The DAV Class 8 Maths Book Solutions Pdf and **DAV Class 8 Maths Chapter 5 Worksheet 2** Solutions of Profit, Loss and Discount offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 5 WS 2 Solutions

Question 1.

The marked price of a pant is ₹ 1250 and the shopkeeper allows a discount of 8% on it. Find the discount and the selling price of the pant.

Solution:

M.P. = ₹ 1250

Discount = 8% of M.P.

∴ Discount offered = \(\frac{8}{100}\) × 1250 = ₹ 100

∴ Selling price = M.P. – Discount

= ₹ 1250 – ₹ 100 = ₹ 1150

Hence, the selling price of the pant = ₹ 1150

and the discount = ₹ 100.

Question 2.

The marked price of a water cooler is ₹ 5400. The shopkeeper offers an off season discount of 20% on it. Find its selling price.

Solution:

M.P. = ₹ 5400

Discount = 20% of M.P.

= \(\frac{20}{100}\) × 5400 = ₹ 1080

∴ Selling price = M.P. – Discount

= ₹ 5400 – ₹ 1080 = ₹ 4320

Hence, the selling price = ₹ 4320.

Question 3.

An almirah of marked price ₹ 4000 is sold for ₹ 3700 after allowing certain discount. Find the rate of discount.

Solution:

M.P. = ₹ 4000

and S.P. = ₹ 3700

∴ Discount = M.P. – S.P.

= ₹ 4000 – ₹ 3700 = ₹ 300

∴ Rate of discount = \(\frac{\text { Discount }}{\text { M.P. }}\) × 100

= \(\frac{300}{4000}\) × 100

= \(\frac{15}{2}\) % or 7\(\frac{1}{2}\) %

Hence, the required rate of discount is 7\(\frac{1}{2}\)%.

Question 4.

Find the rate of discount being given on a ceiling fan whose selling price is ₹ 1175 after allowing a discount of ₹ 75 on its marked price.

Solution:

S.P. = ₹ 1175

and Discount = ₹ 75

∴ M.P. = S.P. + Discount

= ₹ 1175 + ₹ 75 = ₹ 1250

Discount = \(\frac{\text { Discount }}{\text { M.P. }}\) × 100

= \(\frac{75}{1250}\) × 100 = 6%

Hence, the required rate of discount = 6%.

Question 5.

Find the marked price of a washing machine which is sold at ₹ 8400 after allowing a discount of 16%.

Solution:

S.P. = ₹ 8400

and Discount = 16%

Let the M.P. be ₹ x.

∴ Discount = \(\frac{16}{100}\) × x

= ₹ \(\frac{4x}{25}\)

∴ S.P. = M.P. – Discount

= x – \(\frac{4x}{25}\)

= ₹ \(\frac{21 x}{25}\)

∴ \(\frac{21 x}{25}\) = 8400

⇒ x = 8400 × \(\frac{25}{21}\)

⇒ x = 400 × 25 = 10000

Hence, the marked price = ₹ 10,000.

Question 6.

A dinner set was bought for ₹ 2464 after getting a discount of 12% on its marked price. Find the marked price of the dinner set.

Solution:

C.P. = ₹ 2464

Discount = 12% of M.P.

Let the M.P. be ₹ x

∴ Discount = \(\frac{12}{100}\) × x

= ₹ \(\frac{3 x}{25}\)

∴ C.P. = M.P. – Discount

= x – \(\frac{3 x}{25}\)

= ₹ \(\frac{22 x}{25}\)

∴ \(\frac{22 x}{25}\) = 2464

⇒ x = \(\frac{2464 \times 25}{22}\)

⇒ x = 112 × 25 = 2800

Hence the marked price is ₹ 2800.

Question 7.

The marked price of a computer is ₹ 22,000. After allowing a 10% discount, a dealer still makes a profit of 20% . Find the cost price of a computer.

Solution:

M.P. = ₹ 22000

Discount = 10% of M.P.

