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Posts posted by Rob_G

  1. Well

    Assume that 12 are already balanced on the scale. If I switch any gear for the 13th one the scale would still remain balanced. Therefore the 13th gear and the gear I would replace have the same weight. I can take the same 13th gear and repeat this for the other 11 gears for the same result making all 13 of them the same weight.

  2. Assuming we can't just break the goblet and spill it all...

    I get approx 0.606 in which will displace approx 0.908 cubic in of water.

    Oops...I had an extra parenthesis somewhere.

    So what I really got was a radius of approx 0.643 in and displacement of approx 1.02 cubic inches.

  3. So...

    The multiples of 3 are all of the form 3n where n is a natural number. The largest value of n is floor(1000/3) or 333. So we get 3(1), 3(2), 3(3), ..., 3(333). So if we sum them...sum[n=1...333](3n) we get 3* (333*334)/2 or 166,833.

    The same follows for 5 1000/5=200. So sum[n=1...200](5n) or 5*(200*201)/2 or 100,500.

    Finally we add them together to get 267,333.

  4. Assumptions: The other team wont score again, and we will score a second touchdown.

    Go for two. 

    Theoretically, by kicking for the extra point I will score two points. Also, theoretically, I will succeed in one out of the two 2-point conversions. In both situations I get two more points and tie to go to overtime, but if I succeed on the first 2-point conversion, I can kick a field goal after the second touchdown and win. 

    So doing the numbers:

    Extra point: 100% chance of a tie and 50% chance of winning or if you go for one then the second time go for two you still only have a 50% chance of winning. The only difference is what you put your hope for the win in.

    Two Points: You have a 50% chance of converting the first one and when you do you have a 100% chance of winning by kicking a field goal. 

    If you miss the first (50%) you have a 50% chance to make the second so a 25% chance of that outcome. In this case you only tie and go to overtime with a 50% chance of winning or 12.5% overall. Putting this together with the first part gives us a total of 62.5% chance of winning.


    The other part will require paper instead of just my head so I will get back to you with that.

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  5. Well I guess it depends on how the store operates. I could justify 3 different answers.

    $75 because most stores don't allow you to use coupons on bogo sales.

    $67.50 because they take 10% off of the $75 that you would be charged.

    $60 because the store first charges you for both, takes the 10% off the takes off the $75 for the bogo sale.

  6. no.

    There are 'technically' 16 possible outcomes some are just equivalent.

    1,1 2,2 3,3 4,4 all have a 1/16 chance of occuring while the rest have a 1/8 chance.

    1,1 2,2 3,3 4,4

    1,2 2,1 2,3 3,2 3,4 4,3

    1,3 3,1 2,4 4,2

    1,4 4,1

  7. The first three points are always within a semi-circular arc.

    From 4th point onwards, the probability of lying within a semi-circular arc is 1/2. For 4th the probability is 1/2. For 5th, it will be 4th lies within the arc and the 5th also; so 1/4... and so on

    The overall probability is then 1/2(n-3) for n>3. For n<3 it is 1.

    with your first statement. A counter-example would be if the three points formed an inscribed equilateral triangle.

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