Question

50 kg of an alloy of lead and tin contains 60% of lead. How much lead must be
melted into it to make the alloy contain 75% of lead.

Answer 1

Answer:

30 kg

Step-by-step explanation:

Quantity of lead in 50 kg of alloy = 60% of 50 kg

= (50 \times \frac{60}{100})(50×10060) 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 = 7550+x30+x×100=75

⇒ \frac{(30 + x)}{(50 + x)} =\frac{75}{100}(50+x)(30+x)=10075

⇒ \frac{(30+x)}{(50+x)} =\frac{3}{4}(50+x

)(30+x)=43

⇒ 4(30 + x) = 3(50 + x)

⇒ 120 + 4x = 150 + 3x

⇒ 4x - 3x = 150 - 120

⇒ x = 30

∴ quantity of lead to be added = 30 kg.