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QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average [#permalink]
07 Sep 2016, 12:57

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Question Stats:

43% (01:17) correct
56% (01:09) wrong based on 91 sessions

List X: 2, 5, s, t List Y: 2, 5, t

The average (arithmetic mean) of the numbers in list X is equal to the average of the numbers in list Y.

Quantity A

Quantity B

s

0

A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given

Re: QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average [#permalink]
07 Sep 2016, 13:01

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Explanation

Since you are given that the average of the 4 numbers in list X is equal to the average of the 3 numbers in list Y, it follows that \(\frac{(2+5+s+t)}{4} = \frac{(2+5+t)}{3}\). To make it easier to compare Quantity A and Quantity B, you can simplify this equation as follows.

From the equation 3s = 7 + t, if t = –7, then s = 0, but if t = 0, then s > 0. In one case, the quantities are equal, and in the other case, Quantity A is greater. Therefore the correct answer is Choice D. _________________

Sandy If you found this post useful, please let me know by pressing the Kudos Button

Re: QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average [#permalink]
12 Jun 2019, 07:12

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sandy wrote:

List X: 2, 5, s, t List Y: 2, 5, t

The average (arithmetic mean) of the numbers in list X is equal to the average of the numbers in list Y.

Quantity A

Quantity B

s

0

A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given

Another approach is to test some values.

Let t = -7 This means list Y is 2, 5, -7, so list Y has an average of 0 For list X to have an average of 0, s must = 0 In this case, quantity A and quantity B are equal.

Let t = 2 This means list Y is 2, 5, 2, so list Y has an average of 3 For list X to have an average of 3, s must = 3 In this case, quantity A is greater than quantity B

Answer: D

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Re: QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average [#permalink]
02 Oct 2019, 07:56

1

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I got to 3s-1t = 7

Then since A and B have us comparing s and 0, I asked, could s be 0? Sure, if s=0, we are left with -1t = 7, and we can see t could be -7 So, in the case of s=0, our answer is C. Now let's consider if s is ever not equal to 0. Let's try 1. If s=1, then we have 3-1t = 7 Can that ever be true? Sure, let t = -4 Since the A and B could be equal, or not equal, the answer is D.