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Chapter 7

Profit and Loss, Discount and Tax — Exercise 7(E)

Class - 8 RS Aggarwal Mathematics Solutions



Exercise 7E — Multiple Choice Questions

Question 1

By selling an article for ₹ 100, one gains ₹ 10. The gain per cent is

  1. 9%

  2. 10%

  3. 11%

  4. 111911\dfrac{1}{9}%

Answer

S.P. = ₹ 100 and Gain = ₹ 10.

C.P. = S.P. - Gain = ₹ (100 - 10) = ₹ 90.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (1090×100)\Big(\dfrac{10}{90} \times 100\Big)%

= 1009\dfrac{100}{9}%

= 111911\dfrac{1}{9}%

Hence, option 4 is the correct option.

Question 2

By selling an article for ₹ 100, one loses ₹ 10. The loss per cent is

  1. 9%

  2. 91119\dfrac{1}{11}%

  3. 10%

  4. 111911\dfrac{1}{9}%

Answer

S.P. = ₹ 100 and Loss = ₹ 10.

C.P. = S.P. + Loss = ₹ (100 + 10) = ₹ 110.

Loss % = (LossC.P.×100)\Big(\dfrac{\text{Loss}}{\text{C.P.}} \times 100\Big)%

= (10110×100)\Big(\dfrac{10}{110} \times 100\Big)%

= 10011\dfrac{100}{11}%

= 91119\dfrac{1}{11}%

Hence, option 2 is the correct option.

Question 3

A man sold his cow for ₹ 7,920 and gained 10%. The cow was bought for

  1. ₹ 7,000

  2. ₹ 7,200

  3. ₹ 7,128

  4. ₹ 7,840

Answer

S.P. = ₹ 7,920 and Gain % = 10%.

C.P.=100(100+gain percentage)×S.P.=₹(100110×7920)=₹7200\text{C.P.} = \dfrac{100}{(100 + \text{gain percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{110} \times 7920\Big) \\[1em] = ₹ 7200

Hence, option 2 is the correct option.

Question 4

A watch is sold for ₹ 1,080 at a loss of 10%. The cost price of the watch is

  1. ₹ 1,180

  2. ₹ 1,200

  3. ₹ 1,188

  4. ₹ 1,125

Answer

S.P. = ₹ 1,080 and Loss % = 10%.

C.P.=100(100−loss percentage)×S.P.=₹(10090×1080)=₹1,200\text{C.P.} = \dfrac{100}{(100 - \text{loss percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{90} \times 1080\Big) \\[1em] = ₹ 1,200

Hence, option 2 is the correct option.

Question 5

By selling an article for ₹ 240, a trader loses 4%. In order to gain 10% he must sell that article for

  1. ₹ 264

  2. ₹ 273.20

  3. ₹ 275

  4. ₹ 280

Answer

S.P. = ₹ 240 and Loss % = 4%.

C.P.=100(100−loss percentage)×S.P.=₹(10096×240)=₹250\text{C.P.} = \dfrac{100}{(100 - \text{loss percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{96} \times 240\Big) \\[1em] = ₹ 250

Now, C.P. = ₹ 250 and the gain required = 10%.

Required S.P.=(100+gain percentage)100×C.P.=₹(110100×250)=₹275\text{Required S.P.} = \dfrac{(100 + \text{gain percentage})}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{110}{100} \times 250\Big) \\[1em] = ₹ 275

Hence, option 3 is the correct option.

Question 6

A man loses ₹ 20 by selling some articles at the rate of ₹ 3 per piece and gains ₹ 30, if he sells them at ₹ 3.25 per piece. The number of pieces sold by him is

  1. 100

  2. 120

  3. 200

  4. 300

Answer

Let the number of pieces be xx and the cost price of all the pieces be ₹ yy.

When sold at ₹ 3 per piece, there is a loss of ₹ 20.

⇒3x=y−20\Rightarrow 3x = y - 20 .....(1)

When sold at ₹ 3.25 per piece, there is a gain of ₹ 30.

⇒3.25x=y+30\Rightarrow 3.25x = y + 30 .....(2)

Subtracting equation (1) from equation (2),

⇒3.25x−3x=(y+30)−(y−20)⇒0.25x=50⇒x=500.25⇒x=200\Rightarrow 3.25x - 3x = (y + 30) - (y - 20) \\[1em] \Rightarrow 0.25x = 50 \\[1em] \Rightarrow x = \dfrac{50}{0.25} \\[1em] \Rightarrow x = 200

Hence, option 3 is the correct option.

Question 7

The per cent profit made when an article is sold for ₹ 78 is twice as when it is sold for ₹ 69. The cost price of the article is

  1. ₹ 49

  2. ₹ 51

  3. ₹ 57

  4. ₹ 60

Answer

Let the cost price of the article be ₹ xx.

Profit when sold for ₹ 78 = ₹ (78 - xx).

Profit when sold for ₹ 69 = ₹ (69 - xx).

According to the question, the per cent profit in the first case is twice that in the second case.

