Question

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

Answer 1

Answer:

Step-by-step explanation:

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

∴

⇒

⇒

⇒ 4(30 + x) = 3(50 + x)

⇒ 120 + 4x = 150 + 3x

⇒ 4x - 3x = 150 - 120

⇒ x = 30

∴ quantity of lead to be added = 30 kg.

Answer 2

$\huge\rm{Answer:}$

___..

$\huge\sf\blue{Question-}$

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

$\huge\sf\green{Given-}$

• 50 kg of an alloy of lead and tin contains 60% lead.

$\huge\sf\orange{To \: Find-}$

• How much quantity of lead must be added?

$\huge\sf\purple{Solution-}$

Total quantity of lead in 50kg of alloy = 60% of 50 kg

$\sf {= (50 \times \frac{60}{100} )kg = 30kg}$

Now,

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

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

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

Percentage of new alloy =

$\sf {[ \frac{(30 + x)}{(50 + x)} \times 100 ] \%}$

Therefore,

$\sf { \frac{(30 + x)}{(50 + x)} \times 100 = 75}$

$\sf { \longrightarrow \frac{(30 + x)}{(50 + x)} = \frac{75}{100}}$

$\sf { \longrightarrow \frac{(30 + x)}{(50 + x)} = \frac{3}{4}}$

$\sf { \longrightarrow 4(30 + x) = 3(50 + x)}$

[By cross multiplication]

$\sf { \longrightarrow 120 + 4x = 150 + 3x}$

$\sf { \longrightarrow 4x - 3x = 150 - 120}$

$\sf { \longrightarrow x = 30}$

Therefore, the required quantity of lead to be added = 30 kg.