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

$\fbox\orange{Required\:Answer:}$

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

$= (50 \times \dfrac{60}{100} )kg = 30kg.$

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

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

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

Percentage of lead in new alloy

$= 【 \: \: \dfrac{(30 + x)}{(50 + x)} \times 100 \: 】 \%$

$∴ \dfrac{(30 + x)}{(50 + x)} \times 100 = 75$

$= \dfrac{(30 + x)}{(50 + x)} = \dfrac{75}{100} = \dfrac{(30 + x)}{(50 + x)} = \dfrac{3}{4}$

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

$= 120 + 4x = 150 + 3x$

$= 4x - 3x = 150 - 120$

$= x = 30$

∴ quantity of lead to be added = 30kg .

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Answer 2

Step-by-step explanation:

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

= (50 \times \dfrac{60}{100} )kg = 30kg.=(50×

100

60

)kg=30kg.

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

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

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

Percentage of lead in new alloy

= 【 \: \: \dfrac{(30 + x)}{(50 + x)} \times 100 \: 】 \%=【

(50+x)

(30+x)

×100】%

∴ \dfrac{(30 + x)}{(50 + x)} \times 100 = 75∴

(50+x)

(30+x)

×100=75

= \dfrac{(30 + x)}{(50 + x)} = \dfrac{75}{100} = \dfrac{(30 + x)}{(50 + x)} = \dfrac{3}{4}=

(50+x)

(30+x)

=

100

75

=

(50+x)

(30+x)

=

4

3

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

= 120 + 4x = 150 + 3x=120+4x=150+3x

= 4x - 3x = 150 - 120=4x−3x=150−120

= x = 30=x=30

∴ quantity of lead to be added = 30kg .

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