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Quantitative Aptitude > Interest

COMPOUND INTEREST MCQs

Total Questions : 262 | Page 7 of 27 pages
Question 61. The simple interest on a sum of money for 3 years is Rs. 240 and the compound interest on the same sum, at the same rate for 2 years is Rs. 170. The rate of interest is :
  1.    5$5/17$%
  2.    8%
  3.    12$1/2$
  4.    29$1/6$%
 Discuss Question
Answer: Option C. -> 12$1/2$
Answer: (c)Since, S.I. for 3 years = Rs.240S.I. for 2 years = $240/3$× 2 = Rs.160${PR × 2}/100$ = 160PR = 160 × 50 = 8000...(i)Again, C.I. - S.I. = 170 - 160 = Rs.10${PR^2}/10000$ = 10${8000 × R}/10000$ = 10R = $100/8 = 25/2 = 12{1}/2$%
Question 62. If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, the compound interest on the same at the same rate and for the same time is :
  1.    Rs.50.05
  2.    Rs.50.50
  3.    Rs.51.50
  4.    Rs.51.25
 Discuss Question
Answer: Option D. -> Rs.51.25
Answer: (d)Principal = $\text"S.I. × 100"/ \text"Time × Rate"$= ${50 × 100}/{2 × 5}$ = Rs.500C.I. = P$[(1 + R/100)^T - 1]$= 500$[(1 + 5/100)^2 - 1]$= 500$[(1 + 1/20)^2 - 1]$= 500$[(21/20)^2 - 1]$= 500$(441/400 - 1)$= ${500 × 41}/400$ = Rs.51.25
Question 63. A sum of money invested at compound interest amounts in 3 years to Rs.2,400 and in 4 years to Rs.2,520. The interest rate per annum is :
  1.    12%
  2.    5%
  3.    10%
  4.    6%
 Discuss Question
Answer: Option B. -> 5%
Answer: (b)S.I. on Rs.2400 for 1 year= Rs.(2, 520 - 2, 400) = Rs.120Rate = $\text"S.I. × 100"/ \text"Principal × Time"$%= ${120 × 100}/{2400 × 1}$ = 5%Using Rule 7(i),Here, b - a = 4 - 3 = 1B = Rs.2520, A = Rs.2400R% = $(B/A - 1)$ × 100%= $[2520/2400 - 1] × 100%$= $[{2520 - 2400}/2400] × 100%$= $120/2400 × 100%$ = 5%
Question 64. An amount of money at compound interest grows up to Rs.3,840 in 4 years and up to Rs.3,936 in 5 years. Find the rate of interest.
  1.    2.05%
  2.    2.5%
  3.    3.5%
  4.    2%
 Discuss Question
Answer: Option B. -> 2.5%
Answer: (b)A = P$(1 + R/100)^T$3840 = P$(1 + R/100)^4$ ....(i)3936 = P$(1 + R/100)^5$ ...(ii)Dividing equation (ii) by equation (i),$3936/3840 = 1 + R/100$$R/100 = 3936/3840$ - 1= ${3936 - 3840}/3840 = 96/3840$R = $96/3840 × 100$ = 2.5%Using Rule 7(i),Here, b - a = 5 - 4 = 1B = Rs.3,936, A = Rs.3,840R%= $(B/A - 1)$ × 100%= $(3936/3840 - 1)$ × 100%= $({3936 - 3840}/3840) × 100%$= $96/3840 × 100% = 10/4%$ = 2.5%
Question 65. A sum of money is invested at 20% compound interest (compounded annually). It would fetch Rs. 723 more in 2 years if interest is compounded half yearly. The sum is
  1.    Rs.7,500
  2.    Rs.15,000
  3.    Rs.20,000
  4.    Rs.30,000
 Discuss Question
Answer: Option D. -> Rs.30,000
