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Two bells ring at intervals of 9, 15, and 21 minutes. If they last rang together at 11:00 p.m., when did they last ring together?

Mathematics

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  1. Find the prime factorization of each interval.

    The intervals are 9 minutes, 15 minutes, and 21 minutes.

    9=329 = 3^2 15=3×515 = 3 \times 5 21=3×721 = 3 \times 7
  2. Calculate the Least Common Multiple (LCM) of the intervals.

    The LCM represents the time in minutes after which all bells will ring together again.

    LCM(9,15,21)=32×5×7=9×5×7=315 minutes\text{LCM}(9, 15, 21) = 3^2 \times 5 \times 7 = 9 \times 5 \times 7 = 315 \text{ minutes}
  3. Convert the LCM from minutes to hours and minutes.

    315 minutes=31560 hours=5 hours and 15 minutes315 \text{ minutes} = \frac{315}{60} \text{ hours} = 5 \text{ hours and } 15 \text{ minutes}
  4. The bells will next ring together at 11:00 p.m.

    To find when they last rang together, subtract the LCM from 11:00 p.m.

    11:00 p.m.−5 hours 15 minutes11:00 \text{ p.m.} - 5 \text{ hours } 15 \text{ minutes} 11:00 p.m.=23:00 (in 24-hour format)11:00 \text{ p.m.} = 23:00 \text{ (in 24-hour format)} 23:00−5 hours 15 minutes=17:4523:00 - 5 \text{ hours } 15 \text{ minutes} = 17:45

    Converting back to 12-hour format:

    17:45=5:45 p.m.17:45 = 5:45 \text{ p.m.}

    The bells had last rang together at 5:45 p.m.\boxed{\text{5:45 p.m.}}.

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  5. Calculate the nominal area of the rectangle.

    Length L=15 cmL = 15 \text{ cm}
    Width W=10 cmW = 10 \text{ cm}
    Nominal Area A0=L×W=15 cm×10 cm=150 cm2A_0 = L \times W = 15 \text{ cm} \times 10 \text{ cm} = 150 \text{ cm}^2.

  6. Determine the maximum possible dimensions due to the 5% error.

    The error in each measurement is 5%.
    Maximum length Lmax=L+(0.05×L)=15 cm+(0.05×15 cm)=15 cm+0.75 cm=15.75 cmL_{max} = L + (0.05 \times L) = 15 \text{ cm} + (0.05 \times 15 \text{ cm}) = 15 \text{ cm} + 0.75 \text{ cm} = 15.75 \text{ cm}.
    Maximum width Wmax=W+(0.05×W)=10 cm+(0.05×10 cm)=10 cm+0.5 cm=10.5 cmW_{max} = W + (0.05 \times W) = 10 \text{ cm} + (0.05 \times 10 \text{ cm}) = 10 \text{ cm} + 0.5 \text{ cm} = 10.5 \text{ cm}.

  7. Calculate the maximum possible area.

    Amax=Lmax×Wmax=15.75 cm×10.5 cm=165.375 cm2A_{max} = L_{max} \times W_{max} = 15.75 \text{ cm} \times 10.5 \text{ cm} = 165.375 \text{ cm}^2
  8. Calculate the maximum absolute error in the area.

    ΔAmax=Amax−A0=165.375 cm2−150 cm2=15.375 cm2\Delta A_{max} = A_{max} - A_0 = 165.375 \text{ cm}^2 - 150 \text{ cm}^2 = 15.375 \text{ cm}^2
  9. Calculate the maximum percentage error in the area.

    Percentage Error=ΔAmaxA0×100%\text{Percentage Error} = \frac{\Delta A_{max}}{A_0} \times 100\% Percentage Error=15.375 cm2150 cm2×100%=0.1025×100%=10.25%\text{Percentage Error} = \frac{15.375 \text{ cm}^2}{150 \text{ cm}^2} \times 100\% = 0.1025 \times 100\% = 10.25\%

    The maximum percentage error in the area of the rectangle is 10.25%\boxed{10.25\%}.

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  10. Use the property that the sum of an interior angle and its corresponding exterior angle of a polygon is 180∘180^\circ.

    Given interior angle =3x∘= 3x^\circ
    Given exterior angle =(x−20)∘= (x-20)^\circ

    3x+(x−20)=1803x + (x-20) = 180
  11. Solve the equation for xx.

    4x−20=1804x - 20 = 180 4x=180+204x = 180 + 20 4x=2004x = 200 x=2004x = \frac{200}{4} x=50x = 50
  12. Calculate the measure of the exterior angle.

    Exterior angle =(x−20)∘=(50−20)∘=30∘= (x-20)^\circ = (50-20)^\circ = 30^\circ.

  13. Find the number of sides (nn) of the polygon.

    For a regular polygon, the sum of the exterior angles is 360∘360^\circ, and each exterior angle is 360∘n\frac{360^\circ}{n}.

    Exterior angle=360∘n\text{Exterior angle} = \frac{360^\circ}{n} 30∘=360∘n30^\circ = \frac{360^\circ}{n} n=36030n = \frac{360}{30} n=12n = 12

    The number of sides of the polygon is 12\boxed{12}.

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