Class 8: Profit and Loss – Exercise 14C – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: The marked price of a refrigerator is Rs. $\displaystyle 16450$ . The shopkeeper offers on off-season discount of $\displaystyle 16 \%$ on it. Find its selling price.

Answer:

Marked Price $\displaystyle = 16450 \text{ Rs. }$

$\displaystyle \text{Off season discount } = 16 \%$

$\displaystyle \text{Discount } = \frac{16}{100} \times \text{ Market Price } = \frac{16}{100} \times 16450 = 2632 \text{ Rs. }$

$\displaystyle \text{Selling Price } = 16450-2632 = 13818 \text{ Rs. }$

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Question 2: The price of a sweater was slashed down by a shopkeeper from Rs. $\displaystyle 850$ to Rs. $\displaystyle 731$ . Find the rate of discount given by him.

Answer:

$\displaystyle \text{Market Price of the sweater } = 850 \text{ Rs. }$

$\displaystyle \text{Selling Price } = 731 \text{ Rs. }$

$\displaystyle \text{Discount = Marked Price - Selling Price } = 850 - 731 = 119 \text{ Rs. }$

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Question 3: Find the rate of discount being given on mini toy-gun whose selling price is Rs. $\displaystyle 345$ after deducting a discount being given on a mini toy-gun whose selling price is Rs. $\displaystyle 345$ after deducting a discount of Rs. $\displaystyle 30$ on its marked price.

Answer:

$\displaystyle \text{Selling Price of a mini toy gun } = 345 \text{ Rs. }$

$\displaystyle \text{Discount } = 30 \text{ Rs. }$

$\displaystyle \text{Marked Price = Selling Price + Discount } = 345 + 30 = 375 \text{ Rs. }$

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Question 4: After allowing a discount of $\displaystyle 15 \%$ , a baby-suit was sold for Rs. $\displaystyle 1156$ . Find its marked price.

Answer:

Selling Price of a baby suit $\displaystyle = 1156 \text{ Rs. }$

Discount $\displaystyle = 15 \%$

Let Marked Price $\displaystyle = Rs. x$

$\displaystyle \text{Discount } = \frac{15}{100} \times x$

$\displaystyle \text{Selling Price } = x - \frac{15x}{100} = \frac{85}{100} x$

Equating the above with actual selling price

$\displaystyle \frac{85}{100} x = 1156$

$\displaystyle x ( \frac{1156 \times 100}{85} ) = 1360$

Marked Price of the body Suit $\displaystyle = 1360 \text{ Rs. }$

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Question 5: A calculator was bought for Rs. $\displaystyle 435$ after getting a discount of $\displaystyle 13 \%$ . Find the marked price of the calculator.

Answer:

$\displaystyle \text{Purchase Price of the calculator = Selling Price of the calculator = 435} \text{ Rs. }$

Discount $\displaystyle = 13 \%$ on the marked price

Let Marked Price $\displaystyle = x \text{ Rs. }$

$\displaystyle \text{Discount } = \frac{13}{100} x$

M.P. $\displaystyle =$ S.P. $\displaystyle +$ Discount

$\displaystyle x = 435+ \frac{13}{100} x$

$\displaystyle \text{ or } x - \frac{13}{100} x = 435$

$\displaystyle \text{ or } \frac{87}{100} x = 435$

$\displaystyle \text{ or } x = \frac{435 \times 100}{87} = 500$

$\displaystyle \text{Marked price of the calculate } = 500 \text{ Rs. }$

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Question 6: A dealer marked his goods $\displaystyle 35 \%$ above cost price and allowed a discount of $\displaystyle 20 \%$ on the market price. Find his gain or loss per cent.

Answer:

$\displaystyle \text{Let Cost Price } = x \text{ Rs. }$

$\displaystyle \text{Marked Price } = x + \frac{35}{100} x = \frac{135}{100} x$

$\displaystyle \text{Discount } = 20 \% \text{ on M.P.} = ( \frac{20}{100} ) \times M.P. = \ ( \frac{20}{100} ) \times ( \frac{135}{100} )x$

$\displaystyle \text{S.P. = M.P. - Discount} = \frac{135}{100} x- \frac{20}{100} \times \frac{135}{100} x = \frac{80}{100} \times \frac{135}{100} x$

$\displaystyle \text{Profit } = \text{S.P. - Cost \ Price } = \frac{80 \times 135}{100 \times 100} x-x = 1.08x - x = 0.08 x$

$\displaystyle \% \text{ age profit } = \frac{\text{Profit}}{\text{C.P.}} \times 100 = \frac{0.08x}{x} \times 100 = 8 \%$

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Question 7: An article was marked $\displaystyle 40 \%$ above cost price and a discount of $\displaystyle 35 \%$ was given on its marked price. Find the gain or loss per cent made by the shopkeeper.

