an alloy of lead and tin weighs 78 kg and contains 67 of tin how much lead must be melted into to make the alloy contain 60 lead
Answers
Answer:
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.
Step-by-step explanation:
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Answer:
your question is incomplete
Step-by-step explanation:
are 60 and 67 in % or what cantreally understand the question sorry i wasn't able to answer you.