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    Amy is one third the age of her mother. in 12yrs time amy will be half the age of her mother. how old is amy and her mother? Zacken
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    12 and 36
    24 and 48


    This wasn't very hard at all.
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    (Original post by manuel132)
    Amy is one third the age of her mother. in 12yrs time amy will be half the age of her mother. how old is amy and her mother? Zacken
    What have you tried? Let's call the age of her mother m (current age) and Amy's age a (current age) Then you know that (now) a = \frac{m}{3}

    And you also know that (in twelve years time) a + 12 = \frac{m+12}{2}, can you solve this simultaneous equation?

    Moving this to maths.
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    (Original post by Zacken)
    What have you tried? Let's call the age of her mother m (current age) and Amy's age a (current age) Then you know that (now) a = \frac{m}{3}

    And you also know that (in twelve years time) a + 12 = \frac{m+12}{2}, can you solve this simultaneous equation?

    Moving this to maths.
    i don't get it.. can you explain it in detail
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    (Original post by manuel132)
    i don't get it.. can you explain it in detail
    So if Amy's age is a and her mothers age is m, then we know that a is m divided by 3 because Amy's age (a) is a third of her moths age. So a = \frac{m}{3}, do you understand this?

    (Original post by Mayhem™)
    ...
    Full solutions are against forum guidelines.
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    and two positive numbers x and y, where x is the larger of the two numbers, have a difference of one and a product of 72. using algebra and showing your working find the values of x and y Zacken
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    Here is a lovely gcse question my friend sent me a few years ago, and i've never forgot it (and always kept it saved)

    Name:  gcse_question.jpg
Views: 145
Size:  49.0 KB

    Edit: for you to try after of course , so off topic but its too good not to share
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    (Original post by Zacken)
    So if Amy's age is a and her mothers age is m, then we know that a is m divided by 3 because Amy's age (a) is a third of her moths age. So a = \frac{m}{3}, do you understand this?



    Full solutions are against forum guidelines.
    well that sucks
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    (Original post by manuel132)
    and two positive numbers x and y, where x is the larger of the two numbers, have a difference of one and a product of 72. using algebra and showing your working find the values of x and y Zacken
    Well, if their difference is 1 then we know that x-y=1. If their product is 72, then we can write: xy =72.

    Now, remember that x-y=1, which is the same thing as x = y+1, so: xy = 72 can be written as y(y+1)=72, now can you solve this?
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    (Original post by Zacken)
    Well, if their difference is 1 then we know that x-y=1. If their product is 72, then we can write: xy =72.

    Now, remember that x-y=1, which is the same thing as x = y+1, so: xy = 72 can be written as y(y+1)=72, now can you solve this?
    you are genius
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    (Original post by DylanJ42)
    Here is a lovely gcse question my friend sent me a few years ago, and i've never forgot it (and always kept it saved)

    Name:  gcse_question.jpg
Views: 145
Size:  49.0 KB

    Edit: for you to try after of course , so off topic but its too good not to share
    Solution
    Spoiler:
    Show
    x^2 + y^2 = y^2 + 2y + 1 \iff x^2 = 2y+1 so sincem^2 \equiv 1 \pmod{2} \iff m \equiv 1 \pmod{2} then x is odd.
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    (Original post by Zacken)
    Solution
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    x^2 + y^2 = y^2 + 2y + 1 \iff x^2 = 2y+1 so sincem^2 \equiv 1 \pmod{2} \iff m \equiv 1 \pmod{2} then x is odd.
    ahaha, of course you reply with the solution to a gcse question :laugh:
    Spoiler:
    Show
    I half expected a "please make a new thread to share new questions", but nice solution anyway A* gcse maths for you
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    (Original post by Zacken)
    So if Amy's age is a and her mothers age is m, then we know that a is m divided by 3 because Amy's age (a) is a third of her moths age. So a = \frac{m}{3}, do you understand this?



    Full solutions are against forum guidelines.
    can you post the full solution in my inbox then cuz i don't get it still?
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    (Original post by DylanJ42)
    ahaha, of course you reply with the solution to a gcse question :laugh:
    Spoiler:
    Show
    I half expected a "please make a new thread to share new questions", but nice solution anyway A* gcse maths for you
    why have you guys 'hijacked' my thread? do unto others as you want others to do unto you! i saw something like this yesterday and zacken didn't like it. don't be a hypocrite
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    (Original post by manuel132)
    why have you guys 'hijacked' my thread? do unto others as you want others to do unto you!
    no we havent, its 3 posts... plus youre getting help from others, plus that question i posted is for you to try, plus if i posted an alevel question and someone posted me another one to try afterwards id be pretty pleased tbh...

    (Original post by manuel132)
    can you post the full solution in my inbox then cuz i don't get it still?
    ...plus this the worst thing you could request
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    (Original post by DylanJ42)
    no we havent, its 3 posts... plus youre getting help from others, plus that question i posted is for you to try, plus if i posted an alevel question and someone posted me another one to try afterwards id be pretty pleased tbh...



    ...plus this the worst thing you could request
    yes you have. it is 'irrelevant trash' to my question.. remember this yesterday? and this is not a ****ing homework.. i am just solving questions so what do you mean it is the worst thing i could request?

    and getting help someone doesn't mean you should post 'irrelevant trash' to my question..

    you guys are hypocrites.. you said the same to someone yesterday. just because i am looking for help doesn't mean i should keep quite. telling you the truth. idc if i am being helped
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    (Original post by manuel132)
    you are genius
    so long as you do not call him
    awesome genius ....
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    (Original post by manuel132)
    yes you have. it is 'irrelevant trash' to my question.. remember this yesterday? and this is not a ****ing homework.. i am just solving questions so what do you mean it is the worst thing i could request?
    "irrelevant trash"... i do not remember as it was probably zacken who said that yesterday

    I am 90% sure you wont learn from the full solution just from your previous posts, thats all

    also dont get angry please, if you think i have "hijacked" your thread by giving you another question to solve then I wont do it again
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    (Original post by DylanJ42)
    "irrelevant trash"... i do not remember as it was probably zacken who said that yesterday

    I am 90% sure you wont learn from the full solution just from your previous posts, thats all

    also dont get angry please, if you think i have "hijacked" your thread by giving you another question to solve then I wont do it again
    well why did he solve the question then if it was my question..

    the guy said nothing wrong yesterday so why were you being *******s to him? he wasn't even hijacking' but you accused him of
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    (Original post by Zacken)
    What have you tried? Let's call the age of her mother m (current age) and Amy's age a (current age) Then you know that (now) a = \frac{m}{3}

    And you also know that (in twelve years time) a + 12 = \frac{m+12}{2}, can you solve this simultaneous equation?

    Moving this to maths.
    i still don't get this one.. sorry but the truth had to be told. that's how i roll and i know the truth hurts but you gotta deal with it
 
 
 
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