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# If the average (arithmetic mean) of x and y is 60 and the av

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Manager
Joined: 08 Sep 2006
Posts: 50
If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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23 Jul 2007, 19:33
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Difficulty:

5% (low)

Question Stats:

85% (00:55) correct 15% (01:18) wrong based on 314 sessions

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If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z-x?.

A. 70
B. 40
C. 20
D. 10
E. It cannot be determined from the information.
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Joined: 07 Jul 2004
Posts: 4871
Location: Singapore

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23 Jul 2007, 19:37
(x+y)/2 = 60
x+y = 120 ---[1]

(y+z)/2 = 80
y+z = 160 ---[2]

[2]-[1]
z-x = 40
Manager
Joined: 08 Sep 2006
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23 Jul 2007, 19:39
Can you explain why you are subtracting [2] - [1] ?
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23 Jul 2007, 19:42
tinman1412 wrote:
Can you explain why you are subtracting [2] - [1] ?

The question is looking for z-x. If you substract equation 2 from eqaution 1, then you will get z-x on the left hand side of the equation.
Manager
Joined: 08 Jul 2007
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23 Jul 2007, 19:47
Just to expand on the explanation.

x+y=120 means x=120-y
y+z=160 means z=160-y

Thus, z-x=(160-y)-(120-y)=40-y+y=40
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Joined: 21 Jun 2006
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23 Jul 2007, 22:23
x+y/2 = 60
x+y=120
similarly, y+z=160

y+z-x-y=40
z-x=40
Intern
Joined: 08 Sep 2006
Posts: 34

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24 Jul 2007, 07:48
tinman1412 wrote:
If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z-x?.

a)70
b)40
c)20
d)10
e)It cannot be determined from the information.

This question is from the OG 10, but I do not understand their explanation. Hopefully someone can give me their approach.

Thanks,

I did this problem a couple of ways but it is essential the same as other posters.

(x+y)/2 = 60 (from avg formula)
(y+z)/2 = 80 (from avg formula)

set both equal to each other but we need to subtract 20 from 2nd equation to ensure they are equal (60):

(x+y)/2 = (y+z)/2 - 20 ->
x+y = y+z - 40 ->
z - x = 40
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Joined: 08 Jan 2013
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15 Jun 2013, 17:45
briks123 wrote:
Just to expand on the explanation.

x+y=120 means x=120-y
y+z=160 means z=160-y

Thus, z-x=(160-y)-(120-y)=40-y+y=40

Doh! In the absence of multiplication, my feeble eyes discarded the brackets, and I came to the same answer by the wrong route (finding x, y and z).

Substituting equations:
$$z-x = 40-2y; y = 160-z$$
$$z+x = 280$$

Combining equations:
$$(x+y)/2 + (y+z)/2 = 140$$
$$2y+x+z = 280$$

Compare scribbles:
$$y = 0; z = 160; x = 120$$
$$z-x = 40$$

If, like me, you find yourself wandering down the dark tunnel... there can still be luck. I didn't finish the section on time though
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Joined: 10 Oct 2012
Posts: 613
Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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16 Jun 2013, 01:13
2
tinman1412 wrote:
If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z-x?.

A. 70
B. 40
C. 20
D. 10
E. It cannot be determined from the information.

10 second approach :

Mean of x and y is 60. We can assume both x = y = 60. Again, mean of y and z is 80; We can keep the value of y = 60 and assume the value of z = 100. Thus, z-x = 100-60 = 40.

B.
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Joined: 08 Jan 2013
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Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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16 Jun 2013, 02:04
That's brilliant Mau5. Do you know of a guide with general speed tips?
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Joined: 10 Oct 2012
Posts: 613
Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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16 Jun 2013, 02:26
stormbind wrote:
That's brilliant Mau5. Do you know of a guide with general speed tips?

As for general speed tips, as it is with everyone on gmatclub, I have learnt a lot by just observing Bunuel's solutions! In this case particularly, what I did is just one of the many ways of interpreting averages,i.e. if average of 2 things is X, then each of them can be assumed to be X.
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If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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23 Jul 2016, 07:43
tinman1412 wrote:
If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z-x?.

A. 70
B. 40
C. 20
D. 10
E. It cannot be determined from the information.

$$\frac{X+Y}{2}=60 ; X+Y=120$$

$$\frac{Y+Z}{2}=80 ; Y+Z=160$$

$$(Y+Z)-(X+Y)= 160-120$$

$$Y+Z-X-Y=40$$ (+Y AND -Y CANCELS EACH OTHER)

$$Z-X=40$$

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Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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29 Jul 2016, 15:19
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Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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13 Oct 2016, 16:28
y+z - x - y = 160-120
z-x = 40
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Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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16 Oct 2016, 18:20
1
tinman1412 wrote:
If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z-x?.

A. 70
B. 40
C. 20
D. 10
E. It cannot be determined from the information.

We are given that the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80. We can create the following equations:

Equation 1:

(x + y)/2 = 60

x + y = 120

y = 120 - x

Equation 2:

(y + z)/2 = 80

y + z = 160

y = 160 - z

Since y = 120 - x and y = 160 - z, we can equate the two expressions.

120 - x = 160 - z

z - x = 40

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Joined: 31 Oct 2016
Posts: 3
Re: If the average (arithmetic mean) of x and y is 60 and the av  [#permalink]

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18 Oct 2017, 11:46
x+y/2= 60 = 120
y+z/2=80 = 160
x= 120-y
z= 160-y
therefore
z-x= 160-y -( 120-y)
= 40
Re: If the average (arithmetic mean) of x and y is 60 and the av &nbs [#permalink] 18 Oct 2017, 11:46
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