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 – GMATPrepNow.com

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