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Math Help - Year 11 - Worded question

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    Year 11 - Worded question

    I've done the whole exercie of worded questions but I'm struggling with these two.

    6. In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages.

    8. Two children had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an dqual number. How many marbles did each child start with and how many did they finish with?


    I can do it with trial and error but it'd be much more helpful in algebraic terms as this is an algebra topic.
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    8. Two children had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an dqual number. How many marbles did each child start with and how many did they finish with?
    After one child had lost half her marbles and the other had lost 20 they had an equal number.


    1st child=x
    2nd child is=110-x

      x-\frac{x}{2}=110-x-20

    let's say 1/2 is 60

    \frac{2x-x}{2}=110-x-20

    \frac{3}{2}x=90

    x=60

    60-30=110-60-20

    30=30
    do we have the same answer? i'll try the other question.
    Last edited by princess_21; February 10th 2009 at 03:26 AM. Reason: as corrected by Adarsh. I mistook 110 as 100. Thanks Adarsh
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    Quote Originally Posted by princess_21 View Post
    After one child had lost half her marbles and the other had lost 20 they had an equal number.


    1st child=x
    2nd child is=100-x

     x-\frac{1}{2}=100-x-20

    let's say 1/2 is 26

    53-26=100-53-20

    57=57

    do we have the same answer? i'll try the other question.
    Let one Child has x marbles so the other will have 110-x marbles

    So
    according to question
    \frac{x}{2}= 110-x-20
    so \frac{3x}{2} =90
    Thus

    x= \frac{180}{3}
    x = 60
    Go ahead!

    @Princess watch "110" and the second step, otherwise you are completely right
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    6. In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages.

    4yrs ago
    son=x
    mother=5x

    present
    son=x+4
    mother=5x+4

    +4yrs
    son=x+8
    mother=5x+8

    equation:
    5x+8=3(x+8)

    5x+8=3x+24

    5x-3x=24-8

    2x=16

    x=8

    4yrs ago
    son=x-- 8

    mother=5x--- 40

    present
    son=x+4--- 12

    mother=5x+4--- 44

    +4yrs
    son=x+8--- 16

    mother=5x+8 48

    in present time the son is 12 yrs old
    mother is 44yrs old

    do we have the same answer?
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    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by nimrod View Post
    I've done the whole exercie of worded questions but I'm struggling with these two.

    6. In four years time a mother will be three times as old as her son. Four years ago she was five times as old as her son. Find their present ages.
    Let the mathers age is x
    And age of son be y

    so
    (x+4) =3(y+4)
    and
    (x-4) = 5(y-4)

    Get the values of x and y now and tell in case if you still have trouble
    If you want to follow single variable follow Princess
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