{2} \) .....(i)

and 3149.28 = \( \Large P \left(1+ \frac{R}{100}\right)^{3} \) ....(ii)

On dividing Eq. (ii) by Eq. (i), we get

\( \Large 1+ \frac{R}{100} \)=\( \Large \frac{3149.28}{2916} \) => \( \Large \frac{R}{100} \)= \( \Large \frac{3149.28}{2916}-1 \)

=> R=\( \Large \frac{233.28}{2916}\times 100 \)=8%

From Eq. (i),

P=\( \Large \frac{2916\times 100\times 100}{108\times 108} \)=RS.2500

18). At what per cent annual compound interest rate, a certain sum amounts to its 27 times in 3 yr?

A). 100%

B). 150%

C). 75%

D). 200%

View Answer

Correct Answer: 200%

Let sum = P and given that n = 3yr

According to the question

Amount = 27 P

According to the formula,

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

=> 27P= \( \Large P \left(1+ \frac{R}{100}\right)^{3} \)

=> 27=\( \Large \left(1+ \frac{R}{100}\right)^{3} \)

=> \( \Large \left(1+ \frac{R}{100}\right)^{3}=(3)^{3} \)

=> \( \Large 1+ \frac{R}{100} \)=3

=> \( \Large \frac{R}{100} \)=3-1=2

R=200%

19). What will be the principal, if the rate of 5% per annum can yield RS. 410 as compound interest in 2 yr?

A). RS.4000

B). RS.42000

C). RS.8000

D). RS.21000

View Answer

Correct Answer: RS.4000

Let the amount be RS. P

R = 5%,n = 2yr and CI= RS.410

By formula, CI = \( \Large P \left[(1+ \frac{R}{100}\right)^{n}-1] \)

=> 410=\( \Large P \left[(1+ \frac{5}{100}\right)^{2}-1] \)

=> 410=\( \Large P \left[(1+ \frac{1}{20}\right)^{2}-1] \)

=> 410=\( \Large P \left[(\frac{21}{20}\right)^{2}-1] \)

=> 410=\( \Large P\left[ \frac{441}{400}-1 \right] \)

=> 410=\( \Large \left(\frac{441-400}{400}\right)=P\times 41/400 \)

=> P=\( \Large \frac{410\times 400}{41} \)

P=RS.4000

20). Find the compound interest for a sum of RS. 9375 in 2 yr, if the rate of interest for the first year is 2% and for the second year is 4%.

A). RS.570

B). RS.670

C). RS.770

D). RS.760

View Answer

Correct Answer: RS.570

Given, P =RS. 9375, \( \Large R_{1} \) = 2%

and \( \Large R_{2} \) = 4%

According to the formula,

A=\( \Large P \left(1+ \frac{R_{1}}{100}\right) \) \( \Large \left(1+ \frac{R_{2}}{100}\right) \)

=\( \Large 9375 \left(1+ \frac{2}{100}\right) \) \( \Large \left(1+ \frac{4}{100}\right) \)

=\( \Large 9375\times 51/50\times 26/25 \)

=\( \Large 7.5\times 51\times 26 \)=RS.9945

Now, CI=A-P=9945-9375

=RS.570

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