Set \(A\) consists of 8 distinct prime numbers. If \(x\) is

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Set \(A\) consists of 8 distinct prime numbers. If \(x\) is equal to the range of set \(A\) and \(y\) is equal to the median of set \(A\), is the product \(x\cdot y\) even?

1) The smallest integer in the set is 5.
2) The largest integer in the set is 101.

OA A
Source: — Data Sufficiency |

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by Jay@ManhattanReview » Wed Jul 24, 2019 10:09 pm

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AAPL wrote:Manhattan Prep

Set \(A\) consists of 8 distinct prime numbers. If \(x\) is equal to the range of set \(A\) and \(y\) is equal to the median of set \(A\), is the product \(x\cdot y\) even?

1) The smallest integer in the set is 5.
2) The largest integer in the set is 101.

OA A
Given: x = Set A: Consists of 8 distinct prime numbers. x = Range of Set A and y = Median of Set A

We have to determine whether xy is even. Note that for xy to be even, if at least one between x and y must be even.

Let's take each statement one by one.

1) The smallest integer in the set is 5.

We know that except '2' all the prime numbers are odd. Since the smallest prime number in the set is 5, the largest, as well as the smallest number in the set, are odd; thus, x = Range = Odd number - Odd number = Even.

We have x even, thus, xy is even. Sufficient.

2) The largest integer in the set is 101.

Case 1: Say the smallest prime number in the set is 2; thus x = Range = 101 - 2 = 99, and odd number
Again, say the 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 101. We have median = y = (7 + 11)/2 = 9, odd number

Since x and y both are odd, xy would be odd.

Case 2: Say the smallest prime number in the set is 2; thus x = range = 101 - 2 = 99, and odd number
Again, say the 8 prime numbers are 2, 3, 5, 7, 13, 17, 19 and 101. We have median = y = (7 + 13)/2 = 10, even number

Since y is even, xy would be even.

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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