⇒78−xx×100=2×(69−xx×100)⇒78−x=2(69−x)⇒78−x=138−2x⇒2x−x=138−78⇒x=60\Rightarrow \dfrac{78 - x}{x} \times 100 = 2 \times \Big(\dfrac{69 - x}{x} \times 100\Big) \\[1em] \Rightarrow 78 - x = 2(69 - x) \\[1em] \Rightarrow 78 - x = 138 - 2x \\[1em] \Rightarrow 2x - x = 138 - 78 \\[1em] \Rightarrow x = 60

Hence, option 4 is the correct option.

Question 8

A vendor buys oranges at ₹ 2 for 3 oranges and sells them at a rupee each. To make a profit of ₹ 10 he must sell

  1. 10 oranges

  2. 20 oranges

  3. 30 oranges

  4. 40 oranges

Answer

C.P. of 3 oranges = ₹ 2.

C.P. of 1 orange = ₹ 23\dfrac{2}{3}.

S.P. of 1 orange = ₹ 1.

Profit on 1 orange = ₹ (1−23)=₹13\Big(1 - \dfrac{2}{3}\Big) = ₹ \dfrac{1}{3}.

For a profit of ₹ 13\dfrac{1}{3}, he must sell 1 orange.

⇒ For a profit of ₹ 10, he must sell (10×3)(10 \times 3) = 30 oranges.

Hence, option 3 is the correct option.

Question 9

A man gains 10% by selling an article for a certain price. If he sells it at double the price the profit made is

  1. 20%

  2. 60%

  3. 100%

  4. 120%

Answer

Let the cost price of the article be ₹ 100.

Since the gain is 10%, S.P. = ₹ (100 + 10) = ₹ 110.

New S.P. = double of ₹ 110 = ₹ 220.

New profit = ₹ (220 - 100) = ₹ 120.

On a cost price of ₹ 100, the profit is ₹ 120.

Profit % = (ProfitC.P.×100)\Big(\dfrac{\text{Profit}}{\text{C.P.}} \times 100\Big)%

= (120100×100)\Big(\dfrac{120}{100} \times 100\Big)%

= 120%

Hence, option 4 is the correct option.

Question 10

If an article is sold at a gain of 6% instead of at a loss of 6%, then the seller gets ₹ 6 more. The cost price of the article is

  1. ₹ 50

  2. ₹ 94

  3. ₹ 100

  4. ₹ 106

Answer

Let the cost price of the article be ₹ xx.

When the gain is 6%, then S.P. = ₹ (xx + 6% of xx) = ₹ 53x50\dfrac{53x}{50}.

When the loss is 6%, then S.P. = ₹ (xx - 6% of xx) = ₹ 47x50\dfrac{47x}{50}.

Now, the difference between the two S.P.s = ₹ 6.

⇒53x50−47x50=6⇒6x50=6⇒x=6×506⇒x=50\Rightarrow \dfrac{53x}{50} - \dfrac{47x}{50} = 6 \\[1em] \Rightarrow \dfrac{6x}{50} = 6 \\[1em] \Rightarrow x = \dfrac{6 \times 50}{6} \\[1em] \Rightarrow x = 50

Hence, option 1 is the correct option.

Question 11

A radio is sold at a gain of 16%. If it had been sold for ₹ 20 more, 20% would have been gained. The cost price of the radio is

  1. ₹ 350

  2. ₹ 400

  3. ₹ 500

  4. ₹ 600

Answer

Let the cost price of the radio be ₹ xx.

When the gain is 16%, then S.P. = ₹ (xx + 16% of xx) = ₹ 29x25\dfrac{29x}{25}.

When the gain is 20%, then S.P. = ₹ (xx + 20% of xx) = ₹ 6x5\dfrac{6x}{5}.

Now, the difference between the two S.P.s = ₹ 20.

⇒6x5−29x25=20⇒30x−29x25=20⇒x25=20⇒x=500\Rightarrow \dfrac{6x}{5} - \dfrac{29x}{25} = 20 \\[1em] \Rightarrow \dfrac{30x - 29x}{25} = 20 \\[1em] \Rightarrow \dfrac{x}{25} = 20 \\[1em] \Rightarrow x = 500

Hence, option 3 is the correct option.

Question 12

Profit after selling an article for ₹ 425 is the same as loss after selling it for ₹ 355. The cost of the article is

  1. ₹ 385

  2. ₹ 390

  3. ₹ 395

  4. ₹ 400

Answer

Let the cost price of the article be ₹ xx.

Profit when sold for ₹ 425 = ₹ (425 - xx).

Loss when sold for ₹ 355 = ₹ (xx - 355).

According to the question, the profit is equal to the loss.

⇒425−x=x−355⇒425+355=2x⇒2x=780⇒x=7802⇒x=390\Rightarrow 425 - x = x - 355 \\[1em] \Rightarrow 425 + 355 = 2x \\[1em] \Rightarrow 2x = 780 \\[1em] \Rightarrow x = \dfrac{780}{2} \\[1em] \Rightarrow x = 390

Hence, option 2 is the correct option.

Question 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 ₹ 1,092 for it the price paid by A is

  1. ₹ 993.72

  2. ₹ 995.90

  3. ₹ 996

  4. ₹ 1,000

Answer

For B, S.P. = ₹ 1,092 and Gain % = 4%.

