The daily high temperatures were recorded at an Antarctic

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Here's a question I recently created:
The daily high temperatures were recorded at an Antarctic weather station. If a temperature of -38.2 degrees Celsius is 2 units of standard deviation below the mean temperature, and a temperature of -22.6 degrees Celsius is 4 units of standard deviation above the mean, which of the following temperatures is 1 unit of standard deviation above the mean?

A) -35.6 degrees Celsius
B) -33.0 degrees Celsius
C) -30.4 degrees Celsius
D) -27.8 degrees Celsius
E) -25.2 degrees Celsius
NOTE: I can think of at least 3 different ways to solve this question.

Answer: C
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by Brent@GMATPrepNow » Wed Sep 28, 2016 10:50 am
The daily high temperatures were recorded at an Antarctic weather station. If a temperature of -38.2 degrees Celsius is 2 units of standard deviation below the mean temperature, and a temperature of -22.6 degrees Celsius is 4 units of standard deviation above the mean, which of the following temperatures is 1 unit of standard deviation above the mean?

A) -35.6 degrees Celsius
B) -33.0 degrees Celsius
C) -30.4 degrees Celsius
D) -27.8 degrees Celsius
E) -25.2 degrees Celsius
Here's one approach:

Let M = mean
Let d = standard deviation

So, we get: M - 2d = -38.2
and: M + 4d = -22.6

At this point, we COULD solve for the individual values, but we can save some time by simply ADDING the two equations to get:
2M + 2d = -60.8
Divide both sides by 2 to get: M + d = -30.4
Since M + d represents 1 unit of standard deviation above the mean, the correct answer is -30.4

Answer: C

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by Brent@GMATPrepNow » Wed Sep 28, 2016 10:52 am
The daily high temperatures were recorded at an Antarctic weather station. If a temperature of -38.2 degrees Celsius is 2 units of standard deviation below the mean temperature, and a temperature of -22.6 degrees Celsius is 4 units of standard deviation above the mean, which of the following temperatures is 1 unit of standard deviation above the mean?

A) -35.6 degrees Celsius
B) -33.0 degrees Celsius
C) -30.4 degrees Celsius
D) -27.8 degrees Celsius
E) -25.2 degrees Celsius
Here's another approach:
Image
Notice that 1 unit of standard deviation above the mean is an equal distance from 2 units of standard deviation below the mean and 4 units of standard deviation above the mean.

So, 1 unit of standard deviation will be the average of the two outer values
That is, 1 unit of standard deviation above the mean = (-38.2 + -22.6)/2
= 60.8/2
= 30.4

Answer: C
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by Matt@VeritasPrep » Thu Sep 29, 2016 6:26 pm
Let me see if I can win the "laziest solution" ribbon at the Antarctic County Fair.

-38.2 to -22.6 is a six standard deviation swing, so

6*d = 38.2 - 22.6

That'll end in 6, so when I divide it by 6 it'll still end in 6.

Thus d ends in 6. -22.6 - 3*(something that ends in 6) = something that ends in 4, so C.

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by Matt@VeritasPrep » Thu Sep 29, 2016 6:27 pm
No, I'm an idiot, there's a lazier way.

The midpoint of +4 and -2 is +1. So (-22.6 + -38.2) / 2 => exactly what we want.

Touchdown!

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by Matt@VeritasPrep » Thu Sep 29, 2016 6:30 pm
Wait, one more contender.

"Whoever wrote this problem must want to trap me by providing all the standard deviations in between! So the answers are -1 sd, -0 sd, +1 sd, +2 sd, and +3 sd. So it must be C."

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by Brent@GMATPrepNow » Fri Sep 30, 2016 6:47 am
Matt@VeritasPrep wrote:Wait, one more contender.

"Whoever wrote this problem must want to trap me by providing all the standard deviations in between! So the answers are -1 sd, -0 sd, +1 sd, +2 sd, and +3 sd. So it must be C."
When in doubt...C! :-)

All great solutions, Matt!

Cheers,
Brent
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