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by zaarathelab » Tue Feb 07, 2012 6:41 am
Solution X contains only ingredients A and B in a ratio of 2:3. Solution Y contains only ingredients A and B in the ratio 1:2. If solution Z is created by mixing solutions X and Y in the ratio of 3:11, then 630 ounces of solution Z contains how many ounces of a?

a) 68
b) 73
c) 89
d) 219
e) 236

Pls give the shortest way to do this.
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by Jim@StratusPrep » Tue Feb 07, 2012 7:20 am
1st get the ratios into equal groups of 15 parts (multiply solution X by 3 and Solution Y by 5)

X: 2:3 * 3 = 6:9
Y: 1:2 * 5 = 5:10

Then multiply the mixed ratio components by the ingredients in each solution:

Solution x: 6:9 * 3 = 18:27
Solution y: 5:10 * 11 = 55:110

Then sum the results for a 73:137 ratio.

The sum of the ratio components is 210 or 1/3 of 630 so you have to multiply each piece by 3 for the total ratio of 219:411.

Thus the answer is D.
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by vaibhavgupta » Tue Feb 07, 2012 8:07 am
zaarathelab wrote:Solution X contains only ingredients A and B in a ratio of 2:3. Solution Y contains only ingredients A and B in the ratio 1:2. If solution Z is created by mixing solutions X and Y in the ratio of 3:11, then 630 ounces of solution Z contains how many ounces of a?

a) 68
b) 73
c) 89
d) 219
e) 236

Pls give the shortest way to do this.
IMO D. looking for OA
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by LalaB » Tue Feb 07, 2012 9:34 am
X A/B=2/3 or A=2/5*x
y a/b=1/2 or a=1/2*y

z x/y=3/11 It means, that x=3/14*z=3/14*630=135

sol y = total-sol.x=630-135=495

A of sol x+ a of sol Y=2/5*x+1/2*y=2/5*135+1/3*495=219
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by GMATGuruNY » Tue Feb 07, 2012 9:45 am
zaarathelab wrote:Solution X contains only ingredients A and B in a ratio of 2:3. Solution Y contains only ingredients A and B in the ratio 1:2. If solution Z is created by mixing solutions X and Y in the ratio of 3:11, then 630 ounces of solution Z contains how many ounces of a?

a) 68
b) 73
c) 89
d) 219
e) 236

Pls give the shortest way to do this.
Solution Z: 630 ounces
X:Y = 3:11.
This means that of every 14 ounces, 3 ounces are X and 11 ounces are Y.
In other words, X is 3/14 of the total:
X = (3/14)630 = 135 ounces.
Y = 630-135 = 495 ounces.

Solution X: 135 ounces
A:B = 2:3.
This means that of every 5 ounces, 2 ounces are A and 3 ounces are B.
In other words, A is 2/5 of the total:
A = (2/5)135 = 54 ounces.

Solution Y: 495 ounces
A:B = 1:2.
This means that of every 3 ounces, 1 ounce is A and 2 ounces are B.
In other words, A is 1/3 of the total:
A = (1/3)495 = 165 ounces.

Total amount of A = 54+165 = 219 ounces.

The correct answer is D.

Fastest approach:
In the mixture, Y:X = 11:3.
The relative amount of Y (11) is almost 4 times the relative amount of X (3).
Since most of the mixture will come from Y, the fraction of A in the mixture will be very close to the fraction of A in Y:
(1/3)630 = 210.
Since about 1/5 of the mixture will come from X -- and X contains a slightly higher percentage of A -- the actual amount of A in the mixture will be just a bit more than 210.
The correct answer is D.
Last edited by GMATGuruNY on Tue Feb 07, 2012 12:47 pm, edited 3 times in total.
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by LalaB » Tue Feb 07, 2012 10:43 am
one more method -

z=x/y=3/11

lets assume that x=3 y=11 and then z= 14

then A of sol x+a of sol y= (2/5*3+1/3*11)*630/14=73/15*630/14=219

GMATGuruNY wrote:
Fastest approach:
In the mixture, Y:X = 11:3.
The relative amount of Y (11) is almost 4 times the relative amount of X (3).
Since most of the mixture will come from Y, the fraction of A in the mixture will be very close to the fraction of A in Y:
(1/3)630 = 210.
Since about 1/5 of the mixture will come from X -- and X contains a higher percentage of A -- the actual amount of A in the mixture will be just a bit more than 210.
yep, thought this way too.
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