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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 pure copper mixed in the alloy C?
3 kg
4 kg
5 kg
7 kg
Ratio of copper and tin in alloy A = 2:3
Ratio of copper and tin in alloy B = 3:4
20 kg taken from A: Copper = 8 kg and tin = 12 kg 28 kg taken from B: Copper = 12 kg and tin =16 kg
This is mixed with some pure cooper = x kg
Ratio of copper and tin in alloy C = 6:7 Total copper in alloy C/ total tin in alloy C = 6/7
(8 + 12 + x)/ (12 + 16) = 6/7
(20 + x)/ 28 = 6/7
x = 4 kg
By: Munesh Kumari ProfileResourcesReport error
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