soujanyareddy13
asked
in Quantitative Aptitude
Jan 20, 2022
retagged
Mar 23, 2022
by Lakshman Patel RJIT

256 views
1 vote

If a certain weight of an alloy of silver and copper is mixed with $3 \; \text{kg}$ of pure silver, the resulting alloy will have $90 \%$ silver by weight. If the same weight of the initial alloy is mixed with $2 \; \text{kg}$ of another alloy which has $90 \%$ silver by weight, the resulting alloy will have $84 \%$ silver by weight. Then, the weight of the initial alloy, in kg, is

- $4$
- $2.5$
- $3$
- $3.5$

1 vote

Let the alloy contain $a \; \text{kg}$ silver and $b \; \text{kg}$ copper.

Now, when mixed with $3 \; \text{kg}$ pure silver.

$\Rightarrow \frac{a+3}{a+b+3} = \frac{90}{100}$

$\Rightarrow \frac{a+3}{a+b+3} = \frac{9}{10}$

$\Rightarrow 10a + 30 = 9a + 9b + 27$

$\Rightarrow a – 9b = \;– 3 \quad \longrightarrow (1)$

Now, silver in $2^{\text{nd}}$ alloy $ = 2 \left( \frac{90}{100} \right) = \frac{9}{5} = 1.8 \; \text{kg}$

And, $\frac{a+1.8}{a+b+2} = \frac{84}{100}$

$\Rightarrow \frac{a+1.8}{a+b+2} = \frac{21}{25}$

$\Rightarrow 25a + 45 = 21a + 21b + 42$

$\Rightarrow 4a – 21b =\; – 3 \quad \longrightarrow (2)$

From equation $(1)\; \&\; (2).$

$\begin{array} {} (a – 9b = -3) \times 4 \\ (4a – 21b = - 3) \times 1 \\\hline 4a – 36b = -12 \\ 4a – 21b = -3 \\ \; – \quad + \qquad \; +\\\hline \end{array}$

$\Rightarrow -15b = -9$

$\Rightarrow \boxed{b = \frac{3}{5} = 0.6 \; \text{kg}}$

Put the value of $b$ is equal in $(1),$ we get.

$\Rightarrow a – 9 \left(\frac{3}{5} \right) = -3$

$\Rightarrow 5a – 27 = -15$

$\Rightarrow 5a = 12$

$\Rightarrow a = \frac{12}{5}$

$\Rightarrow \boxed{a = 2.4 \; \text{kg}}$

$\therefore$ The weight of the initial alloy $ = a+b = 0.6 + 2.4 = 3 \; \text{kg}.$

Correct Answer $: \text{C}$

Now, when mixed with $3 \; \text{kg}$ pure silver.

$\Rightarrow \frac{a+3}{a+b+3} = \frac{90}{100}$

$\Rightarrow \frac{a+3}{a+b+3} = \frac{9}{10}$

$\Rightarrow 10a + 30 = 9a + 9b + 27$

$\Rightarrow a – 9b = \;– 3 \quad \longrightarrow (1)$

Now, silver in $2^{\text{nd}}$ alloy $ = 2 \left( \frac{90}{100} \right) = \frac{9}{5} = 1.8 \; \text{kg}$

And, $\frac{a+1.8}{a+b+2} = \frac{84}{100}$

$\Rightarrow \frac{a+1.8}{a+b+2} = \frac{21}{25}$

$\Rightarrow 25a + 45 = 21a + 21b + 42$

$\Rightarrow 4a – 21b =\; – 3 \quad \longrightarrow (2)$

From equation $(1)\; \&\; (2).$

$\begin{array} {} (a – 9b = -3) \times 4 \\ (4a – 21b = - 3) \times 1 \\\hline 4a – 36b = -12 \\ 4a – 21b = -3 \\ \; – \quad + \qquad \; +\\\hline \end{array}$

$\Rightarrow -15b = -9$

$\Rightarrow \boxed{b = \frac{3}{5} = 0.6 \; \text{kg}}$

Put the value of $b$ is equal in $(1),$ we get.

$\Rightarrow a – 9 \left(\frac{3}{5} \right) = -3$

$\Rightarrow 5a – 27 = -15$

$\Rightarrow 5a = 12$

$\Rightarrow a = \frac{12}{5}$

$\Rightarrow \boxed{a = 2.4 \; \text{kg}}$

$\therefore$ The weight of the initial alloy $ = a+b = 0.6 + 2.4 = 3 \; \text{kg}.$

Correct Answer $: \text{C}$