C.P. of B=100(100+gain percentage)×S.P.=₹(100104×1092)=₹1050\text{C.P. of B} = \dfrac{100}{(100 + \text{gain percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{104} \times 1092\Big) \\[1em] = ₹ 1050

For A, S.P. = ₹ 1,050 and Gain % = 5%.

C.P. of A=100(100+gain percentage)×S.P.=₹(100105×1050)=₹1000\text{C.P. of A} = \dfrac{100}{(100 + \text{gain percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{105} \times 1050\Big) \\[1em] = ₹ 1000

Hence, option 4 is the correct option.

Question 14

A bicycle is sold at a gain of 16%. If it had been sold for ₹ 60 more, 20% would have been gained. The cost price of the bicycle is

  1. ₹ 1,050

  2. ₹ 1,200

  3. ₹ 1,500

  4. ₹ 1,800

Answer

Let the cost price of the bicycle be ₹ xx.

When the gain is 16%, then S.P. = ₹ (xx + 16% of xx) = ₹ 29x25\dfrac{29x}{25}.

When the gain is 20%, then S.P. = ₹ (xx + 20% of xx) = ₹ 6x5\dfrac{6x}{5}.

Now, the difference between the two S.P.s = ₹ 60.

⇒6x5−29x25=60⇒30x−29x25=60⇒x25=60⇒x=1500\Rightarrow \dfrac{6x}{5} - \dfrac{29x}{25} = 60 \\[1em] \Rightarrow \dfrac{30x - 29x}{25} = 60 \\[1em] \Rightarrow \dfrac{x}{25} = 60 \\[1em] \Rightarrow x = 1500

Hence, option 3 is the correct option.

Question 15

If the cost price of 15 tables be equal to selling price of 20 tables, the loss per cent is

  1. 20%

  2. 25%

  3. 35%

  4. 371237\dfrac{1}{2}%

Answer

Let the cost price of 1 table be ₹ 1.

Then, C.P. of 15 tables = ₹ 15.

S.P. of 20 tables = C.P. of 15 tables = ₹ 15.

⇒ S.P. of 1 table = ₹ 1520=₹34\dfrac{15}{20} = ₹ \dfrac{3}{4}.

Since S.P. of 1 table < C.P. of 1 table, so there is a loss.

Loss on 1 table = ₹ (1−34)=₹14\Big(1 - \dfrac{3}{4}\Big) = ₹ \dfrac{1}{4}.

Loss % = (LossC.P.×100)\Big(\dfrac{\text{Loss}}{\text{C.P.}} \times 100\Big)%

= (14×1001)\Big(\dfrac{1}{4} \times \dfrac{100}{1}\Big)%

= 25%

Hence, option 2 is the correct option.

Question 16

A fruit seller buys lemons at 2 for a rupee and sells them at 5 for ₹ 3. His gain per cent is

  1. 10%

  2. 15%

  3. 20%

  4. 25%

Answer

Suppose the number of lemons bought = L.C.M. of 2 and 5 = 10.

C.P. of 2 lemons = ₹ 1.

C.P. of 10 lemons = ₹ (12×10)\Big(\dfrac{1}{2} \times 10\Big) = ₹ 5.

S.P. of 5 lemons = ₹ 3.

S.P. of 10 lemons = ₹ (35×10)\Big(\dfrac{3}{5} \times 10\Big) = ₹ 6.

Since S.P. > C.P., so there is a gain.

Gain = ₹ (6 - 5) = ₹ 1.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (15×100)\Big(\dfrac{1}{5} \times 100\Big)%

= 20%

Hence, option 3 is the correct option.

Question 17

By selling 45 oranges for ₹ 80 a man loses 20%. How many should he sell for ₹ 48 so as to gain 20% in the transaction?

  1. 15

  2. 18

  3. 20

  4. 25

Answer

S.P. of 45 oranges = ₹ 80 and Loss % = 20%.

C.P. of 45 oranges=100(100−loss percentage)×S.P.=₹(10080×80)=₹100\text{C.P. of 45 oranges} = \dfrac{100}{(100 - \text{loss percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{80} \times 80\Big) \\[1em] = ₹ 100

Now, the gain required = 20%.

S.P. of 45 oranges=(100+gain percentage)100×C.P.=₹(120100×100)=₹120\text{S.P. of 45 oranges} = \dfrac{(100 + \text{gain percentage})}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{120}{100} \times 100\Big) \\[1em] = ₹ 120

So, for ₹ 120 he sells 45 oranges.

⇒ For ₹ 48 he should sell (45120×48)\Big(\dfrac{45}{120} \times 48\Big) = 18 oranges.

Hence, option 2 is the correct option.

Question 18

A trader lists his articles 20% above cost price and allows a discount of 10% on cash payment. His gain per cent is

  1. 5%

  2. 6%

  3. 8%

  4. 10%

Answer

Let the cost price of the article be ₹ 100.

Then, its marked price = ₹ (100+20100×100)\Big(100 + \dfrac{20}{100} \times 100\Big) = ₹ (100 + 20) = ₹ 120.

Discount = 10% of ₹ 120 = ₹ (10100×120)\Big(\dfrac{10}{100} \times 120\Big) = ₹ 12.

