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The sum of five consecutive even integers is 20. What is the product of the median of the sequence and the smallest inte

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by Vincen » Wed Jul 22, 2020 12:16 pm

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The sum of five consecutive even integers is 20. What is the product of the median of the sequence and the smallest integer in the sequence?

(A) 12

(B) 10

(C) 6

(D) 4

(E) 0

[spoiler]OA=E[/spoiler]

Source: Magoosh
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Re: The sum of five consecutive even integers is 20. What is the product of the median of the sequence and the smallest

by sahilaro23 » Fri Jul 24, 2020 11:53 am
Let the 5 numbers be:
a-4, a-2, a, a+2, a+4

Sum of these 5 numbers = 20
5a = 20,
a = 5.

So numbers are - 0, 2, 4, 6, 8

Median = 4

Product of median and smallest number = 4*0 = 0.

Hence E.
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Re: The sum of five consecutive even integers is 20. What is the product of the median of the sequence and the smallest

by Scott@TargetTestPrep » Sat Jul 25, 2020 2:23 pm
Vincen wrote: ↑
Wed Jul 22, 2020 12:16 pm
 The sum of five consecutive even integers is 20. What is the product of the median of the sequence and the smallest integer in the sequence?

(A) 12

(B) 10

(C) 6

(D) 4

(E) 0

[spoiler]OA=E[/spoiler]

Solution:

Let x denote the smallest integer in the sequence. Then, the remaining integers are x + 2, x + 4, x + 6, and x + 8.

We can use the formula sum = average x quantity since we have a consecutive set:

average = (smallest integer + largest integer)/2

average = (x + x + 8)/2 = (2x + 8)/2 = x + 4

Thus:

20 = (x + 4)(5)

4 = x + 4

0 = x

Since the smallest integer is found to be 0, we don’t even need to find the median; the product will be 0.

Alternate Solution:

If the sum of the five consecutive even integers (which is an evenly spaced set) is 20, the median must be the average. Thus, the median = 20/5 = 4. So, the two even integers before 4 are 0 and 2 (and the two even integers after 4 are 6 and 8). Thus, the smallest even integer is 0, and the product of 0 and 4 is 0.

Answer: E

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