Profit and Loss problems
- 1. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is
Options- A. 500 kg
- B. 600 kg
- C. 400 kg
- D. 640 kg Discuss
Correct Answer: 600 kg
Explanation:
Let the sugar sold at 8% gain = P
? Sugar sold at 18% gain = (1000 - P)
Let CP of sugar = ? Q per kg
Total CP = ? 1000 x Q
? [(108/100) x PQ)] + [(118/100) x (1000 - P) x Q]
= (114/100) x 1000 Q
? 108PQ + 118000Q - 118PQ = 114000Q
? 10P = 4000
? P = 400
? Quantity sold at 18% profit = 1000 - 400 = 600 kg
- 2. A dealer buys an article marked at ? 25000 with 20% and 5% off. He spends ? 2000 on its repair and sells it for ? 25000. What is his gain or loss per cent?
Options- A. 21% loss
- B. 10.50% loss
- C. 19.05% gain
- D. 25% gain Discuss
Correct Answer: 19.05% gain
Explanation:
Total CP = (95% of 80% of ? 25000) + (2000)
= (95/100) x (80/100) x 25000 + 2000
= 19000 + 2000 = ? 21000
? CP = ? 21000 and SP = ? 25000
? Gain = 25000 - 21000 = ? 4000
? Gain % = (4000/21000) x 100% = 400/21%
= 19.05% (approx.)
- 3. Cost of a packet of coffee powder and a litre of milk are ? 20 and ? 30 respectively. 10 cups of coffee is made with one packet coffee powder and for each cup 200 ml of milk is used. If coffee is sold at 25% profit, the selling price of each cup of coffee is
Options- A. ? 12.50
- B. ? 6.25
- C. ? 8
- D. ? 10 Discuss
Correct Answer: ? 10
Explanation:
Cost of coffee powder used in one cup = 20/10 = ? 2
Cost of milk used in one cup = (30/1000) x 200 = ? 6
? Cost of each cup coffee = 2 + 6 = ? 8
To gain 25% profit, sale price of each cup of coffee = 125% of 8 = ? 10
- 4. A shopkeeper sells transistors at 15% above its cost price. If he had bought it at 5% more than what he paid for it and sold it for ? 6 more, he would have gained 10%. The cost price of the transistor is?
Options- A. ? 800
- B. ? 1000
- C. ? 1200
- D. ? 1400 Discuss
Correct Answer: ? 1200
Explanation:
Let cost price of transistor = ? N
According to the question,
CP of transistor = (N x 105)/100
And SP of transistor = (115 x N)/100 + 6
? Profit percentage = (SP - CP)/CP x 100
? 10 = [[(115N/100) + 6 - (105N/100)] / (105N/100)] x 100
? 10 = [(10N + 600) x 100]/ 105N
? 105N = 100N + 6000
? 5N = 6000
? N= ? 1200
- 5. A merchant earns a profit of 20% by selling a basket containing 80 apples whose cost is ? 240 but he gives one-fourth of it to his friend at cost price and sells the remaining apples. In order to earn the same profit, at what price must he sell each apple?
Options- A. ? 3.00
- B. ? 3.60
- C. ? 3.80
- D. ? 4.80 Discuss
Correct Answer: ? 3.80
Explanation:
? CP of 80 apples = ? 240
? CP of 1 apples = ? 3
? CP of 20 apples = ? 60
To earn a profit of 20%
SP = 120% of 240 = ? 288
But he sales 1/4 of his apples i.e., 20 apples for ? 60
? SP of remaining 60 apples = 288 - 60 = ? 228
? SP of 1 apple = 228/60 = ? 3.80
- 6. A bookseller sells a book at a gain of 10%. if he had bought it at 4% less and sold it for ? 6 more, he would have gained 118 3/ 4%. The CP of the book is?
Options- A. ? 130
- B. ? 140
- C. ? 150
- D. ? 160 Discuss
Correct Answer: ? 150
Explanation:
Let CP = N
Then, SP = 110N / 100 = ? 11N/10
Now, CP = 96% of N = 96N/100 = ? 24N/25
Now, according to the question,
SP = ? (11N/10) + 6
? (11N/10) + 6 = 1183/4% of 24N/25
? (11N + 60)/10 = (475/400) x (24N/25) = 57N/50
? 550N + 3000 = 570N
? 20N = 3000
? N = 3000/20 = 150
? CP = ? 150
- 7. Assuming that profit of a shopkeeper in a particular commodity is a linear expression of transportation charge (t) and the quantity of commodity (q) . He earns a profit of ? 10000 by selling 20 units at the transport charge of ? 400. He also earn a profit of ? 12000 by selling 25 units at the transport charge of ? 600. What is the linear expression in t and q?
Options- A. 600q - 5t
- B. 500q - 4t
- C. 600q - 4t
- D. 500q - 5t Discuss
Correct Answer: 500q - 5t
Explanation:
By hit and trial ,
Case I t = ? 400 and q = 20 unit
Then, from option (a),
Total profit = 600q - 5t
= 600 x 20 - 5 x 400
= 12000 - 2000 = ? 10000
Case II t ? 600 and q = 25 units
Total profit = 600q - 5t
= 600 x 25 - 5 x 600
= 600(25 - 5)
= 600 x 20
= ? 12000
- 8. A person sold a table at a gain of 15%. Had he bought it for 25% less and sold it for ? 60 less, he would have made a profit of 32%. The cost price of table was
Options- A. ? 300
- B. ? 350
- C. ? 375
- D. ? 400 Discuss
Correct Answer: ? 375
Explanation:
Let the cost price of table = ? N
Then, selling price with 15% gain
= (100 + Gain%) x CP/100
= (100 + 15)% x CP/100
= (115 x N)/100 = ? 115N/100
Now CP = [(100 - 25%) x CP]/100 = ? 75N/100
New SP = ? (115N/100) - 60
Now, according to the question,
[[(115/100 )- 60) - (75N/100)]/ (75N/100)] x 100 = 32
? [[(115N - 6000 - 75N)/100 ]/ (75N/100)] x 100 = 32
? [(40N - 6000)/75N] x 100 =32
? (40N - 6000)/(3N x 4) = 32
? 160N - 24000 = 96N
? 160N - 96N = 24000
? 64N = 24000
? N = 24000/64
? N = ? 375
? The cost price of table is ? 375.
- 9. A person bought two bicycles for ? 1600 and sold the first at 10% profit and the second at 20% profit. If he sold the first at 20% profit and the second at 10% profit, he would get ? 5 more. The difference in the cost price of the two bicycles was
Options- A. ? 25
- B. ? 75
- C. ? 50
- D. ? 40 Discuss
Correct Answer: ? 50
Explanation:
Let the cost price of first bicycle be ? P.
Then, the cost price of second bicycle = ? (1600 - P)
According to the given condition,
[20% of P + 10% of (1600 - P )] - [10% of P + 20% of (1600 - P)] = 5
? [(20P/100) + [10 x (1600 - P)]/100] - [(10P/100) + {20 x (1600 - P)}/100] = 5
? (P/5 + (1600 - P)/10) - (P/10 + (1600 - P)/5) = 5
? P/5 - P/10 + (1600 - P)/10 - (1600 - P)/5 = 5
? 2P = 1600 + 50
? P = 1650/2 = 825
? Cost of second bicycle = (1600 - 825) = ? 775
? Required difference = 825 - 775 = ? 50
- 10. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
Options- A. No profit, no loss
- B. 5%
- C. 8%
- D. 10% Discuss
Correct Answer: 5%
Explanation:
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
Gain =(80/1600*100) % = 5%
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