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problem on ratios

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by uptowngirl92 » Sat Jul 18, 2009 12:49 am
A fruit mixture is made up by 25% fruit A and 75% fruit B. Now if the amount of fruit A is doubled, what is their relative share in the new mixture?

Guys I got confused mid way.Does somebdy have an easy explanation?
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by raghavsarathy » Sat Jul 18, 2009 1:40 am
Let the quantity of the mixture be 4x
Quantity of fruit A = x

Quantity of fruit B = 3x

Amount of fruit A is doubled . Hence it becomes 2x and total quantity of mixture = 2x+ 3x = 5x

Share of A = 2/5 = 40%
Share of B = 3/5 = 60%
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by gmat740 » Sat Jul 18, 2009 6:50 am
Here is another way of solving

Let the mixture be 100ml
A=25ml
B=75ml

now A is doubled,so A=50ml
Total Mixture = A +B = 50+75 = 125

So share of A in 100 ml mixture = (A/A+B)*100
= (50/125)*100 = 40
So, B=60

There can be lot of variations with these kind of mixture questions.
Eg:If the question says, We double A but the volume of the mixture is constant. So in that case, A=50
and B = 50(volume of B will Decrease)

It is very important to break down the problem to its basic form, so that you can exactly answer what is required in the question.
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