Profit Loss Discount and Tax MCQs Class 8 RS Aggarwal Exe-7E Goyal Brothers ICSE Maths Solutions Ch-7. We provide step by step Solutions of council prescribe textbook / publications to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Profit Loss Discount and Tax MCQs Class 8 RS Aggarwal Exe-7E Goyal Brothers ICSE Maths Solutions
| Board | ICSE |
| Publications | Goyal Brothers Prakashan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-7 | Profit Loss Discount and Tax |
| Exe-7E | Multiple Choice Questions |
| Edition | 2024-2025 |
MCQs on Profit Loss Discount and Tax with Answer
The MCQs problems on Profit Loss Discount and Tax is very common now days not only in academic exam but also for various one day exam of NEET and JEE. Hence various type questions with solutions has been given for your complete practice.
Exercise- 7E
(Profit Loss Discount and Tax MCQs Class 8 RS Aggarwal Exe-7E Goyal Brothers ICSE Maths Solutions Ch-7)
Multiple Choice Questions :
Que-1: By selling an article for Rs100, one gains Rs10. The gain per cent is :
(a) 9% (b) 10% (c) 11% (d) 11*(1/9)%
Sol: (d) 11*(1/9)%
Reason: Selling price (S.P) of the article = Rs.100
Profit = Rs.10
Cost price of the article (C.P) will be = Rs.100-Rs.10
= Rs.90
Profit percent = (Profit / Cost Price) × 100
= (10/90) × 100
= 100/9
= 11*(1/9)%.
Que-2: By selling an article for Rs100, one loss Rs10. The loss per cent is:
(a) 9% (b) 9*(1/11)% (c) 10% (d) 11*(1/9)%
Sol: (b) 9*(1/11)%
Reason: Selling price (S.P) of the article = Rs.100
Profit = Rs.10
Cost price of the article (C.P) will be = Rs.100+Rs.10
= Rs.110
Loss percent = (Loss/Cost Price) × 100
= (10/110) × 100
= 100/11
= 9*(1/11)%.
Que-3: A man sold his cow for Rs7920 and gained 10%. The cow was bought for
(a) Rs7000 (b) Rs7200 (c) Rs7128 (d) Rs7840
Sol: (b) Rs7200
Reason: A man sold his cow for rs 7920 and gained 10%
Profit amount will be :
= Purchasing price × 10%
= x × 10%
= x × 10/100
= Rs. x/10
Selling price will be :
= Purchasing price + Profit amount
= x + (x/10)
= (10x + x)/10
= Rs. 11x/10
According to the data mentioned in the question,
11x/10 = 7920
x = 7920 × 10/11
x = 7200
Que-4: A watch is sold for Rs1080 at a loss of 10%. The cost price of the watch is
(a) Rs1180 (b) Rs1200 (c) Rs1188 (d) Rs1125
Sol: (b) Rs1200
Reason: Given information:
– Selling Price = ₹1080
– Loss = 10% of the Cost Price
Loss = 10% of the Cost Price
Loss = 0.10 × Cost Price
₹1080 = Cost Price – 0.10 × Cost Price
₹1080 = 0.90 × Cost Price
Cost Price = ₹1080 / 0.90
Cost Price = ₹1200
Que-5: By selling an article for Rs240, a trader loses 4%. In order to gain 10% he must sell that article for
(a) Rs264 (b) Rs273.20 (c) Rs275 (d) Rs280
Sol: (c) Rs275
Reason: S.P. = Rs. 240; Loss =4%
⇒ 96% of C.P. = Rs. 240
S.,P. with 10% gain
⇒ 110 % of C.P.
Therefore, 110% of C.P.
= Rs. 24096×110
= Rs. 275
Que-6: A man loses Rs20 by selling some articles at the rate of Rs3 per piece and gains Rs30, if he sells them at Rs3.25 per piece. The number of piece sold by him.
(a) 100 (b) 120 (c) 200 (d) 300
Sol: (c) 200
Reason: A man loses Rs 20 by selling some toys at the rate of rs. 3 per piece.
