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

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If 0 < y < x, then which of the following is  [#permalink]

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New post 31 Jan 2017, 07:38
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If 0 < y < x, then which of the following is a possible value of \(\frac{27x + 23y}{3x + 2y}\)?
    I. 8.7
    II. 9.2
    III. 10.8

A) I only
B) II only
C) III only
D) I and II only
E) II and III only

*Kudos for all correct solutions

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Re: If 0 < y < x, then which of the following is  [#permalink]

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New post 31 Jan 2017, 09:13
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GMATPrepNow wrote:
If 0 < y < x, then which of the following is a possible value of \(\frac{27x + 23y}{3x + 2y}\)?
    I. 8.7
    II. 9.2
    III. 10.8

A) I only
B) II only
C) III only
D) I and II only
E) II and III only

*Kudos for all correct solutions


Hi

If you look at a Q like this, where the variable can take any value and you have to find which values can fit in, it will help you if you find the MINIMUM and MAX value..


Let's see now..

\(\frac{27x + 23y}{3x + 2y}=\frac{27x+18y+5y}{3x+2y}=\frac{9(3x+2y)}{3x+2y}+\frac{3y+2y}{3x+2y}\)

1) Min value..
It is 9 + something, so min value is slightly more than 9..
I is out..

2) Max value..
Let's check \(\frac{3y+2y}{3x+2y}\)..
Now x>y so 3x>3y thus 3x+2y>3y+2y....
This means \(\frac{3y+2y}{3x+2y}<1\).
So value will be LESS than 9+1 or <10...
III is also out

ONLY 9.2 is left
B
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Re: If 0 < y < x, then which of the following is  [#permalink]

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New post 31 Jan 2017, 08:10
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GMATPrepNow wrote:
If 0 < y < x, then which of the following is a possible value of \(\frac{27x + 23y}{3x + 2y}\)?
    I. 8.7
    II. 9.2
    III. 10.8

A) I only
B) II only
C) III only
D) I and II only
E) II and III only

*Kudos for all correct solutions


\(A=\frac{27x + 23y}{3x + 2y}=\frac{9(3x+2y)+5y}{3x+2y}=9+\frac{5y}{3x+2y}\)

Since \(0<y<x\), we have \(A > 9\), so (I) is out.

Also \(\frac{5y}{2x+3y}<\frac{5y}{2y+3y}=1 \implies A < 9+1=10\), so (III) is out.

Hence (II) is left. The answer is B.

To check the answer, we have \(A=9.2 \iff \frac{5y}{3x+2y}=0.2=\frac{1}{5} \iff 25y = 3x+2y \iff 3x=23y \iff x=\frac{23y}{3}\)
This result satisfies the condition \(0<y<x\).
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Re: If 0 < y < x, then which of the following is  [#permalink]

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New post 31 Jan 2017, 13:20
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GMATPrepNow wrote:
If 0 < y < x, then which of the following is a possible value of \(\frac{27x + 23y}{3x + 2y}\)?
    I. 8.7
    II. 9.2
    III. 10.8

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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Re: If 0 < y < x, then which of the following is  [#permalink]

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New post 03 Jul 2018, 22:58
I tried it x=2 and y=1 and it solved problem very quickly
Re: If 0 < y < x, then which of the following is &nbs [#permalink] 03 Jul 2018, 22:58
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