RS Aggarwal Solutions for Class 8 Maths Chapter 10- Profit and Loss, are provided here. Our expert faculty team has prepared solutions in order to help you with your exam preparation to acquire good marks in Maths. RS Aggarwal Solutions book for Class 8 Maths comes in very handy at this point. Our solution module utilizes various shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. If you wish to secure an excellent score, solving RS Aggarwal Class 8 Solutions is an utmost necessity. These solutions will help you in gaining knowledge and strong command over the subject. Practicing the textbook questions will help you in analyzing your level of preparation and understanding of the concept.

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## Exercise 10A

**Q1.Find the gain or loss percent when:**

**CP = â‚¹620 and SP = â‚¹713**Solution: we know that SP is more than CP so itâ€™s a gain.As the formula states Gain = SP â€“ CP

713 â€“ 620 = 93

Gain% = (Gain Ã— 100) / CP

= (93 Ã— 100) / 620

= 15%

**CP = â‚¹675 and SP = â‚¹630**Solution: we know that SP is less than CP so itâ€™s a loss.As the formula states Loss = CP â€“ SP

675 â€“ 630 = 45

Loss% = (Loss Ã— 100) / CP

= (45 Ã— 100) / 675

= 6.66%

**CP = â‚¹345 and SP = â‚¹372.60**Solution: we know that SP is more than CP so itâ€™s a gain.As the formula states Gain = SP â€“ CP

372.60 â€“ 345 = 27.60

Gain% = (Gain Ã— 100) / CP

= (27.60 Ã— 100) / 345

= 8%

**CP = â‚¹80 and SP = â‚¹76.80**Solution: we know that SP is less than CP so itâ€™s a loss.As the formula states Loss = CP â€“ SP

80 â€“ 76.80 = 3.20

Loss% = (Loss Ã— 100) / CP

= (3.20 Ã— 100) / 80

= 4%

**Q2.Find the selling price when:**

**CP = â‚¹1650 and Gain = 4%**Solution: As the formula states SP = ((100 + Gain %) /100) Ã— CP= ((100+4)/100) Ã— 1650

= (104/100) Ã— 1650

=1716

âˆ´ The selling price is 1716

**CP = â‚¹915 and Gain = 6 2/3%**Solution: As the formula states SP = ((100 + Gain %) /100) Ã— CP= ((100+(20/3))/100) Ã— 915

= ((320/3)/100) Ã— 915

=976

âˆ´ The selling price is 976

**CP = â‚¹875 and Loss = 12%**Solution: As the formula states SP = ((100 – Loss %) /100) Ã— CP= ((100 – 12)/100) Ã— 875

= (88/100) Ã— 875

=770

âˆ´ The selling price is 770

**CP = â‚¹645 and Loss = 13 1/3**%Solution: As the formula states SP = ((100 – Loss %) /100) Ã— CP= ((100-(40/3))/100) Ã— 645

= ((260/3)/100) Ã— 645

= (260/300) Ã— 645

=559

âˆ´ The selling price is 559

**Q3.Find the cost price when:**

**SP = â‚¹1596 and Gain = 12%**Solution: As the formula states CP = (100 / (100+ Gain %)) Ã— SP= (100/ (100+12)) Ã— 1596

= (100/112) Ã— 1596

=1425

âˆ´ The cost price is 1425

**SP = â‚¹2431 and Loss = 6 Â½ %**Solution: As the formula states CP = (100 / (100 – Loss %)) Ã— SP= (100/ (100-(13/2))) Ã— 2431

= (100/ (187/2)) Ã— 2431

= (200/187) Ã— 2431

=2600

âˆ´ The cost price is 2600

**SP = â‚¹657.60 and Loss = 4%**Solution: As the formula states CP = (100 / (100 – Loss %)) Ã— SP= (100/ (100-4)) Ã— 657.60

