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

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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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New post 23 Jul 2007, 18:37
(x+y)/2 = 60
x+y = 120 ---[1]

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

[2]-[1]
z-x = 40
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New post 23 Jul 2007, 18:39
Can you explain why you are subtracting [2] - [1] ?
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New post 23 Jul 2007, 18:42
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.
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New post 23 Jul 2007, 18:47
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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New post 23 Jul 2007, 21:23
x+y/2 = 60
x+y=120
similarly, y+z=160

y+z-x-y=40
z-x=40
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Re: Average Problem. Please explain [#permalink]

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New post 24 Jul 2007, 06:48
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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New post 15 Jun 2013, 16:45
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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Re: If the average (arithmetic mean) of x and y is 60 and the av [#permalink]

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

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New post 16 Jun 2013, 01:04
That's brilliant Mau5. Do you know of a guide with general speed tips?
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Re: If the average (arithmetic mean) of x and y is 60 and the av [#permalink]

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New post 16 Jun 2013, 01:26
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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Re: If the average (arithmetic mean) of x and y is 60 and the av [#permalink]

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

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New post 23 Jul 2016, 06:43
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)

\(Z-X=40\)

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

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New post 29 Jul 2016, 14:19
X+Y=120
Y+Z=160
Equation 2 - Equation 1
=> Z-X=40
Smash B
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Re: If the average (arithmetic mean) of x and y is 60 and the av [#permalink]

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New post 13 Oct 2016, 15:28
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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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

Answer: B
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Re: If the average (arithmetic mean) of x and y is 60 and the av   [#permalink] 16 Oct 2016, 17:20
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