SELINA Solution Class 9 Compound interest (without using formula) Chapter 2 Exercise 2D

Question 1

What sum will amount of Rs. 6,593.40 in 2 years at C.I. , if the rates are 10 per cent and 11 per cent for the two successive years ?

Sol:

Let principal (p) = Rs. 100
For 1^st year :
P = Rs. 100
R = 10%
T = 1 year

I = ${\displaystyle \frac{100\times 10\times 1}{100}}$ = Rs. 10.

A = 100 + 10 = Rs. 110

For 2^nd year :
P = Rs. 110
R = 11%
T = 1 year

I = ${\displaystyle \frac{110\times 11\times 1}{100}}$ = Rs. 12.10.

A = 110 + 12.10 = Rs. 122.10
If Amount is Rs. 122.10 on a sum of Rs. = 100

If amount is Rs. 1, sum = ${\displaystyle \frac{100}{122.10}}$

If amount is Rs. 6593.40, sum =  ${\displaystyle \frac{100}{122.10}\times 6593.40}$ = Rs. 5400

Question 2

The value of a machine depreciated by 10% per year during the first two years and 15% per year during the third year. Express the total depreciation of the machine, as percent, during the three years.

Sol:

Let the value of the machine in the beginning = Rs. 100

For 1^st year depreciation = 10% of Rs. 100 = Rs. 100
Value of machine for second year = 100 - 10 = Rs. 90

For 2^nd year depreciation = 10% of 90 = Rs. 9
Value of machine for third year = 90 - 9 = Rs. 81

For 3^rd year depreciation = 15% of 81 = Rs. 12.15
Value of machine at the end of third year = 81 - 12.15 = Rs. 68.85

Net depreciation = Rs. 100 - Rs. 68.85 = Rs. 31.15 Or 31.15%.

Question 3

Rachna borrows Rs. 12,000 at 10 percent per annum interest compounded half-yearly. She repays Rs. 4,000 at the end of every six months. Calculate the third payment she has to make at end of 18 months in order to clear the entire loan.

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Sol:

For 1^st half - year :
P = Rs. 12,000; R = 10 % and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{12,000\times 10\times 1}{100\times 2}}$ = Rs. 600

Amount = Rs. 12,000 + Rs. 600 = Rs. 12,600
Money paid at the end of 1^st half year = Rs. 4,000
Balance money for 2^nd half-year = Rs. 12,600 - Rs. 4,000 = Rs. 8,600

For 2^nd half - year :

P = Rs. 8,600; R = 10% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{8,600\times 10\times 1}{100\times 2}}$ =Rs. 430

Amount = Rs. 8,600 + Rs. 430 = Rs. 9,030

Money paid at the end of 2^nd half-year = Rs. 4,000

Balance money for 3^rd half - year = Rs. 9,030 - Rs. 4,000 = Rs. 5,030

For 3^rd half-year

P = Rs. 5,030; R = 10% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{5,030\times 10\times 1}{100\times 2}}$ = Rs. 251.50

Amount = Rs. 5,030 + Rs. 251.50 = Rs. 5,281.50

Question 4

On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is Rs. 2,652. Find the sum.

Sol:

Let Principal = Rs.100
For 1^st year
P = Rs. 100; R = 10 % and T = 1 year

Interest = Rs.${\displaystyle \frac{100\times 10\times 1}{100}}$ = Rs. 10

Amount = Rs. 100 + Rs. 10 = Rs. 110

For 2^nd year

P = Rs. 110; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{110\times 10\times 1}{100}}$ = Rs. 11

Amount = Rs. 110 + Rs. 11 = Rs. 121

For 3^rd year

P = Rs. 121; R = 10 % and T = 1 year

Interest = Rs ${\displaystyle \frac{121\times 10\times 1}{100}}$= Rs. 12.10

Sum of C.I. for 1^st year and 3^rd year = Rs. 10 + Rs. 12.10 = Rs. 22.10

When sum is Rs. 22.10, principal is Rs. 100

When sum is Rs. 2,652, principal = Rs. ${\displaystyle \frac{100\times 2652}{22.10}}$
= Rs. 12,000. 

Question 5

During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by Rs. 2,640 during the second financial year of its purchase.

Sol:

Let original value of machine = Rs. 100
For 1^st year
P = Rs. 100; R = 12% and T = 1 year
Depreciation in 1^st year = Rs ${\displaystyle \frac{100\times 12\times 1}{100}}$ = Rs.12

Value at the end of 1^st year = Rs. 100 - Rs. 12 = Rs. 88

For 2^nd year

P = Rs. 88;  R = 12% and T = 1 year

Depreciation in 2^nd year = Rs.${\displaystyle \frac{88\times 12\times 1}{100}}$ = Rs. 10.56

When depreciation in 2^nd year is Rs.10.56, original cost is Rs.100

When depreciation in 2^nd year is Rs.2,640, original cost = ${\displaystyle \frac{100\times 2640}{10.56}}$ = Rs. 25,000

Question 6

Find the sum on which the difference between the simple interest and compound interest at the rate of 8% per annum compounded annually would be Rs. 64 in 2 years.

