Arithmetic Chapter 2 Profit And Loss
⇒ Cost price: The amount paid to purchase an article is known as its cost price.
⇒ Selling price: The price at which an article is sold is known as its selling price.
⇒ The cost price and selling price are abbreviated as C. P. and S. P. respectively.
⇒ Profit: If S.P. > C. P. then the difference between S. P. and C. P. is called profit.
⇒ Loss: If C. P. > S. P. then the difference between C. P. and S. P. is called loss.
∴ Profit = Selling price (S. P.) – Cost price (C. P.) and Loss Cost price (C. P.) – Selling price (S. P.)
⇒ Marked price: While buying goods we have seen that on every article there is a price marked. This price is known as the marked price of the article.
⇒ Discount = Marked price x Rate of discount
⇒ S. P. = Marked price – Discount
⇒ Profit percentage = \(=\frac{\text { Total profit }}{\text { Cost price }}\) x 100
⇒ Loss percentage = \(=\frac{\text { Total loss }}{\text { Cost price }}\) x 100
⇒ Equivalent discount: On a particular principal the equivalent discount is equal to more than one successive discount on that principal.
⇒ The discount equivalent to successive discounts of a% and b% is (a + b – \(\frac{ab}{100}\))%
Read and Learn More WBBSE Solutions For Class 9 Maths
Arithmetic Chapter 2 Profit And Loss True Or False
Example 1. If selling price of an article is smaller than the cost price, then there will be profit.
Solution: The statement is False.
Example 2. If S.P. and C.P. of an article are equal then there will be no profit or loss.
Solution: The statement is True.
Example 3. Ram purchased a pen for ₹50 and sold it to Shyam at a loss of 20% then the selling price of that pen was ₹40.
Solution: C.P. = ₹50 and S.P. = ₹40
Loss = ₹(50 – 40) = ₹10
Loss percentage = \(\frac{10}{50}\) x 100 = 20
∴ The statement is True.
Arithmetic Chapter 2 Profit And Loss Fill In The Blanks
Example 1. Profit percentage x ________ = Total profit x 100.
Solution: cost price (C. P.).
Example 2. Cost price = \(=\frac{100 \times}{100-\text { Loss percentage }}\)
Solution: Selling price.
Example 3. There is a ________ relation between cost price and selling price
Solution: Direct.
Arithmetic Chapter 2 Profit And Loss Short Answer Type Questions
Example 1. If 20% profit is on cost price, what is profit percentage on selling price?
Solution: If cost price is ₹100 then profit is ₹20
∴ Selling price = ₹(100+ 20) = ₹120
⇒ If selling price is ₹120 then profit is ₹20
⇒ If selling price is ₹1, then profit is ₹\(\frac{20}{120}\)
⇒ If selling price is ₹100, then profit is ₹\(\frac{20}{120}\) x 100
= ₹\(\frac{50}{3}\) = ₹16\(\frac{2}{3}\)
∴ Profit is 16\(\frac{2}{3}\)% on selling price.
Example 2. If 20% profit is on selling price, what is the profit percentage on cost price?
Solution: If S.P. is ₹100 then profit is ₹20
∴ C.P. = ₹(100 – 20) = ₹80
⇒ If C.P. is ₹80, then profit is ₹20
⇒ If C.P. is ₹1, then profit is ₹\(\frac{20}{80}\)
⇒ If C.P. is ₹100, then profit is ₹(\(\frac{20}{80}\) x 100) = ₹25
∴ 25% profit on cost price.
Example 3. By selling 110 mangoes, if the cost price of 120 mangoes has been got, what will be the profit percentage?
Solution: Let S.P. of 110 mango is ₹x
⇒ S.P. of 1 mango is ₹\(\frac{x}{110}\)
⇒ According to condition, C.P. of 120 mangoes is ₹x [x > 0]
∴ C.P. of 1 mango is ₹\(\frac{x}{120}\)
⇒ Profit = \(₹\left(\frac{x}{110}-\frac{x}{120}\right)\)
= \(₹\left(\frac{12 x-11 x}{1320}\right)=₹ \frac{x}{1320}\)
⇒ If C.P. is \(₹ \frac{x}{120}\) then profit is \(₹ \frac{x}{1320}\)
⇒ If C.P. is 1, then profit is \(₹\left(\frac{x}{1320} \times \frac{120}{x}\right)\)
⇒ If C.P. is 100, then profit is \(₹\left(\frac{120 \times 100}{1320}\right)\) = \(\frac{100}{11}\) = 9\(\frac{1}{11}\)
∴ Profit is 9\(\frac{1}{11}\).
Example 4. To submit electricity bill in due time, 15% discount can be obtained. Sumanbabu has got 54 as discount for submission of electricity bill in due time. How much was his electricity bill?
