Christmaths Advent Calendar 2013: Day 15

This year’s Christmaths Advent Calendar will have a question for every day in December up to and including Christmas Day, each one featuring a new mathematical challenge.

The star at the top of this Christmas tree flashes seven times in one minute.

tree

Use your calculation skills to work out the following.

How many flashes will take place in:

  1. 20 minutes
  2. 1 hour
  3. 1 day
  4. 1 week
  5. Between December 10th and December 25th
  6. What if the light remained working for a whole year… how many flashes would there be?
  7. The bulb is guaranteed to last for 50,000 flashes. How many bulbs would you need to keep the star lit for a whole year?

Got an answer? Please leave a comment below.

Mr Pitts currently teaches Computing to the whole school along with some other subjects during teacher’s PPA time. He runs this website and makes sure it all works smoothly, most of the time!

6 Comments

  1. Emma
    Monday, December 16th 2013 @ 12:41

    1 = 400
    2 = 1,200
    3 = 28,800
    4 = 201,600
    5 = 460,800
    6 = 10,483,200
    7 = 210

    • Mr Pitts
      Monday, December 16th 2013 @ 15:53

      I don’t agree with you here. Remember the star flashes seven times in one minute. I think you’ve been working on twenty times in one minute.

  2. Emma
    Tuesday, December 17th 2013 @ 10:57

    Oops your right I read it wrong!
    1 = 140
    2 = 420
    3 = 10,080
    4 = 70,560
    5 = 161,280
    6 = 3,669,120
    7 = 73

    • Mr Pitts
      Tuesday, December 17th 2013 @ 15:36

      1-5 I agree with.

      6: 7 x 60 x 24 x 365 = 3679200

      and 7: 3679200 divided by 50000 = 73.584, which means that 74 bulbs would be needed overall (otherwise you would be 29200 flashes short – or roughly three days without flashing.)

      Clearly, a 50000 flash guarantee is pretty useless here!

  3. Emma
    Thursday, December 19th 2013 @ 09:50

    I worked on the basis that the answer to 6 was 70,560 (a week) X 52 = 3,669,120 obviously my Maths skills aren’t as good as yours Mr Pitts :)

    • Mr Pitts
      Thursday, December 19th 2013 @ 11:04

      Ah… rounding! 52×7=364 therefore a day gets lost from the year that way.