= \(\frac{10}{100}\) × 22000 = ₹ 2200

∴ S.P. = M.P. – Discount

= ₹ 22000 – ₹ 2200 = ₹ 19800

Let C.P. be ₹ x.

Profit = 20% of C.P.

= \(\frac{20}{100}\) × x = ₹ \(\frac{x}{5}\)

∴ S.P. = C.P. + Profit

= x + ₹ \(\frac{x}{5}\) = ₹ \(\frac{6 x}{5}\)

∴ \(\frac{6 x}{5}\) = 19800

⇒ x = 19800 × \(\frac{5}{6}\)

⇒ x = 3300 × 5 = ₹ 16500

Hence, the cost price is ₹ 16500.

Question 8.

The marked price of a double bed is ₹ 9575. A shopkeeper allows a discount of 12% on its marked price and still gains 10%. Find the cost price of the double bed.

Solution:

M.P. = ₹ 9575

Discount = 12% of M.P.

= \(\frac{12}{100}\) × 9575 = ₹ 1149

∴ S.P. = M.P. – Discount

= ₹ 9575 – ₹ 1149 = ₹ 8426

Gain = 10%

∴ S.P. = C.P. (1 + \(\frac{\text { Gain }}{100}\))

⇒ 8426 = C.P. (1 + \(\frac{10}{100}\))

⇒ 8426 = C.P. (\(\frac{11}{10}\))

⇒ C.P. = \(\frac{8426 \times 10}{11}\)

= 766 × 10 = ₹ 7660

Hence the cost price of the double bed = ₹ 7660.

Question 9.

A dealer buys a bicycle for ₹ 1,250 and marks it at 40% above its cost price. If he allows 8% discount, find

(i) Selling price of the bicycle

(ii) Profit percentage

Solution:

(i) C.P. of bicycle = ₹ 1250

M.P. of bicycle = C.P. + 40% of C.P.

= 1250 + \(\frac{40}{100}\) × 1250

= 1250 + 500 = ₹ 1750

S.P. = M.P. (1 – \(\frac{\text { Discount% }}{100}\))

= 1750 (1 – \(\frac{8}{100}\))

= 1750 \(\left(\frac{100-8}{100}\right)\)

= 1750 \(\left(\frac{92}{100}\right)\)

= \(\frac{161000}{100}\) = ₹ 1610

(ii) Profit = S.P. – C.P.

= 1610 – 1250 = ₹ 360

Profit = \(\frac{\text { Profit }}{\text { C.P. }}\) × 100

= \(\frac{360 \times 100}{1250}\)

= \(\frac{144}{5}\) = 28.8%.

Question 10.

Priti allows 8% discount on the marked price of the suits and still makes a profit of 15%. If her gain over the sale of a suit is ₹ 156, find the marked price of the suit.

Solution:

Let M.P. be ₹ x.

∴ Discount = \(\frac{8}{100}\) × M.P.

= \(\frac{8}{100}\) × x

= \(\frac{2 x}{25}\)

∴ S.P. = M.P. – Discount

= x – \(\frac{2 x}{25}\)

= ₹ \(\frac{23 x}{25}\)

∴ S.P. = C.P. (1 + \(\frac{\text { Gain }}{100}\))

⇒ \(\frac{23 x}{25}\) = C.P. (1 + \(\frac{15}{100}\))

∴ \(\frac{23 x}{25}\) = C.P. × \(\frac{23}{20}\)

⇒ C.P. = \(\frac{23 x}{25} \times \frac{20}{23}=₹ \frac{4 x}{5}\)

∴ Gain = S.P. – C.P.

= \(\frac{23 x}{25}-\frac{4 x}{5}=₹ \frac{3 x}{25}\)

∴ \(\frac{3 x}{25}\) = 156

⇒ x = \(\frac{156 \times 25}{3}\)

⇒ 52 × 25 = ₹ 1300

Hence the marked price is ₹ 1300.