Selling price = M.P. - Discount = ₹ (120 - 12) = ₹ 108.

Gain = S.P. - C.P. = ₹ (108 - 100) = ₹ 8.

On a cost price of ₹ 100, the gain is ₹ 8.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (8100×100)\Big(\dfrac{8}{100} \times 100\Big)%

= 8%

Hence, option 3 is the correct option.

Question 19

At what percent above the cost price must an article be marked so as to gain 22.5% after allowing a discount of 2%?

  1. 20%

  2. 24%

  3. 25%

  4. 25.5%

Answer

Let the cost price be ₹ 100.

Since the gain is 22.5%, S.P. = ₹ (100 + 22.5) = ₹ 122.50.

Let the marked price be ₹ xx.

Discount = 2% of ₹ xx, so S.P. = 98% of M.P.

⇒98x100=122.50⇒x=122.50×10098⇒x=125\Rightarrow \dfrac{98x}{100} = 122.50 \\[1em] \Rightarrow x = \dfrac{122.50 \times 100}{98} \\[1em] \Rightarrow x = 125

M.P. = ₹ 125 and C.P. = ₹ 100.

M.P. is above C.P. by ₹ (125 - 100) = ₹ 25.

On a cost price of ₹ 100, the marked price is ₹ 25 above.

Required per cent = (M.P.−C.P.C.P.×100)\Big(\dfrac{\text{M.P.} - \text{C.P.}}{\text{C.P.}} \times 100\Big)%

= (25100×100)\Big(\dfrac{25}{100} \times 100\Big)%

= 25%

Hence, option 3 is the correct option.

Question 20

Arun buys an article with 25% discount on its marked price. He makes a profit of 10% by selling it at ₹ 660. The marked price is

  1. ₹ 600

  2. ₹ 700

  3. ₹ 800

  4. ₹ 685

Answer

For Arun, S.P. = ₹ 660 and Profit % = 10%.

C.P. of Arun=100(100+profit percentage)×S.P.=₹(100110×660)=₹600\text{C.P. of Arun} = \dfrac{100}{(100 + \text{profit percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{110} \times 660\Big) \\[1em] = ₹ 600

Arun bought the article at a discount of 25% on the marked price.

Let the marked price be ₹ xx.

⇒(100−25)100×x=600⇒75x100=600⇒x=600×10075⇒x=800\Rightarrow \dfrac{(100 - 25)}{100} \times x = 600 \\[1em] \Rightarrow \dfrac{75x}{100} = 600 \\[1em] \Rightarrow x = \dfrac{600 \times 100}{75} \\[1em] \Rightarrow x = 800

Hence, option 3 is the correct option.

Question 21

An umbrella marked at ₹ 80 is sold for ₹ 68. The rate of discount is

  1. 12%

  2. 15%

  3. 17111717\dfrac{11}{17}%

  4. 20%

Answer

M.P. = ₹ 80 and S.P. = ₹ 68.

Discount given = M.P. - S.P. = ₹ (80 - 68) = ₹ 12.

Rate of discount = (DiscountM.P.×100)\Big(\dfrac{\text{Discount}}{\text{M.P.}} \times 100\Big)%

= (1280×100)\Big(\dfrac{12}{80} \times 100\Big)%

= 120080\dfrac{1200}{80}%

= 15%

Hence, option 2 is the correct option.

Question 22

A tradesman marks his goods 30% above cost price. If he allows a discount of 6146\dfrac{1}{4}%, then his gain per cent is

  1. 217821\dfrac{7}{8}%

  2. 22%

  3. 233423\dfrac{3}{4}%

  4. 361436\dfrac{1}{4}%

Answer

Let the cost price of the goods be ₹ 100.

Then, its marked price = ₹ (100+30100×100)\Big(100 + \dfrac{30}{100} \times 100\Big) = ₹ (100 + 30) = ₹ 130.

Discount = 6146\dfrac{1}{4}% of ₹ 130 = ₹ (254×1100×130)\Big(\dfrac{25}{4} \times \dfrac{1}{100} \times 130\Big) = ₹ 658\dfrac{65}{8}.

Selling price = M.P. - Discount = ₹ (130−658)=₹9758\Big(130 - \dfrac{65}{8}\Big) = ₹ \dfrac{975}{8}.

Gain = S.P. - C.P. = ₹ (9758−100)=₹1758\Big(\dfrac{975}{8} - 100\Big) = ₹ \dfrac{175}{8}.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (1758×1100×100)\Big(\dfrac{175}{8} \times \dfrac{1}{100} \times 100\Big)%

= 1758\dfrac{175}{8}%

= 217821\dfrac{7}{8}%

Hence, option 1 is the correct option.

Question 23

A discount series of 10%, 20% and 40% is equivalent to a single discount of

  1. 50%

  2. 56.8%

  3. 60%

  4. 70.28%

Answer

Let the marked price be ₹ 100.

First discount = 10% of ₹ 100 = ₹ 10.

Price after the first discount = ₹ (100 - 10) = ₹ 90.

Second discount = 20% of ₹ 90 = ₹ 18.

Price after the second discount = ₹ (90 - 18) = ₹ 72.

