Selina ICSE Solutions for Class 10 Maths – Compound Interest (Using Formula)
Selina ICSE Solutions for Class 10 Maths Chapter 2 Compound Interest (Using Formula)
Exercise 2(A)
Solution 1:
Given : P = ₹ 12,000; n = 3 years and r = 5%
= ₹ 13,891.50
C.I. = ₹ 13,891.50 – ₹ 12,000
= ₹ 1,891.50
Solution 2:
Given : P = ₹ 15,000; n = 2 years ; r1 = 8%; r2 = 10%
= ₹ 17,820
Solution 3:
Given : P = ₹ 6,000; n = 3 years ; r1 = 5%; r2 = 8% and r3 = 10%
= ₹ 7,484.40
C.I. = ₹ 7,484.40 – ₹ 6,000 = ₹ 1,484.40
Solution 4:
Given : Amount = ₹ 5,445; n = 2 years and r = 10%
Solution 5:
Given : C.I = ₹ 768.75; n = 2 years and r = 5%
Solution 6:
Given : C.I = ₹ 1,655; n = 3 years and r = 10%
Solution 7:
Given : Amount = ₹ 9,856; n = 2 years ; r1 = 10%; r2 = 12%
Solution 8:
The sum is ₹ 16,000
Solution 9:
At 5% per annum the sum of ₹ 6,000 amounts to ₹ 6,615 in 2 years when the interest is compounded annually.
Solution 10:
Let Principal = ₹ y
Then Amount = ₹ 1.44y
n = 2 years
Solution 11:
Given : P = ₹ 18,000; C.I. = ₹ 5,958 and n = 3 years
Solution 12:
Given: P = ₹ 5,000; A = ₹ 6,272 and n = 2 years
Solution 13:
Given : P = ₹ 7,000; A = ₹ 9,317 and r = 10%
Solution 14:
Given : P = ₹ 4,000; C.I. = ₹ 630.50 and r = 5%
Solution 15:
Let share of A = ₹ y
share of B = ₹ (28,730 – y)
rate of interest = 10%
According to question
Amount of A in 3 years = Amount of B in 5 years
Therefore share of A = ₹ 15,730
Share of B = ₹ 28,730 – ₹ 15,730 = ₹ 13,000
Solution 16:
Let share of Rohit = ₹ y
share of Rajesh = ₹ (34,522 – y)
rate of interest = 5%
According to question
Amount of Rohit in 12 years = Amount of Rajesh in 9 years
Therefore share of Rohit = ₹ 16,000
Share of Rajesh = ₹ 34,522 – ₹ 16,000 = ₹ 18,522
Solution 17:
(i) Let share of John = ₹ y
share of Smith = ₹ (44,200 – y)
rate of interest = 10%
According to question
Amount of John in 4years = Amount of Smith in 2years
Therefore share of John = ₹ 20,000
Share of Smith = ₹ 44,200- ₹ 20,000 = ₹ 24,200
(ii) Amount that each will receive
Solution 18:
The amount of money in the account = ₹ 22,000
Compound interest for the first year = Simple interest for the first year
Solution 19:
Let’s ₹ x be the sum of the money.
Let A1 be the amount obtained at the end of the 1st year.
Let A2 be the amount obtained at the end of the 2nd year.
Let R be the rate of interest.
The amounts of are in the ratio 20:21.
But rate on interest cannot be negative hence R = 5.
Therefore the rate of interest is 5%.
Solution 20:
Let’s ₹ x be the sum of the money.
The sum of the money is ₹ 30,000.
Exercise 2(B)
Solution 1:
Given: P = ₹ 7,400; r = 5% p.a. and n = 1 year
Since the interest is compounded half-yearly,
Solution 2:
Solution 3:
For the first 2 years
Amount in the account at the end of the two years is ₹ 22,400.
For the remaining one year
The total amount to be paid at the end of the three years is ₹ 27,104.
Solution 4:
The sum of ₹ 24,000 amount ₹ 27,783 in one and a half years at 10% per annum compounded half yearly.
Solution 5:
Solution 6:
The rate of interest is 8%.
Solution 7:
Given: P = ₹ 1,500; C.I.= ₹ 496.50 and r = 20%
Since interest is compounded semi-annually
Solution 8:
Given: P = ₹ 3,500; r = 6% and n = 3 years
Since interest is being compounded half-yearly
Solution 9:
Given: P = ₹ 12,000; n = 1 ½ years and r = 10%
To calculate C.I.