And, he gains rs.30 , if he sells them at 3.25 per piece.
Let,the sold pieces of toy = x
Total selling price in the first case = 3×x = rs. 3x
Cost price = Selling price + loss = 3x + 20 ….(1)
And,
Total selling price in the second case = 3.25 × x = rs. 3.25x
Cost price = Selling price – profit = 3.25x -30 ….(2)
Now, if we compare (1) and (2),we get that,
3x + 20 = 3.25x – 30
3.25x – 3x = 20+30
0.25x = 50
x = 200
Que-7: The per cent profit made when an article is sold for Rs78 is twice as when it is sold for Rs69. The cost price of the article is
(a) Rs49 (b) Rs51 (c) Rs57 (d) Rs60
Sol: (d) Rs60
Reason: Let CP be Rs. x.
According to the question,
(78−x) = 2(69−x)
78−x = 138−2x
⇒ x = 138−78
= 60.
Hence, CP=Rs. 60.
Que-8: A vendor buys oranges at Rs2 for 3 oranges and sells them at a rupee each. To make a profit of Rs10 he must sell
(a) 10 oranges (b) 20 oranges (c) 30 oranges (d) 40 oranges
Sol: (c) 30 oranges
Reason: Suppose he sells x oranges
Then, C.P of x oranges = Rs. 2x/3
S.P. of x oranges = Rs. x
Profit on x oranges = Rs. (x – 2x/3) = Rs. x/3
∵ x/3 = 10
⇒ x = 30
Que-9: A man gains 10% by selling an article for a certain price. If sells it at double the price the profit made is
(a) 20% (b) 60% (c) 100% (d) 120%
Sol: (d) 120%
Reason: Gain% of the man = 10%
Selling price is doubled and we have to find the new profit percent
Let the cost price of article be Rs. 100
According to question
Selling price = Rs. (100 + 100 × 10%)
⇒ Selling price = Rs. (100 + 10)
⇒ Selling price = Rs. 110
According to question
New selling price = 2 × 110 = Rs. 220
New profit = 220 – 100 = Rs. 120
⇒ New profit% = (120/100) × 100%
⇒ New profit% = 120%
Que-10: If an article is sold at a gain of 6% instead of at a loss of 6%, then the seller gets Rs6 more. The cost price of the article is
(a) Rs50 (b) Rs94 (c) Rs100 (d) Rs106
Sol: (a) Rs50
Reason: Cost price of article is “x”
Then sold at 6% gain ⇒ x + 0.06x = 1.06x
Sold at 6% loss ⇒ x – 0.06x = 0.94x
& their difference is Rs. 6
= > 1.06x – 0.94x = 6
= > x = Rs. 50
Que-11: A radio is sold at a gain of 16%. If it had been sold for Rs20 more, 20% would have been gained. The cost price of the radio is
(a) Rs350 (b) Rs400 (c) Rs500 (d) Rs600
Sol: (c) Rs500
Reason: let the C.P. of the bicycle is Rs. x
Selling prices after 16 gain will be 1.16x
20% gain on the cost price is equal to 0.2x
According to the given condition,
1.16x + 20 = 0.2x
0.04x = 20
x = 500
Que-12: Profit after selling an article for Rs425 is the same as loss after selling it for Rs355. The cost of the article is
(a) Rs385 (b) Rs390 (c) Rs395 (d) Rs400
Sol: (b) Rs390
Reason: Let the cost price of the article be Rs. x
⇒ 425 – x = x – 355
