1. A sum of money at 8% per annum compound interest becomes 37,791.36 in 3 years. The sum of money is
(a) Rs. 28000
(b) Rs. 30000
(c) Rs. 32000
(d) Rs. 33000
2. A sum of Rs. 12000 deposited at compound interest becomes double after 5 years. After 20 years, it will become
(a) Rs. 48000
(b) Rs. 96000
(c) Rs. 190000
(d) Rs. 192000
3.The compound interest on a certain sum of money for 2 years at 10% per annum is Rs. 420. The simple interest on the same sum at the same rate and for the same time will be
(a) Rs. 380
(b) Rs. 400
(c) Rs. 350
(d) Rs. 375
4.A principal of Rs. 10,000 after 2 year compound annually, the rate of interest being 10% per annum during the first year and 12% per annum during the second year ( in rupees) will amount to:
(a) Rs. 12,000
(b) Rs. 12,350
(c) Rs. 12,500
(d) Rs. 12,320
5.A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself.
(a) 4
(b) 6
(c) 6.4
(d) 7.5
6. If a sum of money placed at compound interests, compounded annually, doubles itself in 5 years, then the same amount of money will be 8 times of itself in
(a) 25 years
(b) 20 years
(c) 15 years
(d) 10 years
7.The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 48. Then the sum is
(a) Rs. 1000
(b) Rs. 1200
(c) Rs. 1500
(d) Rs. 2000
8.The difference between the compound interest and simple interest for the amount Rs. 5000 in 2 years is Rs. 32. The rate of interest is
(a) 5%
(b) 8%
(c) 10%
(d) 12%
9. The difference between the simple and compound interest on a certain sum of money for 2 years at 4% perannum, is Rs. 1. Find the sum
(a) Rs. 630
(b) Rs. 620
(c) Rs. 625
(d) Rs. 635
10.The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 3 years is Rs. 500. The sum is
(a) Rs. 45000
(b) Rs. 50000
(c) Rs. 55000
(d) None of these
Answers and Solution:
1. (b)
A = p(1+r/100)^n
37791.36 = p(108/100)^3
p = Rs. 30000
2. (d)
sum of Rs.12000 becomes double after 5 years
so 2*12000 = 12000(1+r/100)^5
so 2 = (1+r/100)^5 ........(i)
then after 20 years A = 12000(1+r/100)^20
so A = 12000((1+r/100)^5)^4 = 12000*2^4 = 192000
3. (b)
4. (d)
Amount = P(110/100)*(112/100) = 10000*11/10*112/100 = 12320
5. (b)
In 3 years 2 times
Let in years 4 times = 2^2 times
so n = 3*2 = 6 years
6. (c)
In 5 years 2 times
Let in n years 8 times = 2^3 times
so n = 5*3 = 15 years
7. (b)
Sum of money = difference(100/r)^2
= 48(100/20)^2 = 1200
8. (b)
Difference = pr^2/(100)^2
=> 32 = 5000(r)^2/(100)^2
r = 8%
9. (c)
10. (d)
D = pr^2(300+r)/(100)^3
500 = P(10)^2(300+10)/(100)^3
p = 16129
1. (b)
A = p(1+r/100)^n
37791.36 = p(108/100)^3
p = Rs. 30000
2. (d)
sum of Rs.12000 becomes double after 5 years
so 2*12000 = 12000(1+r/100)^5
so 2 = (1+r/100)^5 ........(i)
then after 20 years A = 12000(1+r/100)^20
so A = 12000((1+r/100)^5)^4 = 12000*2^4 = 192000
3. (b)
4. (d)
Amount = P(110/100)*(112/100) = 10000*11/10*112/100 = 12320
5. (b)
In 3 years 2 times
Let in years 4 times = 2^2 times
so n = 3*2 = 6 years
6. (c)
In 5 years 2 times
Let in n years 8 times = 2^3 times
so n = 5*3 = 15 years
7. (b)
Sum of money = difference(100/r)^2
= 48(100/20)^2 = 1200
8. (b)
Difference = pr^2/(100)^2
=> 32 = 5000(r)^2/(100)^2
r = 8%
9. (c)
10. (d)
D = pr^2(300+r)/(100)^3
500 = P(10)^2(300+10)/(100)^3
p = 16129
Click Here For : Concept and Short Trick on C.I and S.I
No comments:
Post a Comment