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Re: If 0 < y < x, then which of the following [#permalink]
I do not understand this solution , Could you please, write it in other way?
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Re: If 0 < y < x, then which of the following [#permalink]
Is there any way to use plug in number?
If Y=1/3 and X= 1/2 then the result is = 8.5?
using different numbers would give me another solution and none is on the choices
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If 0 < y < x, then which of the following [#permalink]
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GreenlightTestPrep 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


I'm going to add a few extra steps to my original solution.

First notice that \(27x + 18y\) is a multiple of \(3x + 2y\) because \(27x + 18y = 9(3x + 2y)\)

I'm going to use this fact by taking the expression \(27x + 23y\) and rewriting it as \(27x + 18y + 5y\)

When we do this we get: \(\frac{27x + 23y}{3x + 2y} = \frac{27x + 18y + 5y}{3x + 2y}\)

Useful property: \(\frac{a + b}{c} = \frac{a}{c} + \frac{ b}{c}\)

Apply the property to split up our fraction as follows:
\(= \frac{27x + 18y}{3x + 2y} + \frac{5y}{3x + 2y}\)

Rewrite the numerator of the first fraction: \(= \frac{9(3x + 2y)}{3x + 2y} + \frac{5y}{3x + 2y}\)

Simplify the first fraction: \(= 9 + \frac{5y}{3x + 2y}\)

Since x and y are both POSITIVE, the numerator and denominator of \(\frac{5y}{3x + 2y}\) will be POSITIVE, which means \(\frac{5y}{3x + 2y}\) has a POSITIVE value.

This means \(9 + \frac{5y}{3x + 2y}\) will evaluate to be a number that's GREATER THAN 9

So, value I (8.7) is not possible

Now let's take a closer look at \(\frac{5y}{3x + 2y}\)

Notice that \(\frac{5y}{3y + 2y} = \frac{5y}{5y} = 1\) [since the numerator and denominator are EQUAL]

However, since we're told that \(y < x\), we know that \(3y + 2y < 3x + 2y\)
This means \(\frac{5y}{3x + 2y} < 1\) [since the numerator is LESS THAN the denominator]

If \(\frac{5y}{3x + 2y} < 1\), then we can conclude that \(9 + \frac{5y}{3x + 2y} < 10\)

So, value III (10.8) is not possible

This leaves us with value II (9.2), which IS possible.

Answer: B

Cheers,
Brent
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Re: If 0 < y < x, then which of the following [#permalink]
GR3A wrote:
Is there any way to use plug in number?
If Y=1/3 and X= 1/2 then the result is = 8.5?
using different numbers would give me another solution and none is on the choices


No, it would be 9.769. The result will never be less than 9.
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Re: If 0 < y < x, then which of the following [#permalink]
GreenlightTestPrep wrote:
loolah wrote:
I do not understand this solution , Could you please, write it in other way?


I've written another solution. See what you think.




Thank you, it is clear for me :thumbsup:
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Re: If 0 < y < x, then which of the following [#permalink]
4
We can solve this by estimating the maximum and the minimum possible values of the expression
The expression E is
(27x + 23y)/(3x + 2y)

If we divide numerator and denominator by x as it's given x>0
(27 + 23(y/x))/(3+2(y/x))

since we know 0<y<x,
the maximum value y/x can take is 1, or to be more precise 0.99999999999... when y and x are approx equal
and the minimum value it can take is close to 0 when x>>y, like y=1 and x = 10^10.

when we substitute the max value, ie y/x=1, we can the expression E = (27+23)/(3+2) = 10
and the minimum value at y/x=0 is E = 27/3 = 9

Therefore 9<E<10
only option B satisfies this value and hence it is the answer
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If 0 < y < x, then which of the following [#permalink]
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GreenlightTestPrep 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


Calculate the value of \(\frac{27x + 23y}{3x + 2y}\) when x and y are EQUAL.
If x=y=1, we get:
\(\frac{27x+23y}{3x+2y} = \frac{27+23}{3+2} = 10\)

Calculate the value of \(\frac{27x + 23y}{3x + 2y}\) when x and y are FAR APART.
If x=100 and y=0, we get:
\(\frac{27x+23y}{3x+2y} = \frac{27*100+23*0}{3*100+2*0} = \frac{2700}{300} = 9\)

Since x and y cannot actually be equal and y cannot actually be 0, the results above imply the following:
\(9 < \frac{27x + 23y}{3x + 2y} < 10\)
Only option II is possible.

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Re: If 0 < y < x, then which of the following [#permalink]
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• Main logic: Express the given expression in terms of y/x by dividing by x on Nr and Dr.
○ As x>y>0
§ Max value of y/x ≈ to 1
§ And putting y/x value as 1 solves the expression to value 10
□ So option 10.8 is not possible
§ Min value of y/x ≈ 0. It can't be negative.
□ So the expression solves 9 at minimum value.
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Re: If 0 < y < x, then which of the following [#permalink]
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Re: If 0 < y < x, then which of the following [#permalink]
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