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# QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average

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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

Practice Questions
Question: 6
Page: 177
[Reveal] Spoiler: OA

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Retired Moderator
Joined: 07 Jun 2014
Posts: 4804
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 163

Kudos [?]: 2762 [1] , given: 394

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.

$$\frac{(2+5+s+t)}{4} = \frac{(2+5+t)}{3}$$
$$\frac{(7+s+t)}{4} = \frac{(7+t)}{3}$$
$$3s=7+t$$

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.
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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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Expert's post
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

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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
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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.
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Re: QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average [#permalink]  26 May 2020, 13:54
This exercise should be attacked by making assumptions.

we are told that we have the same avg in both sets.

we assume two scenarios.

one where we have an avg of 100 and the other one where we have an avg of -100

with the first assumption we get that s is 93

with the second assumption we would get that s is -100

Re: QOTD#20 List X: 2, 5, s, t List Y: 2, 5, t The average   [#permalink] 26 May 2020, 13:54
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