If the average (arithmetic mean) of seven consecutive

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 am; Followed by:3 members

If the average (arithmetic mean) of seven consecutive

by VJesus12 » Mon Jun 17, 2019 3:03 am

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If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

[spoiler]OA=C[/spoiler]

Source: Magoosh

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Source: Beat The GMAT — Problem Solving | Mon Jun 17, 2019 3:03 am

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by GMATGuruNY » Mon Jun 17, 2019 3:41 am

VJesus12 wrote:If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

For any set of consecutive integers, average = median.
Let the 7 integers be 1, 2, 3, 4, 5, 6, 7.
Since k+2 = average = median = 4, k=2.
Product of the greatest and least = 7*1 = 7.

The correct answer must yield 7 when k=2.
Only C works:
kÂ² + 4k - 5 = 2Â² + (4*2) - 5 = 7

The correct answer is C.

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by Brent@GMATPrepNow » Mon Jun 17, 2019 4:17 am

VJesus12 wrote:If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

Another approach:

For any set of consecutive integers, mean = median.
So, in this case, mean = median = k + 2
So, the three values that come AFTER k + 2 are k+3, k+4, and k+5
And the three values that come BEFORE k + 2 are k-1, k and k+1

The seven values are: k-1, k, k+1, k+2, k+3, k+4, k+5

The product of the greatest and least integer =(k-1)(k+5)
= kÂ² + 4k - 5

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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by Scott@TargetTestPrep » Wed Jun 19, 2019 5:44 pm

VJesus12 wrote:If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

[spoiler]OA=C[/spoiler]

Source: Magoosh

We can let the smallest integer in the set = x and the largest = x + 6; thus:

(x + x + 6)/2 = k + 2

2x + 6 = 2k + 4

2x = 2k - 2

x = k - 1

Thus, the smallest integer is k - 1 and the greatest is k + 5, so the product of those two values is:

(k - 1)(k + 5) = k^2 + 4k - 5

Alternate solution:

Since k + 2 is in the average of the 7 consecutive integers, it's also the median, i.e., the 4th integer. Thus, the largest integer is 3 more than k + 2, i.e., k + 5, and the least integer is 3 less than k + 2, i.e., k - 1. The product of the greatest and least integer is:

(k + 5)(k - 1) = k^2 + 4k - 5

Answer: C

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