For 1 year
P = ₹ 12,000; n = 1 year and r = 10%
For next ½ year
P = ₹ 13,200; n = ½ year and r = 10%
∴ C.I. = ₹ 13,860 – ₹ 12,000 = ₹ 1,860
∴ Difference between C.I. and S.I = ₹ 1,860 – ₹ 1,800 = ₹ 60
Solution 10:
Given: P = ₹ 12,000; n = 1 ½ years and r = 10%
To calculate C.I.(compounded half-yearly)
P = ₹ 12,000; n = 1 ½ years and r = 10%
∴ C.I. = ₹ 13,891.50 – ₹ 12,000 = ₹ 1,891.50
∴ Difference between C.I. and S.I
= ₹ 1,891.50 – ₹ 1,800 = ₹ 91.50
Solution 11:
Solution 12:
Exercise 2(C)
Solution 1:
Initial height(P)= 80 cm
Growth rate = 20%
∴ Growth after 3 months
Solution 2:
Cost of machine in 2008 = ₹ 44,000
Depreciation rate = 12%
(i) Cost of machine at the end of 2009
(ii) Cost of machine at the beginning of 2007(P)
Solution 3:
Value of a machine at the end of 2004(P)= ₹ 27,000
Value of a machine at the beginning of 2007(A)= ₹ 21,870
Time(n)= 2 years
(ii) The value of machine at the beginning of 2004(P)
Solution 4:
Let x be the value of the article.
The value of an article decreases for two years at the rate of 10% per year.
The value of the article at the end of the 1st year is
x – 10% of x = 0.90x
The value of the article at the end of the 2nd year is
0.90x – 10% of (0.90x) = 0.81x
The value of the article increases in the 3rd year by 10%.
The value of the article at the end of 3rd year is
0.81x + 10% of (0.81x) = 0.891x
0.891x = 40,095
⇒ x = 45,000
The value of the article at the end of 3 years is ₹ 40,095.
The original value of the article is ₹ 45,000.
Solution 5:
Population in 2005(P) = 64,000
Let after n years its population be 74,088(A)
Growth rate = 5% per annum
Solution 6:
Let the population in the beginning of 1998 = P
The population at the end of 1999 = 2,85,120(A)
r1 = – 12% and r2 = +8%
Solution 7:
Let sum of money be Rs P and rate of interest= r%
Money after 1 year = ₹ 16,500
Money after 3 years = ₹ 19,965
Solution 8:
Given: P = ₹ 7,500 and Time(n) = 2 years
Let rate of interest = y%
Solution 9:
Let Principal be Rs y and rate= r%
According to 1st condition
Amount in 10 years = Rs 3y
Solution 10:
At the end of the two years the amount is
Mr. Sharma paid ₹ 19,360 at the end of the second year.
So for the third year the principal is A1 – 19,360.
Also he cleared the debt by paying ₹ 31,944 at the end of the third year.
Mr. Sharma borrowed ₹ 40,000.
Solution 11:
Solution 12:
Let ₹ x and ₹ y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.
Pramod will attain the age of 25 years after 25 – 16 = 9 years
Rohit will attain the age of 25 years after 25 -18 = 7 years
Pramod and Rohit should invest in 400:441 ratio respectively such that they will get the same sum on attaining the age of 25 years.
Exercise 2(D)
Solution 1:
Let ₹ 100 be the principal amount.
When the principal is ₹ 100 the compound interest is ₹ 10.25.
Therefore the effective rate per annum is 10.25%.
Solution 2:
Let ₹ x be the value of the property.
Value of the property decreases at the rate of 6 1/4 percent at the beginning of that year.
Value of the property at the end of the 2nd year = Value of the property at the beginning of the 3rd year
Value of the property at the beginning of 2nd year = value of property at the beginning of 1st year
Value of property
Value of the property at the beginning of 2 years is ₹ 2,56,000.
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Solution 7:
(i) Present value of machine(P) = ₹ 97,200
Depreciation rate = 10%
= ₹ 78732
(ii) Present value of machine(A) = ₹ 97,200
Depreciation rate = 10% and time = 2 years
To calculate the cost 2 years ago
Solution 8:
Solution 9:
Solution 10:
Solution 11:
Given: C.I. for the 2nd year = ₹ 4,950 and rate = 15%
Then amount at the end of 2nd year= ₹ 33,000
For first 2 years
A = ₹ 33,000; r1 =10%
Solution 12:
Solution 13:
Solution 14:
P = ₹ 60,000, R = 5%, n = 2 years
The money lender deducts the interest that would be due at the end of the period and handed over the balance to Mrs. Shukla.
Amount received by Mrs. Shukla = ₹ 60,000 – ₹ 6,000 = ₹ 54,000
Now sum deposited in the bank (P) = ₹ 54,000
R = 5% compounded annually, n = 2 years
The amount of money that Mrs. Shukla will have to add to pay the money back to the money lender
= ₹ 60,000 – ₹ 59,535 = ₹ 465
Solution 15:
Let ₹ x be the sum of money.
Rate = 5 % p.a. Simple interest = ₹ 1,200, n = 3years
The amount due and the compound interest on this sum of money at the same rate and after 2 yers
P = ₹ 8,000; rate = 5% p.a., n = 3 years
The amount due after 2 years is ₹ 8,820 and the compound interest is ₹ 820.
Solution 16:
Let x% be the rate of interest.
P = ₹ 6,000, n = 2 years, A = ₹ 6,720
(i) For the first year
The rate of interest is x% = 12%.
(ii) The amount at the end of the second year.
The amount at the end of the second year = ₹ 7,526.40