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vinegar and water

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vinegar and water

by leonswati » Mon Nov 21, 2011 11:51 am
A certain 4-liter solution of vinegar and water consists of x liters of vinegar and y liters of water. How many liters of vinegar does the solution contain?

(1) x/4 = 3/8
(2) y/4 = 5/8

the answer is D, but can anyone plz help me solve it using equations ....... a complete answer... thank you...
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by kanwar86 » Mon Nov 21, 2011 12:07 pm
leonswati wrote:A certain 4-liter solution of vinegar and water consists of x liters of vinegar and y liters of water. How many liters of vinegar does the solution contain?

(1) x/4 = 3/8
(2) y/4 = 5/8

the answer is D, but can anyone plz help me solve it using equations ....... a complete answer... thank you...
Approach - 1) From (1), x xan be found.
2) From (2), y can be found, which when subtracted from total volume of solution gives the value of x.
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by shankar.ashwin » Mon Nov 21, 2011 12:20 pm
Given x + y = 4

1) gives you x -> 3/2 (sufficient)

2) gives you y -> 5/2 (and) x = 4 - 5/2 = 3/2 (sufficient)
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by neelgandham » Mon Nov 21, 2011 12:40 pm
The question can be rephrased to,

If x+y =4, then x = ?
(1) x/4 = 3/8

We get the value of x by simplifying. Hence, sufficient
(2) y/4 = 5/8
y = A fraction and x+y =4 ; x = 4- A fraction. Hence, sufficient

Option D
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by Stuart@KaplanGMAT » Mon Nov 21, 2011 3:23 pm
leonswati wrote:A certain 4-liter solution of vinegar and water consists of x liters of vinegar and y liters of water. How many liters of vinegar does the solution contain?

(1) x/4 = 3/8
(2) y/4 = 5/8

the answer is D, but can anyone plz help me solve it using equations ....... a complete answer... thank you...
Hi!

This question is a great illustration of the most powerful rule to remember for data sufficiency: number of equations vs number of unknowns.

Here's the "official" wording of the rule:
To solve a system with n variables, one requires n distinct, linear, equations.
Put more simply, if you have the same number of linear equations as you have variables, you can answer any question about the system.

Applying the rule to your question:

Q: gives you 1 equation with 2 variables. We're missing 1 more equation.

(1) gives us another linear equation, no new variables: sufficient.

(2) gives us another linear equation, no new variables: sufficient.

Choose (D)!
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