1. Ravi Shankar takes money from the employees’
cooperative society at lower rate of interest and saves in a scheme, which
gives him a compound interest of 20%, compounded annually. Find the least
number of complete years after which his sum will be more than double.
A) 4 years
B) 2 years
C) 6 years
D) 8 years
2. Equal sums of money are deposited in two
different banks by M/s Enterprises, one at compound interest (compounded
annually) and the other at simple interest, both at 5% per annum. If after two
years, the difference in the amounts comes to Rs. 200, what are the amounts
deposited with each bank?
A) Rs.72000
B) Rs.76000
C) Rs.78000
D) Rs.80000
E) Rs.82000
3. Bharat took a loan of Rs. 20000 to purchase one
LCD TV set from a finance company. He promised to make the payment after three
years. The company charges compound interest at the rate of 10% per annum for
the same. But, suddenly the company announces the rate of interest as 15% per
annum for the last one year of the loan period. What extra amount does Rohit
have to pay due to this announcement of the new rate of interest?
A) Rs.4200
B) Rs.1210
C) Rs.1620
D) Rs.3150
E) None of these
4. Maninder puts equal amounts of money in two
schemes: one at 10% per annum compound interest payable half yearly and the
second at a certain per cent per annum compound interest payable yearly. If he
gets equal amounts after 3 yr, what is the value of the second per cent?
A) 10 1/4%
B) 12 1/3%
C) 9 1/2%
D) 8 3/4%
E) Can’t be determined
5. A man borrows of Rs 4000 from a bank compound
interest AT 7.5% rate of interest. At the end of every year, he pays Rs. 1500
as part repayment of loan and interest. How much does he still owe to the bank
after three such installments?
A) 275
B) 325
C) 133.5
D) 127.5
E) 123.45
6. A sum of money invested at compound interest
amounts to Rs. 800 in 3 years and Rs.882 in 5 years. What is the rate of
interest?
A) 2.5%
B) 4%
C) 5%
D) 7.5%
E) 10%
7. The difference between compound interest and
simple interest on a sum for two years at 8% per annum, where the interest is
compounded annually is Rs.16. if the interest were compounded half yearly, the
difference in two interests would be nearly
A) Rs.24.64
B) Rs.21.85
C) Rs.16
D) Rs.16.80
E) Rs.25.85
8. Find the compound interest on Rs. 5000 for 3
years if the rate of interest is 5% for the first year, 6% for the second year
and 10% for the third year.
a) Rs. 120.5
b) Rs. 125.5
c) Rs. 120
d) Rs. 130
e) None of these
9. A man borrows Rs. 4000 at 20% compound rate of
interest. At the end of each year he pays back Rs. 1500. How much amount should
he pay at the end of the third year to clear all his dues?
a) Rs. 2592
b) Rs. 2852
c) Rs. 2952
d) Rs. 2953
e) None of these
10. Divide Rs. 1105 between A and B, so that A’s
share at the end of 5 years may equal B’s share at the end of 7 years, compound
interest being at 10%.
a) Rs. 505, Rs. 600
b) Rs. 605, Rs. 500
c) Rs. 705, Rs. 400
d) Rs. 625, Rs. 480
e) None of these
Answers with Explanations:
1. A) Let us assume the sum of money be S
Since the sum is
compounded annually so it becomes
S (1+20/100)t
≥ 2S
=> (6/5)t ≥
2
=> (1.2)t ≥
2
Hence the least number of
complete years are 4
2. D) Let
the amount be A.
Money in the first bank C.I.
= A (1+5/100)2 – A
= A (21/20)2 –
A = 441A/(400) – A = 41A/400
Money in the 2nd bank
S.I. = (A x R x T)/100 = (A x 5 x 2)/100 = A/10
Therefore (41A/400) –
(A/10) = 200
A = Rs. 80000
3. B) Charges
for three years:
= 20000(1+10/100)3
= 26620
Therefore, the compound
interest that Bharat has to pay = 20000(1+10/100)2(1+15/100)
= Rs. 27830 So, the extra
amount is 27830 – 26620 = Rs. 1210
4. A) Let
the equal amount = A (each)
And the interest = r
Using the equation for
compound interest:
A(1+5/100)6 =
A(1+r/100)3
(1+5/100)2 =
(1+r/100)
(105 x 105) / 100 x 100 =
(100 + r)
r = 10.25%
5. E) After
one year, the amount due will be = 4000(1+7.5/100) – 1500 = Rs. 2800
After the second year,
the amount due will be =2800(1+7.5/100) – 1500 = Rs.1510
After third year, the
amount due will be = 1510(1+7.5/100) – 1500 = 123.25
6. C) Formula
for calculating Amount (A) in C.I
A= P (1+R/100)T
Hence, P [1 + (R/100)]3
= 800 …..(1)
P [1 + (R/100)]5
= 882 ….(2)
Since, (2)/(1) = P [1 +
(R/100)]5 / P [1 + (R/100)]3 = 882/800
[1 + (R/100)]2
= 441/400
[1 + (R/100)] = √(441 /
400) = 21 / 20
∴ R = (1/20) × 100 = 5%
7. A)
8. E)
9. C)
10. B)