I do not understand this solution , Could you please, write it in other way?

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

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

No, it would be 9.769. The result will never be less than 9.

Thank you, it is clear for me

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

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.

• 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.

In the last few steps where u took 5y / 3y + 2y instead of 5y / 3x + 2y, any specific reason for this? how did u know that this trick will get the right ans ?

$A=\frac{27x + 23y}{3x + 2y}=\frac{9(3x+2y)+5y}{3x+2y}=9+\frac{5y}{3x+2y}$
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 $0<y<x$.