The DAV Class 6 Maths Solutions and **DAV Class 6 Maths Chapter 5 Worksheet 4 **Solutions of Percentage and its Applications offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 5 WS 4 Solutions

Question 1.

Fill in the following blanks:

(a) Principal = ₹ 600, Time = 2 years, Rate of interest = 10% p.a., simple interest = _____.

Solution:

Here, P = ₹ 600,

T = 2 years,

R = 10% p.a.

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 120

Hence, the required S.I. = ₹ 120.

(b) Principal = ₹ 6500, Time = 1½ years, Rate of interest = 10% p.a., simple interest = _____.

Solution:

Here, P = ₹ 6500,

T = 1 \(\frac{1}{2}\) years

= \(\frac{3}{2}\) years,

R = 10 % p.a.

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 975

Hence, the required Simple interest S.I. = ₹ 975.

(c) Principal = ₹ 1280, Time = 2½ years, Rate of intere.st = 2% %, Simple interest = _____, Amount = _____.

Solution:

Here, P = ₹ 1280,

T = 2 \(\frac{1}{2}\) years

= \(\frac{5}{2}\) years,

R = 2 \(\frac{3}{4}\) %

= \(\frac{11}{4}\) % p.a.

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 88

Amount = P + S.I.

= ₹ 1280 + ₹ 88

= ₹ 1368

Amount = ₹ 1368.

(d) Principal = ₹ 2600, time = 3 years, Rate of interest = 5½ % p.a. Simple interest = _____ , Amount = ____ .

Solution:

Here, P = ₹ 2600,

T = 3 years

R = 5 \(\frac{1}{2}\) %

= \(\frac{11}{2}\) % p.a.

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 429

Amount = P + S.I.

= ₹ 2600 + ₹ 429

= ₹ 3029

Amount = ₹ 3029.

Question 2.

Hari borrowed ₹ 450 from his friend at 6% per annum. Fie returned the amount after 8 months. How much money did he pay?

Solution:

Here, P = ₹ 450,

R = 6% p.a.,

T = 8 months

= \(\frac{8}{12}=\frac{2}{3}\) years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 18

∴ Amount = P + S.I.

= ₹ 450 + ₹ 18

= ₹ 468

Hence, the money to be returned = ₹ 468.

Question 3.

Find the simple interest on ₹ 2000 for 6 months at the rate of 4 % per annum.

Solution:

Here, P = ₹ 2000,

R = 4 \(\frac{1}{2}\)% p.a.

= \(\frac{9}{2}\) % p.a.,

T = 6 months

= \(\frac{6}{12}=\frac{1}{2}\) years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 45

Hence, the S.I. = ₹ 45.

Question 4.

A farmer borrowed ₹ 5300 at 8% p.a. At the end of 2 years, he cleared his account by paying ₹ 4600 and watch.

Find the cost of watch.

Solution:

Here, P = ₹ 5300,

R = 8% p.a.,

T = 2\(\frac{1}{2}\) years

= \(\frac{5}{2}\) years Z Z

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 1060

Amount = P + S.I.

= ₹ 5300 + ₹ 1060

= ₹ 6360

Cost of watch = ₹ 6360 – ₹ 4600 = ₹ 1760

Hence, the required cost = ₹ 1760.

Question 5.

Harish borrows ₹ 2500 at 3% p.a. and ₹ 1000 at 5% p.a. interest. What interest will Harish have to pay after 2 years?

Solution:

Case I:

Here, P = ₹ 2500,

R = 3% p.a.,

T = 2 years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 150

Case II:

Here, P = ₹ 1000,

R = 5% p.a.,

T = 2 years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 100

Total simple interest = ₹ 150 + ₹ 100 = ₹ 250

Hence, the required simple interest = ₹ 250.

Question 6.

Rohan deposited ₹ 5000 at 8% p.a. for 3 \(\frac{1}{2}\) years and Sunil deposited ₹ 5000 at 6% p.a. for 4 years. Who will get more interest? What amount will each get?

Solution;

Case I:

Here, P = ₹ 5000,

R = 8% p.a.,

T = 3 \(\frac{1}{2}\)

= \(\frac{7}{2}\) years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 1400

Case II:

Here, P = ₹ 5000,

R = 6% p.a.,

T = 4 years

∴ Simple interest, S.I. = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)

=

= ₹ 1200

Amount = P + S.I.

= ₹ 5000 + ₹ 1200

= ₹ 6200

∴ Rohan gets S.I. = ₹ 1400

and Sunil gets S.I. = ₹ 1200

∴ Rohan gets more S.I.

Amount got by Rohan = ₹ 6400

and by Sunil = ₹ 6200

So, Rohan gets more amount.