= (100/96) Ã— 657.60

=658

âˆ´ the cost price is 658

**SP = â‚¹34.40 and Gain = 7 Â½ %**Solution: As the formula states CP = (100 / (100+ Gain %)) Ã— SP= (100/ (100+ (15/2))) Ã— 34.40

= (100/ (215/2)) Ã— 34.40

= (200/215) Ã— 34.40

=32

âˆ´ the cost price is 32

**Q4. Manjit bought an iron safe for â‚¹12160 and paid â‚¹340 for its transportation. Then, he sold it for â‚¹12875. Find his gain percent.**

Solution: Lets solve by using, the total cost of iron safe = purchase cost + transportation

= 12160 + 340 = 12500

CP of iron safe = 12500

SP of iron safe = 12875

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

Gain = SP â€“ CP = 12875 â€“ 12500 = 375

Gain % = (Gain Ã— 100) / CP

= (375 Ã— 100) / 12500

= 3%

âˆ´ The Gain percent is 3%

**Q5. Robin purchased an old car for â‚¹73500. He spent â‚¹10300 on repairs and paid â‚¹2600 for its insurance. Then he sold it to a mechanic for â‚¹84240. What was his percentage gain or loss? Hint: overheads =â‚¹10300+â‚¹2600= â‚¹12900**

Solution: We know that,

Actual price of old car= purchase price + overheads

=73500 + 12900

=86400

âˆ´ the CP =â‚¹86400

SP = â‚¹84240

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

Loss = CP â€“ SP

=86400 â€“ 84240

= 2160

Loss% = (Loss Ã— 100) / CP

= (2160 Ã— 100) / 86400

= 2.5%

âˆ´ The Loss percent is 2.5%

**Q6.Hari bought 20kg of rice at â‚¹36 per kg and 25kg of rice at â‚¹32 per kg. He mixed the two varieties and sold the mixture at â‚¹38 per kg. Find his gain percent in the whole transaction.**

Solution: We know that,

Total weight of rice = 20 + 25 = 45

So, total cost of both varieties = (20Ã—36) + (25Ã—32) = 720 + 800 = 1520

âˆ´ The CP =â‚¹1520

SP = weight Ã— Rate = 45Ã—38 = 1710

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

Gain = SP â€“ CP = 1710 â€“ 1520 = â‚¹190

Gain % = (Gain Ã— 100) / CP

= (190 Ã— 100) / 1520

= 12.5%

âˆ´ The Gain percent is 12.5%

**Q7. Coffee costing â‚¹250 per kg was mixed with chicory costing â‚¹75 per kg in the ratio 5:2 for a certain blend. If the mixture was sold at â‚¹230 per kg, find the gain or loss percent. Hint: let 5kg of coffee be mixed with 2kg of chicory**

Solution: Let us consider x as the common multiple

Cost of 5kg of coffee= 5x = 5 Ã— 250 = â‚¹1250

Cost of 2kg of coffee= 2x = 2 Ã— 75 = â‚¹150

âˆ´ the cost of the mixture is 5x + 2x = 1250 + 150

7x = 1400

x = 1400/7 = 200

So, CP of mixture = 200

SP of mixture = 230

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

Gain = SP â€“ CP = 230 â€“ 200 = â‚¹30

Gain % = (Gain Ã— 100) / CP

= (30 Ã— 100) / 200

= 15%

âˆ´ The Gain percent is 15%

**Q8. If the selling price of 16 water bottles is equal to the cost price of 17 water bottles, find the gain percent earned by the dealer.**