⇒ x + x = 425 + 355
⇒ 2x = 780
⇒ x = 780/2
⇒ x = 390
Que-13: A sold a watch to B at a gain of 5% and B sold it to C at a gain of 4%. If C paid Rs1092 for it the price paid by A is
(a) Rs993.72 (b) Rs995.90 (c) Rs996 (d) Rs1000
Sol: (d) Rs1000
Reason: Let the cost price of watch for A = 100
Given that,
Gain percentage = 5%
Then,
The selling price of watch for A = the cost price of watch for B = 100 + 5 = 105
Given that,
Gain percentage = 4 %
Then,
The selling price of watch for B = the cost price of watch for C = 105 + 4 % of 105 = 105 + 4.2 = 109.2
Given that,
109.2 = Rs 1092
1 = Rs 10
100 = Rs 1000
Hence, the price paid by A = Rs 1000
Que-14: A bicycle is sold at a gain of 16%. If it had been sold for Rs60 more, 20% would have been gained. The cost price of the bicycle is
(a) Rs1050 (b) Rs1200 (c) Rs1500 (d) Rs1800
Sol: (c) Rs1500
Reason: (20% of C.P.) – (16% of C.P.) = Rs. 60
⇒ 4% of C.P. = 60
⇒ (4/100 x C.P.) = 60
⇒ C.P. = Rs. (100 x 60/4)
∴ C.P. of the bicycle = 1500
Que-15: If the cost price of 15 tables be equal to selling price of 20 tables, the loss per cent is
(a) 20% (b) 25% (c) 35% (d) 37*(1/2)%
Sol: (b) 25%
Reason: Loss percentage ={(CP-SP)/CP}*100
here,
CP of 15 tables = SP of 20 tables
there fore,
CP of 20 tables = SP of 15 tables
Loss percentage = {(20-15)/20}*100
. = {5/20}*100
= {1/4}*100
= 0.25*100
= 25%
Que-16: A fruit seller buys lemons at 2 for a rupee and sells them at 5 for Rs3. His gain per cent is
(a) 10% (b) 15% (c) 20% (d) 25%
Sol: (c) 20%
Reason: Gain% = (Gain/C.P.)× 100
C.P. of 2 lemons = Rs. 1
⇒ C.P. of 1 lemons = Rs. 1/2 = 0.5
S.P. of 5 lemons = Rs. 3
⇒ S.P. of 1 lemon = Rs. 3/5 = 0.6
Using above formula-
⇒ Gain% = {(0.6-0.5)/0.5} x 100
= 1/5 x 100
= 20%.
Que-17: By selling 45 oranges for Rs80 a man loss 20%. How many should he sell for Rs48 so as to gain 20% in the transaction ?
(a) 15 (b) 18 (c) 20 (d) 25
Sol: (b) 18
Reason: Cost Price of 45 – Oranges
= ₹ 80 x 100/(100–20) = ₹100.
So to have a profit or gain of 20% , the man has to sell 45- oranges for ₹120.
For ₹120 he has to sell = 45- oranges
So for ₹48 he has to sell
= (48/120) x 45
= 18 oranges
Que-18: A trader lists his articles 20% above cost price and allows a discount of 10% on cash payment. His gain per cent is
(a) 5% (b) 6% (c) 8% (d) 10%
Sol: (c) 8%
Reason: Marked price = 20% above the cost price
Discount = 10%
Profit% = ((S.P – C.P)/100) × 100
Let cost price of item = 100 Rs.
Marked price = 120 Rs
Discount = 120 × 10/100 = 12 Rs.
Net S.P = 120 – 12 = 108 Rs.
Profit% = (8/100) × 100 = 8%
Que-19: At what percent above the cost price must an article be marked as so to gain 22.5% after allowing a discount of 2% ?
