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# If the range of the set containing the numbers x, y, and z

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If the range of the set containing the numbers x, y, and z [#permalink]

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05 Feb 2012, 15:31
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If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.
[Reveal] Spoiler: OA

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05 Feb 2012, 15:58
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If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

Let the numbers in ascending order be {x, y, z} (it really doesn't matter how we group them).

The range of a set is the difference between the largest and smallest elements in the set.
Given: $$z-x=8$$. Question: $$x=?$$

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5 --> $$x+y+z+8=4*12.5=50$$ --> $$x+y+z=42$$: 2 distinct linear equations 3 with unknowns. Not sufficient.

(2) The mean and the median of the set containing the numbers x, y, and z are equal --> $$mean=\frac{x+y+z}{3}=y=median$$ --> $$x+z=2y$$: 2 distinct linear equations with 3 unknowns. Not sufficient.

(1)+(2) 3 distinct linear equations with 3 unknowns --> we can solve for $$x$$. Sufficient.

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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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14 Jul 2012, 02:38
hello sir..

here when you have calculated the mean .. as per the data in 1st option mean should include 8 in the set and its value is 12.5

but you have excluded it...
can you just let me know where i am goin wrong

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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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14 Jul 2012, 02:45
mohan514 wrote:
hello sir..

here when you have calculated the mean .. as per the data in 1st option mean should include 8 in the set and its value is 12.5

but you have excluded it...
can you just let me know where i am goin wrong

I don't understand what you mean.

(1) says: The average of the set containing the numbers x, y, z, and 8 is 12.5. So, $$\frac{x+y+z+8}{4}=12.5$$ --> $$x+y+z+8=4*12.5=50$$ --> $$x+y+z=42$$.
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if the range of the set containing the numbers x, y, and z [#permalink]

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15 Sep 2012, 06:47
if the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.

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Re: if the range of the set containing the numbers x, y, and z [#permalink]

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15 Sep 2012, 07:00
if the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.

Merging similar topics. Please refer to the solution above.
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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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15 Sep 2012, 07:10
I had a doubt regarding option A:

In that we are given that x + y+ z = 42 and also that range is 8.

Now lets say smallest number is x and largest is z then equation is (x) + (x+n) + (x+8). Where 0<n<8 .

isnt this constraint enough to give us the required answer? Since we are just looking for the smallest number, for finding the middle number we will need another equation.

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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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15 Sep 2012, 07:20
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I had a doubt regarding option A:

In that we are given that x + y+ z = 42 and also that range is 8.

Now lets say smallest number is x and largest is z then equation is (x) + (x+n) + (x+8). Where 0<n<8 .

isnt this constraint enough to give us the required answer? Since we are just looking for the smallest number, for finding the middle number we will need another equation.

From (1) we can have many options, for example: {10, 14, 18} or {11, 12, 19}, therefore this statement is not sufficient.

Hope it's clear.
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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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21 Sep 2014, 17:41
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If the range of the set containing the numbers x, y, and z [#permalink]

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22 Sep 2014, 08:26
calreg11 wrote:
If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5.
(2) The mean and the median of the set containing the numbers x, y, and z are equal.

C.

did it like this

1) x+y+z+8 = 50
=> x+y+z = 42
and |x-z| = 8
so, sets can be: {8,16,16}, {9,15,17}, {10,14,18}, {11,13,19}, and {12,12,20}
A is insufficient.

2) (x+y+z)/3 = y
=> x+z = 2y
and |x-z| = 8
B is insufficient.

(1)+(2)
among the sets in (1) the sets which have x+z = 2y is {10,14,18}
hence C.
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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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17 May 2016, 05:05
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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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29 Jul 2017, 06:45
Hello from the GMAT Club BumpBot!

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Re: If the range of the set containing the numbers x, y, and z [#permalink]

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04 Aug 2017, 17:38
Bunuel wrote:
If the range of the set containing the numbers x, y, and z is 8, what is the value of the smallest number in the set?

Let the numbers in ascending order be {x, y, z} (it really doesn't matter how we group them).

The range of a set is the difference between the largest and smallest elements in the set.
Given: $$z-x=8$$. Question: $$x=?$$

(1) The average of the set containing the numbers x, y, z, and 8 is 12.5 --> $$x+y+z+8=4*12.5=50$$ --> $$x+y+z=42$$: 2 distinct linear equations 3 with unknowns. Not sufficient.

(2) The mean and the median of the set containing the numbers x, y, and z are equal --> $$mean=\frac{x+y+z}{3}=y=median$$ --> $$x+z=2y$$: 2 distinct linear equations with 3 unknowns. Not sufficient.

(1)+(2) 3 distinct linear equations with 3 unknowns --> we can solve for $$x$$. Sufficient.

Hey Bunuel ,
It doesn't say that x,y,z are the ONLY numbers in the set, right?
I found this question quite ambiguous.
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Re: If the range of the set containing the numbers x, y, and z   [#permalink] 04 Aug 2017, 17:38
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