Solution: Let us consider the CP of 17 water bottles = â‚¹100

âˆ´ The CP of 17 water bottles = SP of 16 water bottles = â‚¹100

SP of 17 water bottles = (100/16) Ã— 17 = â‚¹106.25

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

Gain = SP â€“ CP = 106.25 â€“ 100 = â‚¹6.25

Gain % = (Gain Ã— 100) / CP

= (6.25 Ã— 100) / 100

= 6.25%

âˆ´ The Gain percent is 6.25%

**Q9. The cost price of 12 candles is equal to the selling price of 15 candles. Find the loss percent.**

Solution: Let us consider the SP of 15 candles = â‚¹100

âˆ´ The CP of 12 candles = SP of 15 candles = â‚¹100

CP of 15 candles = (100/12) Ã— 15 = â‚¹125

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

Loss = CP â€“ SP = 125 â€“ 100 = â‚¹25

Loss % = (loss Ã— 100) / CP

= (25 Ã— 100) / 125

= 20%

âˆ´ The Loss percent is 20%

**Q10. By selling 130 cassettees, a man gains an amount equal to the selling price of 5 cassettes. Find the gain percent.**

Solution: Let us consider the cassettes price as x

The SP of 5 cassettes = 5x

SP of 130 cassettes = 130x

âˆ´ The CP of 130 cassettes = 130x â€“ 5x = 125x

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

Gain = SP â€“ CP = 130x â€“ 125x = 5x

Gain % = (Gain Ã— 100) / CP

= (5x Ã— 100) / 125x

=500x/125x

= 4%

âˆ´ The Gain percent is 4%

**Q11. By selling 45 lemons, a vendor loses a sum equal to the selling price of 3 lemons. Find the loss percent.**

Solution: Let us consider the lemons price as x

The SP of 3 lemons = 3x

SP of 45 lemons = 45x

âˆ´ The CP of 45 lemons = 45x + 3x = 48x

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

Loss = CP â€“ SP = 48x â€“ 45x = 3x

Loss % = (loss Ã— 100) / CP

= (3x Ã— 100) / 48x

=300x/48x

= 6.25%

âˆ´ The Loss percent is 6.25%

**Q12.Oranges are bought at 6 for â‚¹20 and sold at 4 for â‚¹18. Find the gain or loss percent.**

Solution: We know that the CP of 6 oranges = â‚¹20

âˆ´ The CP of 1 orange = â‚¹20/6

The SP of 4 oranges = â‚¹18

The SP of 1 orange = â‚¹18/4

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

Gain = SP â€“ CP = 18/4 â€“ 20/6

By taking LCM of 4 and 6 we get, (54-40)/12 = 7/6

Gain % = ((7/6) Ã— 100) / (20/6)

= (700/6) / (20/6)

=35%

âˆ´ The Gain percent is 35%

**Q13. A vendor purchased bananas at â‚¹40 per dozen and sold them at 10 for â‚¹36. Find the gain or loss percent. Hint: let the number of bananas bought be LCM (12, 10) = 60**

Solution: We know that the SP of 10 bananas = â‚¹36

SP of 1 banana = â‚¹36/10 = â‚¹3.6

SP of 1 dozen bananas = â‚¹3.6 Ã— 12 = â‚¹43.20

CP of 1 dozen bananas = â‚¹40

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

Gain = SP â€“ CP = 43.20 â€“ 40 = â‚¹3.2

Gain % = (3.2 Ã— 100) / 40

=8%

âˆ´ The Gain percent is 8%

**Q14. A man bought apples at 10 for â‚¹75 and sold them at â‚¹75 per dozen. Find his loss percent.**

Solution: We know that the CP of 10 apples = â‚¹75

CP of 1 apple = â‚¹75/10 = â‚¹7.5

CP of 1 dozen apples = â‚¹7.5 Ã— 12 = â‚¹90

SP of 1 dozen bananas = â‚¹75

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

Loss = CP â€“ SP = 90 â€“ 75 = 15

Loss % = (loss Ã— 100) / CP

= (15 Ã— 100) / 90

= 16.66%

âˆ´ The Loss percent is 16.66%

**Q15. A man purchased some eggs at 3 for â‚¹16 and sold them at 5 for â‚¹36. Thus he gained â‚¹168 in all. How many eggs did he purchase? Hint: let the number of eggs bought be x. then, CP = â‚¹ ((16/3) Ã— x) = â‚¹ (16x/3) and SP = â‚¹ ((36/5) Ã— x) = â‚¹ (36x/5).**

Solution: Let us consider the number of eggs as x

The CP of 10 eggs = â‚¹16x

CP of 1 egg = â‚¹16x/3

SP of an egg = â‚¹36x/5

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

Gain = SP â€“ CP = (36x/5) â€“ (16x/3) = 168

By taking the LCM 5 and 3 we get, (108x-80x)/15 = 168

âˆ´ 28x =168 Ã— 15

x = (2520) /28

x = 90

âˆ´ the number of eggs is 90.

