Mixture Problem

Question

The volume of a substance A is half the sum of the volumes of substances B & C, and the volume of substance B is one fifth of the sum of the volumes of A & C. Find the ratio of the volume of C to the sum of the volumes of A & B.

Could anyone offer a fast solution?

Answer: [highlight]1:1[/highlight]

Could anyone offer a fast solution?

Answer:   1:1

arzad · Answer

From the argument we have:
2A=B+C
5B=A+C
Subtraction of the equations results in:
3A=6B
A=2B
Lets substitute in the ratio we need to find:
 C/(A+B)
=(2A-B)/(A+B)
=(4B-B)/(2B+B)
=3B/3B
=1/1 OR 1:1

KarishmaB · Answer

The question is simple and straight forward but it has a great takeaway - always keep the end in mind.

Given: A = 1/2(B + C) ->  2A = B + C 
B = 1/5(A + C) ->  5B = A + C

Required: C/(A+B)
Makes me think that I need to get x (the required ratio) in the following equation: C = (A+B)*x
So basically, the above two highlighted equations have to be modified to get the same co-efficient for A and B. 
(2A - B = C) * 2
-A + 5B = C
When we add these two, we get 3A + 3B = 3C
or C = A+B
The required ratio is 1/1 or 1:1

SonyGmat · Answer

Thats a good strategy to follow from now on. Thanks Karishma!

And +1 for your blog! Really interesting articles that can help you solve problems a lot faster. Great Job! Though, I have a few questions:P

Mindreko · Answer

Hi Karishma,

I'm not sure why you did this: 
(2A - B = C) * 2

could you please explain why?

My own working for this qsn:
2A = B +C
5B = A + C

So 3A = 6B, and A = 2B

we need ratio of volumes of c to a+b

or c / a + b

We know A = 2B
We also know C = 2A - B

So C/A+B

2A-B/2B+B
4B-B/3B = 3B/3B = 1/1 or 1:1

Hope that helps.

KarishmaB · Answer

The reason I multiplied the first equation by 2 is that I wanted to make the co-efficient of a and b equal.
I have two equations: 
2A = B + C or 2A - B = C
5B = A + C or -A + 5B = C
I need C in terms of (A +B) so I need to make the coefficient of A and B equal.
If I add these two directly, I get A + 4B = 2C. 
But if I multiply the first equation by 2 and then add, I get 3A + 3B = 3C. I tried to think of a way to make the co-efficients same. Just like you do when you solve equations simultaneously. It saves me the extra steps needed to find each variable in terms of another variable (the way you have done and got A and C in terms of B)
Here the co-efficient of A and B is same so 3(A+B) = 3C. I get C in terms of (A+B)
Obviously there is nothing wrong in what you have done. I just tried to save myself some steps and some time.

LalaB · Answer

@VeritasPrepKarishma, thanks. what I like in ur solutions is that u always try to use ur logic.
it is much more interesting

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