Question

50 kg of an alloy of lead and tin contains 60 percent of lead. how much lead must be melted into it to make the alloy contain 75 percent of lead? (please solve it by unitary method)​

Answer 1

Step-by-step explanation:

Quantity of lead in 50 kg of alloy = 60% of 50 kg = (50 \times \frac{60}{100})(50×

100

60

) kg = 30 kg.

Let the required quantity of lead to be added be x kg.

Then, weight of lead = (30 + x) kg.

And, weight of new alloy = (50 + x) kg.

Percentage of lead in new alloy = [\frac{(30 + x)}{(50 + x)} \times 100][

(50+x)

(30+x)

×100] %

∴ \frac{30 + x}{50 + x} \times 100 = 75

50+x

30+x

×100=75

⇒ \frac{(30 + x)}{(50 + x)} =\frac{75}{100}

(50+x)

(30+x)

=

100

75

⇒ \frac{(30+x)}{(50+x)} =\frac{3}{4}

(50+x)

(30+x)

=

4

3

⇒ 4(30 + x) = 3(50 + x)

⇒ 120 + 4x = 150 + 3x

⇒ 4x - 3x = 150 - 120

⇒ x = 30

∴ quantity of lead to be added = 30 kg

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