Compound Interest Practice Problems –
Set 2
1. The population of town is decreasing by a certain rate of interest
(Compounded annually). If the current population be 29160 and ratio of in
population second year and 3rd year be 10 : 9. When was the population 3 years
ago?
a) 40,000
b) 35,000
c) 55,000
d) 33,000
Answer:
A)
Explanation:
P = 29160
Population after 2 years = 29160 (1 - R/100)^2 …..
(i)
Population after 2 years = 29160 (1 - R/100)^3 ……
(ii)
ATQ, Eq (i)/Eq (ii) = 10/9
⇒ 1/(1 - R/100) = 10/9 ⇒ R = 10%
Population 3 years ago-
⇒ x (1 – 10/100)^3 = 29160 => x = 29160/[(9/10)^3]
= (29160 × 1000)/(9 × 9 × 9) = 40000
2. Irfan borrows a sum of Rs. 64000 at 5% pa compound interest. He
repays a certain amount at the end of one year and the balance amount of Rs.
35700 at the end of the second year. What amount does he repay in the first
year?
a) Rs.34000
b) Rs.37200
c) Rs.36400
d) Rs.35700
e) Rs.33200
Answer:
E)
Explanation:
Sum = Rs. 64000
CI for 1st year= [(64000×5)/100] =
Rs. 3200
:. A= (64000+3200) = Rs. 67200
Let the amount repaid be Rs. X.
Then, the sum at the beginning of the 2nd year = 67200-x
Amount at the 2nd year = (67200-x)
×1.05 = 35700
Or, x=Rs. 33200
3. A man gave 50% of his savings of 84100 to his wife and divided the
remaining sum among his two sons A and B of 15 and 13 years of age
respectively. He divided it in such a way that each of his sons, when they
attain the age of 18 years would receive the same amount at 5% compound
interest per annum, the share of B was
a) 20,550
b) 20,000
c) 20,500
d) 22,000
e) 22,500
Answer: B)
Explanation:
Total savings = Rs. 84100.
Wife got = 50% of 84100 = Rs. 42050.
Let P1 to A and P2 B as principals.
P1[1+5/100]3 = P2[1+5/100]5
P1/P2 = [1+5/100]2=> P1/P2 = 441/440
P2 = 440/881 * 42050
P2 = 20,000.
So, B got Rs. 20,000.
4. Manoj invests Rs. x in insurance which gives her returns at 21%
annually and Rs. y in mutual funds which gives her returns of 10% compounded
half yearly. If Manoj gets the same returns from both the investments after 1
year, then what is the square root of the ratio of x to y?
a) 1 : 2
b) 11 : 21
c) 21 : 22
d) 21 : 25
e) None of these
Answer:
C)
Explanation:
Amount earned from insurance after one year;
A1 = (100 + Interest) × Principal = 121% of x
Applying net% effect in the 2nd scenario to get the
effective rate of interest compound half-yearly, we get
Net % effect = x + y + xy/100 % = 5 + 5 + (5 *
5)/100 = 10.25%
∴ Amount
earned from mutual funds
A2 = (100 + interest) × Principal = (100 + 10.25)% =
110.25% of y
Given, A1 = A2
121% of x = 110.25% of y
∴ x/y
= 110.25/121 = 441/484
√(x/y) = √(441/484) = 21 : 22
5. Sanjay purchased a hotel worth rupees 10 lakhs and Anita purchased a
car worth Rs. 16 lakh. The value of hotel every year increase by 20% of the
previous value and the value of car every depreciates by 25%. What is the
difference between the price of hotel and car after 3 years?
a) 10,53,000
b) 10,63,000
c) 11,53,000
d) 10,43,000
e) 11,43,000
Answer: A)
Explanation:
Amount of the hotel after 3 years =10 lakh (1 +
20/100)^3 = 10,00,000 × 216/125 = 1728000
Amount of the hotel after 3 years = 16 lakh (1 – 25/100)^3
= 16,00,000 × 27/64 = 6,75,000
Difference = 17,28,000 - 6,75,000 = 10,53,000.
6. A has lent some money to B at 6% p.a. and C at 10% at the end of the
year he has gain the over all interest at 8% p.a. in what ratio has he lent the
money to A and B?
a) 1 : 2
b) 2 : 1
c) 1 : 1
d) 2 : 3
e) 3 : 2
Answer: C)
Explanation:
By allegation and mixture:
6
10
8
2
2
Therefore, the ratio is- 2 : 2 = 1:1.
7. The SI on a certain sum of money for 3 yr at 8% pa is half the CI on
Rs. 8000 for 2 yr at 10% pa. Find the sum placed on simple interest?
a) Rs.3500
b) Rs.3800
c) Rs.4000
d) Rs.4200
e) Rs.4500
Answer:
A)
Explanation:
Applying the net% effect formula, we get = 10 + 10 +
(10 * 10)/100 = 21%
Now, 21% of 8000 = 1680
Sum of SI is half of CI = 1680/2 = 840
∴ Sum =
(840 * 100)/(8 * 8) = Rs.3500
8. If the compound interest on certain sum at 16 2/3% for 3 years is Rs. 1270. Find the simple interest on the same
sum at the same rate for the same period.
a) Rs.1202
b) Rs.1104
c) Rs.1080
d) Rs.1432
e) Rs.1252
Answer:
C)
Explanation:
Let the sum be Rs.x, then,
CI = [x * (1 + 50/[3*100])^3 – x) = 127x/216
So, 127x/216 = 1270 => x = (1270 × 126)/127 =
2160
Thus, the sum is Rs.2160
So, SI = Rs.(2160 × 50/3 × 3 × 1/100) = Rs.1080
9. Arvind takes a loan of Rs. 10500 at 10% p.a. compounded annually
which is to be repaid in two equal annual installments. First at the end of one
year and other at the end of the second year. The value of each installment.
a) 3032
b) 6050
c) 4500
d) 5630
e) 5120
Answer:
B)
Explanation:
Let the installments be x. Then,
According to the question,
10500 = [x/(1 + 10/100)] + [x/(1 + 10/100)^2]
From formula, A = P(1 + R/100)^n => P = A/[(1 +
R/100)^n]
=> 10500 = [10/100 + 100/121]x
=> 10500 = (110x + 100x)/121
=> x = 10500 * 121/210 = 6050
10. A man borrows Rs. 5100 to be paid back with compound interest at the
rate of 4% pa by the end of 2 yr in two equal yearly investments. How much will
each installment be?
a) Rs.2704
b) Rs.2800
c) Rs.3000
d) Rs.2500
e) Rs.2809
Answer: A)
Explanation:
Let the installments be x. Then,
According to the question,
=> x/(1 + 4/100) + x/(1 + 4/100)^2 = 5100
=> x/(25/26) + x/[(26/25)^2] = 5100
=> [25x × 26 + 652x]/676 = 5100
=> 650x + 652x = 5100 × 676
=> x = 5100 * 676/1275 = Rs.2704
