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For three consecutive odd integers, the product of seven...

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For three consecutive odd integers, the product of seven...

by BTGmoderatorLU » Fri Mar 09, 2018 2:52 pm
For three consecutive odd integers, the product of seven and the median of the integers is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

A. 9.0
B. 10.5
C. 11.0
D. 12.5
E. 13.0

The OA is B.

I'm really confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
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by [email protected] » Sat Mar 10, 2018 4:50 pm
Hi LUANDATO,

We're told that we have three CONSECUTIVE ODD integers and the product of 7 and the median of the integers is thirty-five greater than the sum of the other two integers. We're asked for the average of Set A (if Set A contains the original three consecutive odd integers as well as the three products involving each of those integers and the number 2). This question is certainly wordy, but it essentially comes down to some basic arithmetic and figuring out what the 3 consecutive odd integers are. We can beat this question with a bit of 'brute force'

To start, let's see what happens when the consecutive odd integers are the simplest ones we can come up with:
IF... they're 1, 3 and 5, then...
the median = 3
the product described = (7)(3) = 21
the sum of the other two = 1+5 = 6
The difference is 21 - 6 = 15. This is TOO SMALL (it's supposed to be 35), so the three consecutive odds must be LARGER than these three odds

IF... they're 5, 7 and 9, then...
the median = 7
the product described = (7)(7) = 49
the sum of the other two = 5+9 = 14
The difference is 49 - 14 = 35
This is an exact match for what we were told, so these MUST be the odd integers.

Thus, Set A contains:
5, 7, 9, (5)(2) = 10, (7)(2) = 14 and (9)(2) = 18
The average of those 6 values is (5+7+9+10+14+18)/6 = 63/6 = 10.5

Final Answer: B

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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