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I was preparing for Quantitative aptitude exams and I came across this question of ratios

An alloy contains copper and zinc in ratio 5:2 and another alloy contains zinc and tin in the ratio 3:2. If 2 parts of 1st alloy and one part of second alloy are melted together to form a new alloy of copper, zinc and tin, the ration of the metals will be?

What I've understood is that 2 parts of 1st ratio(lets call it A) means 2*A=10:4 and 1 part of the second ratio(lets call it B) means 1*B=3:2. Now when these parts are melted together the ratios will be copper:zince:tin=10:4+3:2 Did I get this right or not?

Masroor
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  • The ratio will be $\frac 2 3 \times \frac 5 7 : (\frac 2 3 \times \frac 2 7 + \frac 1 3 \times \frac 3 5) : \frac 1 3 \times \frac 2 5.$ – little o Apr 19 '19 at 10:20
  • the question had following 4 choices a)5:5:4 b)10:7:4 c)5:6:4 d)10:7:8 – Masroor Apr 19 '19 at 10:25
  • @Dbchatto67 won't it be $\frac{2}{3}\times\frac{5}{7}:\frac{2}{3}\times\frac{2}{7} + \frac{1}{3}\times\frac{3}{5}:\frac{1}{3}\times\frac{2}{5}$ – Vizag Apr 19 '19 at 10:26
  • I have done exactly that @Vizag.I first think the two alloys are mixed in the ratio $1:1.$ – little o Apr 19 '19 at 10:29
  • The ratio will be $$\frac 2 3 \times \frac 5 7 : \left (\frac 2 3 \times \frac 2 7 + \frac 1 3 \times \frac 3 5 \right ) : \frac 1 3 \times \frac 2 5 = \frac {10} {21} : \frac {41} {105} : \frac {2} {15} =50:41:14.$$ – little o Apr 19 '19 at 10:30
  • can anyone please tell what x part of some ratio A:B means? – Masroor Apr 19 '19 at 10:32
  • In the context of your question you can think that we are mixing 2 Kg of Alloy 1 with 1 Kg of Alloy 2 – Vizag Apr 19 '19 at 10:37
  • That's what "part" essentially means. You can just simply think that we are mixing x litres of this liquid with y litres of this liquid or x Kgs of this metal with y Kgs of that metal. – Vizag Apr 19 '19 at 10:40
  • so if for example I've got a ratio lets call it A=3:7 and I've to determine x parts of A out of y. What I would is (x/x+y)*(3/10+7/10). Is this correct? – Masroor Apr 19 '19 at 10:44
  • You're thinking about it wrong. Take for example two solutions A and B. You mix 2 parts of A with 1 part of B to make a new solution C. Now if I take 3 litres of C, it will have 2 litres of A and 1 litres of B. So let's complicate it a little. Suppose A is 50% water, and B is 75% water and we mix 2 parts of A with 1 part of B to make a new solution C. Now since C is 2/3 A and 1/3 B, what %age of C is water? – Vizag Apr 19 '19 at 10:56
  • water in C=(50% of 2/3) + (75% of 1/3) – Masroor Apr 19 '19 at 11:04
  • That is exactly right. – Vizag Apr 19 '19 at 11:18

2 Answers2

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So $A_1$ has the following composition:

$$\frac{5}{7} \text{ Cu, } \frac{2}{7} \text{ Zn } $$

$A_2$ has the following composition:

$$\frac{3}{5} \text{ Zn, } \frac{2}{5} \text{ Pb } $$

Now when they are mixed in the ratio $2:1$, the overall mixture will have Cu:Zn:Pb in the ratio :

$$ \frac{2}{3}\times\frac{5}{7}:\frac{2}{3}\times\frac{2}{7} + \frac{1}{3}\times\frac{3}{5}:\frac{1}{3}\times\frac{2}{5}$$

Vizag
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  • the question had following 4 choices a)5:5:4 b)10:7:4 c)5:6:4 d)10:7:8 is it possible that we're doing something wrong related to 2 parts and 1 parts thing? – Masroor Apr 19 '19 at 10:46
  • There surely is a problem with those options. The approach is correct. – Vizag Apr 19 '19 at 10:50
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In the first alloy, copper : zinc = 5:2 in the 2nd , zinc : tin = 3:2 Let the 70x of first alloy is mixed with 35x of 2nd alloy ( 35 being LCM of (5+2 = 7 and 3+2 =5)) so amount of zinc and copper in the mixture , 70x = 50x -copper , 20x -zinc 35 x of 2nd will contain 21x - zinc and 14x tin so required ratio of copper : zinc : tin = 50x :(20x+21x):14x = 50:41 : 14

for more concepts-https://www.handakafunda.com/ratio-and-proportion-concepts-properties-and-cat-questions/

Kouts Km
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