## Problem of the day 25.06.09

**The real numbers**, *x, y, z, w* **satisfy**

*2x+y+z+w=1*

*x+3y+z+w=2*

*x+y+4z+w=3*

*x+y+z+5w=25*

Find *w*.

**Options**

1. 2

2. 7/2

3. 11/2

4. 5

5. None of these

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congrats!!!

i followed and enjoyed every problem of this blog last year. i used to persevere with those problems. hoping to get problems of that level and sheen this year also.

i m getting w=19/7

i.e. (5)

correct me if i m wrong

yodhaJune 26, 2009 at 5:22 pm

Thanks!

The answer is incorrect

RahulJune 29, 2009 at 12:27 pm

now i got it!

x=-2 , z=0 , y=-1/2 and w=11/2

yodhaJuly 3, 2009 at 8:34 am

right!! post the process 🙂

RahulJuly 3, 2009 at 9:10 am

subtracting eqn. 2 from 1 , 3 from 2 and 4 from 3,we get

x-2y=-1 , 2y-3z=-1 , 3z-4w=-22

=> x = 4w+24.

now, multipling eqn. 1 by 2 and adding it with eqn. 2

5(x+y) + 3z + 3w = 4.

substituting the value of x+y from eqn. 3, we get

17z+2w=11

now solving the two eqns. in z and w. we get z and x

and hence can obtain the values of other variables also.

yodhaJuly 4, 2009 at 2:39 pm

the symmetry is too close to neglect, multiplying the equations by 12, 6, 4, 3 resp. and adding will give the value of x+y+z+w

anonymousJuly 19, 2009 at 2:39 pm