Question

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

Answer 1

Answer:

Answer is 30 kg ,

Step-by-step explanation:

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Answer 2

_________________________

Quantity of lead in 50 kg of alloy

=> 60% of 50 kg

\begin{gathered} \lgroup 50 \times \frac{60}{100} \rgroup \: kg \\ \\ \lgroup 5 \cancel0 \: \times \frac{6 \cancel0}{1 \cancel0\cancel0 }\rgroup \\ \\ = > \bold{ 30kg}\end{gathered}

⟮50×

100

60

⟯kg

⟮5

0

×

1

0

0

6

0

⟯

=>30kg

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

\begin{gathered} = \lbrace \frac{(30 + x)}{(50 + x)} \times 100 \rbrace \% \\ \\ \therefore \: \frac{(30 + x)}{(50 + x)} \times 100 = 75 \\ \\ = > \frac{(30 + x)}{(50 + x)} = \frac{ \cancel{75} \: {}^{3} }{ \cancel{100} \: {}^{4} } \\ \\ = > \frac{(30 + x)}{(50 + x)} = \frac{3}{4} \\ \\ = > 4(30 + x) = 3(50 + x) \\ \\ = > 120 + 4x = 150 + 3x \\ \\ = > 4x - 3x = 150 - 120 \\ \\ = > x = 30\end{gathered}

={

(50+x)

(30+x)

×100}%

∴

(50+x)

(30+x)

×100=75

=>

(50+x)

(30+x)

=

100

4

75

3

=>

(50+x)

(30+x)

=

4

3

=>4(30+x)=3(50+x)

=>120+4x=150+3x

=>4x−3x=150−120

=>x=30

Therefore,

Quantity of lead to be added = 30 kg.

__________________________

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