Third discount = 40% of ₹ 72 = ₹ 28.80.

Price after the third discount = ₹ (72 - 28.80) = ₹ 43.20.

Net discount = ₹ (100 - 43.20) = ₹ 56.80.

On a marked price of ₹ 100, the discount is ₹ 56.80.

Discount % = (DiscountM.P.×100)\Big(\dfrac{\text{Discount}}{\text{M.P.}} \times 100\Big)%

= (56.80100×100)\Big(\dfrac{56.80}{100} \times 100\Big)%

= 56.8%

Hence, option 2 is the correct option.

Question 24

The difference between a discount of 40% on ₹ 500 and two successive discounts of 36% and 4% on the same amount is

  1. Nil

  2. ₹ 1.93

  3. ₹ 2

  4. ₹ 7.20

Answer

For a single discount of 40%:

Discount = 40% of ₹ 500 = ₹ (40100×500)\Big(\dfrac{40}{100} \times 500\Big) = ₹ 200.

Price after discount = ₹ (500 - 200) = ₹ 300.

For two successive discounts of 36% and 4%:

First discount = 36% of ₹ 500 = ₹ (36100×500)\Big(\dfrac{36}{100} \times 500\Big) = ₹ 180.

Price after the first discount = ₹ (500 - 180) = ₹ 320.

Second discount = 4% of ₹ 320 = ₹ (4100×320)\Big(\dfrac{4}{100} \times 320\Big) = ₹ 12.80.

Price after the second discount = ₹ (320 - 12.80) = ₹ 307.20.

Required difference = ₹ (307.20 - 300) = ₹ 7.20.

Hence, option 4 is the correct option.

Question 25

On an article with marked price ₹ 20,000, 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

  1. nothing

  2. ₹ 690

  3. ₹ 715

  4. ₹ 785

Answer

Marked price of the article = ₹ 20,000

If d1d_1%, d2d_2% and d3d_3% are three successive discounts, then :

Price = (100−d1)100×(100−d2)100×(100−d3)100×M.P.\dfrac{(100 - d_1)}{100} \times \dfrac{(100 - d_2)}{100} \times \dfrac{(100 - d_3)}{100} \times \text{M.P.}

For the successive discounts of 20%, 20% and 10%:

Price=₹[(1−20100)(1−20100)(1−10100)×20000]=₹(80100×80100×90100×20000)=₹(0.8×0.8×0.9×20000)=₹11520\text{Price} = ₹ \Big[\Big(1 - \dfrac{20}{100}\Big)\Big(1 - \dfrac{20}{100}\Big)\Big(1 - \dfrac{10}{100}\Big) \times 20000\Big] \\[1em] = ₹ \Big(\dfrac{80}{100} \times \dfrac{80}{100} \times \dfrac{90}{100} \times 20000\Big) \\[1em] = ₹ (0.8 \times 0.8 \times 0.9 \times 20000) \\[1em] = ₹ 11520

For the successive discounts of 40%, 5% and 5%:

Price=₹[(1−40100)(1−5100)(1−5100)×20000]=₹(60100×95100×95100×20000)=₹(0.6×0.95×0.95×20000)=₹10830\text{Price} = ₹ \Big[\Big(1 - \dfrac{40}{100}\Big)\Big(1 - \dfrac{5}{100}\Big)\Big(1 - \dfrac{5}{100}\Big) \times 20000\Big] \\[1em] = ₹ \Big(\dfrac{60}{100} \times \dfrac{95}{100} \times \dfrac{95}{100} \times 20000\Big) \\[1em] = ₹ (0.6 \times 0.95 \times 0.95 \times 20000) \\[1em] = ₹ 10830

Since ₹ 10,830 < ₹ 11,520, the second offer is better.

Saving = ₹ (11,520 - 10,830) = ₹ 690.

Hence, option 2 is the correct option.

Exercise 7E — Mental Maths

Question 1

Fill in the blanks:

(i) ......... = (100+gain percentage)100\dfrac{(100 + \text{gain percentage})}{100} × ......... .

(ii) Discount is always allowed on the ......... price.

(iii) ......... = M.P. - Discount.

(iv) If the S.P of an article is 65\dfrac{6}{5} of its cost price, then the gain per cent is ......... .

(v) A shopkeeper incurs a ......... if he sells 15 articles at the cost price of 12 articles.

Answer

(i) When an article is sold at a gain, the selling price is obtained from the cost price by the formula S.P.=(100+gain percentage)100×C.P.\text{S.P.} = \dfrac{(100 + \text{gain percentage})}{100} \times \text{C.P.}

Hence, S.P. = (100+gain percentage)100\dfrac{(100 + \text{gain percentage})}{100} × C.P.

(ii) Discount is a reduction given by the shopkeeper on the price marked on the article.

Hence, discount is always allowed on the marked price.

(iii) The price at which an article is finally sold is obtained by subtracting the discount from the marked price.

Hence, S.P. = M.P. - Discount.

(iv) S.P.=65\text{S.P.} = \dfrac{6}{5} of C.P.

Since S.P. > C.P., so there is a gain.

