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 65 of 118 pages

Question 641. A sum becomes Rs.1,352 in 2 years at 4% per annum compound interest. The sum is

Rs.1,250
Rs.1,225
Rs.1,245
Rs.1,270

Discuss Question

Answer: Option A. -> Rs.1,250
Answer: (a)Using Rule 1,Let the sum be Rs.x.1352 = $x(1 + 4/100)^2$1352 = $x(1 + 1/25)^2$$1352 = x(26/25)^2$$x = {1352 × 25 × 25}/{26 × 26}$ = Rs.1250

Question 642. The compound interest on Rs.8,000 at 15% per annum for 2 years 4 months, compounded annually is:

Rs.3100
Rs.2980
Rs.3109
Rs.3091

Discuss Question

Answer: Option C. -> Rs.3109
Answer: (c)Using Rule 1,Amount = P$(1 + R/100)^t$= 8000$(1 + 15/100)^{2{1}/3}$= 8000$(1 + 3/20)^2(1 + 3/{20 × 3})$= $8000 × 23/20 × 23/20 × 21/20$ = Rs.11109Compound Interest= Rs.(11109 - 8000) = Rs.3109.

Question 643. The compound interest on Rs.6,000 at 10% per annum for 1$1/2$ years, when the interest being compounded annually, is

Rs.900
Rs.910
Rs.930
Rs.870

Discuss Question

Answer: Option C. -> Rs.930
Answer: (c)Amount = 6000$(1 + 10/100) × (1 + {{1/2} × 10}/100)$= $6000 × 11/10 × 21/20$ = Rs.6930Using Rule 4,If the time is in fractional form i.e.,t = nF, thenA = P$(1 + r/100)^n(1 + {rF}/100)$e.g. t =3$5/7$ yrs, thenA = P$(1 + r/100)^3(1 + r/100 × 5/7)$

Question 644. At what rate per annum will Rs.32000 yield a compound interest of Rs.5044 in 9 months interest being compounded quarterly ?

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Answer: Option B. -> 20%
Answer: (b)Using Rule 1,Let the rate of CI be R per cent per annum.CI = P$[(1 + R/100)^T - 1]$5044 = 32000$[(1 + R/400)^3 - 1]$[Since, Interest is compounded quarterly]$5044/32000 = (1 + R/400)^3 - 1$$(1 + R/400)^3 - 1 = 1261/8000$$(1 + R/400)^3 = 1 + 1261/8000$$(1 + R/400)^3 = 9261/8000 = (21/20)^3$1 + $R/400 = 21/20$$R/400 = 21/20 - 1 = 1/20$R = $400/20$ = 20

Question 645. The compound interest on Rs.16,000 for 9 months at 20% per annum, interest being compounded quarterly, is

Rs.2,518
Rs.2,520
Rs.2,522
Rs.2,524

Discuss Question

Answer: Option C. -> Rs.2,522
Answer: (c)Using Rule 1,The interest is compounded quarterly.R = $20/4$ = 5%Time = 3 quartersC.I. = P$[(1 + R/100)^T - 1]$= 16000$[(1 + 5/100)^3 - 1]$= $16000[(21/20)^3 - 1]$= $16000({9261 - 8000}/8000)$= $16000 × 1261/8000$ = Rs.2522

Question 646. In what time will Rs.1000 becomes Rs.1331 at 10% per annum compounded annually ?

3$1/2$ years
3 years
2 years
2$1/2$ years

Discuss Question

Answer: Option B. -> 3 years
Answer: (b)Using Rule 1,Let the required time be n years.Then,1331 = 1000$(1 + 10/100)^n$[$P_1 = P(1+ r/100)^n$]$1331/1000 = ({10 + 1}/10)^n$$(11/10)^n = (11/10)^3$n = 3

Question 647. A sum of Rs.8000 will amount to Rs.8820 in 2 years if the interest is calculated every year. The rate of compound interest is

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Answer: Option A. -> 5%
Answer: (a)Using Rule 1,If the rate of C.I. be r% per annum, thenA = P$(1 + R/100)^T$8820 = 8000$(1 + r/100)^2$$8820/8000 = (1 + r/100)^2$$441/400 = (21/20)^2 = (1 + r/100)^2$$1 + r/100 = 21/20$$r/100 = 21/20 - 1 = 1/20$r = $1/20$ × 100r = 5% per annum

Question 648. A sum of money on compound interest amounts to Rs.10648 in 3 years and Rs.9680 in 2 years. The rate of interest per annum is :

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Answer: Option D. -> 10%
Answer: (d)Using Rule 1,Let the sum be P and rate of interest be R% per annum. Then,$P(1 + R/100)^2 = 9680$ ...(i)$P(1 + R/100)^3 = 10648$ ...(ii)On dividing equation (ii) by (i)$1 + R/100 = 10648/9680$$R/100 = 10648/9680$ -1= ${10648 - 9680}/9680$$R/100 = 968/9680 = 1/10$R = $1/10 × 100$ = 10%

Question 649. An amount of Rs.6,000 lent at 5% per annum compound interest for 2 years will become

Rs.6,615
Rs.600
Rs.6,610
Rs.6,600

Discuss Question

Answer: Option A. -> Rs.6,615
Answer: (a)Using Rule 1,A = P$(1 + R/100)^T$= 6000$(1 + 5/100)^2$= 6000 × $21/20 × 21/20$ = Rs.6615

Question 650. The sum of money that yields a compound interest of Rs.420 during the second year at 5% p.a is

Rs.21,000
Rs.4,000
Rs.8,000
Rs.42,000

Discuss Question

Answer: Option C. -> Rs.8,000
Answer: (c)Using Rule 1,CI = P$[(1 + R/100)^T –1] - {PR}/100$420 = P$[(1 + 5/100)^2 - 1] - {P × 5}/100$420 = P$[(21/20)^2 - 1] - {5P}/100$420 = ${41P}/400 - {5P}/100 = {21P}/400$P = ${420 × 400}/21$ = Rs.8000