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PR TEST-1 Math DS --Average Weight

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PR TEST-1 Math DS --Average Weight

by iikarthik » Sun Jul 04, 2010 8:01 am
The average weigt of 3 people is 148 pounds.If none of the three weigh the same,
what is the weight of the middle person?

1) The lightest person weighs 120 pounds

2) The differnce between the weights of the lightest person and the middle person is the same as the
difference between the weights of the middle and the heaviest person.

OA is : B

Hi all,
My ans is [spoiler]C. My Solution is as follows:

1)Nothing can be concluded - not sufficient

Let the lightest person weigh x kg
since the diff in weights is the same between the lightest & the middle as well as the middle & the heaviest,use 'y; to denote that difference.

so the 3 different weights are X -y , X , X+Y Bt the value fo Y can be modified to arrive at different combination of weights.so need option 1) to have a fixed value for the weights.So the final answer is C[/spoiler]..I dont understand where i went wrong..Assiatance required.
Thanks
karthik
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by kvcpk » Sun Jul 04, 2010 8:05 am
Let the lightest person weigh x kg
since the diff in weights is the same between the lightest & the middle as well as the middle & the heaviest,use 'y; to denote that difference.

so the 3 different weights are X -y , X , X+Y Bt the value fo Y can be modified to arrive at different combination of weights.so need option 1) to have a fixed value for the weights.So the final answer is C..I dont understand where i went wrong..Assiatance required.
Hi,

The mistake is that, you started with Let the lightest person weigh x kg
and then you changed it to be the middle person's weight.

Let the middle person weight be x.
Then,
x-y,x,x+y are the weights of all three.
so x = 148

Done!!
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by outreach » Sun Jul 04, 2010 8:04 pm
let the wt of lightest person be 'a'
let the wt of middle person be 'm'
let the wt of lightest person be 'b'

given
m-a=b-m
2m=a+b

now
a+m+b=544(3*148)
3m=544
m=148
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by san2009 » Mon Jul 05, 2010 2:14 am
i went with B as well -- with statement 2 you get to two equations, that get you the middle number

however, how can we DEFINITIVELY say that there is NO other set of three evenly spaced integers that will sum to 444?

thanks
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by kvcpk » Mon Jul 05, 2010 2:52 am
san2009 wrote:i went with B as well -- with statement 2 you get to two equations, that get you the middle number

however, how can we DEFINITIVELY say that there is NO other set of three evenly spaced integers that will sum to 444?

thanks
There might exist more than one set with the three numbers evenly spaced. But the middle person weight should be only 148. weights of the other two can vary.

it says the weights are evenly spaced. So, let x be the middle persons weight and y be the difference.
then, weights are x-y,x,x+y
so average is 3x/3 = x = 148. We are done because we are asked only for middle person weight.

The following sets are possible:
147,148,149
146,148,150
144,148,152
143,148,153
...................
......

Hope this helps!!
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by san2009 » Mon Jul 05, 2010 3:54 am
got it. thanks
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by Fiver » Mon Jul 05, 2010 6:24 am
iikarthik wrote:The average weigt of 3 people is 148 pounds.If none of the three weigh the same,
what is the weight of the middle person?

1) The lightest person weighs 120 pounds

2) The differnce between the weights of the lightest person and the middle person is the same as the
difference between the weights of the middle and the heaviest person.
St2 implies that the set is in AP, and in an AP the mean = median.
No need to calculate, with this principle the mid person's weight (median) = 148 (mean)
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