Gain=S.P.−C.P.=65C.P.−C.P.=C.P.5\text{Gain} = \text{S.P.} - \text{C.P.} \\[1em] = \dfrac{6}{5}\text{C.P.} - \text{C.P.} \\[1em] = \dfrac{\text{C.P.}}{5}

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (C.P.5×100C.P.)\Big(\dfrac{\text{C.P.}}{5} \times \dfrac{100}{\text{C.P.}}\Big)%

= 20%

Hence, the gain per cent is 20%.

(v) Let the cost price of 1 article be ₹ 1.

Then, C.P. of 15 articles = ₹ 15.

S.P. of 15 articles = C.P. of 12 articles = ₹ 12.

Since S.P. < C.P., the shopkeeper incurs a loss.

Hence, a shopkeeper incurs a loss if he sells 15 articles at the cost price of 12 articles.

Question 2

Write true (T) or false (F):

(i) Gain or loss is always reckoned on the S.P. .........

(ii) If a shopkeeper buys an article for ₹ 75 and sells it for ₹ 100, then his gain is 25%. .........

(iii) C.P. = 100(100−loss percentage)\dfrac{100}{(100 - \text{loss percentage})} × S.P. .........

(iv) If no discount is given, then cost price becomes equal to selling price. .........

Answer

(i) Gain or loss is always reckoned on the S.P.

Gain or loss is always reckoned on the cost price, and not on the selling price.

Hence, the statement is False.

(ii) If a shopkeeper buys an article for ₹ 75 and sells it for ₹ 100, then his gain is 25%.

C.P. = ₹ 75 and S.P. = ₹ 100.

Gain = ₹ (100 - 75) = ₹ 25.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (2575×100)\Big(\dfrac{25}{75} \times 100\Big)%

= 1003\dfrac{100}{3}%

= 331333\dfrac{1}{3}%

Since the gain per cent is 331333\dfrac{1}{3}% and not 25%, the statement is false.

Hence, the statement is False.

(iii) C.P. = 100(100−loss percentage)\dfrac{100}{(100 - \text{loss percentage})} × S.P.

When an article is sold at a loss, S.P.=(100−loss percentage)100×C.P.\text{S.P.} = \dfrac{(100 - \text{loss percentage})}{100} \times \text{C.P.}, and so C.P.=100(100−loss percentage)×S.P.\text{C.P.} = \dfrac{100}{(100 - \text{loss percentage})} \times \text{S.P.}

Hence, the statement is True.

(iv) If no discount is given, then cost price becomes equal to selling price.

If no discount is given, the article is sold at its marked price, so the selling price becomes equal to the marked price and not to the cost price.

Hence, the statement is False.

Exercise 7E — Assertion-Reason Questions

Question 1

Assertion (A): Aman bought a new motorcycle for ₹ 2,50,000. Next year, its price decreased by 10% and further next year, it decreased by 12%. The overall decrease in the price of the motorcycle is 20.8%.

Reason (R): SP = (100−loss percentage)100\dfrac{(100 - \text{loss percentage})}{100} × CP.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

  2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

  3. Assertion (A) is true but Reason (R) is false.

  4. Assertion (A) is false but Reason (R) is true.

Answer

Original price of the motorcycle = ₹ 2,50,000.

Price after a decrease of 10% = ₹ (100−10100×250000)=₹(90100×250000)\Big(\dfrac{100 - 10}{100} \times 250000\Big) = ₹ \Big(\dfrac{90}{100} \times 250000\Big) = ₹ 2,25,000.

Price after a further decrease of 12% = ₹ (100−12100×225000)=₹(88100×225000)\Big(\dfrac{100 - 12}{100} \times 225000\Big) = ₹ \Big(\dfrac{88}{100} \times 225000\Big) = ₹ 1,98,000.

Total decrease in the price = ₹ (2,50,000 - 1,98,000) = ₹ 52,000.

Overall decrease % = (Total decreaseOriginal price×100)\Big(\dfrac{\text{Total decrease}}{\text{Original price}} \times 100\Big)%

= (52000250000×100)\Big(\dfrac{52000}{250000} \times 100\Big)%

= 5200000250000\dfrac{5200000}{250000}%

= 20.8%

So, Assertion (A) is true.

The formula SP=(100−loss percentage)100×CP\text{SP} = \dfrac{(100 - \text{loss percentage})}{100} \times \text{CP} is a correct statement, so Reason (R) is true.

However, this formula applies to a single transaction only. The overall decrease of 20.8% arises because the second decrease is calculated on the already reduced price, and this successive effect is not explained by the formula in Reason (R).

So, Reason (R) is true but it is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

Question 2

Assertion (A): Selling price of 9 articles is equal to the cost price of 15 articles. In this transaction, there is a profit of 662366\dfrac{2}{3}%.

Reason (R): Gain% = Actual gain100\dfrac{\text{Actual gain}}{100} × CP.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

  2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

  3. Assertion (A) is true but Reason (R) is false.

  4. Assertion (A) is false but Reason (R) is true.

Answer

Let the cost price of 1 article be ₹ 1.

Then, C.P. of 15 articles = ₹ 15.

S.P. of 9 articles = C.P. of 15 articles = ₹ 15.

