Mathematics Thread

[Ender]

what standard is set here? I only post those questions I got stucked with my son's primary 5 questions.

[Ender]

correct

 

 

other way is list out

 

eee

eoe

eoo

ooo

oee

ooe

ooo

oeo

Tarzan teach one is it? should be o,e,o,e,o............fade

[1fast1]

Easy one. But let's see who has the most elegant (in my view) solution:

 

Your doctor gives you 5 bottles filled with pills. The pills contained in each bottle are identical in appearance and size. Four of the five bottles hold pills that weigh 10 grams each, and the remaining bottle holds pills that weigh 9 grams each. However, you don't know which bottle is the one holding the lightweight pills. Keep in mind that all of the pills in all bottles are identical in appearance and size. You are given a weighing scale that can only be used once. Using this scale (only once), how do you determine which bottle is holding the 9 gram pills? Note: assume the bottles are quite large and that you have an unlimited quantity of pills from each bottle.

[Wind30]

I think the classic question for maths is

 

1) you have 12 stones and a balance scale. One of the stones is different in weight (either heavier or lighter) than the rest of the 11 stones (all identical).

How many weighings do you need to find out the special stone?

 

 

[Wind30]

deletef

Edited July 28, 2013 by Wind30

[Wind30]

Here's the most elegant way of almost immediately "seeing" the answer. Although it lessens the computation time, the time taken to get the insight may be longer.

 

Anyway, here it is:

 

There are an equal number of odd and even balls in the box (call it box 1). If we switch all the odds to evens and evens to odds (call that box 2), the final probabilities will not change with regard to the sum of any selection with replacement.

 

Consider box 1: Let the prob. of an odd-sum from be p. Then the prob. of an even-sum is (1-p). Now, in the drawn balls, switch every even ball to odd and odd ball to even. The parity of the sum (even or odd) in a drawn lot of 3 will always change. Hence an odd sum will become even and an even sum will become odd. Hence the prob. of getting an even sum now is p. However, the situation is now equivalent to drawing an even sum from box 2, and the prob. of that is (1-p).

 

Hence p = 1-p and p = 1/2.

 

this solution is good as you can extend it to even like 20 balls....

[1fast1]

I think the classic question for maths is

 

1) you have 12 stones and a balance scale. One of the stones is different in weight (either heavier or lighter) than the rest of the 11 stones (all identical).

How many weighings do you need to find out the special stone?

 

This one I solved as a kid.

 

Spoiler:

 

3 weighings. I believe I found more than one algorithm to do the weighings.

 

The more interesting question is how many coins can be disposed of with w weighings, in general. The answer to that exists (or existed) on the web, in a truly elegant inductive proof which I found very impressive.

[1fast1]

this solution is good as you can extend it to even like 20 balls....

 

Careful. If you have an even number draw, the parity does not change with an even-odd switch (e.g. if I draw 4 balls, and they are odd-even-even-even, the parity of the sum is odd. If I switch them all around to even-odd-odd-odd, the parity remains odd). So the same reasoning doesn't quite apply. It is possible to modify the reasoning to accommodate this, but it's an even more subtle argument than the one in my post. It's much easier to simply prove that exactly half the possibilities have odd parity (and half, even parity) no matter how many balls are drawn (odd or even), and since there are an equal number of odd and even balls at any one time, the probability is always half.

Edited July 29, 2013 by Turboflat4

[VellfireS]

I'm starting a mathematics thread since there seem to be mathematics enthusiasts in here (myself included). Go ahead and post interesting and challenging questions.

 

A box contains 100 balls numbered from 1 to 100. It 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd?

 

Remember to time (max 2 mins) yourself because this was taken from a timed exam

wah. muz go back home.. sit down.. do..

[Porker]

No need. It's simple but some people think even plus even plus even equal odd though

 

Too much theory but still logic error. Love you deep deep brother

[VellfireS]

No need. It's simple but some people think even plus even plus even equal odd though

 

Too much theory but still logic error. Love you deep deep brother

hahaha.. bobian la.. my maths sibei jialat wan..

 

wah.. you say like that.. from you.. sounds too too too wong liao..

[1fast1]

Since this thread exists to irritate people (after all, it's from my beloved Porker, who loves people with itchy backsides so he can personally scratch them, but not with his fingers ), here's something to irritate everyone.

 

See diagram (not to scale, hor).

 

 

In triangle ABC, you're given:

 

angle BDC = 90 degrees (so BD is the height and AC is the base)

 

angle ABC = 110 degrees

 

side AC = 100cm

 

height BD = 40cm

 

 

You're asked to find all the other sides and angles, i.e. find lengths of AB and BC and angles BAC and BCA.

 

OK, now go do it.

[Aaronlkl]

I learned something today!   

 

 

https://www.facebook.com/youandimagazine/videos/1202518693114534/

[Vid]

Math question - If the letters of the word RESCUE are laid out in a random order, what is the probability they will spell SECURE?

[Datsun366]

Six per mute six?

[Ender]

Math question - If the letters of the word RESCUE are laid out in a random order, what is the probability they will spell SECURE?

2/720 = > 1/360

Edited May 16, 2017 by Ender

[Kusje]

Six per mute six?

 

2!/6!

[Wind30]

my kid took 30mins to finish her SA2 (1hour 45mins) primary 3. 

 

Got full marks :) sorry for the blatant yaya...