If w+x+y+z=8 what is sqrt(w^2+x^2+y^2+z^2) ? (1)

Question

If w+x+y+z=8 what is sqrt(w^2+x^2+y^2+z^2) ?

(1) |w|=|x|=|y|=|z|

(2) wx=yz=wz

yessuresh · Answer

Let me try this one....

From (1) w^2 = x^2 = y^2 = z^2
 So, w^2 + x^2 + y^2 + z^2 = 4w^2
 Also, from the stem w+x+y+z = 8. Combining this with (1)
 We can say each w=x=y=z=2.
 With this we can find sqrt(w^2 + x^2 + y^2 + z^2). SUFF
From (2) we can say y=w and x=z
 therefore w+x+y+z = 2(w+x) = 8 => w+x = 4
 w^2 + x^2 + y^2 + z^2 = 2w^2 + 2x^2
 there fore sqrt(w^2 + x^2 + y^2 + z^2) = sqrt(2(w^2 + z^2))
 INSUFF

So, the answer is A.

2times · Answer

Satement 1)  - |w|=|x|=|y|=|z| 

a) if all nos are +ve then w=x=y=z=2
b) if one no is -ve then 3 nos are 4 and one is -4
c) if 2 nos are -ve ... not possible
d) if 3 nos are negative ... not possible

There are 2 possible combinations of numbers that will give w+x+y+z=8;
2+2+2+2 =8 & 4+4+4-4 =8 - INSUFF

 statement 2) 

X&Z are same sign & Y &W are same sign 
but says nothing about the values - INSUFF

Combining - we get value of w=x=y=z = 2 

we get W^2 +X^2 + Y^2 +Z^2 = 16
but the question asks sqrt (W^2 +X^2 + Y^2 +Z^2 )......

so I am slightly confused that does sqrt (16) give us ans as 4 (in which case option C) or ans as +/-4 (in which case and E)

 Thus ans is either E or C 

Can anyone please clarify my last point?

Vel · Answer

I got C. All numbers = 2, not 4 and -4. If it were 4 and -4 the second ST would't work. Am I right?

2times · Answer

Vel,

Maybe I ddin't put across my question clearly

Once we get the final value as sqrt(2^2 +2^2+2^2+2^2) = sqrt (16)

does that mean the value of the expreresion (& not of w,x,y,z)= 4 thus C
or does that mean the value of the expreresion (& not of w,x,y,z)= +/-4.. thus E

anyone?

kevincan · Answer

I'm forced to use sqrt(n) to mean the positive square root of n, as I can't 

..........................______
use the symbol V n

If n is the square root of 25, |n|=5

but sqrt(25)=5

Professor · Answer

you got it and it should be C since sqrt(16) is only 4   not -4  .


from i, either they all are 2 or three 4s and one -4.
from ii, w=y and x=z and (w+x) = (y+z) = 4

from i and ii, w=x=y=z=2

If w+x+y+z=8 what is sqrt(w^2+x^2+y^2+z^2) ? (1)

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