Kiss those Math Headaches GOODBYE!

Fun Math Problem #2


Here is the second in my series of “Fun Math Problems.”

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.

To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.

The Problem:  Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

Rusting clock face

Image by The Hidaway (Simon) via Flickr

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Comments on: "Fun Math Problem #2" (6)

  1. Diana M. said:

    I left at 12:05 and returned at 1:00.

    Like

  2. Diana M. said:

    ::sigh:: RTP RTP (Read the Problem) – my answer should have said I was gone 55 minutes.

    Like

  3. ZeroSum Ruler said:

    Hep! I know this isn’t the best place for my question, but I need help with a probability problem and I really stink, and always have, at probability. Here goes…

    If the probability of wining on a scratch ticket is 1/3, what is the chance of losing on all of 30 scratch tickets? (2/3)^30? (2/3)^10? Ugh!

    Like

    • Hi ZS, assuming that whether or not one wins or loses on one scratch ticket (what is that, anyhow?) is independent from winning or losing on any other scratch ticket, you treat each event as an independent event. Laws of probability tell us to multiply the various probabilities of independent events. It appears that the probability of winning on any particular scratch ticket must be 2/3. So then the probability of winning on 30 scratch tickets in a row (if that is what your problem is asking) must be your first answer: (2/3)^30 = approximately 5.2 x 10^–6, which is about .0000052, or 52 out of 10 million, which boils down to 1 chance out of 192,307. Not great chances, but in the world of lottery chances, it could be far worse!

      Like

      • ZeroSum Ruler said:

        I can’t believe it! The reason I am trying to figure out this problem is because I gave my friend 30 $1 scratch tickets for her 30th birthday. Each ticket stated that “one in three tickets is a winner” and she lost on all 30! So the chance of that happening is very, very slim. Thank you for your help. You add further evidence to my belief that the Massachusetts State lottery is rigged!

        Like

  4. Man, you definitely did the research with this post! I have not ever thought of most of that before.
    Wonderful!

    Like

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