⇒ S.P. of 1 article = ₹ 159=₹53\dfrac{15}{9} = ₹ \dfrac{5}{3}.

Since S.P. of 1 article > C.P. of 1 article, so there is a profit.

Profit on 1 article = ₹ (53−1)=₹23\Big(\dfrac{5}{3} - 1\Big) = ₹ \dfrac{2}{3}.

Profit % = (ProfitC.P.×100)\Big(\dfrac{\text{Profit}}{\text{C.P.}} \times 100\Big)%

= (23×1001)\Big(\dfrac{2}{3} \times \dfrac{100}{1}\Big)%

= 2003\dfrac{200}{3}%

= 662366\dfrac{2}{3}%

So, Assertion (A) is true.

The correct formula is Gain percentage=Actual gainCP×100\text{Gain percentage} = \dfrac{\text{Actual gain}}{\text{CP}} \times 100, and not Actual gain100×CP\dfrac{\text{Actual gain}}{100} \times \text{CP} as given.

So, Reason (R) is false.

Hence, option 3 is the correct option.

Exercise 7E — Competency Focused Questions

Question 1

Which of the following does mean a 100% profit?

  1. Buying at ₹ 100 and selling at ₹ 50

  2. Buying at ₹ 100 and selling at ₹ 100

  3. Buying at ₹ 50 and selling at ₹ 100

  4. Buying at ₹ 50 and selling at ₹ 150

Answer

A profit of 100% means that the profit is equal to the cost price, i.e. S.P. = 2 × C.P.

Let us check each option:

Option 1: C.P. = ₹ 100 and S.P. = ₹ 50. Here S.P. < C.P., so there is a loss of 50%.

Option 2: C.P. = ₹ 100 and S.P. = ₹ 100. Here S.P. = C.P., so there is neither profit nor loss.

Option 3: C.P. = ₹ 50 and S.P. = ₹ 100. Profit = ₹ 50 and Profit % = (5050×100)\Big(\dfrac{50}{50} \times 100\Big)% = 100%.

Option 4: C.P. = ₹ 50 and S.P. = ₹ 150. Profit = ₹ 100 and Profit % = (10050×100)\Big(\dfrac{100}{50} \times 100\Big)% = 200%.

Hence, option 3 is the correct option.

Question 2

A shopkeeper buys a ball and a bat for ₹ 200 and ₹ 500 respectively. For selling, he marks the bat at ₹ 525. What price should he ask for the ball, if he wants to make a profit of 5% on the whole transaction?

  1. ₹ 220

  2. ₹ 215

  3. ₹ 210

  4. ₹ 200

Answer

Total C.P. of the ball and the bat = ₹ (200 + 500) = ₹ 700.

For a profit of 5% on the whole transaction,

Total S.P.=(100+profit percentage)100×C.P.=₹(100+5100×700)=₹(105100×700)=₹735\text{Total S.P.} = \dfrac{(100 + \text{profit percentage})}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{100 + 5}{100} \times 700\Big) \\[1em] = ₹ \Big(\dfrac{105}{100} \times 700\Big) \\[1em] = ₹ 735

S.P. of the bat = ₹ 525.

S.P. of the ball = ₹ (735 - 525) = ₹ 210.

Hence, option 3 is the correct option.

Question 3

A shopkeeper bought an item and plans to sells it in two ways:

(i) The shopkeeper gives a discount of xx% on the marked price and adds xx% tax on the discounted price of the item.

(ii) The shopkeeper hikes the marked price by xx% and then gives xx% discount on the hiked price of the item.

Which option is better for the shopkeeper?

  1. (i)

  2. (ii)

  3. both will give the same revenue

  4. none of these

Answer

Let the marked price of the item be ₹ MM.

For option (i):

Price after a discount of xx% = ₹ M(1−x100)M\Big(1 - \dfrac{x}{100}\Big).

Price after adding xx% tax on it = ₹ M(1−x100)(1+x100)M\Big(1 - \dfrac{x}{100}\Big)\Big(1 + \dfrac{x}{100}\Big) = ₹ M(1−x210000)M\Big(1 - \dfrac{x^2}{10000}\Big).

For option (ii):

Price after a hike of xx% = ₹ M(1+x100)M\Big(1 + \dfrac{x}{100}\Big).

Price after a discount of xx% on it = ₹ M(1+x100)(1−x100)M\Big(1 + \dfrac{x}{100}\Big)\Big(1 - \dfrac{x}{100}\Big) = ₹ M(1−x210000)M\Big(1 - \dfrac{x^2}{10000}\Big).

The final price is the same in both the options.

Hence, option 3 is the correct option.

Question 4

Amazon offers you 10% cashback (discount) on an article of ₹ 500 and adds 15% delivery charges on the discounted amount of the article. How much do you need to pay?

  1. ₹ 500 + 5% of ₹ 500

  2. ₹ 500 + 15% of ₹ 500 - 10% of ₹ 575

  3. ₹ 500 - (10% of ₹ 500 + 15% of ₹ 500)

  4. ₹ 500 - 10% of ₹ 500 + 15% of ₹ 450

Answer

Marked price of the article = ₹ 500.

Cashback (discount) = 10% of ₹ 500 = ₹ 50.

