Compound Interest - Aptitude MCQ Questions and Solutions with Explanations

51

A certain some of money and Rs. 2420 in 2 years and Rs. 2662 in 3 years at same rate of compound interest, compounded annually. The rate of interest per annum is = ?

A) 6% B) 8% C) 9% D) 10%

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Amount after three years}} \cr & {\text{ = Rs. 2662}} \cr & {\text{Amount after two years}} \cr & {\text{ = Rs. 2420}} \cr & {\text{Net interest earned in the }}{{\text{3}}^{{\text{rd}}}}{\text{ year}} \cr & {\text{ = }}\,{\text{2662}} - {\text{2420}} \cr & {\text{ = Rs}}{\text{. 242}} \cr & {\text{Rate of interest (r)}} \cr & {\text{ = }}\frac{{242}}{{2420}} \times {\text{100 = 10% }} \cr} $$
(∴ 2^nd year's amount is principal for 3^rd year)

52

Kamal took Rs. 6800 as a loan which along with interest is to be repaid in two equal annual installment. If the rate of interest is $$12\frac{1}{2}$$ % compounded annually, then the value of each installment is = ?

A) Rs. 8100 B) Rs. 4150 C) Rs. 4050 D) Rs. 4000

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Rate of interest}} \cr & {\text{r}} = {\text{12}}\frac{1}{2}\% = \frac{1}{8} \cr} $$

Year	Principal	Installment
⇒ I	8_×9 →	9_×9	......(i)
⇒ II	64 →	81	......(ii)

Since, installment is equal hence multiply equation (i) by 9
⇒ Total principal = 72 + 64 = 136 units
136 units → 6800
1 units → 50
81 units → 4050
⇒ Each installment = Rs. 4050

53

A man invests Rs 4000 for 3 years at compound interest. After one year the money amounts to Rs. 4320. What will be the amount (to the nearest rupee) due at the end of 3 years ?

A) Rs. 4939 B) Rs. 5039 C) Rs. 5789 D) Rs. 6129

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Le the rate be R }}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Then,}} \cr & {\text{4000}}\left( {1 + \frac{{{\text{R }}}}{{100}}} \right) = 4320 \cr & \Rightarrow 1 + \frac{{{\text{R }}}}{{100}} = \frac{{4320}}{{4000}} = \frac{{108}}{{100}} \cr & \Rightarrow \frac{{{\text{R }}}}{{100}} = \frac{8}{{100}} \cr & \Rightarrow {\text{R }} = 8 \cr & \therefore {\text{Amount after 3 yeras}} \cr & {\text{ = Rs}}{\text{. }}\left[ {4000 + {{\left( {1 + \frac{8}{{100}}} \right)}^3}} \right] \cr & {\text{ = Rs}}{\text{. }}\left( {4000 \times \frac{{27}}{{25}} \times \frac{{27}}{{25}} \times \frac{{27}}{{25}}} \right) \cr & {\text{ = Rs}}{\text{. }}\left( {\frac{{629856}}{{125}}} \right) \cr & {\text{ = Rs}}{\text{. }}5038.848 \approx 5039 \cr} $$

54

A sum of Rs. 13360 was borrowed at $${\text{8}}\frac{3}{4}$$ % per annum compound interest and paid back in two years in two equal annual installments. What was the amount of each installment ?

A) Rs. 5769 B) Rs. 7569 C) Rs. 7009 D) Rs. 7500

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Rate of interest (r)}} \cr & {\text{ = 8}}\frac{3}{4}\% = \frac{7}{{80}} = \frac{{87 \to {\text{ Installment}}}}{{80 \to {\text{Principal}}}} \cr} $$

⇒	I	80_×87	→	87_×87	......(i)
⇒	II	6400	→	7569	......(ii)

Since, installment is equal hence multiply equation (i) by 87
⇒ Total principal = 6960 + 6400 = 13360
⇒ 13360 units = Rs. 13360
⇒ 1 units = Rs. 1
⇒ 7569 units = Rs. 7569
∴ Each installment = Rs. 7569

55

An amount of Rs. 10000 becomes Rs. 14641 in 2 years if the interest is compounded half yearly. What is the rate of compound interest p.c.p.a. ?

