# Moderate Alligations or Mixtures Solved QuestionAptitude Discussion

 Q. An alloy contains zinc, copper and tin in the ratio $2:3:1$ and another contains copper, tin and lead in the ratio $5:4:3$. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:
 ✖ A. $\dfrac{1}{10} \text{ kg}$ ✔ B. $\dfrac{1}{8} \text{ kg}$ ✖ C. $\dfrac{1}{4} \text{ kg}$ ✖ D. $\dfrac{1}{2} \text{ kg}$

Solution:
Option(B) is correct

In the first alloy, ratio of Zinc, Copper and Tin is given as,

$Z : C : T = 2 : 3 : 1$

Similarly, In the second alloy, ratio of Copper, Tin and Lead is given as,

$C : T : L = 5 : 4 : 3$

The trick here is to arrive at a quantity where calculation becomes easy.

To do that, we take LCM of $6 (=2+3+1)$, taken as 2 kg Zinc, 3 kg Copper and 1 Kg Lead),  and $12 (=5+4+3)$, taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead, which is $12$.

Thus we assume that both the alloys are being mixed at 12 Kgs each.

Alloys are mixed together to form third alloy. Then the ratio of content in it,

$Z : C : T : L = 4 : (6+5) : (2+4) : 3$

Weight of the third alloy,

$T= 12+12 = 24$ Kg.

$L = \dfrac{3}{24}$

$=\dfrac{1}{8} \text{ kg}$

## (4) Comment(s)

Kirthana
()

The total LCM is taken as 12. In the first ratio each of the numbers is multiplied by 2 so that the total of it gives 12. The second ratio need not be multiplied,as already the total is 12.

so zinc = 2*2=4

Copper=(3*2)+5

tin=(1*2)+4

Since equal quantities of weight are taken from both the alloys,

12+12 =24

Out of this 24kg, 3 parts is lead, hence 1/8 kg is the answer.

Juhi Chopra
()

quantity of lead in 1kg of first alloy=0

quantity of lead in 1kg of second alloy=3/12=1/4kg

therefore quantity of lead in 1kg of new alloy=1/8(1/4/1/2+1/2=1/8)

Snehal
()

Plz anyone explain how the ratio Z : C : T : L is calculated??

Snehal
()

Z : C : T : L = 4 : (6+5) : (2+4) : 3 how??????