**Q16. A dealer sold a camera for â‚¹1080 gaining 1/8 of its cost price. Find (i) the cost price of the camera, and (ii) the gain percent earned by the dealer.**

**Hint: let CP = â‚¹x. then, gain = â‚¹x/8. Therefore, SP= â‚¹ (x + (x/8)) = â‚¹9x/8**

Solution:

(i) Let us consider the cost price of camera be x

SP of camera = x + 1x/8 =1080

x + x/8 = 1080

9x/8 = 1080

9x = 1080 Ã— 8

x = 960

âˆ´ The CP of the camera = â‚¹960

(ii) Gain = SP â€“ CP = 1080 – 960 = 120

Gain % = (120 Ã— 100) / 960

=12.5%

âˆ´ The Gain percent is 12.5%

**Q17. Meenakshi sells a pen for â‚¹54 and loss 1/10 of her outlay. Find (i) the cost cost price of the pen, and (ii) the loss percent**

Solution: (i) Let us consider the cost price of pen be x

SP of the pen = x – 1x/10 =54

x – x/10 = 54

9x/10 = 54

9x = 54 Ã— 10

x = 60

âˆ´ the CP of the camera = â‚¹60

(ii) Loss = CP â€“ SP = 60 – 54 = 6

Loss % = (6 Ã— 100) / 60

=10%

âˆ´ The Loss percent is 10%

## 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%

## Exercise 10C

**Q1. The list price of a refrigerator is â‚¹14650. If 6% is charged as sales tax, find the cost of the refrigerator.**

Solution: we know that the list price =â‚¹14650

Sales tax = 6%

So, Sales tax amount = 6% of â‚¹14650

= 6% Ã— â‚¹14650

= â‚¹879

âˆ´ Final price of the refrigerator = list price + sales tax

= 14650 + 879

= â‚¹15529

**Q2. Reena bought the following articles from a general store:**

**i. 1 tie costing â‚¹250 with ST @ 6%**

**ii. Medicines costing â‚¹625 with ST @ 4%**

**iii. Cosmetics costing â‚¹430 with ST @ 10%**

**iv. Clothes costing â‚¹1175 with ST @ 8%
calculate the total amount to be paid by Reena**

Solution: we know that the cost of a tie = â‚¹250, ST on tie = 6%

ST amount on tie = 6% of â‚¹250

= â‚¹15

âˆ´ Final cost of tie = 250 + 15 = â‚¹265

We know that the cost of medicines = â‚¹625, ST on medicines = 4%

ST amount on medicines = 4% of â‚¹625

= â‚¹25

âˆ´ Final cost of medicines = 625 + 25 = â‚¹650

We know that the cost of cosmetics = â‚¹430, ST on medicines = 10%

ST amount on cosmetics = 10% of â‚¹430

= â‚¹43

âˆ´ Final cost of cosmetics = 430 + 43 = â‚¹473

We know that the cost of clothes = â‚¹1175, ST on medicines = 8%

ST amount on clothes = 8% of â‚¹1175

= â‚¹94

âˆ´ Final cost of clothes = 1175 + 94 = â‚¹1269

So, the total amount to be paid by reena is â‚¹265 + â‚¹650 + â‚¹473 + â‚¹1269 = â‚¹2657

