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22 Jul 2021, 07:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (03:04) correct
30%
(03:20)
wrong
based on 207
sessions
History
Date
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One ounce of Solution X contains only ingredients a and b in a ratio of 2 : 3. One ounce of Solution Y contains only ingredients a and b in a ratio of 1 : 2. If Solution Z is created by mixing solutions X and Y in a 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
A. 68
B. 73
C. 89
D. 219
E. 236
22 Jul 2021, 09:18
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Several ways to handle this one.
The one that came to mind to me first was to use a sort of 'teeter totter' weighted average approach. For one, the mixture can't be less than 1/3(a) or more than 2/5(a) (imagine if it were ALL x or ALL y...It would be 40% or 33.3%, so a mixture must fall between these). So everything except D and E are out.
Solution X is 40% a, solution Y is 33.3% a.
Solution Z is 3/14 X and 11/14 Y.
(33.3% a)_____________________________(40%a)
11/14 3/14
Y X
Since X has 3/14 the weight, it will 'pull' the average 3/14 the distance from Y's average. The distance between 33.3 and 40 is ~7, and 3/14*(~7) = 1.5. So solution Z is ~(33.3%+1.5% = 34.8%) a.
34.8% of 630 is a little more than 1/3 630, which is about D (If you don't trust that: ~35%(630) = 63*3 + 31.5 = 188 + 31.5 ~ 219).
You could also say a ratio of 3:11 with a total of 630 gives a ratio 'multiplier' of 45. (Since 3*45 + 11*45 = 14*45 = 630)
This means: 135 ounces of X and 495 ounces of Y.
In 135 ounces of X, there will be 54 ounces of a. (40% of 135 = 20% of 270 = 27*2)
In 495 ounces of Y there will be 165 ounces of a. ((1/3)*495 = 495/3 = (480 + 15)/3 = 160 + 5)
Together that is 219 ounces of a.
One could also say that, since X is in '5' parts and Y is in '3' parts, it might be nice to make the Z ratio into pieces that fit this nicely.
3:11 could be written as 15:55 (to make the 'x' portion nice) and then scaled up to 45:165 (to make the '3' parts nice for the 'y').
The 45 and the 165 could then be broken up into the a:b ratios, respective to what is written:
So 45 will be 2:3 as 18:27 and 165 will be 1:2 as 55:110
This gives an
a : b : a : b
18:27:55:110
which can then combine into:
a:b
73:137
a:b:total =
73:137:210
Since total is actually 630, the ratio multiplier is '3,' and 73*3 = 219.
The one that came to mind to me first was to use a sort of 'teeter totter' weighted average approach. For one, the mixture can't be less than 1/3(a) or more than 2/5(a) (imagine if it were ALL x or ALL y...It would be 40% or 33.3%, so a mixture must fall between these). So everything except D and E are out.
Solution X is 40% a, solution Y is 33.3% a.
Solution Z is 3/14 X and 11/14 Y.
(33.3% a)_____________________________(40%a)
11/14 3/14
Y X
Since X has 3/14 the weight, it will 'pull' the average 3/14 the distance from Y's average. The distance between 33.3 and 40 is ~7, and 3/14*(~7) = 1.5. So solution Z is ~(33.3%+1.5% = 34.8%) a.
34.8% of 630 is a little more than 1/3 630, which is about D (If you don't trust that: ~35%(630) = 63*3 + 31.5 = 188 + 31.5 ~ 219).
You could also say a ratio of 3:11 with a total of 630 gives a ratio 'multiplier' of 45. (Since 3*45 + 11*45 = 14*45 = 630)
This means: 135 ounces of X and 495 ounces of Y.
In 135 ounces of X, there will be 54 ounces of a. (40% of 135 = 20% of 270 = 27*2)
In 495 ounces of Y there will be 165 ounces of a. ((1/3)*495 = 495/3 = (480 + 15)/3 = 160 + 5)
Together that is 219 ounces of a.
One could also say that, since X is in '5' parts and Y is in '3' parts, it might be nice to make the Z ratio into pieces that fit this nicely.
3:11 could be written as 15:55 (to make the 'x' portion nice) and then scaled up to 45:165 (to make the '3' parts nice for the 'y').
The 45 and the 165 could then be broken up into the a:b ratios, respective to what is written:
So 45 will be 2:3 as 18:27 and 165 will be 1:2 as 55:110
This gives an
a : b : a : b
18:27:55:110
which can then combine into:
a:b
73:137
a:b:total =
73:137:210
Since total is actually 630, the ratio multiplier is '3,' and 73*3 = 219.
08 Aug 2021, 10:52
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There is a very useful formula in these types of problems:
W1/W2 = (Ay - Avg) / (Avg - Ax) Where W is weight of mixture X and Y and A is the concentration of X and Y. Avg is the average concentration of the resulting solution.
In our case the equation would become: 3/11 = (1/3 - x)/(x-2/5)
x = 73/(15*14) => the amount of x in 630 grams of mixture is 630*73/(15*14) = 219
Hence D
More on this formula: https://gmatclub.com/forum/weighted-ave ... 06999.html (Thanks Bunuel)
W1/W2 = (Ay - Avg) / (Avg - Ax) Where W is weight of mixture X and Y and A is the concentration of X and Y. Avg is the average concentration of the resulting solution.
In our case the equation would become: 3/11 = (1/3 - x)/(x-2/5)
x = 73/(15*14) => the amount of x in 630 grams of mixture is 630*73/(15*14) = 219
Hence D
More on this formula: https://gmatclub.com/forum/weighted-ave ... 06999.html (Thanks Bunuel)
