Quantitative Aptitude > Interest
COMPOUND INTEREST MCQs
Total Questions : 262
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Answer: Option C. -> Rs.1,575.20
Answer: (c)Using Rule 3,If there are distinct 'rates of interest' for distinct time periods i.e.,Rate for 1st year → $r_1$%Rate for 2nd year → $r_2$%Rate for 3rd year → $r_3$% and so onThen A = P$(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)$...C.I. = A - P
Answer: (c)Using Rule 3,If there are distinct 'rates of interest' for distinct time periods i.e.,Rate for 1st year → $r_1$%Rate for 2nd year → $r_2$%Rate for 3rd year → $r_3$% and so onThen A = P$(1 + r_1/100)(1 + r_2/100)(1 + r_3/100)$...C.I. = A - P
Answer: Option B. -> 1$1/2$ years
Answer: (b)Using Rule 1 and 2,Time = t half yearand R = 5% per half yearA = P$(1 + R/100)^T$$92610/80000 = (1 + 5/100)^T$$9261/8000 = (21/20)^T$T = 3 half years or 1$1/2$ years$(21/20)^3 = (21/20)^T$
Answer: (b)Using Rule 1 and 2,Time = t half yearand R = 5% per half yearA = P$(1 + R/100)^T$$92610/80000 = (1 + 5/100)^T$$9261/8000 = (21/20)^T$T = 3 half years or 1$1/2$ years$(21/20)^3 = (21/20)^T$
Answer: Option B. -> 1$1/2$ years
Answer: (b)Using Rule 1 and 2,Let the required time be t years. Interest is compounded half yearly.Time = 2t half years and rate= $20/2$ = 10%1000$(1 + 10/100)^{2t}$ = 1331$(11/10)^{2t} = 1331/1000$$(11/10)^{2t} = (11/10)^3$ ⇒ 2t = 3t = $3/2$ years or 1$1/2$ years
Answer: (b)Using Rule 1 and 2,Let the required time be t years. Interest is compounded half yearly.Time = 2t half years and rate= $20/2$ = 10%1000$(1 + 10/100)^{2t}$ = 1331$(11/10)^{2t} = 1331/1000$$(11/10)^{2t} = (11/10)^3$ ⇒ 2t = 3t = $3/2$ years or 1$1/2$ years
Answer: Option B. -> 5%
Answer: (b)Using Rule 1,A = P$(1 + R/100)^T$Let rate be 'r'${1102.50}/1000 = (1 + r/100)^2$$11025/10000 = (1 + r/100)^2$$(105/100)^2 = (1 + r/100)^2$1 + $r/100 = 105/100$$r/100 = 5/100$ = 5%
Answer: (b)Using Rule 1,A = P$(1 + R/100)^T$Let rate be 'r'${1102.50}/1000 = (1 + r/100)^2$$11025/10000 = (1 + r/100)^2$$(105/100)^2 = (1 + r/100)^2$1 + $r/100 = 105/100$$r/100 = 5/100$ = 5%
Answer: Option B. -> 1$1/2$ years
Answer: (b)Using Rule 1 and 2,Rate = 10% per annum = 5% half yearlyA = P$(1 + R/100)^T$926.10 = 800$(1 + 5/100)^T$$9261/8000 = (21/20)^T$$(21/20)^3 = (21/20)^T$Time = 3 half years = 1$1/2$ years
Answer: (b)Using Rule 1 and 2,Rate = 10% per annum = 5% half yearlyA = P$(1 + R/100)^T$926.10 = 800$(1 + 5/100)^T$$9261/8000 = (21/20)^T$$(21/20)^3 = (21/20)^T$Time = 3 half years = 1$1/2$ years
Answer: Option C. -> 2 years
Answer: (c)Using Rule 1,A = P$(1 + R/100)^T$30000 + 4347 = $30000(1 + 7/100)^T$$34347/30000 = (107/100)^T$$11449/10000 = (107/100)^2 = (107/100)^T$Time = 2 years
Answer: (c)Using Rule 1,A = P$(1 + R/100)^T$30000 + 4347 = $30000(1 + 7/100)^T$$34347/30000 = (107/100)^T$$11449/10000 = (107/100)^2 = (107/100)^T$Time = 2 years
Answer: Option C. -> Rs.3109
Answer: (c)Using Rule 1,Amount = P$(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.
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.
Answer: Option B. -> Rs.5,000
Answer: (b)Using Rule 1,5832 = P$(1 + 8/100)^2$5832 = P$(1 + 2/25)^2$5832 = P $× 27/25 × 27/25$P = ${5832 × 25 × 25}/{27 × 27}$ = Rs.5000
Answer: (b)Using Rule 1,5832 = P$(1 + 8/100)^2$5832 = P$(1 + 2/25)^2$5832 = P $× 27/25 × 27/25$P = ${5832 × 25 × 25}/{27 × 27}$ = Rs.5000
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$(1 + r/100)^3(1 + r/100 × 5/7)$
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$(1 + r/100)^3(1 + r/100 × 5/7)$
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
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
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