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 72 of 118 pages

Question 711. The income of a company increases 20% per year. If the income is Rs. 26,64,000 in the year 2012, then its income in the year 2010 was :

Rs.21,20,000
Rs.28,55,000
Rs.28,20,000
Rs.18,50,000

Discuss Question

Answer: Option D. -> Rs.18,50,000
Answer: (d)Using Rule 1,Let the income of company in 2010 be Rs.PAccording to the question,A = P$(1 + R/100)^T$2664000 = P$(1 + 20/100)^2$2664000 = P$(1 + 1/5)^2$2664000 = P × $(6/5)^2$P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000

Question 712. If the compound interest on a sum of money for 3 years at the rate of 5% per annum is Rs.252.20, the simple interest on the same sum at the same rate and for the same time is

Rs.250
Rs.220
Rs.245
Rs.240

Discuss Question

Answer: Option D. -> Rs.240
Answer: (d)Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$

Question 713. The compound interest on a certain sum of money at a certain rate per annum for two years is Rs.2,050, and the simple interest on the same amount of money at the same rate for 3 years is Rs.3, 000. Then the sum of money is

Rs.25, 000
Rs.20,000
Rs.21,000
Rs.18,000

Discuss Question

Answer: Option B. -> Rs.20,000
Answer: (b)Using Rule 6,The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will beC.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$ For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$

Question 714. The compound interest on a certain sum of money at a certain rate for 2 years is Rs.40.80 and the simple interest on the same sum is Rs.40 at the same rate and for the same time. The rate of interest is

5% per annum
2% per annum
4% per annum
3% per annum

Discuss Question

Answer: Option C. -> 4% per annum
Answer: (c)Let the principal be P and rate of interest be r per cent per annum. Then,C. I = P$[(1 + r/100)^2 - 1]$40.80 = P$[(1 + r/100)^2 - 1]$...(i)S.I. = ${P.r.t}/100 ⇒ 40 = {Pr × 2}/100$ ...(ii)${40.80}/40 = P[(1 + r/100)^2 - 1]/{{2Pr}/100}$ ⇒ 1.02= $100/{2r}[1 + r^2/10000 + {2r}/100 - 1]$1.02 = $r/200$ +1$r/200$ = 1.02 - 1r = 0.02 × 200 = 4% per annum.Using Rule 10,Here, C.I. = Rs.40.80, S.I. = Rs.40, R = ?C.I.= S.I.$(1 + R/200)$40.80 = 40$(1 + R/200)$$4080/4000 = 1 + R/200$$408/400 = {200 + R}/200$408 = 400 + 2R2R = 8 ⇒ R = 4%

Question 715. On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. If it was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned ?

Rs.35
Rs.3.50
Rs.14
Rs.7

Discuss Question

Answer: Option D. -> Rs.7
Answer: (d)Principal = $\text"S.I. × 100"/\text"Time × Rate"$= ${350 × 100}/{2 × 4}$ = Rs.4375Difference = ${PR^2}/10000$= ${4375 × 4 × 4}/10000$ = Rs.7

Question 716. The compound interest on a certain sum of money invested for 2 years at 5% per annum is Rs.328. The simple interest on the sum, at the same rate and for the same period will be

Rs.287
Rs.320
Rs.300
Rs.308

Discuss Question

Answer: Option B. -> Rs.320
Answer: (b)Let the principal be P.C.I. = P$[(1 + R/100)^T - 1]$328 = P$[(1 + 5/100)^2 - 1]$328 = P$[441/400 -1]$328 = P$[{441 - 400}/400]$P = ${328 × 400}/41$ = Rs.3200S.I. = ${PRT}/100 = {3200 × 5 × 2}/100$ = Rs.320Using Rule 10,Here, C.I. = Rs.328, R = 5%, S.I. = ?C.I.= S.I.$(1 + R/200)$328 = S.I.$(1 + 5/200)$328 = S.I.$(1 + 1/40)$S.I. = ${328 × 40}/41$S.I. = 8 x 40 = Rs.320

Question 717. If the compound interest on a certain sum for 2 years at 3% per annum is Rs.101.50, then the simple interest on the same sum at the same rate and for the same time will be

Rs.98.25
Rs.90.00
Rs.100.00
Rs.95.50

Discuss Question

Answer: Option C. -> Rs.100.00
Answer: (c)Let the sum be P.101.50 = P$[(1 + 3/100)^2 - 1]$[Since, C.I. = P$[(1 + r/100)^n - 1]$]101.50 = P$[(103/100)^2 - 1]$=P$({10609 - 10000}/10000)$P = Rs.${101.50 × 10000}/609 = Rs.1015000/609$S.I. = ${1015000 × 2 × 3}/{609 × 100}$ = Rs.100 Using Rule 10,The simple interest for a certain sum for 2 years at an annual rate interest R% is S.I., thenC.I. = S.I.$(1 + R/200)$

Question 718. There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs.8000 after 2 years at the same rate of interest.

Rs.2125
Rs.2500
Rs.2250
Rs.2000

Discuss Question

Answer: Option A. -> Rs.2125
Answer: (a)Using Rule 1,Let S.I. = Rs.100, & Principal = Rs.100Rate = $\text"S.I. × 100"/\text"Principal × Time"$= ${100 × 100}/{100 × 8} = 25/2$%C.I. = P$[(1 + r/100)^T - 1]$= 8000$[(1 + 25/200)^2 - 1]$= 8000 $(81/64 - 1)$= ${8000 × 17}/64$ = Rs.2125

Question 719. 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 :

5$5/17$%
8%
12$1/2$
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 720. 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 :

Rs.50.05
Rs.50.50
Rs.51.50
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