OG 2016 DATA Sufficiency #135

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OG 2016 DATA Sufficiency #135

by AshB » Sat Dec 26, 2015 8:51 am
I am not a quant whiz but I think the solution for statement 2 that OG came up with is not right!

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by [email protected] » Sat Dec 26, 2015 10:19 am
Hi AshB,

When posting questions in the Forums, you should post the entire question.

For this particular prompt, you should also explain why you disagree with the solution.

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by AshB » Sat Dec 26, 2015 3:02 pm
A list contains 3 different numbers. Does the median of the 3 numbers equal the average(arithmetic mean) of the 3 numbers?
1) The range of 3 numbers is equal to twice the difference between the greatest number and the median
2) The sum of the 3 numbers is equal to 3 times one of the numbers

OG solves 2) by deriving that all the 3 numbers in the list are equal, which contradicts the question stem information: "different numbers"? Does this happen often on GMAT?

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by GMATGuruNY » Sat Dec 26, 2015 4:21 pm
AshB wrote:A list contains 3 different numbers. Does the median of the 3 numbers equal the average(arithmetic mean) of the 3 numbers?
1) The range of 3 numbers is equal to twice the difference between the greatest number and the median
2) The sum of the 3 numbers is equal to 3 times one of the numbers
When numbers are EVENLY SPACED, the median is equal to the average.
Question stem, rephrased:
Are the 3 numbers evenly spaced?

Statement 1:
Let the 3 numbers, in ascending order, be a, b and c.
Range = biggest - smallest = c-a.
Difference between the greatest number and the median = c-b.

Since the range is equal to twice the difference between the greatest number and the median, we get:
c-a = 2(c-b)
c-a = 2c - 2b
2b = a + c
b = (a+c)/2.

Since b is equal to the average of a and c, b must be HALFWAY between a and c, implying that a, b and c are EVENLY SPACED.
SUFFICIENT.

Statement 2:
Case 1: One of the numbers is 1.
If the sum is equal to three times 1, then the sum is equal to 3.
Options for the 3 numbers:
0, 1, 2 --> 0+1+2 = 3.
-1, 1, 3 --> -1+1+3 = 3
-10, 1, 12 --> -10+1+12 = 3.
In every case, the 3 numbers are EVENLY SPACED.

Case 2: One of the numbers is 10.
If the sum is equal to three times 10, then the sum is equal to 30.
Options for the 3 numbers:
9, 10, 11 --> 9+10+11 = 30.
0, 10, 20 --> 0+10+20 = 30.
-10, 10, 30 --> -10+10+30 = 30.
In every case, the 3 numbers are EVENLY SPACED.

As Cases 1 and 2 illustrate, for the sum to be equal to three times one of the numbers, the 3 numbers must be EVENLY SPACED.
SUFFICIENT.

The correct answer is D.
Last edited by GMATGuruNY on Sun Dec 27, 2015 10:36 pm, edited 1 time in total.
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by GMATGuruNY » Sat Dec 26, 2015 4:43 pm
AshB wrote:OG solves 2) by deriving that all the 3 numbers in the list are equal, which contradicts the question stem information: "different numbers"? Does this happen often on GMAT?
The OG's explanation for Statement 2 is faulty.
Since the prompt requires that the 3 numbers be DIFFERENT, the explanation for Statement 2 should not discuss cases in which the 3 numbers are EQUAL.
As my solution above shows, statement 2 can be satisfied by 3 different numbers.
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