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is z – y = y – x?

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by mehravikas » Sat Aug 01, 2009 9:05 pm
If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
(2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.


Source: MGMAT
OA: C
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Source: — Data Sufficiency |
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by real2008 » Mon Aug 03, 2009 12:32 am
i think answer is E
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by zeenab » Mon Aug 03, 2009 11:44 am
Hi,

1)
Mean of {x,y,z,4} > Mean of {x,y,z}

This means that atleast 2 numbers in x y z are less than 4. This condition can only bring the mean higher by adding 4 to the set.
Also solving this
(x+y+z+4)/4 > (x+y+z)/3
gives x+y+z < 12

however the set can be anything in the range
x,y,z = 1,3,5 , or 1,2,8
Insufficient

2)
Median of {x,y,z,4} < {x,y,z}
This does not restrict the values of x,y,z
4 can be anywhere, so the sorted order can be anything
4,x,y,z
x,4,y,z
x,y,4,z
x,y,z,4

Median of {x,y,z} = y
for the median of {x,y,z,4} to be less than y,
these two are only possible
4,x,y,z
x,4,y,z

still it doesnt give us much information about how the numbers are spread.
Insufficient.

Now lets try to combine the two.

x+y+z < 12 from 1)

from 2) firstchoice is out, 4,x,y,z as we cannot have the above condition with this.

So the choice is x,4,y,z and x+y+z < 12

I dont think its possible for the values of x,y,z to satisfy both the conditions. So we can know that z-y not equal to y-x.

Can someone confirm?
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by mehravikas » Mon Aug 03, 2009 12:25 pm
Your explanation is in sync with MGMAT's explanation. However, I am looking for a shorter way to solve this problem.

Thanks,
Vikas
zeenab wrote:Hi,

1)
Mean of {x,y,z,4} > Mean of {x,y,z}

This means that atleast 2 numbers in x y z are less than 4. This condition can only bring the mean higher by adding 4 to the set.
Also solving this
(x+y+z+4)/4 > (x+y+z)/3
gives x+y+z < 12

however the set can be anything in the range
x,y,z = 1,3,5 , or 1,2,8
Insufficient

2)
Median of {x,y,z,4} < {x,y,z}
This does not restrict the values of x,y,z
4 can be anywhere, so the sorted order can be anything
4,x,y,z
x,4,y,z
x,y,4,z
x,y,z,4

Median of {x,y,z} = y
for the median of {x,y,z,4} to be less than y,
these two are only possible
4,x,y,z
x,4,y,z

still it doesnt give us much information about how the numbers are spread.
Insufficient.

Now lets try to combine the two.

x+y+z < 12 from 1)

from 2) firstchoice is out, 4,x,y,z as we cannot have the above condition with this.

So the choice is x,4,y,z and x+y+z < 12

I dont think its possible for the values of x,y,z to satisfy both the conditions. So we can know that z-y not equal to y-x.

Can someone confirm?
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by tohellandback » Mon Aug 03, 2009 9:22 pm
I don't think there is any shorter way to solve this:
best you can do is substitute numbers for 1 and 2 but you still need to do consider a few cases when you combine both:
my solution:

is z – y = y – x or 2y=x+z. basically asking whether y is the arithmetic mean of the numbers x,y and z

1) not sufficient.
take the set (2,3,4) and then take (1,2,5)
in (2,3,4) 3 is the AM. while in (2,2,5), 2 is not.

2) take the numbers (4,6,8) and then take the numbers (4,5,10)
not sufficient.

combining
from 2,
x+y+z/3<x+y+z+4/4
x+y+z/3-x+y+z/4<1
x+y+z/12<1
x+y+z<12

from 1,
the only form will be (x,4,y,z) because
if you take (4,x,y,z). X is atleast 4 y atleast 5 and z atleast 6 and the sum will be >12

if you take x,y,z,4- the new median will be Y+z/2 which is greater than y so not possible
so the form is (x,y,4,z)

new median less than Y

4+y/2< y
4+y<2y
y>4

take Y values >4, z>Y and sum of X Y and Z <12.
there will be no set where Y is the mean.
Answer C
The powers of two are bloody impolite!!
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