## Selina ICSE Solutions for Class 10 Maths – Compound Interest (Without using formula)

**Selina ICSE Solutions for Class 10 Maths Chapter 1 Compound Interest (Without using formula)**

**Exercise 1(A)**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Solution 10:**

**Solution 11:**

**Solution 12:**

Let Rs.*x* be the sum.

Compound interest

The difference between the simple interest and compound interest at the rate of 8% per annum compounded annually should be ₹ 64 in 2 years.

⇒ ₹ 0.08x – ₹ 0.0864x = ₹ 64

⇒ ₹ 0.0064x = ₹ 64

⇒ x = ₹ 10000

Hence the sum is ₹ 10000.

**Solution 13:**

**Solution 14:**

**Solution 15:**

**Solution 16:**

**Solution 17:**

**Solution 18:**

**Solution 19:**

Amount = ₹ 30,400 + ₹ 3,040 + ₹ 4,000 = ₹ 37,440

The amount in Mrs. Kapoor’s account on 01/01/2012 is ₹ 37,440.

**Solution 20:**

(i) Let x% be the rate of interest charged.

The compound interest for the second year is ₹ 920

Rs. (80x + 1.20x^{2}) = ₹ 920

⇒ 1.20x^{2} + 80x – 920 = 0

⇒ 3x^{2} + 200x – 2300 = 0

⇒ 3x^{2} + 230x – 30x – 2300 = 0

⇒ x(3x + 230) -10(3x + 230) = 0

⇒ (3x + 230)(x – 10) = 0

⇒ x = -230/3 or x = 10

As rate of interest cannot be negative so x = 10.

Therefore the rate of interest charged is 10%.

(ii) For 1^{st} year:

Interest = ₹ 120x = ₹ 1200

For 2^{nd} year:

Interest = ₹ (80x + 1.20x^{2}) = ₹ 920

The amount of debt at the end of the second year is equal to the addition of principal of the second year and interest for the two years.

Debt = ₹ 8,000 + ₹ 1200 + Rs.920 = ₹ 10,120

**Exercise 1(B)**

**Solution 1:**

**Solution 2:**

Difference between the C.I. of two successive half-years

= ₹ 760.50 – ₹ 650 = ₹ 110.50

₹ 110.50 is the interest of one half-year on ₹ 650

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Solution 10:**

(i) The interest charged is compounded because if the interest charged is simple, then the interest for two years will be double of interest for one year which is not given.

(ii) C.I. for 1^{st} year= ₹ 720

C.I. for two years= ₹ 1,497.60

C.I. for 2^{nd} year = ₹ 1,497.60 – ₹ 720 = ₹ 777.60

Difference between the C.I. of two successive years

= ₹ 777.60 – ₹ 720

= ₹ 57.60

₹ 57.60 is the interest for one year on ₹ 720.

**Solution 11:**

(i) C.I. for second year = ₹ 864

C.I. for third year = ₹ 933.12

Difference between the C.I. of two successive years= ₹ 933.12 – ₹ 864= ₹ 69.12

₹ 69.12 is the interest of one year on ₹ 864

(ii) Let the sum of money= ₹ 100

Interest on it for 1st year= 8% of ₹ 100= ₹ 8

Amount in one year= ₹ 100+ ₹ 8= ₹ 108

Similarly, C.I. for 2nd year= 8% of ₹ 108 = ₹ 8.64

When C.I. for 2nd year is ₹ 8.64, sum = ₹ 100

Principal for 4th year= ₹ 10,000+₹ 800+₹ 864+₹ 933.12 = ₹ 12,597.12

Interest for 4th year= 8% of ₹ 12,597.12 = ₹ 1,007.77

**Solution 12:**

(i) Amount in three years = ₹ 20,160

Amount in four years = ₹ 24,192

Difference between the amounts of two successive years

= ₹ 24,192 – ₹ 20,160= ₹ 4,032

⇒ ₹ 4,032 is the interest of one year on ₹ 20,160

(ii) Let amount in two years = ₹ 100

And amount in three years = ₹ 100+ 20% of ₹ 100

= ₹ 100+ ₹ 20

= ₹ 120

When amount in 3 years is ₹ 120, amount in two years = ₹ 100

(iii) Amount in 5 years = ₹ 24,192+ 20% of ₹ 24,192

= ₹ 24,192 +₹ 4,838.40

= ₹ 29,030.40

**Solution 13:**

(ii)The total interest paid in two years = ₹ 350 + ₹ 560 = ₹ 910

(iii) The total amount of money paid in two years to clear the debt

= ₹ 8,000+ ₹ 910

= ₹ 8,910

**Solution 14:**

**Solution 15:**

**Solution 16:**

**Solution 17:**

(i) Difference between depreciation in value between the first and second years = ₹ 4,000 – ₹ 3,600 = ₹ 400

⇒ Depreciation of one year on ₹ 4,000 = ₹ 400

(ii) Let ₹ 100 be the original cost of the machine.

Depreciation during the 1st year = 10% of ₹ 100 = ₹ 10

When the values depreciates by ₹ 10 during the 1st year, Original cost = ₹ 100

⇒ When the depreciation during 1st year = ₹ 4,000

The original cost of the machine is ₹ 40,000.

