Hello Aspirants. Welcome to Online Maths in AffairsCloud.com. Here we are creating question sample in Compound Interest, which is common for all the IBPS, SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in exams !!
- On a certain sum of money, after 2 years the simple interest and compound interest obtained are Rs 800 and Rs 960 respectively. What is the sum of money invested?
A) Rs 1420
B) Rs 1325
C) Rs 1000
D) Rs 1405
E) Rs 1375C) Rs 1000
Explanation:
Diff = 960-800 = 160
r = 2*Diff*100/SI
So r = 2*160*100/800 = 40%
Now 160 = Pr2/1002 - Rs 6000 becomes Rs 7200 in 3 years at a certain rate of compound interest. What will be the amount received after 9 years?
A) Rs 11,498
B) Rs 10,352
C) Rs 9,368
D) Rs 10,368
E) None of theseD) Rs 10,368
Explanation:
6000[1 + r/100]3 = 7200
So [1 + r/100]3 = 6/5
So 6000[1 + r/100]9 = 6000*(6/5)*(6/5)*(6/5) - A man borrows Rs 4000 at 8% compound interest for 3 years. At the end of each year he paid Rs 500. How much amount should he pay at the end of 3rd year to clear the debt?
A) Rs 4254.5
B) Rs 3465.2
C) Rs 3485.2
D) Rs 4345.4
E) Rs 3915.6E) Rs 3915.6
Explanation:
Amount after 1 yr = 4000[1 + 8/100] = 4320
Paid 500, so P = 4320 – 500 = 3820
Amount after 2nd yr = 3820[1 + 8/100] = 4125.6
So P= 4125.6-500 = 3625.6
Amount after 3rd yr = 3625.6[1 + 8/100] = 3915.6 - A sum of money is lent for 2 years at 20% p.a. compound interest. It yields Rs 482 more when compounded semi-annually than compounded annually. What is the sum lent?
A) Rs 25,600
B) Rs 20,000
C) Rs 26,040
D) Rs 40,500
E) None of theseB) Rs 20,000
Explanation:
P[1 + (r/2)/100]4 – P[1 + r/100]2 = 482
P[1 + 10/100]4 – P[1 + 20/100]2 = 482
Solve, P = 20,000 - The compound interest obtained after 1st and 2nd year is Rs 160 and Rs 172.8 respectively on a certain sum of money invested for 2 years. What is the rate of interest?
A) 10%
B) 8%
C) 8.5%
D) 9%
E) 9.2%B) 8%
Explanation:
Difference in interest for both yrs = 172.8 – 160 = 12.8
So (r/100)*160 = 12.8 - A sum of money becomes Rs 35,280 after 2 years and Rs 37,044 after 3 years when lent on compound interest. Find the principal amount.
A) Rs 32,000
B) Rs 28,000
C) Rs 31,500
D) Rs 32,500
E) None of theseA) Rs 32,000
Explanation:
P[1 + r/100]3 = 37,044, and P[1 + r/100]2 = 35,280
Divide both equations, [1 + r/100] = 37044/35280 = 21/20
So P[21/20]2 = 35280 - The difference between compound interest earned after 3 years at 5% p.a. and simple interest earned after 4 years at 4% p.a. is Rs 76. Find the principal amount.
A) Rs 32,000
B) Rs 28,000
C) Rs 31,500
D) Rs 32,500
E) None of theseA) Rs 32,000
Explanation:
[P[1 + 5/100]3 – P] – P*4*4/100 = 76
P [9261/8000 – 1 – 16/100] = 76 - A sum of money is lent at simple interest and compound interest. The ratio between the difference of compound interest and simple interest of 3 years and 2 years is 35 : 11. What is the rate of interest per annum?
A) 20 3/4%
B) 17 2/5%
C) 18 2/11%
D) 22 1/5%
E) 24 5/6%C) 18 2/11%
Explanation:
Difference in 3 yrs = Pr2(300+r)/1003
Difference in 2 yrs = Pr2/1002
So Pr2(300+r)/1003 / Pr2/1002 = 35/11 (300+r)/100 = 35/11 - A sum of money borrowed at 5% compound interest is to paid in two annual installments of Rs 882 each. What is the sum borrowed?
A) Rs 1650
B) Rs 2340
C) Rs 2630
D) Rs 1640
E) Rs 2640D) Rs 1640
Explanation:
P = 882/[1 + 5/100] + 882/[1 + 5/100]2 - Rs 3903 is to be divided in a way that A’s share at the end of 7 years is equal to the B’s share at the end of 9 years. If the rate of interest is 4% compounded annually, find A’s share.
A) Rs 2475
B) Rs 1875
C) Rs 2175
D) Rs 1935
E) Rs 2028B) Rs 1875
Explanation:
A’s share = (1 + 4/100)7
B’s share = (1 + 4/100)9
Divide both, B/A = (1 + 4/100)2 = 676/625
So A’s share = 625/(676+625) * 3903
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