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In the sets above, which of the numbers in set Y, if added to set X, w

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In the sets above, which of the numbers in set Y, if added to set X, w  [#permalink]

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New post 04 Jul 2018, 07:37
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62% (01:40) correct 38% (01:44) wrong based on 40 sessions

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x = {10, 3, 6, 5}
Y = {5, 6, 10}
In the sets above, which of the numbers in set Y, if added to set X, will
be equal to both the average (arithmetic mean) and the median of the
changed set X?
I. 5
II. 6
III 10

A. Only I
B. only II
C. only III
D. Only II and III
E. None of the numbers

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Re: In the sets above, which of the numbers in set Y, if added to set X, w  [#permalink]

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New post 04 Jul 2018, 09:06
Given
X -(10,3,6,5)
Y- (5,6,10)

If we add 5 to X
X- (3,5,5,6,10)
number added =5
median - 5
mean 3+5+5+6+10 /5 =29/5 not equal to 5
5 doesnt satisfy the conditions

If we add 6 to X
X - (3,5,6,6,10)
median - 6
mean - 3+5+6+6+10 /5 = 30/5 =6
clearly 6 satisfies.

We dont need to solve for 10 as it cannot be the median of X.

So answer is only 6
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Re: In the sets above, which of the numbers in set Y, if added to set X, w  [#permalink]

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New post 04 Jul 2018, 09:14
gmatbusters wrote:
x = {10, 3, 6, 5}
Y = {5, 6, 10}
In the sets above, which of the numbers in set Y, if added to set X, will
be equal to both the average (arithmetic mean) and the median of the
changed set X?
I. 5
II. 6
III 10

A. Only I
B. only II
C. only III
D. Only II and III
E. None of the numbers


set X = {3,5,6,10}
by adding any number the median can never be the extremes, so discard 10 in y..

MEDIAN : addition of either of 5 and 6 will make it the median of set X.
so lets check on MEAN..
Mean of X is \(\frac{3+5+6+10}{4}=\frac{24}{4}=6\)......
so if you add 5, the mean cannot be 5 of the set
But when we add 6, the mean will remain 6

so answer 6
B
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Re: In the sets above, which of the numbers in set Y, if added to set X, w &nbs [#permalink] 04 Jul 2018, 09:14
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