If 0 < y < x, then which of the following is a possible value of

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

I tried it x=2 and y=1 and it solved problem very quickly

Hi Brent BrentGMATPrepNow, is below calculation correct? Thanks Brent
(0 < y < x) > x=2 , y = 1

(27x + 23y)/(3x + 2y) = (27*2 + 23*1)/ 3*2 + 2*1 = 77/8 = 9.6 > therefore close to answer choice II.

Yes, your calculations are correct (a calculator will also confirm your answer)

Let, y=k and x=k+d (where d≥+1)
(27x+23y)/(3x+2y)
=(50k+27d)/(5k+3d)
={9(5k+3d)+5k}/(5k+3d)
=9+{(5k)/(5k+3d)}

{(5k)/(5k+3d)}<1
[because 5k+3d>5k]

So, 9<(27x+23y)/(3x+2y)<10
Hence, ii. Is only available option.

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