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What is the average (arithmetic mean) of a list of

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What is the average (arithmetic mean) of a list of

by M7MBA » Thu Mar 01, 2018 6:57 am
What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?

(1) The remainder when the fourth integer is divided by 5 is 3.
(2) The ratio of the largest integer to the smallest integer is 5:4.

The OA is the option B .

Experts, may you show me what is the best approach to solve this DS question? I am confused about the ratio. <i class="em em-confused"></i>
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by Jay@ManhattanReview » Mon Mar 05, 2018 3:42 am
M7MBA wrote:What is the average (arithmetic mean) of a list of 6 consecutive two-digit integers?

(1) The remainder when the fourth integer is divided by 5 is 3.
(2) The ratio of the largest integer to the smallest integer is 5:4.

The OA is the option B .

Experts, may you show me what is the best approach to solve this DS question? I am confused about the ratio. <i class="em em-confused"></i>
We have to find out the average (arithmetic mean) of a list of 6 consecutive two-digit integers.

Let's take each statement one by one.

(1) The remainder when the fourth integer is divided by 5 is 3.

Case 1: Say the 6 consecutive two-digit integers are 20, 21, 22, 23, 24, and 25. The average = 22.5.
Case 2: Say the 6 consecutive two-digit integers are 30, 31, 32, 33, 34, and 35. The average = 32.5.

No unique answer. Insufficient.

(2) The ratio of the largest integer to the smallest integer is 5 : 4.

Say the largest integer = 5k and the smallest is 4k, thus 5k - 4k = k. The value of k must be 5 since there 6 consecutive integers.

Thus, the smallest integer = 4k = 5*4 = 20. Thus, the 6 consecutive two-digit integers are 20, 21, 22, 23, 24, and 25. And the average = 22.5. Sufficient.

The correct answer: B

Hope this helps!
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