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M23-37

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M23-37  [#permalink]

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New post 16 Sep 2014, 01:20
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A
B
C
D
E

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  5% (low)

Question Stats:

84% (01:07) correct 16% (01:01) wrong based on 111 sessions

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Re M23-37  [#permalink]

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New post 16 Sep 2014, 01:20
Official Solution:


(1) The sum of numbers in each of the rows is equal. So, \(x+2+1=2+1+y=1+z+2\), which gives that \(x=y=z\). Not sufficient.

(2) The sum of numbers in each of the columns is equal. So, \(x+2+1=2+1+z=1+y+2\), which again gives that \(x=y=z\). Not sufficient.

(1)+(2) All we know is that \(x=y=z\), which is not sufficient to get the value of \(x+y+z\). Not sufficient.


Answer: E
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Re: M23-37  [#permalink]

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New post 15 Feb 2016, 01:01
i thought this question was fair and good.
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Re: M23-37  [#permalink]

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New post 04 Jul 2017, 19:56
Where it is mentioned that x, y and z should be distinct numbers?
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Re: M23-37  [#permalink]

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Re: M23-37  [#permalink]

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New post 05 Jul 2017, 22:21
Bunuel wrote:
x 2 1

2 1 y

1 z 2

If in this table above \(x\), \(y\), and \(z\) stand for numbers, what is the value of \(x + y + z\) ?



(1) The sum of numbers in each of the rows is equal.

(2) The sum of numbers in each of the columns is equal.



(1) The sum of numbers in each of the rows is equal

This does not give us the values of x, y and z, only gives information of \(x =y = z\), which is not sufficient to get the values of \(x+y+z\) as x, y and z can be any number and hence values will change.

Hence, (1) =====> is NOT SUFFICIENT

(2) The sum of numbers in each of the columns is equal

Once again, this does not give us the values of x, y and z, only gives information of \(x =y = z\), which is not sufficient to get the values of \(x+y+z\) as x, y and z can be any number and hence values will change.

Hence, (2) =====> is NOT SUFFICIENT

Combining (1) & (2)

Even after combing we will know that x = y = z, and as the values are not provided we will not be able to find the value of \(x + y + z\)

(1) & (2) =====> is NOT SUFFICIENT

Hence, Answer is E
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Re: M23-37  [#permalink]

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New post 05 Apr 2018, 07:15
+1 for option E.

Statement 1 : We get x=y=z ; NS
Statement 2 : We get x=y=z ; NS

Both together : No additional info ! ; NS

Hence option E.
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Re: M23-37  [#permalink]

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New post 07 Dec 2018, 10:11
mchoudh7 wrote:
Where it is mentioned that x, y and z should be distinct numbers?


You true man! it is not mentioned anywhere , consider

X=y=z=2 then value of x+y+z =6


if , x=y=z=3 then value of x+y+z=9

clearly we are getting different answer each time
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Re: M23-37  [#permalink]

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New post 14 Oct 2019, 11:17
Yes, its not mentioned in the question that XYZ have a distinct from each other. Why should we take this as an assumption that they cant be equal to each other. Can anybody please explain ?
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Re: M23-37  [#permalink]

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New post 14 Oct 2019, 12:26
anshuliitd wrote:
Yes, its not mentioned in the question that XYZ have a distinct from each other. Why should we take this as an assumption that they cant be equal to each other. Can anybody please explain ?


From Statement (1) and from Statement (2) we are getting that x=y=z

we are not assuming anywhere that they can not be equal.
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Re: M23-37   [#permalink] 14 Oct 2019, 12:26
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