Students can refer and download RS Aggarwal Solutions for Class 8 Maths Chapter 10- Exercise 10B, Profit and Loss from the links provided below. Our experts have solved the RS Aggarwal Solutions to ensure that the students are thorough with their basic concepts by practising the solutions. In Exercise 10B of RS Aggarwal Class 8 Maths, we shall see more about discount related questions.

**Download PDF of RS Aggarwal Solutions for Class 8 Maths Chapter 10- Exercise 10B**

### Access Answers to RS Aggarwal Solutions for Class 8 Maths Chapter 10- Profit and Loss Exercise 10B

**Q1. The marked price of a water cooler is â‚¹4650. The shopkeeper offers an off-season discount of 18**%** on it. Find its selling price.**

Solution: we know that the marked price = â‚¹4650

The discount = 18%

And the discount in amount = 18% of the marked price

= (18/100) Ã— 4650

= â‚¹ 837

By using the formula SP = marked price – discount

= 4650 â€“ 837 = â‚¹ 3813

**Q2. The price of a sweater was slashed from â‚¹960 to â‚¹816 by a shopkeeper in the winter season. Find the rate of discount given by him.**

Solution: we know that the marked price = â‚¹960

The SP = â‚¹816

By using the formula Discount = marked price – SP

= 960 â€“ 816 = â‚¹144

âˆ´ The Discount% = (Discount/market price) Ã— 100

= (144/960) Ã— 100 = 15%

**Q3. Find the rate of discount being given on a shirt whose selling price is â‚¹1092 after deducting a discount of â‚¹208 on its marked price. Hint. MP = (SP) + (discount).**

Solution: we know that the SP = â‚¹1092

The Discount = â‚¹208

By using the formula market price = SP + Discount

= 1092 + 208 = â‚¹1300

âˆ´ The Discount% = (Discount/market price) Ã— 100

= (208/1300) Ã— 100 = 16%

**Q4. After allowing a discount of 8% on a toy, it is sold for â‚¹216.20. Find the marked price of the toy.**

Solution: we know that the SP = â‚¹216.20

The Discount = 8%

Now letâ€™s consider the market price as x

By using the formula market price â€“ Discount = SP

x â€“ (x Ã— 8/100 ) = SP

(100x â€“ 8x/100) = 216.20

92x/100 = 216.20

x = (216.20 Ã— 100) / 92

x = 235

âˆ´ The Market price of a toy = â‚¹235

**Q5. A tea set was bought for **â‚¹**528 after getting a discount of 12% on its marked price. Find the marked price of the tea set. **

Solution: we know that the SP = â‚¹528

The Discount = 12%

Now letâ€™s consider the market price as x

By using the formula market price â€“ Discount = SP

x â€“ (x Ã— 12/100) = SP

(100x â€“ 12x/100) = 528

88x/100 = 528

x = (528 Ã— 100) / 88

x = 600

âˆ´ The Market price of a tea set = â‚¹600

**Q6. A dealer marks his goods at 35% above the cost price and allows a discount of 20% on the marked price. Find his gain or loss per cent.**

Solution: let us consider the CP of goods as x

Market price of the goods when goods marked above the 35% CP is

Market price = x + (35x/100) = 135x/100

So the Discount = 20%

Discount amount = 20% of 135x/100 = 27x/100

âˆ´ The SP = Market price â€“ Discount

= (135x/100) â€“ (27x/100) = 108x/100 = 1.08x

Since, SP is more than CP itâ€™s a gain

We know that, Gain = SP â€“ CP

= 1.08x â€“ x = 0.08x

âˆ´ Gain % = (Gain Ã— 100)/ CP

= (0.08x Ã— 100)/ x

= 8%

**Q7. A cellphone was marked at 40% above the cost price and a discount of 30% was given on its marked price. Find the gain or loss per cent made by the shopkeeper.**

Solution: let us consider the CP of goods as x

Market price of the goods when goods marked above the 40% CP is

Market price = x + (40x/100) = 140x/100 = 1.4x

So the Discount = 30%

Discount amount = 30% of 1.40x = 0.42x

âˆ´ The SP = Market price â€“ Discount

= 1.4x â€“ 0.42x= 0.98x

Since, SP is less than CP itâ€™s a loss

We know that, Loss = CP â€“ SP

= x â€“ 0.98x = 0.02x

âˆ´ Loss % = (Loss Ã— 100)/ CP

= (0.02x Ã— 100)/ x

= 2%

### Access other Exercises of RS Aggarwal Solutions for Class 8 Maths Chapter 10- Profit and Loss

Exercise 10A Solutions 34 Questions

Exercise 10C Solutions 11 Questions

Exercise 10D Solutions 21 Questions

## RS Aggarwal Solutions for Class 8 Maths Chapter 10- Exercise 10B

Exercise 10B of RS Aggarwal Class 8, Profit and Loss. This exercise mainly deals with discount- In order to increase sales or clear the old stock, sometimes the shopkeepers offer a certain percentage of rebate on the marked price. This rebate is known as a discount.

Students are suggested to practice the problems on a regular basis which will help them excel in their exams and increase their overall percentage. Practicing as many times as possible helps in building time management skills and also boosts the confidence level to achieve high marks.