The DAV Class 7 Maths Solutions and **DAV Class 7 Maths Chapter 5 Worksheet 4** Solutions of Application of Percentage offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 5 WS 4 Solutions

Question 1.

Find the unknown quantity in each of the following.

Answer:

(i) P = ₹ 400, R = 5%, T = 3 years, S.I. = ?

S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{400 \times 5 \times 3}{100}\) = 60

Amount = P + S.I. = ₹ 400 + ₹ 60

= ₹ 460

P = ₹ 450, R = 4\(\frac{1}{2}\)%, = \(\frac{9}{2}\)%, T = 3 years 4 months, S.I. = ?

= 3 + \(\frac{4}{12}\) = 3 + \(\frac{1}{3}=\frac{10}{3}\) years

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{450 \times 9 \times 10}{2 \times 3 \times 100}=\frac{135}{2}\) = 67.50

(iii) P = ₹ 500, R = 15\(\frac{1}{2},=\frac{31}{2}\)%, = T = 146 days = \(\frac{146}{365}=\frac{2}{5}\) years, S.I. = ?

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{500 \times 31 \times 2}{2 \times 5 \times 100}\) = ₹ 31

Amount = P + S.I.

= 500 + 31

= ₹ 531

(iv) P = ₹ 1200, R = 15%, T = 22 days of April + 31 days for May + 20 days for June = 73 days

= \(\frac{73}{365}\) years = \(\frac{1}{5}\)year

S.I. = \(\frac{P \times R \times T}{100}=\frac{1200 \times 15 \times 1}{100 \times 5}\)

Amount = P + S.I.

= ₹ 1200 + ₹ 36

= ₹ 1236

Question 2.

Find the amount from the investment ofX 4500for 2 years at 5 paise per rupee interest.

Answer:

P = ₹ 4500, R = 5 paise per rupee = 5%, T = 2 years

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{4500 \times 5 \times 2}{100}\) = ₹ 450

∴ Amount = P + S.I. = ₹ 4500 + ₹ 450

= ₹ 4950

Question 3.

Ramesh took a loan oft 80,000 from a bank at 12% per annum and paid it back after 7 months together with interest. Find the amount he paid to the bank.

Answer:

P = ₹ 80,000, R = 12% P.a., T = \(\frac{7}{12}\) year

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{80,000 \times 12 \times 7}{100 \times 12}\) = ₹ 5600

∴ Amount = P + S.I. = ₹ 80,000 + ₹ 5600

= ₹ 85,600

Question 4.

Rahul deposited ₹ 7000 at 7% P.a. for 4y years and Rohan deposited ₹ 7000 at 6% p.a. for 5 years. Who will get more interest? What amount will each get?

Answer:

For Rahul,

P = ₹ 7000, R = 7 % p.a., T = 4y years = y years

∴ S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{7000 \times 7 \times 9}{100 \times 2}\) = ₹ 2205

∴ Amount = ₹ 7000 + ₹ 2205

= ₹ 9205

For Rohan,

P = ₹ 7000, R = 6%, T = 5 years

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{7000 \times 6 \times 5}{100}\) = ₹ 2100

∴ Amount = ₹ 7000 + ₹ 2100 = ₹ 9100

Hence the amount got by Rahul is more than Rohan.

Question 5.

Rawit deposited ₹ 80,000 in a bank which pays him 6% interest. After 3 years he withdraws the money and buys a car for ₹ 90,000. How much money is left with him?

Answer:

Here, P = ₹ 80,000, R = 6%, T = 3 years.

S.I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{80,000 \times 6 \times 3}{100}\) = ₹ 14400

Amount that he receives from the bank

= ₹ 80,000 + ₹ 14,400

= ₹ 94,400

Money spent on buying the Car = ₹ 90,000

Money left with him = ₹ 94,400 – ₹ 90,000

= ₹ 4400