**Q3. Tanvy bought a watch for **â‚¹**1980 including VAT at 10%. Find the original price of the watch.**

Solution: we know that the VAT = 10%

SP = â‚¹1980

So letâ€™s consider the original price of watch as x

âˆ´ VAT amount = 10% of x

= x/10

âˆ´ x + x/10 = 1980

(10x + x)/10 = 1980

11x /10 = 1980

x = (1980 Ã— 10) / 11

x = 1800

âˆ´ the original price of the watch excluding VAT is â‚¹1800

**Q4. Mohit bought a shirt for **â‚¹**1337.50 including VAT at 7%. Find the original price of the shirt.**

Solution: we know that the VAT = 7%

SP = â‚¹1337.50

So letâ€™s consider the original price of shirt as x

âˆ´ VAT amount = 7% of x

= 7x/100

âˆ´ x + 7x/100 = 1337.50

(100x + 7x)/100 = 1337.50

107x /100 = 1337.50

x = (1337.50 Ã— 100) / 107

x = 1250

âˆ´ the original price of the shirt excluding VAT is â‚¹1250

**Q5. Karuna bought 10 g of gold for **â‚¹**15756 including VAT at 1%. What is the rate of gold per 10 g?**

Solution: we know that the VAT = 1%

SP = â‚¹15756

So letâ€™s consider the original price of shirt as x

âˆ´ VAT amount = 1% of x

= 1x/100

âˆ´ x + x/100 = 15756

(100x + x)/100 = 15756

101x /100 = 15756

x = (15756 Ã— 100) / 101

x = 15600

âˆ´ the original price of the 10g Gold excluding VAT is â‚¹15600

## Exercise 10D

**Select the correct answer in each of the following:**

**1. Rajan buys a toy for â‚¹75 and sells it for â‚¹100. His gain percent is**

**25%****20%****33 1/3%****37 Â½%**

Solution: we know that the CP = 75

SP = 100

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

Gain = SP â€“ CP

= 100 â€“ 75

= 25

âˆ´ Gain% = (Gain Ã— 100) /CP

= (25 Ã— 100) / 75

= 33.33%

**2. A bat is bought for â‚¹120 and sold for â‚¹105. The loss percent is**

**15%****12 Â½%****16 2/3%****14 1/5%**

Solution: we know that the CP = â‚¹120

SP = â‚¹105

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

Loss = CP â€“ SP

= 120 â€“ 105

= 15

âˆ´ Loss% = (loss Ã— 100) /CP

= (15 Ã— 100) / 120

= 12.5%

**3. A bookseller sells a book for â‚¹100, gaining â‚¹20. His gain percent is**

**20**%**25**%**22**%**None of these**

Solution: we know that the SP = â‚¹100

Gain = â‚¹20

By using the formula CP = SP â€“ Gain

= 100 â€“ 20

= 80

âˆ´ Gain% = (Gain Ã— 100) /CP

= (20 Ã— 100) / 80

= 25%

**4. On selling an article for â‚¹48, a shopkeeper loses 20%. In order to gain 20%, what would be the selling price?**

**â‚¹52****â‚¹56****â‚¹68****â‚¹72**

Solution: we know that the SP = â‚¹48

Loss percent = 20%

âˆ´ We calculate, CP = (100 / (100 – Loss %)) Ã— SP

= (100 / 100-20) Ã— 48

= 100/80 Ã— 48

=â‚¹60

We calculate, SP = ((100 + Gain %) /100) Ã— CP

= ((100+20) /100) Ã— 60

= 120/100 Ã— 60

=â‚¹72

**5. On selling an article at a certain price a man gains 10%. On selling the same article at double the price, gain percent is**