Discounted amount = ₹ (500 - 50) = ₹ 450, which is ₹ 500 - 10% of ₹ 500.

The delivery charges are added on the discounted amount, so

Delivery charges = 15% of ₹ 450.

Amount to be paid = ₹ 500 - 10% of ₹ 500 + 15% of ₹ 450.

Hence, option 4 is the correct option.

Question 5

A shopkeeper sells an item for 25% discount and still makes 25% profit. The discount he needs to offer to achieve a profit of 36% is:

  1. 10%

  2. 15%

  3. 18.4%

  4. 22.6%

Answer

Let the marked price of the item be ₹ 100.

Discount = 25% of ₹ 100 = ₹ 25.

S.P. = ₹ (100 - 25) = ₹ 75, and on this the profit is 25%.

C.P.=100(100+profit percentage)×S.P.=₹(100125×75)=₹60\text{C.P.} = \dfrac{100}{(100 + \text{profit percentage})} \times \text{S.P.} \\[1em] = ₹ \Big(\dfrac{100}{125} \times 75\Big) \\[1em] = ₹ 60

For a profit of 36%,

Required S.P.=(100+36)100×C.P.=₹(136100×60)=₹81.60\text{Required S.P.} = \dfrac{(100 + 36)}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{136}{100} \times 60\Big) \\[1em] = ₹ 81.60

Required discount = ₹ (100 - 81.60) = ₹ 18.40.

On a marked price of ₹ 100, the discount is ₹ 18.40.

Discount % = (DiscountM.P.×100)\Big(\dfrac{\text{Discount}}{\text{M.P.}} \times 100\Big)%

= (18.40100×100)\Big(\dfrac{18.40}{100} \times 100\Big)%

= 18.4%

Hence, option 3 is the correct option.

Question 6

A teacher buys a book of cost price ₹ 100. As she goes to school next day, her friend asks her to sells that book to her. As a friend, she sells the book at 20% loss to her. And then her friend sells that book at 20% gain to some other colleague. The final selling price of the book is:

  1. ₹ 80

  2. ₹ 90

  3. ₹ 96

  4. ₹ 100

Answer

For the teacher, C.P. = ₹ 100 and Loss % = 20%.

S.P.=(100−loss percentage)100×C.P.=₹(100−20100×100)=₹80\text{S.P.} = \dfrac{(100 - \text{loss percentage})}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{100 - 20}{100} \times 100\Big) \\[1em] = ₹ 80

For the friend, C.P. = ₹ 80 and Gain % = 20%.

S.P.=(100+gain percentage)100×C.P.=₹(100+20100×80)=₹96\text{S.P.} = \dfrac{(100 + \text{gain percentage})}{100} \times \text{C.P.} \\[1em] = ₹ \Big(\dfrac{100 + 20}{100} \times 80\Big) \\[1em] = ₹ 96

Hence, option 3 is the correct option.

Question 7

A dealer allows a discount of 10% on the marked price. How much above the cost price must he mark his goods to make a profit of 8%?

  1. 8%

  2. 10%

  3. 15%

  4. 20%

Answer

Let the cost price of the goods be ₹ 100.

Since the profit is 8%, S.P. = ₹ (100 + 8) = ₹ 108.

Let the marked price be ₹ xx.

Discount = 10% of ₹ xx, so S.P. = 90% of M.P.

⇒90x100=108⇒x=108×10090⇒x=120\Rightarrow \dfrac{90x}{100} = 108 \\[1em] \Rightarrow x = \dfrac{108 \times 100}{90} \\[1em] \Rightarrow x = 120

M.P. = ₹ 120 and C.P. = ₹ 100.

M.P. is above C.P. by ₹ (120 - 100) = ₹ 20.

On a cost price of ₹ 100, the marked price is ₹ 20 above.

Required per cent = (M.P.−C.P.C.P.×100)\Big(\dfrac{\text{M.P.} - \text{C.P.}}{\text{C.P.}} \times 100\Big)%

= (20100×100)\Big(\dfrac{20}{100} \times 100\Big)%

= 20%

Hence, option 4 is the correct option.

Question 8

A dishonest seller sells fruits with a 960 g weight instead of a 1 kg weight. If he sold the fruits at the cost price, his gain is:

  1. 4%

  2. 4154\dfrac{1}{5}%

  3. 4124\dfrac{1}{2}%

  4. 4164\dfrac{1}{6}%

Answer

Let the cost price of the fruits be ₹ 1 per gram.

The seller uses a weight of 960 g in place of 1 kg = 1000 g.

So, he actually gives 960 g of fruits but charges the price of 1000 g.

C.P. of the fruits sold = ₹ 960.

S.P. of the fruits sold = ₹ 1,000.

Since S.P. > C.P., so there is a gain.

Gain = ₹ (1,000 - 960) = ₹ 40.

Gain % = (GainC.P.×100)\Big(\dfrac{\text{Gain}}{\text{C.P.}} \times 100\Big)%

= (40960×100)\Big(\dfrac{40}{960} \times 100\Big)%

= 4000960\dfrac{4000}{960}%

= 256\dfrac{25}{6}%

= 4164\dfrac{1}{6}%

Hence, option 4 is the correct option.

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