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:

Given:-

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

To Find:-

How much lead must be melted into it to make the alloy contain 75% of lead.

Solution:-

Lead contents in 100 kg alloy = 60 kg

$\rm \therefore \: Lead \: contents \: in \: 50 \: kg \: alloy \\ \looparrowright \: \bigg( \frac{60}{100} \times 50 \: kg \bigg) = 30 \: kg$

Let x kg of lead be melTed into the alloy to make it an alloy containing 75% lead.

♠ Weight of the new alloy = (50 + x) kg

♠ Weight of lead in the new alloy =(30 + x) kg

Now,

Lead contents in (50 + x) kg alloy = (30 + x) kg

$\rm \: \bigstar \: Lead \: contents \: in \: one \: kg \: alloy \\ \hookrightarrow \: \rm \bigg( \frac{30 + x}{ 50 + x} \bigg ) \\ \rm \: \bigstar Lead \: contents \: in \: 100 \: kg \: alloy \\ \twoheadrightarrow \rm \bigg( \frac{30 + x}{50 + x} \bigg) \times 100 \: kg \\ \rm \bigstar \: Percentage \: of \: lead \: in \: new \: alloy \\ \rm \leadsto \: \bigg( \frac{30 + x}{5 0 + x} \times 100 \bigg)\%$

But, The percentage of lead in the new alloy is given as 75%

So,

ACQ

$\therefore \: \rm \frac{30 + x}{50 + x} \times 100 = 75 \\ \dashrightarrow \rm \: 100(30 + x) = 75(50 + x) \\ \dashrightarrow \rm \: 300 + 100x = 3750 + 75x \\ \dashrightarrow \rm \: 100x - 75x = 3750 - 3000 \\ \dashrightarrow \rm \: 25x = 750 \\\dashrightarrow \rm \: x = \cancel\frac{750}{25} \\ \dashrightarrow \rm \: x = 30 \: kg$

♣ Hence, Required lead to be added = 30 kg