Question

Of the two kinds of alloys, silver and copper are contained in the ratio of 5 : 1 in the first and 7 : 2 in the second, what weight of the second alloy should be melted with first alloy to make up a 5 kg mass with 80% of silver?

Answer 1

Therefore the of second alloy should be = 7.5 gram.

Explanation:

Let the weight of second alloy be x

The amount of silver in 5 kg of first alloy is$=5\times \frac{5}{6}$ $=\frac{25}{6}$ Kg

The amount of silver in x Kg of second alloy is $=x\times \frac{7}{9}$  $=\frac{7x}{9}$ Kg

According to the problem,

$\frac{\frac{25}{6} +\frac{7x}{9} }{x+5} =\frac{80}{100}$

$\Rightarrow \frac{150+28x}{36(x+5)} =\frac{4}{5}$

$\Rightarrow 750 +140x=144x+720$

⇒4x=30

⇒x=7.5

Therefore the of second alloy should be = 7.5 gram.

Answer 2

Explanation:

Let the weight of second alloy be x

The amount of silver in 5 kg of first alloy is=5\times \frac{5}{6}=5×

6

5

=\frac{25}{6}=

6

25

Kg

The amount of silver in x Kg of second alloy is =x\times \frac{7}{9}=x×

9

7

=\frac{7x}{9}=

9

7x

Kg

According to the problem,

\frac{\frac{25}{6} +\frac{7x}{9} }{x+5} =\frac{80}{100}

x+5

6

25

+

9

7x

=

100

80

\Rightarrow \frac{150+28x}{36(x+5)} =\frac{4}{5}⇒

36(x+5)

150+28x

=

5

4

\Rightarrow 750 +140x=144x+720⇒750+140x=144x+720

⇒4x=30