**20**%**100**%**120**%**140**%

Solution: let us consider the CP be â‚¹100

Gain = 10%

We calculate, SP = ((100 + Gain %) /100) Ã— CP

= ((100+10) /100) Ã— 100

= 110/100 Ã— 100

=â‚¹110

Now, double the SP = 110 Ã— 2 = â‚¹220

Now calculating Gain = SP â€“ CP

= 220 â€“ 100 = â‚¹120

âˆ´ Gain% = (Gain Ã— 100) /CP

= (120 Ã— 100) / 100

= 120%

**6. Bananas are bought at 3 for â‚¹2 and sold at 2 for â‚¹3. The gain percent is**

**25%****50%****75%****125%**

Solution: we know that the CP for 3 bananas = â‚¹2

CP for 1 banana = â‚¹2/3

SP for 2 bananas = â‚¹3

SP for 1 banana = â‚¹3/2

Since SP is more than CP its Gain

Gain = SP â€“ CP

= 3/2 â€“ 2/3

= 5/6

âˆ´ Gain% = (Gain Ã— 100) /CP

= ((5/6) Ã— 100) / (2/3)

= 5/4 Ã— 100

= 125%

**7. If the selling price of 10 pens is the same as the cost price of 12 pens then gain percent is**

**2**%**12**%**20**%**25**%

Solution: let us consider the CP as x

SP of 1 pen = x/10

CP of 1 pen = x/12

Since SP is more than CP its Gain

Gain = SP â€“ CP

= x/10 â€“ x/12

= x/60

âˆ´ Gain% = (Gain Ã— 100) /CP

= ((x/60) Ã— 100) / (x/12)

= 20%

**8. On selling 100 pencils a man gains the selling price of 20 pencils. His gain percent is **

**20%****25%****22 Â½ %****16 2/3 %**

Solution: let us consider the CP of a pencil be x

SP of 100 pencils = 100x

Gain of 20 pencils = 20x

âˆ´ we calculate CP = SP â€“ Gain

= 100x â€“ 20x

= 80x

âˆ´ Gain% = (Gain Ã— 100) /CP

= ((20x) Ã— 100) / 80x

= 25%

**9. Ravi buys some toffees at 5 for a rupee and sells them at 2 for a rupee. His gain percent is**

**30%****40%****50%****150%**

Solution: we know that the CP of 1 toffee = â‚¹1/5

SP of 1 toffee = â‚¹1/2

Since SP is more than CP itâ€™s a Gain

Gain = SP â€“ CP

= Â½ – 1/5

= 3/10

âˆ´ Gain% = (Gain Ã— 100) /CP

= ((3/10) Ã— 100) / (1/5)

= 150%

**10. Oranges are bought at 5 for â‚¹10 and sold at 6 for â‚¹15. His gain percent is**

**50%****40%****30%****25%**

Solution: we know that the CP of 1 orange = â‚¹10/5 = â‚¹2

SP of 1 orange = â‚¹15/6 = â‚¹2.5

Since SP is more than CP itâ€™s a Gain

Gain = SP â€“ CP

=2.5 – 2

= 0.5

âˆ´ Gain% = (Gain Ã— 100) /CP

= ((0.5) Ã— 100) / 2

= 25%

## RS Aggarwal Solutions for Class 8 Maths Chapter 10 – Profit and Loss

Chapter 10- Profit and Loss contain four exercises and the RS Aggarwal solutions present on this page provide the solutions for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Calculating the Cost price.
- Calculating the Selling price.
- Calculating the Profit or Gain ï¼….
- Calculating the Loss or Lossï¼….
- Calculating the Discount.
- Calculating the Sales Tax or Value added Tax(VAT).

### Also, Access RS Aggarwal Solutions for Class 8 Chapter 10 Exercises

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 10 – Profit and Loss

RS Aggarwal Solutions for Class 8 Maths Chapter 10 – Profit and Loss, ensures that the students are thorough and familiar with the concepts. Regular revision of important concepts and formulas over time to time is the best way to strengthen your ability to solve the problems. The solutions are designed in such a way that they are easy to understand and solve.

As the Chapter is about Profit and Loss, now let us understand the concept of profit and loss in our daily life, as and when we purchase goods we generally use the statements such as cost price, selling price, profit, loss, and discount. These statements are financial statements that summarise the revenues, cost and expenses incurred during a specified period, usually a fiscal quarter or a year. Here, in the RS Aggarwal Solutions book, many such exercise problems are given which enhances familiarity with the concepts.