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# If John's average (arithmetic mean) score for three games of pinball

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Manager
Joined: 06 Jul 2013
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If John's average (arithmetic mean) score for three games of pinball  [#permalink]

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26 Jun 2018, 18:16
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25% (medium)

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85% (01:04) correct 15% (01:11) wrong based on 30 sessions

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If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.

Spoiler: :: Doubt
This question is #90 from Jeff Sackmann's Statistics-Sets 100 Challenge Q's.

I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:

a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient

b) Clearly not sufficient alone.

Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?
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Joined: 02 Sep 2009
Posts: 51229
Re: If John's average (arithmetic mean) score for three games of pinball  [#permalink]

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26 Jun 2018, 19:15
TippingPoint93 wrote:
If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.

Spoiler: :: Doubt
This question is #90 from Jeff Sackmann's Statistics-Sets 100 Challenge Q's.

I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:

a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient

b) Clearly not sufficient alone.

Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?

If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

The median of a set with odd number of terms is the middle term, when arranged in ascending/descending order.

Say, the three scores are x, y, and z, where $$x \leq y \leq z$$. We are told that x + y + z = 3*112 and asked to find the value of y.

(1) The average (arithmetic mean) of John's highest and lowest scores was 118 --> x + z = 2*118. Substitute this in the above equation: 2*118 + y = 3*112 --> y = 100. Sufficient.

(2) John's highest score was 154 --> z = 154. Not sufficient.

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If John's average (arithmetic mean) score for three games of pinball  [#permalink]

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26 Jun 2018, 20:22
Bunuel wrote:
TippingPoint93 wrote:
If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.

Spoiler: :: Doubt
This question is #90 from Jeff Sackmann's Statistics-Sets 100 Challenge Q's.

I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:

a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient

b) Clearly not sufficient alone.

Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?

If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

The median of a set with odd number of terms is the middle term, when arranged in ascending/descending order.

Say, the three scores are x, y, and z, where $$x \leq y \leq z$$. We are told that x + y + z = 3*112 and asked to find the value of y.

(1) The average (arithmetic mean) of John's highest and lowest scores was 118 --> x + z = 2*118. Substitute this in the above equation: 2*118 + y = 3*112 --> y = 100. Sufficient.

(2) John's highest score was 154 --> z = 154. Not sufficient.

Hi Bunuel,

Thanks for responding. If the average of the lowest and highest is 118, then the third term will always be 100. [I understand this part]

But couldn't the largest term be 140 and the smallest be 96? making 100 the median? And couldn't the largest term also be 120 with other term 116, still averaging to 118, thus, making 100 the smallest term and hence not the median?
Math Expert
Joined: 02 Sep 2009
Posts: 51229
Re: If John's average (arithmetic mean) score for three games of pinball  [#permalink]

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26 Jun 2018, 20:30
1
TippingPoint93 wrote:
Bunuel wrote:
TippingPoint93 wrote:
If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

(1) The average (arithmetic mean) of John's highest and lowest scores was 118.
(2) John's highest score was 154.

Spoiler: :: Doubt
This question is #90 from Jeff Sackmann's Statistics-Sets 100 Challenge Q's.

I'm a bit stumped by OA on this question. I don't understand why it's not C. My thought process:

a) Say we have X, Y, Z. If X is 140, Z could be 96 to average 118, and then Y would be 100, making Y the median because X>Y>Z. If X is 120, and Z is 116, they average to 118, Y would still be 100 and Z would be the median because X>Z>Y. Not Sufficient

b) Clearly not sufficient alone.

Combined - If you know what the highest (X) is, and you know what the average of X and Z is and the average of X, Y, Z, then one could determine the median. Am I missing something?

If John's average (arithmetic mean) score for three games of pinball was 112, what was his median score?

The median of a set with odd number of terms is the middle term, when arranged in ascending/descending order.

Say, the three scores are x, y, and z, where $$x \leq y \leq z$$. We are told that x + y + z = 3*112 and asked to find the value of y.

(1) The average (arithmetic mean) of John's highest and lowest scores was 118 --> x + z = 2*118. Substitute this in the above equation: 2*118 + y = 3*112 --> y = 100. Sufficient.

(2) John's highest score was 154 --> z = 154. Not sufficient.

Hi Bunuel,

Thanks for responding. If the average of the lowest and highest is 118, then the third term will always be 100. [I understand this part]

But couldn't the largest term be 140 and the smallest be 96? making 100 the median? And couldn't the largest term also be 120 with other term 116, still averaging to 118, thus, making 100 the smallest term and hence not the median?

The highlighted case violates the statement (1). (1) says: The average (arithmetic mean) of John's highest and lowest scores was 118. In the case when the scores are 100, 116 and 120, the highest and lowest scores are 100 and 120, and their average is 110, not 118.
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Joined: 06 Jul 2013
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Re: If John's average (arithmetic mean) score for three games of pinball  [#permalink]

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26 Jun 2018, 20:37
Bunuel Right, that makes sense. Thanks
Re: If John's average (arithmetic mean) score for three games of pinball &nbs [#permalink] 26 Jun 2018, 20:37
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