Practice MCQ Questions and Answer on Compound Interest

The difference between CI and SI for 3 years Rs. 992. If rate of interest is 10%. Find the Principal ?

(A) Rs. 22000
(B) Rs. 30000
(C) Rs. 28000
(D) Rs. 32000

Solution: $\begin{aligned} Rate = 10 %, \\ Let Principal = P \\ S .I . = \frac{P \times 10 \times 3}{100} = \frac{3 P}{10} \\ C .I . = P {{(1 + \frac{1}{10})}^{3} - 1} \\ \Rightarrow C .I . - S .I . = 992 \\ \Rightarrow P {{(1 + \frac{1}{10})}^{3} - 1} - \frac{3 P}{10} = 992 \\ \Rightarrow P {{(\frac{11}{10})}^{3} - 1 - \frac{3}{10}} = 992 \\ \Rightarrow P {\frac{(1331 - 1000 - 300)}{1000}} = 992 \\ \Rightarrow P (\frac{31}{1000}) = 992 \\ \Rightarrow P = 32000 \end{aligned}$ Alternate: Rate = 10% = $\frac{1}{10}$ Let principal ⇒ (10)3 = 1000 Interest 1st year → 100 Interest 2nd year → 100 + 10 Interest 3rd year = 100 + 10 + 10 + 1 C.I – S.I = 331 – 300 = 31 31 → 992 1 → 32 P → 32000

A loan of Rs. 12300 at 5% per annum compound interest, is to be repaid in two equal annual installments at the end of every year. Find the amount of each installment ?

(A) Rs. 6651
(B) Rs. 6615
(C) Rs. 6516
(D) Rs. 6156

Solution: $5 % = \frac{1}{20} = \frac{21 \to Installment}{20 \to Principal}$ Year Principal Installment ⇒ I 20×21 → 21×21 ......(i) ⇒ II 400→ 441 .....(ii) Since, installment is equal, hence multiply equation (i) by 21 ⇒ Total principal = 420 + 400 = 820 ⇒ 820 units = Rs. 12300 ⇒ 1 units = Rs. 15 ⇒ 441 units = Rs. 6615 ∴ Each installment = Rs. 6615

At what rate percent per annum will a sum of Rs. 15,625 amount to Rs. 21,952 in three years, if the interest is compounded annually?

(A) 12%
(B) 8%
(C) 9%
(D) 10%

Solution: $\begin{aligned} 15625 : 21952 \to in three years \\ \sqrt[3]{15625} : \sqrt[3]{21952} \to in one year \end{aligned}$ $Rate = \frac{3}{25} \times 100 = 12 %$

A sum of Rs. 12,000 amounts to Rs. 20,736 in 3 years at a certain rate percent per annum interest compounded annually. What will amount of the same sum to in 2 years at the same rate on compound interest?

(A) Rs. 15,640
(B) Rs. 17,820
(C) Rs. 17,280
(D) Rs. 14,520

Solution: Amount in 2 years $\to 12000 \times \frac{6}{5} \times \frac{6}{5} = 17280$

What is the compound interest on Rs. 5,000 for 2 years at 10%, per annum, compounded half-yearly?

(A) Rs. 1,025.00
(B) Rs. 2,320.50
(C) Rs. 1,050.00
(D) Rs. 1,077.50

A sum amount to Rs. 7,562 in 4 years and to Rs. 8,469.44 in 5 years at a certain rate percent per annum, when the interest is compounded yearly. The rate of interest is:

(A) 8%
(B) 12%
(C) 20%
(D) 15%

Solution: $\begin{aligned} 7562________8469.44 \\ Rate of interest = \frac{907.44}{7562} \times 100 = 12 % \end{aligned}$

Shashi had a certain amount of money. He invested $\frac{2}{3}$ of the total money in scheme A for 6 years and rest of the money he invested in scheme B for 2 years. Scheme A offers simple interest at a rate of 12% p.a. and scheme B offers compound interest ( compound annually) at a rate of 10% p.a. If the total interest obtained from both the schemes is Rs. 2750. What was the total amount invested by him in scheme A and scheme B together ? (Approximate value)

(A) Rs. 4500
(B) Rs. 4200
(C) Rs. 4050
(D) Rs. 5000

Solution: Let the total sum of money invested by Shashi be Rs. x In scheme A money invested at simple interest for 6 years at a rate of 12% p.a. $∴ \frac{2}{3} of x \times \frac{12 \times 6}{100} = \frac{48 x}{100} . . . . (i)$ In scheme B money at compound interest for 2 year at a rate of 10% p.a. $\begin{aligned} \frac{x}{3} {(1 + \frac{10}{100})}^{2} - \frac{x}{3} \\ \Rightarrow \frac{x}{3} {(1 + \frac{10}{100})}^{2} - \frac{x}{3} = \frac{7 x}{100} \end{aligned}$ According to given information we get $\begin{aligned} \Rightarrow \frac{48 x}{100} + \frac{7 x}{100} = 2750 \\ \Rightarrow 55 x = 2750 \times 100 \\ \Rightarrow x = \frac{2750 \times 100}{55} \\ \Rightarrow x = R s . 5000 \end{aligned}$

At what percent is the annual compound interest rate taken such that a certain sum amounts to its 16 times in 4 years?

(A) 175%
(B) 100%
(C) 200%
(D) 75%

Solution: $\Rightarrow Rate = \frac{1}{1} \times 100 = 100 Answer$

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

(A) 6%
(B) 6.5%
(C) 7%
(D) 7.5%

Solution: $\begin{aligned} Let the rate be R % p . a . \\ Then, 1200 \times {(1 + \frac{R}{100})}^{2} = 1348.32 \\ \Rightarrow {(1 + \frac{R}{100})}^{2} = \frac{134832}{120000} = \frac{11236}{10000} \\ ∴ {(1 + \frac{R}{100})}^{2} = {(\frac{106}{100})}^{2} \\ \Rightarrow 1 + \frac{R}{100} = \frac{106}{100} \\ \Rightarrow R = 6 % \end{aligned}$

10.

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

(A) Rs. 2160
(B) Rs. 3120
(C) Rs. 3972
(D) Rs. 6240

Solution: $\begin{aligned} Let P = Rs . 100 Then, \\ S .I . = Rs . 60 and \\ T = 6 years \\ ∴ R = \frac{100 \times 60}{100 \times 6} = 10 % p . a . \\ Now, P = R s . 12000 \\ T = 3 year and \\ R = 10 % p . a . \\ ∴ C .I . = R s . [12000 \times {{(1 + \frac{10}{100})}^{3} - 1}] \\ = R s . (12000 \times \frac{331}{1000}) \\ = R s . 3972 \end{aligned}$

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Practice MCQ Questions and Answer on Compound Interest

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