Sol:

Let Rs. X be the sum.
Simple Interest (I) = ${\displaystyle \frac{\text{X}\times 8\times 1}{100}}$ = 0.08X

Compound interest
For 1^st year :
P = Rs. X, R = 8% and T = 1
⇒ Interest (I) = ${\displaystyle \frac{\text{X}\times 8\times 1}{100}}$ = 0.08X

For 2^nd year :
P = Rs. X + Rs. 0.08X = Rs.1.08X
⇒ Interest (I) = ${\displaystyle \frac{\text{1.08X}\times 8\times 1}{100}}$ = 0.0864X

The difference between the simple interest and compound interest at the rate of 8% per annum compounded annually should be Rs. 64 in 2 years.
⇒ Rs. 0.08X - Rs. 0.0864X = Rs. 64
⇒ Rs. 0.0064X = Rs. 64
⇒ x = Rs.10,000
Hence the sum is Rs.10,000

Question 7

A sum of Rs. 13,500 is invested at 16% per annum compound interest for 5years. Calculate :
(i) the interest for the first year.
(ii) the amount at the end of first year.
(iii) the interest for the second year, correct to the nearest rupee.

Sol:

For 1^st year :

P = Rs. 13,500; R = 16% and T = 1 year

Interest = Rs. ${\displaystyle \frac{13,500\times 16\times 1}{100}}$= Rs. 2,160

Amount = Rs. 13,500 + Rs. 2,160= Rs. 15,660

For 2^nd year :

P = Rs. 15,660; R = 16% and T = 1 year

Interest = Rs. ${\displaystyle \frac{15,660\times 16\times 1}{100}}$= Rs. 2,505.60 = Rs. 2,506

Question 8

Saurabh invests Rs. 48,000 for 7 years at 10% per annum compound interest. Calculate:
(i) the interest for the first year.
(ii) the amount at the end of second year.
(iii) the interest for the third year.

Sol:

For 1^st year :
P = Rs. 48,000; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{48,000\times 10\times 1}{100}}$ = Rs. 4,800

Amount = Rs. 48,000 + Rs. 4,800 = Rs. 52,800

For 2^nd year :

P = Rs. 52,800; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{52,800\times 10\times 1}{100}}$= Rs. 5,280

Amount = Rs. 52,800 + Rs. 5,280 = Rs. 58,080

For 3^rd year :

P = Rs. 58,080; R = 10% and T = 1 year

Interest = Rs.${\displaystyle \frac{58,080\times 10\times 1}{100}}$ = Rs. 5,808

Question 9

Ashok borrowed Rs. 12,000 at some rate on compound interest. After a year, he paid back Rs.4,000. If the compound interest for the second year is Rs. 920, find:

1. The rate of interest charged
2. The amount of debt at the end of the second year

Sol:

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

For 1^st year :
P = Rs.12,000, R = X% and T = 1
⇒ Interest (I) = ${\displaystyle \frac{12,000\times \text{X}\times 1}{100}}$ = 120X

For 2^nd year:
After a year, Ashok paid back Rs. 4,000.
P = Rs.12,000 + Rs. 120X - Rs. 4,000 = Rs. 8,000 + Rs.120X
⇒ Interest (I) = ${\displaystyle \frac{(8000+120\text{X})\times 1}{100}}$ = ( 80X + 1.20X² )

The compound interest for the second year is Rs. 920.
Rs. ( 80X + 1.20X² ) = Rs. 920
⇒ 1.20X² + 80X - 920 = 0
⇒ 3X² + 200X - 2300 = 0
⇒ 3X² + 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 = Rs.120X = Rs.1200
For 2^nd year :
Interest = Rs.( 80X + 1.20X² ) = Rs.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 = Rs. 8,000 + Rs. 1200 + Rs. 920 = Rs. 10,120

Question 10

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

For 1^st year :
P = Rs.12,000, R = X% and T = 1
⇒ Interest (I) = ${\displaystyle \frac{12,000\times \text{X}\times 1}{100}}$ = 120X

For 2^nd year:
After a year, Ashok paid back Rs. 4,000.
P = Rs.12,000 + Rs. 120X - Rs. 4,000 = Rs. 8,000 + Rs.120X
⇒ Interest (I) = ${\displaystyle \frac{(8000+120\text{X})\times 1}{100}}$ = ( 80X + 1.20X² )

The compound interest for the second year is Rs. 920.
Rs. ( 80X + 1.20X² ) = Rs. 920
⇒ 1.20X² + 80X - 920 = 0
⇒ 3X² + 200X - 2300 = 0
⇒ 3X² + 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 = Rs.120X = Rs.1200
For 2^nd year :
Interest = Rs.( 80X + 1.20X² ) = Rs.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 = Rs. 8,000 + Rs. 1200 + Rs. 920 = Rs. 10,120

Sol:

Total interest obtained in the first year = Rs, 1500
lnterest for the second year - Total interest obtained in the first year
= Rs. 1725 - Rs. 1500
= Rs. 225

Rate of interest for the second year = ${\displaystyle \frac{225}{1500}}$ x 100 = 15%

Interest for the third year - Interest for the second year
= Rs. 2,070 - Rs. 1,725
= Rs. 345

Rate of interest for the third year
= ${\displaystyle \frac{345}{1,725}}$ x 100 = 20%

So, rate of interest for the second year and third year are 15% and 20% respectively.