GreenlightTestPrep 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
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 possibleNow 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 possibleThis leaves us with value II (9.2), which IS possible.
Answer: B
Cheers,
Brent