Download RS Aggarwal Solutions for Class 8 Maths Chapter 10- Exercise 10A, Profit and Loss from the links provided below. BYJUâ€™S expert team have solved the RS Aggarwal Solutions to ensure that the students are thorough with their fundamentals by practising the solutions. In Exercise 10A of RS Aggarwal Class 8 Maths, we shall see about the cost price, selling price, profit or gain and loss related questions.

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

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

**i 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%

**ii 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%

**iii 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%

**iv 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:**

**i 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

**ii CP = â‚¹915 and Gain = \(6\frac{2}{3}\)%**

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

**iii 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

**iv CP = â‚¹645 and Loss = \(13\frac{1}{3}\)%**

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:**

**i 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

**ii 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

**iii 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

**iv 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%

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

Exercise 10B Solutions 14 Questions

Exercise 10C Solutions 11 Questions

Exercise 10D Solutions 21 Questions

## RS Aggarwal Solutions for Class 8 Maths Chapter 10- Profit and Loss Exercise 10A

Exercise 10A of RS Aggarwal Class 8, Profit and Loss. This exercise mainly deals with the cost price- The amount for which an article is bought; selling price- The amount for which an article is sold; profit- when selling price is more than the cost price then its a gain; loss- when selling price is less than the cost price then its a loss.

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