A) 10% B) 12% C) 16% D) 20% E) None of these

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Let the rate be R% p}}{\text{.a}}{\text{. }} \cr & {\text{Then,}} \cr & {\text{10000}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^4} = 14641 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{200}}} \right)^4} = \frac{{14641}}{{10000}} = {\left( {\frac{{11}}{{10}}} \right)^4} \cr & \Rightarrow 1 + \frac{{\text{R}}}{{200}} = \frac{{11}}{{10}} \cr & \Rightarrow \frac{{\text{R}}}{{200}} = \frac{1}{{10}} \cr & \Rightarrow {\text{R}} = {\text{20% }} \cr} $$

56

In how many years will a sum of Rs. 800 at 10% per annum compounded semi annually become Rs. 926.10?

A) $$1\frac{1}{3}$$ years B) $$1\frac{1}{2}$$ years C) $$2\frac{1}{3}$$ years D) $$2\frac{1}{2}$$ years

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Let the time be }}n{\text{ year}} \cr & {\text{Then,}} \cr & {\text{800}} \times {\left( {1 + \frac{5}{{100}}} \right)^{2n}} = 926.10 \cr & \Leftrightarrow {\left( {1 + \frac{5}{{100}}} \right)^{2n}} = \frac{{9261}}{{8000}} \cr & \Leftrightarrow {\left( {\frac{{21}}{{20}}} \right)^{2n}} = {\left( {\frac{{21}}{{20}}} \right)^3} \cr & \Leftrightarrow 2n = 3 \cr & \Leftrightarrow n = \frac{3}{2} \cr & \therefore n = 1\frac{1}{2}{\text{years}} \cr} $$

57

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

Discuss

Answer & Solution

Answer: Option B

Solution:

$$5\% = \frac{1}{{20}} = \frac{{21 \to {\text{ Installment}}}}{{20 \to {\text{ Principal}}}}$$

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

58

An amount of Rs 6000 lent at 5% per annum compounded interest for 2 years will become = ?

A) Rs. 600 B) Rs. 6600 C) Rs. 6610 D) Rs. 6615

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Amount = 6000}}{\left( {1 + \frac{5}{{100}}} \right)^2} \cr & \Rightarrow {\text{Amount = 6000}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \cr & \Rightarrow {\text{Amount = Rs. 6615}} \cr} $$

59

The simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is ?

A) Rs. 1550 B) Rs. 1650 C) Rs. 1750 D) Rs. 2000

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{C}}{\text{.I}}{\text{.}} \cr & {\text{ = Rs}}{\text{.}}\left[ {4000 \times {{\left( {1 + \frac{{10}}{{100}}} \right)}^2} - 4000} \right] \cr & = {\text{ Rs}}{\text{.}}\left( {4000 \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} - 4000} \right) \cr & = {\text{Rs}}{\text{. 840}} \cr & \therefore {\text{Sum = Rs}}{\text{.}}\left( {\frac{{420 \times 100}}{{3 \times 8}}} \right) \cr & = {\text{Rs}}{\text{. }}1750 \cr} $$

60

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12000 after 3 years at the same rate ?

A) Rs. 2160 B) Rs. 3120 C) Rs. 3972 D) Rs. 6240 E) None of these

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Let P}} = {\text{Rs}}.100 \cr & {\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{. = Rs}}.60{\text{ and}} \cr & {\text{T = 6 years}} \cr & {\text{R = }}\frac{{100 \times 60}}{{100 \times 6}}{\text{ = 10% p}}{\text{.a}}{\text{.}} \cr & {\text{Now}} \cr & {\text{P = Rs 12000,}} \cr & {\text{T = 3 years and}} \cr & {\text{R = 10% p}}{\text{.a}}{\text{.}} \cr & \therefore {\text{C}}{\text{.I}}{\text{. = Rs}}{\text{.}}\left[ {12000 \times \left\{ {{{\left( {1 + \frac{{10}}{{100}}} \right)}^3} - 1} \right\}} \right] \cr & = {\text{Rs}}{\text{.}}\left( {12000 \times \frac{{331}}{{1000}}} \right) \cr & = {\text{Rs}}{\text{. }}3972 \cr} $$