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Profit & Loss
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CAT 2025 Lesson : Profit & Loss - Successive Mark-up & Discount

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5. Successive Mark ups or discounts

This is an extension of 4.7 Successive Changes section in Percentages lesson. The multiplication factors for the respective percentage changes can be successively multiplied.

Example 8

A manufacturer sells a product at a profit of 25%25 \% to the stockist, who in turn sells the product at a profit of 20%20 \% to a wholesaler, who in turn sells it at a profit of 33.33%33.33 \% to a retailer, who in turn sells it at a profit of 10%10 \% to the end customer. By what percentage is the customer's purchase price greater than the cost of manufacturing the product?

Solution

As all the numbers in the question are in percentage terms, we can assume a number. Let the cost of manufacturing the product be Rs. 100100.

At each stage, the value of the product is increasing by the said percentage. As the growths are successive, we can multiply them one after the other.

Customer’s purchase price=100×(1+25100)×(1+20100)×(1+33.33100)×(1+10100)\text{Customer's purchase price} = 100 \times \left(1 + \dfrac{25}{100} \right) \times \left(1 + \dfrac{20}{100} \right) \times \left(1 + \dfrac{33.33}{100} \right) \times \left(1 + \dfrac{10}{100} \right)

=100×54×65×43×1110== 100 \times \dfrac{5}{4} \times \dfrac{6}{5} \times \dfrac{4}{3} \times \dfrac{11}{10} = Rs. 220220

Customer's purchase price was higher by 220−100100×100%=120%\dfrac{220 - 100}{100} \times 100 \% = 120 \%

Answer: 120%120 \%

Example 9

A trader purchased an item at Rs. 500500. The trader then increased the price by 25%25 \% and reduced it by 4%4 \%. The trader once again increased the price by 25%25 \% and reduced it by 4%4 \% and then sold the item. What was the overall profit of the trader?

Solution

Like in the previous example, the mark-ups and discount factors can be successively multiplied to arrive at the final selling price.

Selling price =500×(1+25100)×(1−4100)×(1+25100)×(1−4100)= 500 \times \left(1 + \dfrac{25}{100} \right) \times \left(1 - \dfrac{4}{100} \right) \times \left(1 + \dfrac{25}{100} \right) \times \left(1 - \dfrac{4}{100} \right)

=500×54×2425×54×2425=500×3625=720= 500 \times \dfrac{5}{4} \times \dfrac{24}{25} \times \dfrac{5}{4} \times \dfrac{24}{25} = 500 \times \dfrac{36}{25} = 720

Profit =720−500== 720 - 500 = Rs. 220220

Answer: Rs. 220220

6. Common Types

For the following common types, applicable formulae and their derivation are provided. If you are clear with the concepts detailed, you will be able to derive these formulae in a matter of seconds. Therefore, you needn't memorise them.

6.1 When Discount and Mark up percentages are equal

Let x%x \% be the discount and marked up percentage of the product.

SP=CP(1+x100)(1−x100)\text{SP} = \text{CP} \left(1 + \dfrac{x}{100} \right) \left(1 - \dfrac{x}{100} \right)

⇒ SP=CP(1−x21002)\text{SP} = \text{CP}\left(1 - \dfrac{x^{2}}{100^{2}} \right)

In this scenario, a loss will always be occurred and the loss would equal x2100%\dfrac{x^{2}}{100} \%

Example 10

John buys a product at Rs. 500500 and marks it up by 15%15 \%. After extensive bargaining with a customer, he sells it to her at a discount of 15%15 \% on the marked up price. What is the overall percentage of profit or loss incurred by John?

(1) 1.50%1.50 \%            (2) 1.75%1.75 \%            (3) 2.25%2.25 \%            (4) 3.00%3.00 \%           

Solution

Directly applying the formula, we get Loss %=152100%=\% = \dfrac{15^{2}}{100} \% = 2.25%\bm{2.25 \%}

Alternatively

SP =500×(1+15100)×(1−15100)=500×(1−1521002)= 500 \times \left(1 + \dfrac{15}{100} \right) \times \left(1 - \dfrac{15}{100} \right) = 500 \times \left(1 - \dfrac{15^{2}}{100^{2}} \right)

⇒ SP=500×(1−0.0225)⟶(1) \text{SP} = 500 \times (1 - 0.0225) \longrightarrow (1)

As SP=CP×(1−Loss%)\text{SP} = \text{CP} \times (1 - \text{Loss} \%), from equation (1)(1) we get

Loss %=0.0225×100%=\% = 0.0225 \times 100 \% = 2.25%\bm{2.25 \%}

Answer: (3) 2.25%2.25 \%

6.2 Two products with same SP and same Profit/Loss percentage

When the selling price of two products is the same and one is sold at a profit of x%x \% and the other at a loss of x%x \%, then the two transactions combined would have resulted in a loss of x2100%\bm{\dfrac{x^{2}}{100} \%}

Example 11

Pratap trades in motorcycles. On a certain day he sells two motorcycles for Rs. 480480 each. If one was sold at a profit of 20%20 \% and the other at a loss of 20%20 \%, what was his overall profit/loss from the transaction?

Solution

Total Selling Price (SP) =480×2=960= 480 \times 2 = 960

Overall loss =202100%CP=4%= \dfrac{20^{2}}{100} \% \text{CP} = 4 \% of CP =0.04CP= 0.04 \text{CP}

CP−0.04CP=960\text{CP} - 0.04 \text{CP} = 960
⇒0.96CP=960⇒ 0.96 \text{CP} = 960
⇒CP=1000⇒ \text{CP} = 1000

Overall loss =0.04CP=0.04×1000== 0.04 \text{CP} = 0.04 \times 1000 = 4040

Answer: Loss of Rs. 4040

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