(a) 20% (b) 24% (c) 25% (d) 25.5%
Sol: (c) 25%
Reason: Since a discount of 2% is given on the marked price:
S = M×(1−(2/100)) = M×0.98
The profit is 22.5%, so:
S = C×(1+(22.5/100)) = C×1.225
M×0.98 = C×1.225
M = (C×1.225)/0.98
Percentage above cost price = {(M−C)/C}×100
Percentage above cost price = [{(C×1.225)/0.98}−C]/C × 100
Percentage above cost price = ((1.225/0.98)−1)×100
1.225/0.98 = 1.25
Percentage above cost price
= (1.25−1) × 100
= 0.25×100
= 25%
Que-20: Arun buys an article with 25% discount on its marked price. He makes a profit of 10% by selling it at Rs660. The marked price is
(a) Rs600 (b) Rs700 (c) Rs800 (d) Rs685
Sol: (c) Rs800
Reason: Selling price of article = Rs. 660
Profit = 10%
Profit% = (S.P – C.P)/C.P × 100
According to the question;
10 = (660 – CP)/CP × 100
⇒ CP = 660 × 100/110 = Rs. 600
Now,
Ravi buys an article with a discount of 25% on its marked price’
∴ Marked price = 600/0.75 = Rs. 800
Que-21: An umbrella marked at Rs80 is sold for Rs68. The rate of discount is
(a) 12% (b) 15% (c) 17*(11/17)% (d) 20%
Sol: (b) 15%
Reason: Discount = Marked Price – Selling Price
⇒ Discount = 80 – 68 = 12
Discount % = Discount/Marked Price × 100
⇒ Discount % = 12/80 × 100 = 15%
Que-22: A tradesman marks his goods 30% above cost price. If he allows a discount of 6*(1/4)%, then his gain per cent is
(a) 21*(7/8)% (b) 22% (c) 23*(3/4)% (d) 36*(1/4)%
Sol: (a) 21*(7/8)%
Reason: let C.P. be Rs. 100. Then, marked price
= Rs. 130.
S.P. = (100−(25/4))% of Rs.130
= ((375/400)×130)
= 121.875
Profit% = 21.875
= 21875/1000
= 21*(7/8)%
Que-23: A discount series of 10%, 20% and 40% is equivalent to a single discount of
(a) 50% (b) 56.8% (c) 60% (d) 70.28%
Sol: (b) 56.8%
Reason: Successive discounts of 40%, 20% and 10% are provided on an article.
M.P. = S.P. + Discount
Discount % = Discount/M.P. × 100
Let M.P. be Rs. x.
⇒ First discount = x × 40% = 2x/5
⇒ Successive discount = {x – (2x/5)} × 20% = 3x/25
⇒ Second successive discount = (3x/5 – 3x/25) × 10% = 12x/250
⇒ Equivalent single discount = (2x/5) + (3x/25) + (12x/250) = 142x/250
⇒ Single discount % = (142x/250)/(x)} × 100 = 56.8%
Que-24: The difference between a discount of 40% on Rs500 and two successive discounts of 36% and 4% on the same amount is
(a) Nil (b) Rs1.93 (c) Rs2 (d) Rs7.20
Sol: (d) Rs7.20
Reason: Amount given = Rs 500
Discount percentage given = 40%
Successive discounts given = 36% and 4%
Successive discount = (-x) + (-y) + (-x) × (-y)/100
where x and y are discounts.
Two successive discounts of 36% and 4%
= -36 – 4 + (36 × 4)/100 = -38.56%
Difference between both discount
= -40% + 38.56% = -1.44%
Difference in rupees = 500 × 1.44%
= 500 × 1.44/100
= 7.20 rupees
Que-25: On an article with marked price Rs20000, a customer has a choice between three successive discounts of 20%, 20% and 10% and three successive discounts of 40%, 5% and 5%. By choosing the better offer he can save
(a) nothing (b) Rs690 (c) Rs715 (d) Rs785
Sol: (b) Rs690
Reason: M.P. = Rs 20000
S.P. after choosing 1st set of successive discounts = 80% of 80% of 90% of Rs 20000
= (80/100) x (80/100) x (90/100) x Rs 20000 = Rs 11520
S.P. after choosing 2nd set of successive discounts = 60 % of 95% of 95% of Rs 20000
= (60/100) x (95/100)x (95/100) x Rs 20000 = Rs 10830
The second offer is better and the customer can save (Rs 11520 – Rs 10830)
= Rs 690.
–: Profit Loss Discount and Tax MCQs Class 8 RS Aggarwal Exe-7E Goyal Brothers ICSE Maths Solutions Ch-7 :–
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