Answer:

$\displaystyle \text{Let Cost Price } = Rs. x$

$\displaystyle \text{Marked Price } = x + \frac{40}{100} x = \frac{140}{100} x$

$\displaystyle \text{Discount } = 35 \% \text{ of the M.P. } = \frac{35}{100} \times \frac{140}{100} x$

$\displaystyle \text{S.P. = M.P. - Discount} = \frac{140x}{100} - \frac{35}{100} \times {} \frac{140}{100} x = \frac{65}{100} \times \frac{140}{100} x$

$\displaystyle \text{ Gain = Cost - S.P. }= x - \frac{65}{100} \times \frac{140}{100} x = x - 0.91 x = 0.09x$

$\displaystyle \% \text{Gain } = \frac{0.09x}{x} \times 100 = 9 \%$

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Question 8: A dealer purchased a washing machine for Rs. $\displaystyle 7660$ . after allowing a discount of $\displaystyle 12 \%$ on its marked price, he gains $\displaystyle 10 \%$ . Find the marked price.

Answer:

The cost price $\displaystyle = 7660 \text{ Rs. }$

Let M.P. $\displaystyle = x \text{ Rs. }$

$\displaystyle \text{Discount } = \frac{12}{100}$

$\displaystyle \text{S.P. = M.P. - Discount = Rs. } x - \frac{12}{100} x = Rs. \frac{88}{100} x$

$\displaystyle \text{Gain is 10 \% of the cost Price = Rs. } \frac{10}{100} \times 7660 = 766 \text{ Rs. }$

$\displaystyle \text{Gain = S.P. - Cost Price = } \frac{88}{100} x - 7660$

$\displaystyle 766 = \frac{88}{100} x-7660$

$\displaystyle \frac{88}{100} x = 7660+766 = 8426$

$\displaystyle x = ( \frac{8426 \times 100}{88} ) = 9575 \text{ Rs. }$

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Question 9: A shopkeeper bought a sewing machine for Rs. $\displaystyle 3750$ . after allowing a discount of $\displaystyle 10 \%$ on its marked price, he gains $\displaystyle 26 \%$ . Find the marked price of the sewing machine.

Answer:

$\displaystyle \text{Cost of the sewing machine } = 3750 \text{ Rs. }$

$\displaystyle \text{Let Mark Price. } = x \text{ Rs. }$

$\displaystyle \text{Discount } = \frac{10}{100} x$

$\displaystyle \text{Selling Price } = x - \frac{10}{100} x = \frac{90}{100} x$

$\displaystyle \text{Gain = Selling Price - Cost Price } = ( \frac{90}{100} x - 3750)$ … … … … … i)

$\displaystyle \text{Gain is 26 \% of the CP =} \frac{26}{100} \times 3750 = 975$ … … … … … ii)

Equating i) and ii)

$\displaystyle \frac{90}{100} x-3750 = 975$

$\displaystyle \text{ or } x = (975 +3750) \times \frac{100}{90} = 5250$

$\displaystyle \text{Marked Price } = 5250 \text{ Rs. }$

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Question 10: After allowing a discount of $\displaystyle 10 \%$ on the marked price, a trader still makes a profit of $\displaystyle 17 \%$ . By what per cent is the marked price above cost price?

Answer:

$\displaystyle \text{ Let Marked Price } = x \text{ Rs. }$

$\displaystyle \text{S.P. } = x - \text{Discount of 10 \% of } x = x- \frac{10}{100} x = \frac{90}{100} x$

$\displaystyle \text{ \% Gain }= \frac{S.P-C.P}{C.P} \times x$

$\displaystyle \text{ (\% gain)(C.P.) = S.P. } \times 100 - C.P.\times 100$

$\displaystyle \text{ (\% gain) (C.P.) +C.P. } \times 100 = S.P.100$

$\displaystyle \text{C.P. } = \frac{\text{S.P. } \times 100}{ \text{\% gain }+100} = \frac{x\frac{90}{100} \times 100}{17+100} = \frac{90}{117} x$

$\displaystyle \text{Difference between M.P. and C.P. } = x - \frac{90}{117} x = \frac{27}{117} x$

$\displaystyle \text{ \% age of difference in marked price over cost price } =\frac{ \frac{27}{117}x}{ \frac{90}{117}x} \times 100 = 30 \%$

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Question 11: After allowing a discount of $\displaystyle 12 \%$ on the marked price, a shopkeeper still gains $\displaystyle 21 \%$ . By what per cent is the marked price above cost price?

