# Thread: PSAT Practice Question with Inequalities

1. ## PSAT Practice Question with Inequalities

Hello

While working on some practice PSAT questions, I came across this question:

The various averages (arithmetic means) of three of the four numbers c, d, e, and f are calculated, and are arranged from greatest to least as follows

The average of c, d, and e
The average of d, e, and f
The average of e, f, and c
The average of f, c, and d

Which of the following correctly orders c, d, e, and f from greatest to least?
(A) c > d > f > e
(B) d > f > e > c
(C) d > e > c > f
(D) e > c > f > d
(E) e > d > c > f

My answer was (A), but the correct answer was (E). I do not see how my work was wrong, other than I might have perhaps used the wrong method. I set it up like you do when you solve equations with two variables by subtracting one whole equation from another. Of course, I'm not totally positive if you can do this with inequalities, but I honestly don't see why not. Here's my work for clarity's sake:

c + d + e > d + e + f
-(d+ e + f > f + c + d)
-----------------------
f > e

d + e + f > e + f + c
-(e + f + c > f + c + d)
------------------------
c > e

So, I found that both c and f are greater than e and picked the only choice that had this written, (A).

Anyone want to tell me if I'm doing this wrong, and, if so, how it is actually supposed to be done? Thanks.

2. this is probably the wrong way to solve this but i would just use a simple and fast point system:

"The various averages (arithmetic means) of three of the four numbers c, d, e, and f are calculated, and are arranged from greatest to least as follows

The average of c, d, and e *** anything in this row 4 points)
The average of d, e, and f *** anything in this row 3 points)
The average of e, f, and c** " " " " 2 points)
The average of f, c, and d**** " " " " 1 point)

Which of the following correctly orders c, d, e, and f from greatest to least?
(A) c > d > f > e
(B) d > f > e > c
(C) d > e > c > f
(D) e > c > f > d
(E) e > d > c > f"

and it works out to (E) e>d>c>f
e=9 pnts
d= 8 pnts
c= 7 pnts
f= 6 pnts

again thats how i usually solve things like that its not really the right way but it works most of the time

3. Hello, seuzy13!

It's much easier than you think . . .

The various arithmetic means of three of the four numbers $c, d, e, f$ are calculated,
and are listed from greatest to least as follows:

. . $\begin{array}{c}\text{mean of }c, d, e \\
\text{mean of }d, e, f \\ \text{mean of }c, e, f \\ \text{mean of }c,d,f \end{array}$

Which of the following correctly orders $c, d, e, f$ from greatest to least?

$\begin{array}{c}(A)\;c > d > f > e\qquad (B)\;d > f > e > c \qquad (C)\;d > e > c > f \\ \\ (D)\;e > c > f > d \qquad (E)\;e > d > c > f \end{array}$

From the list, we have: . $\frac{c+d+e}{3} \;>\;\frac{d+e+f}{3} \;>\:\frac{c+e+f}{3} \;>\;\frac{c+d+f}{3}$

$\text{Muliply by 3: }\;\underbrace{c+d+e}_{{\color{blue}[1]}} \;>\;\underbrace{d+e+f}_{{\color{blue}[2]}} \;>\;\underbrace{c+e+f}_{{\color{blue}[3]}} \;>\;\underbrace{c+d+f}_{{\color{blue}[4]}}$

$\text{From }{\color{blue}[1] > [2]}\!:\;\;c+d+e \:>\:d+e+f \quad\Rightarrow\quad c\:>\:f \;\;{\color{blue}[4]}$

$\text{From }{\color{blue}[2] > [3]}\!:\;\;d+e+f \:>\:c+e+f \quad\Rightarrow\quad d \:>\:c\;\;{\color{blue}[5]}$

$\text{From }{\color{blue}[3] > [4]}\!:\;\;c+e+f \:>\:c+d+f \quad\Rightarrow\quad e \:>\:d\;\;{\color{blue}[6]}$

$\text{So we have: }\;\underbrace{(e \:>\:d)}_{{\color{blue}[6]}} \;\;\wedge\;\; \underbrace{(d \:>\:c)}_{{\color{blue}[5]}} \;\; \wedge\;\;\underbrace{(c \:>\:f)}_{{\color{blue}[4]}}$

. . $\text{Therefore: }\:e \:>\:d\:>\:c\:>\:f\quad\hdots\;\;\text{Answer (E)}$

4. Yes, I realized a long time ago how to get the answer correctly. And I figured out why the way I was doing it wasn't working. It is very simple. You can pretty much see it just by looking at it. I never thought of doing it GooooZ's way. It's strange how that works.

Anyway, I did take my PSAT about a week ago, and I think I did very well. There were no problems like this one, but it did help me to get the basic idea of how the questions would be asked and the ways they would try to trick you.