## ML Aggarwal Class 9 Solutions Chapter 2 Chapter Test

Question 1.

₹ 10000 were lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half-yearly ?

Solution:

Principal = ₹10000

Rate of interest (r) = 10% p.a.

Period = 1 year

Question 2.

A man invests ₹3072 for two years at compound interest. After one year the money amounts to ₹3264. Find the rate of interest and the amount due at the end of 2nd year.

Solution:

Principal (P) = ₹3072

Amount (A) = ₹3264

Period (n) = 1 year

We know that

Question 3.

What sum will amount to ₹28090 in two years at 6% per annum compound interest ? Also find the compound interest.

Solution:

Amount (A) = ₹28090

Rate (r) = 6% p.a.

Period (n) = 2 year

Question 4.

Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹422 find :

(i) the equal sums

(ii) compound interest for each sum.

Solution:

Let each equal sum = ₹100

In first case, rate (r) = 5%

Period (n) = 2 years

Question 5.

The compound interest on a certain sum of money for 2 years is ₹1331·20 and the simple interest on the same sum for the same period and at the same rate is ₹1280. Find the sum and the rate of interest per annum.

Solution:

C.I. for 2 years = ₹1331·20

and S.l. for 2 years = ₹1280·00

Difference = ₹1331·20 – ₹1280·00

= ₹51·20

Question 6.

On what sum will the difference between the simple and compound interest for 3 years at 10% p.a. is ₹232·50 ?

Solution:

Let sum (P) = ₹100

Rate (r) = 10% p.a.

Period (n) = 3 years

Question 7.

The simple interest on a certain sum for 3 years is ₹1080 and the compound interest on the same sum at the same rate for 2 years is ₹741·60. Find

(i) the rate of interest,

(ii) the principal.

Solution:

Simple interest for 3 years = ₹ 1080

Question 8.

In what time will ₹2400 amount to ₹2646 at 10% p.a. compounded semiannually ?

Solution:

Amount (A) = ₹2646

Principal (P) = ₹2400

Rate (r) = 10% p.a. or 5% semi-annually Let period = n half-years

We know that

Hence time = 2 half-years or 1 year

Question 9.

Sudarshan invested ₹60000 in a finance company and received ₹79860 after \(1 \frac{1}{2}\) years. Find the rate of interest per annum compounded half-yearly.

Solution:

Principal (P) = ₹60000

Amount (A) = ₹79860

Period (n) = \(1 \frac{1}{2}\)years = 3 half-years

We know that

Question 10.

The population of a city is 320000. If the annual birth rate is 9·2% and the annual death rate is 1·7%, calculate the population of the town after 3 years.

Solution:

Birth rate = 9·2%

and Death rate = 1·7%

∴ Net growth rate = 9·2 – 1·7 = 7·5%

Present population (P) = 320000

Period (n) = 3 years

∴ Population after 3 years (A)

Question 11.

The cost of a car purchased 2 years ago, depriciates at the rate of 20% every year. If its present worth is ₹315600, find

(i) its purchased price

(ii) its value after 3 years.

Solution:

Present value of car = ₹315600

Rate of depreciation (r) = 20%

Question 12.

Amar Singh started a business with an initial investment of ₹400000. In the first year, he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in the third year rose to 10%. Calculate his net profit for the entire period of 3 years.

Solution:

Investment (P) = ₹400000

Loss in the first year = 4%

Profit in the second year = 5%

Profit in the third year = 10%

Question 13.

The cost of a washing machine depreciates by ₹720 during the second year and by ₹648 during the third year. Calculate :

(i) the rate of depreciation per annum.

(ii) the original cost of the machine.

(iii) the value of the machine at the end of third year.

Solution:

Amount of depreciation for the second year = ₹720

and amount of depreciation for third year = ₹648

Difference = ₹720 – ₹648 = ₹72

∴ ₹72 is depreciation on ₹720 for 1 year.