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The average of 10 consecutive integers is 12. Then, 9 . . .

Expert replies
by M7MBA » Sat Jan 13, 2018 11:29 am
The average of 10 consecutive integers is 12. Then, 9 is deducted from the first consecutive number, 8 is deducted from the second, 7 is deducted from the third, and so on until the last number which remains unchanged. What is the new average?
A. 55
B. 50
C. 6
D. 7.5
E. 7

The OA is the option D.

How can I solve this PS question? I think is difficult. Experts, could you give me some help? I would be thankful.
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Source: — Problem Solving |

by [email protected] » Sat Jan 13, 2018 12:14 pm
Hi M7MBA,

We're told that the average of 10 consecutive integers is 12. Then, 9 is deducted from the first consecutive number, 8 is deducted from the second, 7 is deducted from the third, and so on until the last number (which remains unchanged). We're asked for the new average. You can approach this question in a number of different ways - depending on how you want to do the math.

To start, since the average of the 10 number is 12, the SUM of the numbers is (10)(12) = 120.

We're going to subtract 9, then 8, then 7, etc.... down to 0 from that total. The sum of the integers from 9 to 0, inclusive, is 45.

Thus, the 'new' sum is 120 - 45 = 75, so the 'new' average is 75/10 = 7.5

Final Answer: D

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by Scott@TargetTestPrep » Mon Aug 05, 2019 4:16 pm
M7MBA wrote:The average of 10 consecutive integers is 12. Then, 9 is deducted from the first consecutive number, 8 is deducted from the second, 7 is deducted from the third, and so on until the last number which remains unchanged. What is the new average?
A. 55
B. 50
C. 6
D. 7.5
E. 7
The initial sum is 10 x 12 = 120.

The total removed amount is 10 x (9 + 0)/2 = 45.

So the new average is (120 - 45)/10 = 75/10 = 7.5.

Answer: D

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