(iii) Total depreciation during all the three years

= Depreciation in value during(1st year + 2nd year + 3rd year)

= ₹ 4,000 + ₹ 3,600 + 10% of (₹ 40,000 – ₹ 7,600)

= ₹ 4,000 + ₹ 3,600 + ₹ 3,240

= ₹ 10,840

The cost of the machine at the end of the third year

= ₹ 40,000 – ₹ 10,840 = ₹ 29,160.

**Exercise 1(C)**

**Solution 1:**

Let the sum of money be ₹ 100

Rate of interest = 10% p.a.

Interest at the end of 1st year = 10% of ₹ 100 = ₹ 10

Amount at the end of 1st year = ₹ 100 + ₹ 10 = ₹ 110

Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11

Amount at the end of 2nd year = ₹ 110 + ₹ 11 = ₹ 121

Interest at the end of 3rd year = 10% of ₹ 121 = ₹ 12.10

Sum of interest of 1st year and 3rd year = ₹ 10 + ₹ 12.10 = ₹ 22.10

When sum of both interest is ₹ 22.10, principal is ₹ 100

When sum of both interest is ₹ 1,768, principal = ₹ = ₹ 8,000

**Solution 2:**

**Solution 3:**

**Solution 4:**

Cost of machine= ₹ 32,000

Depreciation rate every year = 5%

∴ Cost of machine after one year = ₹ 32,000- 5% of ₹ 32,000

= ₹ 32,000- ₹ 1,600

= ₹ 30,400

Cost of machine after two year = ₹ 30,400- 5% of ₹ 30,400

= ₹ 30,400- ₹ 1,520

= ₹ 28,880

∴ Total depreciation in two years = ₹ 32,000 – ₹ 28,880 = ₹ 3,120.

**Solution 5:**

Let the sum of money be ₹ 100.

Rate of interest= 10%p.a.

Interest at the end of 1st year = 10% of ₹ 100= ₹ 10

Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110

Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11

Amount at the end of 2nd year = ₹ 110 + ₹ 11= ₹ 121

Interest at the end of 3rd year = 10% of ₹ 121= ₹ 12.10

∴ Difference between interest of 3rd year and 1st year = ₹ 12.10 – ₹ 10 = ₹ 2.10

When difference is ₹ 2.10, principal is ₹ 100.

When difference is ₹ 252, principal = = ₹ 12,000.

**Solution 6:**

(i) C.I. for 2nd year = ₹ 9,680

C.I. for 3rd year = ₹ 10,648

Difference in both interests = ₹ 10,648 – ₹ 9,680 = ₹ 968

(ii) Interest for 4th year = ₹ 10,648+ 10% of ₹ 10,648

= ₹ 10,648 + ₹ 1,064.80

= ₹ 11,712.80

(iii) Let principal be ₹ 100

Rate of interest= 10% p.a.

Interest at the end of 1st year = 10% of ₹ 100= ₹ 10

Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110

Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11

When C.I. for 2nd year is ₹ 11, principal is ₹ 100

When C.I. for 2nd year is ₹ 9,680, principal = ₹ = ₹ 88,000

Interest for 1st year = 10% of ₹ 88,000 = ₹ 8,800.

**Solution 7:**

(i) Amount in two years = ₹ 9,680

Amount in three years = ₹ 10,648

∴ Difference in both amounts = ₹ 10,648 – ₹ 9,680 = ₹ 968

(ii) Amount in 4 years = ₹ 10,648+ 10% of ₹ 10,648

= ₹ 10,648 + ₹ 1,064.80

= ₹ 11,712.80

(iii) Let principal be ₹ 100

Rate of interest= 10%p.a.

Interest at the end of 1st year = 10% of ₹ 100= ₹ 10

Amount at the end of 1st year = ₹ 100 + ₹ 10= ₹ 110

Interest at the end of 2nd year = 10% of ₹ 110 = ₹ 11

Amount at the end of 2nd year = ₹ 110 +₹ 11= ₹ 121

When amount at the end of 2nd year is ₹ 121, principal is ₹ 100

When amount at the end of 2nd year is ₹ 9,680, principal

= ₹

= ₹ 8,000

∴ Amount in one year = ₹ 8,000+10% of ₹ 8,000

= ₹ 8,000 + ₹ 800

= ₹ 8,800

**Solution 8:**

**Solution 9:**

**Solution 10:**

(i) The population of a town increases by 10% every 3 years.

The population of the town after 3 years

= 72,600 + 10% of 72,600

= 72,600 + 7,260

= 79,860

The population of the tower after 6 years

= 79,860 + 10% of 79,860

= 79,860 + 7,986

= 87,846

The population of the town after 6 years is 87,846.

(ii) Let x be the population of the town 6 years ago.

The present population of the town is 72,600.

The population of the town 3 years ago

= x + 10% of x

= x + 0.10x

= 1.10x

The present population of the town

= 1.10x + 10% of 1.10x

⇒ 72,600 = 1.10x + 0.110x

⇒ 72,600 = 1.210x

⇒ x = 60,000

The population of the town before 6 years ago was 60,000.