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08 Dec 2021, 03:45
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There are two alloys, A and B, of copper and zinc. The ratio (by weight) of copper and zinc in alloy A is 8 : 1 and that in alloy B is 2 : 7. It is found that, if alloy A and alloy B are mixed in a certain ratio, the weights of copper and zinc in the resultant alloy are also in that ratio. What is that ratio?
(A) 4 : 3
(B) 3 : 2
(C) 5 : 3
(D) 2 : 1
(E) 3 : 1
Are You Up For the Challenge: 700 Level Questions
(A) 4 : 3
(B) 3 : 2
(C) 5 : 3
(D) 2 : 1
(E) 3 : 1
Are You Up For the Challenge: 700 Level Questions
08 Dec 2021, 04:51
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Let, the ratio is K:1
SO, Copper : Zinc = 8k+2 : k+7
Now,
8k+2/k+7= k/1
solve for k, k=2
so the ratio is k:1=2:1
SO, Copper : Zinc = 8k+2 : k+7
Now,
8k+2/k+7= k/1
solve for k, k=2
so the ratio is k:1=2:1
16 Nov 2024, 09:26
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Bunuel
Given,
Alloy A: Copper - 8 Parts; Zinc - 1 Part --> 8x : 1x
Alloy B: Copper - 2 Parts; Zinc - 7 Parts --> 2y : 7y
Given,
If alloy A and alloy B are mixed in a certain ratio, the weights of copper and zinc in the resultant alloy are also in that ratio.
Writing the equation ->
\(\frac{Copper }{ Zinc }\) -> \(\frac{(8x + 2y) }{ (x + 7y)} =\frac{ x}{y } \)
Expanding further,
\(8xy + 2y^2 = x^2 + 7xy\)
\(x^2 - xy - 2y^2 = 0\)
Finding the roots of quadratic equation ->
\((x+y)(x-2y) = 0\)
For the above to hold true (x+y) or (x-2y) or both the terms need to be 0.
(x+y) cannot be 0 here because the ratio is positive.
Therefore, \((x-2y)=0\) -> \(x=2y\) -> \(\frac{x}{y}=\frac{2}{1}\)
Ratio is therefore 2:1!