Answer: (d)Let the principal be Rs. x.When the interest is compounded annually,C.I. = P$[(1 + R/100)^T - 1]$= P$[(1 + 20/100)^2 - 1]$= P$[(6/5)^2 - 1]$= P$(36/25 - 1)$ = Rs.${11P}/25$When the interest is compounded half–yearly,C.I. = P$[(1 + 10/100)^4 - 1]$= P$[(11/10)^4 - 1]$= P$(14641/10000 - 1)$= Rs.${4641P}/10000$${4641P}/10000 - {11P}/25$ = 723${4641P - 4400P}/10000$ = 723${241P}/10000$ = 723P = ${723 × 10000}/241$ = Rs.30000
Question 66. A sum becomes Rs.4500 after two years and Rs.6750 after four years at compound interest. The sum is
  1.    Rs.3050
  2.    Rs.4000
  3.    Rs.3000
  4.    Rs.2500
 Discuss Question
Answer: Option C. -> Rs.3000
Answer: (c)P$(1 + r/100)^2$ = 4500 ...(i)P$(1 + r/100)^4$ = 6750 .....(ii)On dividing equation (ii) by equation (i), we get$(1 + r/100)^2 = 6750/4500$From equation (i),P × $6750/4500$ = 4500P = ${4500 × 4500}/6750$ = Rs.3,000Using Rule 7(ii),Here, b - a = 4 - 2 = 2B = Rs.6750, A = Rs.4500R% = $[(B/A)^{1/2} - 1]$ × 100%= $[(6750/4500)^{1/2} - 1] × 100%$= $[(3/2)^{1/2} - 1] ×$ 100%$(3/2)^{1/2} = 1 + R/100$$3/2 = (1 + R/100)^2$A = P$(1 + R/100)^2$4500 = P × $3/2$ ⇒ P = Rs.3000
Question 67. On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned?
  1.    Rs.35
  2.    Rs.3.50
  3.    Rs.14
  4.    Rs.7
 Discuss Question
Answer: Option D. -> Rs.7
Answer: (d)Principal = $\text"S.I. × 100"/ \text"Time × Rate"$= ${350 × 100}/{2 × 4}$ = Rs.4375C.I. = P$[(1 + R/100)^T - 1]$= 4375$[(1 + 4/100)^2 - 1]$= 4375$[(1 + 1/25)^T - 1]$= 4375$[(26/25)^2 - 1]$= 4375$(676/625 - 1)$= ${4375 × 51}/625$ = Rs.357Required difference= Rs.(357 - 350) = Rs.7
Question 68. The difference between the compound interest and simple interest for the amount 5,000 in 2 years is 32. The rate of interest is
  1.    12%
  2.    5%
  3.    10%
  4.    8%
 Discuss Question
Answer: Option D. -> 8%
Answer: (d)Using Rule 6,Difference of 2 years= ${p × r^2}/10000$32 = ${5000 × r^2}/10000$$r^2 = {32 × 10000}/5000$ = 64r = $√{64}$ = 8%
Question 69. On what sum of money will the difference between S.I and C.I for 2 years at 5% per annum be equal to 25 ?
  1.    9000
  2.    10,000
  3.    9,500
  4.    10,500
 Discuss Question
Answer: Option B. -> 10,000
Answer: (b)Using Rule 6,Difference = ${PR^2}/10000$25 = ${P × 5 × 5}/10000$P = Rs.10000
Question 70. On what sum does the difference between the compound interest and the simple interest for 3 years at 10% is 31 ?
  1.    1000
  2.    1500
  3.    1100
  4.    1200
 Discuss Question
Answer: Option A. -> 1000
Answer: (a)Let the sum be xr = 10%, n = 3 yearsS.I. = ${x × r × n}/100$S.I.= ${x × 10 × 3}/100 = 3/10x$C.I.= $[(1 + r/100)^n - 1]x$= $[(1 + 10/100)^3 - 1]x$= $[(11/10)^3 - 1]x$$(1331/1000 - 1)x = 331/1000x$$331/1000x - 3/10x$ = 31or $({331 - 300})/1000x = 31$or $31/1000x$ = 31or x = 1000Sum = Rs.1000Using Rule 6,Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$31 = P × $(10/100)^2(3 + 10/100)$31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000

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