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An alloy A contains two elements, copper and tin in the ratio of 2 ∶ 3, whereas an alloy B contains the same elements in the ratio of 3 ∶ 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 ∶ 7, then what is the quantity of the pure copper mixed in the alloy C?

An alloy A contains two elements, copper and tin in the ratio of 2 ∶ 3, whereas an alloy B contains the same elements in the ratio of 3 ∶ 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 ∶ 7, then what is the quantity of the pure copper mixed in the alloy C? Correct Answer 4 kg

GIVEN :

An alloy A contains two elements, copper and tin in the ratio of 2 ∶ 3

whereas an alloy B contains the same elements in the ratio of 3 ∶ 4

If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C which now contains copper and tin in the ratio of 6 ∶ 7

 

ASSUMPTION :

Let the amount of pure copper added be ‘x’ kg.

 

CALCULATION :

Amount of copper in 20 kg of alloy A = 2/ (2 + 3) × 20 = 8 kg

Amount of copper in 28 kg of alloy B = 3/ (3 + 4) × 28 = 12 kg

Total amount of copper in alloy C = 8 + 12 + x = 20 + x

Total amount of alloy C produced = 20 + 28 + x = 48 + x

⇒ 20 + x = 6/ (6 + 7) × (48 + x)

⇒ 20 + x = 6/13 × (48 + x)

⇒ 260 + 13x = 288 + 6x

⇒ 7x = 28

∴ x = 4 kg

 

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