$\frac{1}{2}$ of 4 kg of an alloy is lead and the rest of it is tin. $\frac{1}{6}$ of 5 kg of another alloy is lead and the rest of it is tin. Find the ratio of lead and tin in the mixture of the given quantities of these two alloys.
Answer & explanation
Correct answer: option 3
\(\frac{1}{2}\) of 4 Kg of an alloy is lead and rest is tin.
⇒ Quantity of lead in first alloy = \(\frac{1}{2}\) of 4 Kg = 2 Kg .. (1.)
⇒ Quantity of Tin in first alloy = 4 - 2 = 2 Kg ..(2.)
⇒ \(\frac{1}{6}\) of 5 Kg of another alloy is lead and the rest of it is tin.
⇒ Quantity of lead in second alloy = \(\frac{1}{6}\) of 5 Kg = \(\frac{5}{6}\) ..(3.)
⇒ Quantity of Tin in first alloy = 5 - \(\frac{5}{6}\)
⇒\(\frac{30 - 5}{6}\) = \(\frac{25}{6}\) ..(4.)
The ratio of lead and tin in the mixture of the given quantities of these two alloys.
Eq.(1) + Eq.(2) + Eq(3) + Eq(4)
⇒ 2 + \(\frac{5}{6}\) : 2 = \(\frac{25}{6}\)
⇒ \(\frac{17}{6}\) : \(\frac{37}{6}\)
⇒ 17 : 37