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Mixture problem

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Mixture problem

by vishwas.arora » Mon Aug 08, 2011 4:24 am
Milk costs $7 per gallon and water costs $3 per gallon. A mixture contains 'x' gallons of milk and 'y' gallons of water. Is y>x?

(1) x>3
(2) The cost of the 5 gallons of the mixture is less than $30.

Any expert replies please.

OA after discussion.

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by kmittal82 » Mon Aug 08, 2011 4:59 am
from the stem:

Cost of x+y gallons of mixture = 7x + 3y

(1) Clearly not enough

(2) We can have 5 gallons of mixture in a variety of combinations, but we also need to make sure that the cost is less than $30. For instance:

x = 5, y = 0, cost = 35 -> Not valid
x = 4, y = 1 , cost = 28 + 3 = 31 -> Not valid
x =3, y = 2, cost = 21+6 = 27 -> Valid
x =2 , y =3, cost = 14 + 9 = 23 -> Valid
x=1 , y=4 , cost = 7+12 = 19 -> Valid
x=0, y = 5, cost = 15 -> Valid

Clearly, the combination is valid for both y>x and y<x, so not sufficient

Combine (1) + (2)
We know that x > 3, but we also know that x can't be 4. So, lets put x = 3.5 and y = 1.5, total cost = 29. We can probably stretch the value of x and y a little more to get even closer to 30, but in each case, y will be less than x.

Hence, (C)

OA please?
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by gmatboost » Mon Aug 08, 2011 8:06 am
This seems like a strange question.

Essentially, the statements are:
1. x > 3
2. x + y = 5

So the price really has nothing to do with answering the question.
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by vishwas.arora » Mon Aug 08, 2011 8:23 am
kmittal82 wrote:from the stem:

Cost of x+y gallons of mixture = 7x + 3y

(1) Clearly not enough

(2) We can have 5 gallons of mixture in a variety of combinations, but we also need to make sure that the cost is less than $30. For instance:

x = 5, y = 0, cost = 35 -> Not valid
x = 4, y = 1 , cost = 28 + 3 = 31 -> Not valid
x =3, y = 2, cost = 21+6 = 27 -> Valid
x =2 , y =3, cost = 14 + 9 = 23 -> Valid
x=1 , y=4 , cost = 7+12 = 19 -> Valid
x=0, y = 5, cost = 15 -> Valid

Clearly, the combination is valid for both y>x and y<x, so not sufficient

Combine (1) + (2)
We know that x > 3, but we also know that x can't be 4. So, lets put x = 3.5 and y = 1.5, total cost = 29. We can probably stretch the value of x and y a little more to get even closer to 30, but in each case, y will be less than x.

Hence, (C)

OA please?
Fine that alone both statements don't give the answer.
But while combining both (1) and (2), IMO we can't assume that x+y = 5. It is just stating that for certain 5 gallons of mixture the cost is less than $30.

OA is E
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by gmatboost » Mon Aug 08, 2011 8:26 am
But while combining both (1) and (2), IMO we can't assume that x+y = 5. It is just stating that for certain 5 gallons of mixture the cost is less than $30.
I agree that the wording is ambiguous. It could go either way. The GMAT will not present that issue.
Just make sure you understand the result with each of the interpretations.
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by GmatKiss » Mon Aug 08, 2011 8:31 am
IMO:E

1. x > 3 (Not sufficient)
2. x + y = 5 (Not sufficient)

1+2,

x can be (4,5,6,.....n)

Minimum possible combination:4(7)+1(3) = 31 -> violates condition (cost less than 30)
Not sufficient!
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by samir2562 » Thu Oct 20, 2011 3:06 pm
Any ideas why this is not C? If x+y=5, then knowing that x>3, it must be that x>y
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by samir2562 » Thu Oct 20, 2011 3:06 pm
Any ideas why this is not C? If x+y=5, then knowing that x>3, it must be that x>y
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by bpdulog » Fri Oct 21, 2011 7:57 am
C
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by rishimaharaj » Tue Nov 01, 2011 8:09 am
In my attempt I came up with the following table:

| x | y | $x | $y | $$ |
------------------------
| 0 | 5 | 00 | 15 | 15 |
| 1 | 4 | 07 | 12 | 19 |
| 2 | 3 | 14 | 09 | 23 |
| 3 | 2 | 21 | 06 | 27 |
| 4 | 1 | 28 | 03 | 31 |
| 5 | 0 | 35 | 00 | 35 |

Statement 1 gives no info on Y so it is insufficient.

Statement 2 says that the price is less than $30, which could mean:
(x,y) = (0,5) (1,4) (2,3) or (3,2).
In some cases Y > X and in others X > Y, so insufficient.

Statements 1 and 2 together clash...
If x > 3, then x has to be either 4 or 5.
But this directly conflicts with Statement 2, where the cost is either $31 or $35.


Because of this, I'd go with answer choice E.
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