The value of an alloy varies directly with the concentration of metal

Hi Nisha0987,

Please find the solution for the problem below.

Let the,
value of alloy be = v
concentration of metal A = a
concentration of metal B = b

\(v \propto \frac{a}{b^2}\)

\(v = \frac{ka}{b^2}\) (where 'k' is a constant)

Given:
b = 1.3b

∴ \(v' = \frac{ka}{(1.3b)^2}\)

\(v' = \frac{ka}{1.69b^2}\)

We are asked by approximately what percentage should the concentration of metal A be increased to avoid a drop in the value of the alloy:

since, \(v = \frac{ka}{b^2}\)

In order to make v' = v, if 'a' gets increased by 69% i.e a = 1.69a, then:

\(v' = \frac{k*1.69a}{1.69b^2} = \frac{ka}{b^2} = v\)

∴ if concentration of metal A is increased by 69% i.e approximately by 70% then the value of the alloy will not drop. (D)

I hope this helps.

Link to EG video solution-

From the question we can form the following equation -

­V = \(\frac{K(CA)}{CB^2}\)

V = Value of alloy
CA = Concentration of metal A
CB = Concentration of metal B
K = Constant­

Let' assume the value for CA and CB as '100'

V = \(\frac{K(100)}{100^2}\)
V = k(0.01)

Now, the question states that the concentration of metal B has been increased by 30%, so value of CB will be = 100+30=130. But as the value of V needs to remain the same, we can form the following equation and calculate the value of CA -

V = k(0.01) = \(\frac{K(CA)}{130^2}\)

0.01 = \(\frac{(CA)}{130^2}\)

CA = 0.01 x 130 x 130 = 169 ~ 170

Now we need to calculate the percentage change for CA, to check how much it has increased to maintain the value of V.

% Change for CA = \(\frac{FV-IV}{IV}\) x 100
= \(\frac{170-100}{100}\) x 100
= 70%

­

KarishmaB Bunuel could you please explain how to solve this question logically? Thanks

As per given direct and inverse relations,

\(\frac{V * Cb^2}{Ca} = Constant\)

If Cb increases by 30%, it becomes 1.3Cb. When it is squared it becomes 1.69Cb^2. To keep the value same, we should multiply Ca by 1.69 too.

\(\frac{V * Cb^2}{Ca} = \frac{V * 1.69Cb^2}{1.69Ca}\)

Hence Ca increases by 69%. Answer (D)

Direct and Inverse Variation is discussed here: https://youtu.be/AT86tjxJ-f0

I understand now. thank you!