Important word problems on Linear Equations in two variables 3

Question for practice
The following questions are more helpful for class tenth students for all boards.

Questions 3.  A sum of Rs. 41 was divided among 50 boys and girls. Each boy gets 65 paise and each girl 90 paise. Find the number of boys and girls.


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Comments

  1. Unknown17 August 2012 at 03:09

    Ans. Let no.of boys = x
    Let no. of girls = y
    x+y = 50 ......(1)
    0.65x+0.9y = 41 .....(2)
    by elimination from (1) and (2),
    x = 16
    y = 34
    therefore, No. of boys = 16
    No. of girls = 34

    ReplyDelete
  2. Unknown17 August 2012 at 04:19

    This comment has been removed by the author.

    ReplyDelete
  3. Unknown17 August 2012 at 04:20

    Let numbers of boys = b
    Let numbers of girls = g
    (1)...... b+g = 50
    (2)...... 0.65b+0.9g = 41 [1 rs = 100 p ]
    by elimination from (1) and (2),
    b = 16
    g = 34
    therefore, Numbers of boys = 16
    Numbers of girls = 34

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  4. Unknown17 August 2012 at 04:46

    let the number of boys be x
    let the number of girls be y

    x+y = 50 (1)

    65x+90y = 4100
    => 13x+18y = 820 (2)
    from (1) and (2) , by elimination y = 34
    therefore, x = 16

    Hence, the number of boys = 16 and number of girls = 34

    ReplyDelete
  5. Unknown17 August 2012 at 05:41

    let the number of boys be x and the number of girls be y.
    therefore,
    x+y=50(1)
    and
    0.65x+0.90y=41
    or
    65x+90y=4100
    13x+18y=820(2)
    from (1) and (2) by elimination we get
    x=16
    y=34
    therefore,
    the number of boys=16
    the number of girls=34

    ReplyDelete
  6. Mahima Batheja17 August 2012 at 07:48

    let no. of boys be x
    let no.of girls be y
    x+y = 50 ----(1)
    65x + 90y = 4100 -------(2)[Re.1=100P]
    by elimination method
    y=34 (no. of girls)
    x=16 (no. of boys)

    ReplyDelete
  7. Unknown18 August 2012 at 05:29

    let the number of boys be b
    and number of girls be g
    we know, b+g=50 ( number of boys and number of girls) equation 1

    0.65b + 0.90g=41
    OR
    0.65b + 0.90g=41*100
    =>65b + 90g=4100 equation 2

    subtracting eq. 1 from eq.2 i.e.
    b + g = 50*90
    65b + 90g = 4100
    - - - ( sign changes)

    => 90b + 90g = 4500

    so, 90b + 90g =4500
    65b + 90g = 4100
    (-) (-) (-)
    _____________________
    25b = 400 ( since 90g and 90g got cancelled , we left with)
    b = 400/25 = 80/5 = 16
    = b = 16
    putting this value in equation 2
    65(16) + 90g = 4100
    1040 + 90g = 4100
    90g = 4100 - 1040 = 3060
    90g = 3060
    g = 3060/90 = 102/3 = 34
    => g = 34
    so , number of girls = 34
    and number of boys = 16

    ReplyDelete
  8. Manan Raheja19 August 2012 at 22:55

    Let the no. of boys = x
    Let the no. of girls = y

    So, x + y = 50
    => 13x + 13 y = 650........(1)

    Also, 65x/100 + 90y/100 = 41
    => 65x + 90y = 4100
    => 13x + 18y = 820........(2)

    Eliminating (1) and (2), we get

    y = 34 and x = 16!!!

    THANKS!!!

    ReplyDelete

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