Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.2 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

## Practice Set 14.2 8th Std Maths Answers Chapter 14 Compound Interest

Compound Interest class 8 practice set 14.2 Question 1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.

Solution:

Here, P = Initial number of workers = 320

R = Increase in the number of workers per year = 25%

N = 2 years

A = Number of workers after 2 years

∴ The number of workers after 2 years would be 500.

Question 2.

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

Solution:

Here, P = Present number of sheeps = 200

R = Increase in number of sheeps per year = 8%

N = 3 years

A = Number of sheeps after 3 years

= \(\frac{0.32}{25} \times 27 \times 27 \times 27\)

= 0.0128 × 27 × 27 × 27

= 251.9424

= 252

∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).

8th Class Math Practice Set 14.2 Question 3.

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.

Solution:

Here, P = Present number of trees in the forest = 40,000

R = Increase in the number of trees per year = 5%

N = 3 years

A = Number of trees after 3 years

= 5 × 21 × 21 × 21

= 5 × 9261

= 46,305

∴ The expected number of trees in the forest after 3 years is 46,305.

Std 8 Maths Practice Set 14.2 Question 4.

The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.

Solution:

Here, P = Cost price of machine = Rs 2,50,000

R = Rate of depreciation per year = 10%

N = 2 years

A = Depreciated price of the machine after 2 years

= 2,500 × 81

= Rs 2,02,500

Depreciation in price = Cost price (P) – Depreciated price (A)

= 2,50,000 – 2,02,500

= Rs 47,500

∴ The depreciation in price of the machine after 2 years would be Rs 47,500.

Question 5.

Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.

Solution:

Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years

i. \(\mathbf{A}=\mathbf{P}\left[1+\frac{\mathbf{R}}{100}\right]^{N}\)

ii. Interest = Amount (A) – Principal (P)

= 4036.80 – 3000

= Rs 1036.80

∴ The compound interest after 2 years would be Rs 1036.80.

Question 6.

A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.

Solution:

Here, P = Rs 15,000, R = 12 p.c.p.a, and

N = 3 years

∴ The amount required to settle the loan after 3 years is Rs 21,073.92.

Practice Set 14.2 Class 8 Question 7.

A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

Solution:

Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years

∴ P = 4 x 50 x 50

∴ P = Rs 10,000

∴ The principal is Rs 10,000.

Question 8.

The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.

Solution:

Here, P = Population of a suburb = 16,000

N = 2 years

A = Increase in the population after 2 years = 17,640

R = Rate of increase in population

∴5 = R

i.e., R = 5%

∴The rate of increase in the population is 5 p.c.p.a.

Compound Interest Practice Set 14.2 Question 9.

In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.

Solution:

Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847

∴Rs 700 will amount to Rs 847 in 2 years.

Practice Set 14.2 Question 10.

Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.

Solution:

Here, P = Rs 20,000, R = 8 p.c.p.a.,

N = 2 years

i. Simple interest (I)

Simple interest (I) = Rs 3200

ii. Compound Interest (I):

= 32 × 27 × 27

= Rs 23,328

Compound interest (I)

= Amount (A) – Principal (P)

= 23,328 – 20,000

= Rs 3328 ,..(ii)

iii. Difference

= Compound interest – Simple interest

= 3328 – 3200 … [Form (i) and (ii)]

= Rs 128

∴ The difference between compound interest and simple interest is Rs 128.

[Note: The question is modified as per the answer given in the textbook.]

**Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities**

8th Standard Maths Practice Set 14.2 Question 1.

Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90)

Solution:

(Students should attempt this activity at their own.)