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.

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

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

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.

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.