Simple & Compound Interest(Quantitative Aptitude > Interest ) Questions and answers for exam Preparation

Quantitative Aptitude > Interest

SIMPLE & COMPOUND INTEREST MCQs

Compound Interest, Simple Interest, Interest (combined)

Total Questions : 1171 | Page 81 of 118 pages

Question 801. A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a different rate becomes three times in 12 years. The lower rate of interest per annum is

20%
15%
16$2/3$%
15$3/4$%

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Answer: Option C. -> 16$2/3$%
Answer: (c)The sum gets doubled in 5 years and tripled in 12 years.Clearly rate of interest for 12 years will be lower.Let Principal be x.then, Rate = ${SI × 100}/\text"Principal × Time"$= ${2x × 100}/{x × 12} = 50/3 = 16{2}/3%$Using Rule 3,$R_1 = {(2 - 1)}/5 × 100%$ = 20%$R_2 = {(3 - 1)}/12 × 100% = 16{2}/3%$Lower rate of interest =16$2/3%$

Question 802. A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?

5 years
6 years
8 years
4 years

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Answer: Option D. -> 4 years
Answer: (d)Let the principal be Rs.1.A = P$(1 + R/100)^T$8 = 1$(1 + R/100)^3$$2^3 = 1(1 + R/100)^3$2 = 1$(1 + R/100)^1$$2^4 = (1 + R/100)^4$Time = 4 yearsUsing Rule 11,Here, $x = 8, n_1 = 3, y = 16, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(8)^{1/3} = (16)^{1/n_2}$$(2^3)^{1/3} = (2^4)^{1/n_2}$$2^1 = 2^{4/n_2}$1= $4/n_2$$n_2$ = 4 years

Question 803. If a sum of money compounded annually becomes 1.44 times of itself in 2 years, then the rate of interest per annum is

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Answer: Option A. -> 20%
Answer: (a)A = P$(1 + R/100)^T$1.44P = P$(1 + R/100)^2$$(1.2)^2 = (1 + R/100)^2$$1 + R/100$ = 1.2R = 0.2 × 100 = 20%Using Rule 8,Here, n = 1.44, t = 2 yearsR% = $(n^{1/6} - 1) × 100%$= $[(1.44)^{1/2} - 1] × 100%$= [(1.2) - 1] × 100%= 0.2 × 100% ⇒ R% = 20%

Question 804. A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?

16 years
12 years
8 years
13 years

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Answer: Option C. -> 8 years
Answer: (c)A = P$(1 + R/100)^T$Let P be Rs.1, then A = Rs.22 = 1$(1 + R/100)^4$$2^2 = (1 + R/100)^8$Time = 8 yearsUsing Rule 11,Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(2)^{1/4} = (4)^{1/n_2}$$(2)^{1/4} = (2^2)^{1/n_2}$$2^{1/4} = 2^{1/n_2}$$1/4 = 2/n_2$$n_2$ = 8 years

Question 805. A sum of money placed at compound interest doubles itself in 5 years. In how many years, it would amount to eight times of itself at the same rate of interest ?

20 years
10 years
7 years
15 years

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Answer: Option D. -> 15 years
Answer: (d)Let the Principal be P and rate of interest be r%.2 P = P$(1 + r/100)^2$2 = $(1 + r/100)^5$ ...(i)On cubing both sides,8 = $(1 + r/100)^15$Time = 15 yearsUsing Rule 5,Here, m = 2, t = 5 yearsIt becomes 8 times = $2^3$ timesin t × n = 5 × 3 = 15 years

Question 806. A sum of money invested at compound interest doubles itself in 6 years. At the same rate of interest it will amount to eight times of itself in :

10 years
15 years
18 years
12 years

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Answer: Option C. -> 18 years
Answer: (c)Let the sum be x. Then,$2x = x(1 + r/100)^6$2 = $(1 + r/100)^6$Cubing both sides,8 =$((1 + r/100)^6)^3$8 = $(1 + r/100)^18$$8x = x(1 + r/100)^18$The sum will be 8 times in 18 years.i.e., Time = 18 yearsUsing Rule 5,Here, m = 2, t = 6 yearsIt will becomes 8 times of itself= $2^3$ times of it selfin t × n years = 6 × 3 = 18 years

Question 807. If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is :

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Answer: Option A. -> 50%
Answer: (a)Suppose P = Rs.100and amount A = Rs.225A = P$(1 + r/100)^t$or 225 = $100(1 + r/100)^2$or $225/100 = [1 + r/100]^2$or 1 + $r/100 = 15/10$or ${100 + r}/100 = 15/10$or 100 + r = 150or r = 50% Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$

Question 808. A sum of money becomes eight times of itself in 3 years at compound interest. The rate of interest per annum is

10%
100%
20%
80%

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Answer: Option B. -> 100%
Answer: (b)Let the principal be x and the rate of compound interest be r% per annum. Then,$8x = x(1 + r/100)^3$8 = $(1 + r/100)^3$$2^3 = (1 + r/100)^3$2 = $1 + r/100 ⇒ r/100 = 1 ⇒ r = 100%$Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$

Question 809. A sum borrowed under compound interest doubles itself in 10 years. When will it become fourfold of itself at the same rate of interest ?

40 years
15 years
24 years
20 years

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Answer: Option D. -> 20 years
Answer: (d)Let the sum be x which becomes 2x in 10 years.Hence, 4x in 20 yearsMethod 2 : Unitary Method can also be used.Using Rule 5,Here, m = 2, t = 10Time taken to become 4 times = $2^2$ times= t × n = 10 × 2 = 20 years

Question 810. A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself ?

7.5 years
4 years
6.4 years
6 years

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Answer: Option D. -> 6 years
Answer: (d)A = P$(1 + R/100)^T$Let P. Rs., A = Rs.22 = 1$(1 + R/100)^3$On squaring both sides.4 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 2, n_1 = 3, y = 4, n_2$ = ?$x^{1/n_1} = y^{1/n_2}$$2^{1/3} = 4^{1/n_2}$$2^{1/3} = (2^2)^{1/n_2}$$2^{1/3} = 2^{2/n_2}$$1/3 = 2/n_2$$n_2$ = 6 Years