February 24th, 2010

At what time will the swimming pool be filled?

storm asked:


A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am?

Gala
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This entry was posted on Wednesday, February 24th, 2010 at 8:56 am and is filed under Swimming Pool. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.

4 Responses to “At what time will the swimming pool be filled?”

  1. Davis P Says:
    February 27th, 2010 at 1:53 am

    Frank

    A rate = 1/3
    B rate = 1/6

    A rate( t - 9) + B rate (t-10) = 1
    1/3 (t-9) + 1/6(t-10) = 1
    2t-18 + t - 10 = 6
    3 t = 34
    t = 34/3
    time is 11:20.

  2. Ted Says:
    February 28th, 2010 at 5:51 am

    Chong

    At 9am the pool will be 1/3 full. ( 1hour of pumping out of the 3 hours required using pumpA)

    At 10:00, rate of filling will be 1/3 + 1/6 = 2/6 + 1/6 = 3/6 per hour

    After x hours, pool will be 3/6 x full

    So to fill the remaining 2/3, 3/6 x = 2/3

    x = 2/3 times 6/3 = 12/9 = 1 1/3 hours

    So the pool will be full at 11:20

  3. Waheed Says:
    March 2nd, 2010 at 1:58 am

    Cherry

    At what time will the swimming pool be filled?
    A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am?

    let swimming pool is x cubic units.
    Rate of flow from pipe A is x/3 units per hour.
    Rate of flow from pipe B is x/6 units per hour.

    form 9 am to 10 am rom pipe A x/3 units of water is collected.
    remaining quantity required to fill the swimming pool is x - x/3 = 2x/3 cubic units.
    When A and B both pipes are working total rate of flow will be x/3 + x/6= x/2 cubic unit per hour.

    to fill 2x/3 cubic units, at rate of x/2 units per hour we need ( 2x/3) / (x/2) = 4/3 = 1.33 hours.

    10 am + 1 hour 20 min = 11 am , 20 min ans

  4. Lord Says:
    March 3rd, 2010 at 10:39 pm

    Leonore

    the rates of the two pumps are

    pump A: 1 / 3 , pump B: 1 / 6
    Working together, If pump A works for t hours then pump B works t - 1 hours since it started 1 hour late. Hence
    t * (1 / 3) + (t - 1) * (1 / 6) = 1

    Solve for t

    t = 7 / 3 hours = 2.3 hours = 2 hours 20 minutes.
    The swimming pool will be filled at

    9 + 2:20 = 11:20