Solution: Let, the electricity bill of Sumanbabu was ₹x
⇒ Discount = \(₹\left(x \times \frac{15}{100}\right)=₹ \frac{3 x}{20}\)
⇒ According to question, \(₹ \frac{3 x}{20}\) = 54
⇒ x = \(\frac{54 \times 20}{3}\)
⇒ x = 360
∴ The electricity bill was ₹360
Example 5. A commodity is sold at ₹480 with a loss of 20% on selling price, what is the cost price of the commodity?
Solution: If selling price of a commodity is ₹100 then loss is ₹20.
⇒ C.P. = ₹(100+20) = ₹120
⇒ If S.P. is ₹100, then C.P. is ₹120
⇒ If S.P. is ₹1 then C.P. is \(₹ \frac{120}{100}\)
⇒ If S.P is ₹480 then C.P. is ₹\(\frac{120 \times 480}{100}\) = ₹576
∴ Cost price of the commodity is ₹576.
Example 6. If a commodity is sold with successive discounts of 20% and 10%, what will be the equivalent discount?
Solution: Let the marked price of the commodity is ₹100
⇒ Then first discount is ₹20
⇒ The net price after 1st discount = ₹(100 – 20) = ₹80
⇒ Second discount = 10% of ₹80 = \(₹ \left(80 \times \frac{10}{100}\right)\) = ₹8
⇒ Total discount = ₹(20 + 8) = ₹28
∴ The equivalent discount is 28%.
Example 7. By selling a clock for 180, Rohit loses 10%, for what amount should be sell it as to gain 10%. [By proportion]
Solution: The loss is 10%
If C.P. of the clock is ₹100, then S.P. will be ₹(100 – 10) = ₹90
In mathematical language, the problem is,
⇒ S.P.(₹)
90
180
⇒ C.P.(₹)
100
?
⇒ The relation between S.P. and C.P. is direct.
∴ The direct proportion is, 90: 180 100: ? (The required C.P.)
∴ The required cost price = \(₹ \frac{180 \times 100}{90}=₹ 200\)
⇒ Rohit wants to make 10% profit
⇒ In mathematical language, the problem is
C.P(₹)
100
200
S.P. (₹)
100+ 10 = 110
?
⇒ The relation between S.P. and C.P. is direct
∴ The direct proportion is, 100: 200 : : 100:? (The required C.P.)
∴ The required selling price = \(₹ \frac{200 \times 110}{100}\) = ₹220
∴ To get 10% profit, Rohit has to sell the clock at ₹220
Example 8. Some toffees are bought at 15 for a rupee and the same number at 10 a rupee. Find the gain or loss percent.
Solution: C.P. of 15 toffees is ₹1
⇒ C.P. of 1 toffee is ₹\(\frac{1}{5}\)
⇒ S.P. of 10 toffees is ₹1
⇒ S.P. of 1 toffee is ₹\(\frac{1}{5}\)
⇒ Profit = \(₹\left(\frac{1}{10}-\frac{1}{15}\right)=₹ \frac{1}{30}\)
⇒ Profit % = \(\frac{\frac{1}{30}}{\frac{1}{15}} \times 100=\frac{15}{30} \times 100=50\)
∴ The percentage profit of toffees is 50.
Example 9. Akash sells a shirt at a loss of 20%. Had he sold the shirt for 200 more, he would have earned a profit of 5%. Determine the cost price of the shirt.
Solution: Let, the C.P. of the shirt is ₹x [x > 0]
⇒ Loss = 20% of ₹x = \(₹\left(x \times \frac{20}{100}\right)=₹ \frac{x}{5}\)
⇒ S.P = \(₹\left(x-\frac{x}{5}\right)=₹ \frac{4 x}{5}\)
⇒ Had Akash sold the shirt for ₹200 more, i.e. \(₹\left(\frac{4 x}{5}+200\right)\), he would have earned a profit of 5%
∴ x + x x \(\frac{5}{100}\) = \(\frac{4x}{5}\) + 200
⇒ x + \(\frac{x}{20}\) – \(\frac{4x}{5}\) = 200
⇒ \(\frac{20 x+x-16 x}{20}=200\)
⇒ 5x = 200 x 20
⇒ x = \(\frac{200 \times 20}{5}\)
⇒ x = 800
∴ Cost price of the shirt is ₹800
Example 10. A dishonest businessman defrauds by false balance, to the extent of 10% both in buying and in selling his goods. Find the actual gain percent of the businessman.
Solution: Since the businessman defrauds to the extent of 10% in buying.
⇒ So he takes goods of ₹110 in exchange of ₹100
⇒ Again, the businessman sells goods of ₹100 at ₹110
⇒ the businessman sells goods of ₹1 at ₹\(\frac{110}{100}\)
⇒ the businessman sells goods of ₹110 at \(₹ \frac{110 \times 110}{100}\) = ₹121
∴ The businessman finally gets ₹121 in exchange of ₹100
∴ The actual gain percent = (121 – 100) = 21.