Answer:

$\displaystyle \text{Let marked price in } x \text{ Rs. }$

$\displaystyle \text{S.P. } = x-12 \% \text{ discount on } x = x - \frac{12}{100} x = \frac{88}{100} x$

$\displaystyle \text{C.P. } = \frac{S.P. \times 100}{ \% gain\ +100} = \frac{\frac{88}{100}x \times 100}{21+100} = \frac{88}{121} x$

$\displaystyle \text{Difference between M.P. and C.P. } = x - \frac{88}{121} x = \frac{33}{121} x$

$\displaystyle \text{\% of difference in marked price over cost price } = \frac{\frac{33}{121}x}{\frac{88}{121}x} \times 100 = \frac{300}{8} = 37.5 \%$

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Question 12: Find a single discount equivalent to two successive discounts of $\displaystyle 20 \%$ and $\displaystyle 10 \%$

Answer:

Let the M.P. of the article be $\displaystyle 100 \text{ Rs. }$

$\displaystyle 1^{st}$ Discount is $\displaystyle 20 \%$ , Discount $\displaystyle = 20 \text{ Rs. }$

Reduced Price after the first discount $\displaystyle = Rs. (100-20) = 80 \text{ Rs. }$

$\displaystyle 2^{nd} \text{ discount is 10 \% of } 80 = \frac{10}{100} \times 80 = 8 \text{ Rs. }$

Price after the second discount $\displaystyle = 80 - 8 = 72 \text{ Rs. }$

S.P. of the article $\displaystyle 72 \text{ Rs. }$

Net discount $\displaystyle = 100 - 72 = 28$ Rs.

Thus, net discount on M.P. of Rs. $\displaystyle 100$ is $\displaystyle 28 \text{ Rs. }$

Single discount equivalent to given successive discounts $\displaystyle = 28 \%$

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Question 13: Find a single discount equivalent to two successive discounts of $\displaystyle 40 \%$ and $\displaystyle 5 \%$ .

Answer:

$\displaystyle \text{Let the M.P. of an article is Rs. } 100$

$\displaystyle \text{Discount on it is } = 40 \% \ of \ 100 = 40 \text{ Rs. }$

$\displaystyle \text{Reduced Price after first discount } = (100-40) = 60 \text{ Rs. }$

$\displaystyle \text{Next Discount is } = 5 \% \ of \ 60$

$\displaystyle ( \frac{5}{100} \times 60) = 3 \text{ Rs. }$

$\displaystyle \text{Price after second Discount } = 60-3 = 57 \text{ Rs. }$

$\displaystyle \text{That is S.P. } = 57 \text{ Rs. }$

$\displaystyle \text{Net Discount } = M.P. -S.P. = 100 -57 = 43 \text{ Rs. }$

Thus, net discount on MP. of $\displaystyle 100 \text{ Rs. }$ is $\displaystyle 43. \text{ Rs. }$

Single discount equivalent to given successive discounts = 43 \% $

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Question 14: Find a single discount equivalent to three successive discounts of $\displaystyle 20 \% , 5 \%$ and $\displaystyle 1 \%$

Answer:

$\displaystyle \text{Let the M.P. of an article in Rs. } 100$

$\displaystyle \text{It discount on it is } = 20 \%$ of $\displaystyle 100 = 20 \text{ Rs. }$

$\displaystyle \text{Reduced price after the its discount } = 100 - 20 = 80 \text{ Rs. }$

$\displaystyle \text{Second discount on it } = 5 \% of \ 80 = \frac{5}{100} \times 80 = 4 \text{ Rs. }$

$\displaystyle \text{Price after second discount } = 80-4 = 76 \text{ Rs. }$

$\displaystyle \text{Third discount on it is 1 \% of } 76 = \frac{1}{100} \times 76 = 0.76 \text{ Rs. }$

$\displaystyle \text{Price after Third discount } = 76-0.76 = 75.24 \text{ Rs. }$

$\displaystyle \text{ S.P. of the article } = 75.24 \text{ Rs. }$

$\displaystyle \text{ Thus, net discount on M.P. of Rs. } 100$ in Rs. $\displaystyle 24.76$

$\displaystyle \text{Single discount equivalent to given successive discounts } = 24.76 \%$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 8: Profit and Loss – Exercise 14C

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