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Maths help??

  • 12-03-2013 8:54pm
    #1
    Hector122
    Closed Accounts Posts: 58 ✭✭


    can anyone help me with this:
    Two taps running together take 6 minutes to fill a bath.
    The hot tap, working by itself takes 100 seconds longer to fill the bath than the cold tap working by itself.
    To the nearest second, how long would it take each tap to fill the bath by itself?


Comments

  • #2
    bpb101
    Registered Users, Registered Users 2 Posts: 2,827 ✭✭✭bpb101


    have you got an answer?
    i think ive got an answer but i am not 100% sure


  • #3
    Hector122
    Closed Accounts Posts: 58 ✭✭Hector122


    bpb101 wrote: »
    have you got an answer?
    i think ive got an answer but i am not 100% sure
    No i dont if you want to throw it up there fire away! :P


  • #4
    bpb101
    Registered Users, Registered Users 2 Posts: 2,827 ✭✭✭bpb101


    forgive me if it is wrong, im not best chaper.

    Right So.
    6 x60 = 360 seconds

    let x= cold tap, let x+100 = hot

    in 360 sec = x + (x+100)
    > 360 = x + (x+ 100)

    360=2x +100
    360-100 =2x

    260/2=x

    x=130 +orginial time to fill bath = 490
    x+100= 590

    cold=490
    hot=590

    i am not 100% sure. Can you get back with the right answer.?


  • #5
    Leaving Cert Student
    Registered Users, Registered Users 2 Posts: 1,026 ✭✭✭Leaving Cert Student


    Kind of a confusing one alright... that answer ^^ looks too simple I would be amazed if that was it but sure it might be


  • #6
    Sensor
    Closed Accounts Posts: 102 ✭✭Sensor


    Forget mine, its all balls


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  • #7
    Sensor
    Closed Accounts Posts: 102 ✭✭Sensor


    OUT OF OUR MYSERY!

    Let V = volume of bath in litres
    Combined rate = V/360 litres/sec

    Let time for cold tap = t sec
    Rate for cold tap = V / t litres /sec

    Time for hot tap = t + 100 sec
    Rate for hot tap = V / (t + 100) litres/sec

    Combined rate = V/t + V / (t + 100) litres/sec

    V/t + V/(t + 100) = V/360

    1/t + 1/(t + 100) = 1/360

    360 (t + 100) + 360 t = t (t + 100)

    720 t + 36000 = t² + 100 t

    t² - 620 t - 36000 = 0

    t = 673 sec

    Tap times are thus 673 sec and 773 sec


  • #8
    bpb101
    Registered Users, Registered Users 2 Posts: 2,827 ✭✭✭bpb101


    Sensor wrote: »
    OUT OF OUR MYSERY!

    Let V = volume of bath in litres
    Combined rate = V/360 litres/sec

    Let time for cold tap = t sec
    Rate for cold tap = V / t litres /sec

    Time for hot tap = t + 100 sec
    Rate for hot tap = V / (t + 100) litres/sec

    Combined rate = V/t + V / (t + 100) litres/sec

    V/t + V/(t + 100) = V/360

    1/t + 1/(t + 100) = 1/360

    360 (t + 100) + 360 t = t (t + 100)

    720 t + 36000 = t² + 100 t

    t² - 620 t - 36000 = 0

    t = 673 sec

    Tap times are thus 673 sec and 773 sec

    Thanks, I didnt think mine was right . Great help for me as well, i very possible question in june


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