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06 Dec 2019, 06:05
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Question Stats:
62% (02:07) correct
38%
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If \(w < x < y < z\), is the average (arithmetic mean) of w, x, y, and z greater than y ?
(1) \(w + z = x + y\)
(2) \(w = 2\) and \(z = 18\)
Are You Up For the Challenge: 700 Level Questions
(1) \(w + z = x + y\)
(2) \(w = 2\) and \(z = 18\)
Are You Up For the Challenge: 700 Level Questions
06 Dec 2019, 10:43
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If w<x<y<z, is the average (arithmetic mean) of w, x, y, and z greater than y ?
\(\frac{w + x+ y + z}{4} > y\)?
(1) w+z=x+y
Now adding x + y on both sides
w + x + y + z = x + y + x + y
\(\frac{w + x + y + z}{4} = \frac{x + y}{2} \)
As \(\frac{x + y}{2}\) < y always
\(\frac{w + x + y + z}{4}\) < y
SUFFICIENT.
(2) w=2 and z=18
Clearly, INSUFFICIENT.
As x = 3 y = 4 then \(\frac{w + x+ y + z}{4} = \frac{27}{4}\) > y
OR
x = 3 y = 15 \(\frac{w + x+ y + z}{4} = \frac{38}{4}\) < y
Answer A.
\(\frac{w + x+ y + z}{4} > y\)?
(1) w+z=x+y
Now adding x + y on both sides
w + x + y + z = x + y + x + y
\(\frac{w + x + y + z}{4} = \frac{x + y}{2} \)
As \(\frac{x + y}{2}\) < y always
\(\frac{w + x + y + z}{4}\) < y
SUFFICIENT.
(2) w=2 and z=18
Clearly, INSUFFICIENT.
As x = 3 y = 4 then \(\frac{w + x+ y + z}{4} = \frac{27}{4}\) > y
OR
x = 3 y = 15 \(\frac{w + x+ y + z}{4} = \frac{38}{4}\) < y
Answer A.
General Discussion
23 Jan 2021, 01:31
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Is: (W + X + Y + Z) / 4 > Y?
Is: W + X + Y + Z > 4Y?
Is: W + X + Z > 3Y?
Given: W < X < Y < Z
S1: W + Z = X + Y
W = X + Y - Z
Inserting S1 into the question stem
Is: (X + Y - Z) + X + Z > 3Y?
Is: 2X + Y > 3Y?
Is: X > Y?
We are given that X < Y
Definite answer: s1 sufficient alone.
S2: again plug in the given values into the question stem.
Question becomes:
Is: 20 + X > 3Y?
Even given the constraint that X < Y, you can easily find two cases that give you a Yes and No.
If X = -2 and Y = 1 ——-Yes
If X = 2 and Y = 20 ——- NO
A
S1 alone is sufficient
Posted from my mobile device
Is: W + X + Y + Z > 4Y?
Is: W + X + Z > 3Y?
Given: W < X < Y < Z
S1: W + Z = X + Y
W = X + Y - Z
Inserting S1 into the question stem
Is: (X + Y - Z) + X + Z > 3Y?
Is: 2X + Y > 3Y?
Is: X > Y?
We are given that X < Y
Definite answer: s1 sufficient alone.
S2: again plug in the given values into the question stem.
Question becomes:
Is: 20 + X > 3Y?
Even given the constraint that X < Y, you can easily find two cases that give you a Yes and No.
If X = -2 and Y = 1 ——-Yes
If X = 2 and Y = 20 ——- NO
A
S1 alone is sufficient
Posted from my mobile device
