Question

a shopkeeper bought a car for 1,50,000. He spent 10,000 for its painting and repairing and then sold it for 2,00,000. find his profit or loss

Answer 1

Answer:

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Answer 2

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Given :-

• Purchase price of Car = ₹ 1, 50, 000
• Amount spent on painting and repairing = ₹ 10, 000
• Selling Price of Car = ₹ 2, 00, 000

To find :-

• Profit or Loss

Formula used :-

● When SP > CP, then there is a profit in transaction and Profit is evaluated as

• ◇ Profit = Selling Price - Cost Price

● When CP > SP, then there is a loss in transaction and Loss is evaluated as

• ◇ Loss = Cost Price - Selling Price

Solution:-

Purchase price of Car = ₹ 1, 50, 000

Amount spent on painting and repairing of car = ₹ 10, 000

⟹ Total Amount spent on car = ₹ 1, 50, 000 + ₹ 10, 000 = ₹ 1, 60, 000

⟹ Cost Price of Car = ₹ 1, 60, 000

& Selling Price of Car = ₹ 2, 00, 000

■ Since SP > CP

∴ In this case, there is profit

So, Profit = Selling Price - Cost Price

⟹ Profit = 2, 00, 000 - 1, 60, 000 = ₹ 40, 000

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Enhance your knowledge

Profit and Loss Basic Concepts :-

• Profit, P = SP – CP; if SP>CP

• Loss, L = CP – SP; if CP>SP

• P% = (P/CP) x 100

• L% = (L/CP) x 100

• SP = {(100 + P%)/100} x CP

• SP = {(100 – L%)/100} x CP

• CP = {100/(100 + P%)} x SP

• CP = {100/(100 – L%)} x SP

• Discount = MP – SP

• SP = MP -Discount

• For false weight, profit percentage will be P% = (True weight – false weight/ false weight) x 100.

• When there are two successful profits say m% and n%, then the net percentage profit equals to (m+n+mn)/100

• When the profit is m% and loss is n%, then the net % profit or loss will be: (m-n-mn)/100

• If a product is sold at m% profit and then again sold at n% profit then the actual cost price of the product will be: CP = [100 x 100 x P/(100+m)(100+n)]. In case of loss, CP = [100 x 100 x P/(100-m)(100-n)]

• If P% and L% are equal then, P = L and loss % = P²/100