GMATPrepNow wrote:
If 0 < y < x, then which of the following is a possible value of \(\frac{27x + 23y}{3x + 2y}\)?
A) I only
B) II only
C) III only
D) I and II only
E) II and III only
*Kudos for all correct solutions
The two approaches above are great, so I won't duplicate them
Instead, I'll show you another approach.
Let's check to see whether (27x + 23y)/(3x + 2y) can equal any of the 3 given values (8.7, 9.2 and 10.8)
Start with I.
Can (27x + 23y)/(3x + 2y) = 8.7?
To make our work easier (without tons of decimals), let's rewrite 8.7 as 87/10
So, we have: (27x + 23y)/(3x + 2y) = 87/10
Cross multiply to get: 10(27x + 23y) = 87(3x + 2y)
Expand: 270x + 230y = 261x + 174y
Rearrange to get: 9x = -56y
Divide both sides by 9 to get: x = (-56/9)y
This is a problem, since x and y are both SUPPOSED to be positive. However, we can see by this equation that, if y is positive then x is NEGATIVE. Likewise, if x is positive then y is NEGATIVE.
This tells us that it's IMPOSSIBLE to find x- and y-values that satisfy the given condition (0 < y < x) so that (27x + 23y)/(3x + 2y) = 8.7
So, statement I is not possible
Now try II.
Can (27x + 23y)/(3x + 2y) = 9.2?
To make our work easier, notice that 9.2 = 92/10 = 46/5
So, we have: (27x + 23y)/(3x + 2y) = 46/5
Cross multiply to get: 5(27x + 23y) = 46(3x + 2y)
Expand: 135x + 115y = 138x + 92y
Rearrange to get: 23y = 3x
Divide both sides by 3 to get: (23/3)y = x
One possible solution to this equation is x = 23 and y = 3
Since these x- and y-values satisfy the given condition (0 < y < x), we can see that it IS POSSIBLE for (27x + 23y)/(3x + 2y) to equal 9.2
So, statement II IS possible
Now try III.
Can (27x + 23y)/(3x + 2y) = 10.8?
To make our work easier, notice that 10.8 = 108/10 = 54/5
So, we have: (27x + 23y)/(3x + 2y) = 54/5
Cross multiply to get: 5(27x + 23y) = 54(3x + 2y)
Expand: 135x + 115y = 162x + 108y
Rearrange to get: 7y = 27x
Divide both sides by 7 to get: y = (27/7)x
This is a problem, since x and y are both SUPPOSED to be such that 0 < y < x.
However, we can see by the equation, y = (27/7)x, that y will always be greater than x.
For example, if x = 7, then y = 27. Likewise, if x = 14, then y = 54, and so on.
This tells us that it's IMPOSSIBLE to find x- and y-values that satisfy the given condition (0 < y < x) so that (27x + 23y)/(3x + 2y) = 10.8
So, statement III is not possible
This is a TIME-CONSUMING approach, but it's better than guessing!
Answer: B
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Brent Hanneson – Creator of gmatprepnow.com
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