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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
Carcass wrote:
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.


I am unable to understand (Quantity A)
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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
2
Allegation Method:

A----------------------------------B
32--------------23-----------------17

(32-23)------------------------(23-17)
9--------------------------------6

In cross we simplify as:
A/B = 6/9 = 2/3

How much in liquid A: part of A/Total = 2/(2+3)
2/5 = 0.4 or 40% Option A
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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
\([m][m][m]Solve it applying weight average formula

\frac{A}{B} = \frac{ (AVERAGE-B)}{(A-AVERAGE)}

=> \frac{A}{B} = \frac{ (23-17)}{(32-23}
=> \frac{A}{B} = \frac{6}{9} = \frac{2}{3}

SO, AMOUNT OF A IS \frac{2}{(3+2)}= \frac{2}{5} X 100 = 40 PERCENT \)[/m][/m][/m]
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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
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I always make the same chart for all mixture questions:

Liquid Iodine Water Total
A .32x .68x x
B .17y .83y y

total liquid is x+y
total iodine is .32x+.17y

so

iodine is 23% of total or
.32x+.17y=(.23)(x+y)
.32x+.17y=.23x+.23y
.09x=.06y
or 9x=6y
the questions is asking what is x/(x+y)
so since x=6y/9 you can plug in (6y/9)/((6y/9)+y))
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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Liquid A is 32 percent iodine and the rest water. Liquid B i [#permalink]
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