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Given: N1 = 3 taps, T1 = 54 minutes.

Mathematics

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Answers

2 parts
a

18 minutes\text{18 minutes}

b

13.5 minutes\text{13.5 minutes}

Step-by-step solution

6 steps

Here's how to solve the problem using inverse proportion:

The total amount of "work" (filling the tank) is constant. This means the number of taps multiplied by the time taken is constant.
Let NN be the number of taps and TT be the time in minutes.
So, N×T=constantN \times T = \text{constant}.

Given: N1=3N_1 = 3 taps, T1=54T_1 = 54 minutes.
The constant work is 3 taps×54 minutes=162 tap-minutes3 \text{ taps} \times 54 \text{ minutes} = 162 \text{ tap-minutes}.

Part a· 3 steps

9 taps

  1. Set up the inverse proportion equation.

    Let T2T_2 be the time taken by 9 taps.

    N1T1=N2T2N_1 T_1 = N_2 T_2 3 taps×54 minutes=9 taps×T23 \text{ taps} \times 54 \text{ minutes} = 9 \text{ taps} \times T_2

  2. Calculate the total work units.

    3×54=162 tap-minutes3 \times 54 = 162 \text{ tap-minutes}

  3. Solve for T2T_2.

    162=9×T2162 = 9 \times T_2 T2=1629T_2 = \frac{162}{9} T2=18 minutesT_2 = 18 \text{ minutes}

    It will take 9 taps 18 minutes\boxed{\text{18 minutes}} to fill the tank.

Part b

13.5 minutes\text{13.5 minutes}

12 taps

  1. Set up the inverse proportion equation.

    Let T3T_3 be the time taken by 12 taps.

    N1T1=N3T3N_1 T_1 = N_3 T_3 3 taps×54 minutes=12 taps×T33 \text{ taps} \times 54 \text{ minutes} = 12 \text{ taps} \times T_3

  2. Use the total work units calculated previously.

    162 tap-minutes=12 taps×T3162 \text{ tap-minutes} = 12 \text{ taps} \times T_3

  3. Solve for T3T_3.

    T3=16212T_3 = \frac{162}{12} T3=816T_3 = \frac{81}{6} T3=272T_3 = \frac{27}{2} T3=13.5 minutesT_3 = 13.5 \text{ minutes}

    It will take 12 taps 13.5 minutes\boxed{\text{13.5 minutes}} to fill the tank.

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