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A metal of 50000/77 kg/m3 is used to make the weights for measuring. But the dishonest shopkeeper purchases a tampered weight. The original weight should be a cylinder of 7 cm radius and 5 cm height. The tampered weight also looks same but inside the cylinder there is a hollow cylindrical region of 2.8 cm radius and 2.5 cm height. If honestly he could make 20% profit, what is the profit (%) he makes by dishonest method?

A metal of 50000/77 kg/m3 is used to make the weights for measuring. But the dishonest shopkeeper purchases a tampered weight. The original weight should be a cylinder of 7 cm radius and 5 cm height. The tampered weight also looks same but inside the cylinder there is a hollow cylindrical region of 2.8 cm radius and 2.5 cm height. If honestly he could make 20% profit, what is the profit (%) he makes by dishonest method? Correct Answer 30 10/23

Original volume of standard weight = πr2h

⇒ 22/7 × 72 × 5

⇒ 770 cm3

Volume of metal in the tampered weight = Original volume – Volume of cavity

⇒ 770 – 22/7 × 2.82 × 2.5

⇒ 708.4 cm3

% volume filled in tampered weight = 708.4/770 × 100

⇒ 92%

That is if he sells 100 grams, he actually sells only 92 grams

Let cost price of one gram be Rs. x.

Selling price = (100 + Profit%)/100 × Cost price

⇒ (100 + 20)/100 × x

⇒ 1.2x

∴ Profit% = (Selling price of 100 grams – Cost price of 92 grams)/Cost price of 92 grams × 100

⇒ (1.2x × 100 – x × 92) / (x × 92) × 100

⇒ 2800/92

⇒ 30 10/23%

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