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26 Aug 2019, 08:29
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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54% (02:13) correct
46%
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Consider three positive numbers \(x\), \(y\), and \(z\) such that \(x>y>z\). If \(M\) is the median and \(A\) is the average, is \(A>M\)?
1) \(M=20\)
2) \(xz=M^2\)
1) \(M=20\)
2) \(xz=M^2\)
26 Aug 2019, 12:30
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Asad
There's some math jargon in this one, so let's translate first.
We have three POSITIVE numbers, x, y, and z, from smallest to largest.
The median, M, is the middle number. That means M is equal to y. So, if the problem tells you anything about M, it's actually just telling you about y.
A is the average, but the variable A isn't used in the statements. It's just used in the question. So, get rid of the variable, and just translate like this:
"Is the average of x, y, and z greater than y?"
Okay, when does that happen? There are two approaches: you could either use algebra to figure it out, or you could use a back-of-the-napkin approach.
Here's the back-of-the-napkin way: If you have three numbers, and they're evenly spaced, the average will be equal to the middle number, like this:
z --------------- y = A --------------- x
If the numbers aren't evenly spaced, the two numbers that are closer together will be on the same side of the average, like this:
z ---- y ----- A ----------------------- x
or
z ----------------------- A -----y ----- x
So, when is the average greater than y? The average is greater than y whenever y is closer to z than it is to x. That is, whenever y is less than the halfway point between z and x. (That's this scenario:
z ---- y ----- A ----------------------- x)
Or, you can use algebra to come to the exact same conclusion. Algebraically, the average is (x + y + z)/3. So, the question asks:
Is (x + y + z)/3 > y?
Simplify it:
Is x + y + z > 3y?
Is x + z > 2y?
Is y < (x + z)/2?
Is y smaller than the average of x and z?
----------------
Now we know exactly what the question is saying. Let's focus on the statements.
Statement 1: Remember that we were going to read anything with an M in it, as if it just said 'y'. Therefore, this statement tells us that y = 20. However, y could be closer to z:
z = 19, y = 20, x = 1000000000
Or, y could be closer to x:
z = 1, y = 20, x = 21
So, the answer to the question might be either 'yes' or 'no'. This statement is insufficient.
Statement 2: According to this statement, \(xz = y^2\). It's not totally obvious how this relates to what we're looking for, so let's test a simple case first to help us understand. Remember that z is the smallest number, x is the largest, and all three numbers are positive.
Case 1: z = 1, y = 2. In this case, x(1) = 4, so x = 4. y is closer to z than to x (lower than the average of x and z), so the answer is "yes".
Next, our job is to try to find a case where y is closer to x. There are two types of cases that tend to cause weird things to happen with exponents and multiplication: negative numbers, and fractions. We can't use negative numbers, because the question said everything has to be positive. Let's try some fractions.
Case 2: z = 1/4, y = 1/2. In this case, x(1/4) = 1/4, so x = 1. This still doesn't do it: y is still closer to z and the answer is still "yes".
Case 3: z = 1/10, y = 1/5. In this case, x(1/10) = 1/25, so x = 10/25 = 2/5. That STILL doesn't do it: y is STILL closer to z.
Case 4: Let's try to get z and y really far apart. z = 1/10, y = 9/10. In this case, x(1/10) = 81/100, so x = 81/10 = 8.1. y is definitely closer to z.
You can stop here, but if you know a bit about geometric sequences, you can also rationalize this by saying that if xz = y^2, then x/y = y/z, so z, y, and x form a geometric sequence. In a geometric sequence, the gaps between the terms get bigger as you keep going up. Therefore, y will be closer to the term before it (z) than to the term after it (x).
The answer will always be "yes," and this statement is Sufficient. The answer is B.
27 Aug 2019, 03:27
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This is a Yes-No DS question. Plugging in values in the statements for getting to the answer works most of the times.
We know that x, y, z are positive numbers such that x > y > z.
Question: Is Average > Median?
Remember that y is the median.
So, technically speaking, the question is asking if Average > y?
Statement 1: M = 20 = y
Do not waste too much time. We only know the middle value. We don't the value of x and y. So, we can't determine the Average. It could be more, less, or equal.
Insufficient
Statement 2: xz = \(M^2\)
OR xz = \(y^2\)
Let's plug in values.
x = 1
y = 2 = Median
z = 4
Average = 7/3
Yes
x = 2
y = 4 = Median
z = 8
Average = 14/3
Yes
x = 1/8
y = 1/4 = Median
z = 1/2
Average = 7/8
Yes
Sufficient
Thus, the answer is B.
Deepti Singh
Master Trainer - GMAT
Manya - The Princeton Review
We know that x, y, z are positive numbers such that x > y > z.
Question: Is Average > Median?
Remember that y is the median.
So, technically speaking, the question is asking if Average > y?
Statement 1: M = 20 = y
Do not waste too much time. We only know the middle value. We don't the value of x and y. So, we can't determine the Average. It could be more, less, or equal.
Insufficient
Statement 2: xz = \(M^2\)
OR xz = \(y^2\)
Let's plug in values.
x = 1
y = 2 = Median
z = 4
Average = 7/3
Yes
x = 2
y = 4 = Median
z = 8
Average = 14/3
Yes
x = 1/8
y = 1/4 = Median
z = 1/2
Average = 7/8
Yes
Sufficient
Thus, the answer is B.
Deepti Singh
Master Trainer - GMAT
Manya - The Princeton Review
