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Problem of the day 25.06.09

with 6 comments

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


Written by Rahul

June 25, 2009 at 4:57 am

Posted in Algebra, Problems

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6 Responses

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  1. 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

    yodha

    June 26, 2009 at 5:22 pm

    Reply

  2. Thanks!

    The answer is incorrect

    Rahul

    June 29, 2009 at 12:27 pm

    Reply

  3. now i got it!

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

    yodha

    July 3, 2009 at 8:34 am

    Reply

  4. right!! post the process :)

    Rahul

    July 3, 2009 at 9:10 am

    Reply

  5. 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.

    yodha

    July 4, 2009 at 2:39 pm

    Reply

    • 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

      anonymous

      July 19, 2009 at 2:39 pm

      Reply

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