The worlds first riddle!

[size=7]but of course i would say write it, at least a little, discreetly[/size]

No assumption, the problem stated it:

"You bought m red plums and n yellow plums from each store"

Markr

I quote the original problem for my fallible memory not yours



m red plums and n yellow plums at each store spending a total of $10

so

n/2+m/3+n/2+m/6=$10
combining
n/2+m./2=10
and n+m=20

so then you bought n+m plums at each store, and there are two stores, so from both stores you bought 2n+2m plums, or 40 plums altogether.

I made the problem harder than stated, but by blind accidental stupidity I still stand by my conclusion. The answer is d (40).

Rap

You're right. I stopped at n+m=20, forgetting to double it.

Hi guy's. My post of an hour ago seems to be missing. I will see if I have a backup.

Good to see you have been busy.

Great answers Rap. It's not often one can see a flurry of confessions and a correct answer - good fun.

However, please remind me not to go on a heist with you guys, confession may be good for the sole but not for a jury.


PS I think 3 is over doing things Mark. It was only a double error.


Back later.

Check out Rap's answers. (However, ignore the woman bumps analysis if you are under 21)


McRonalds
The answer is 43.

For any desired number if it is divisible by 3 it can easily be made with 6 and 9 packs, except if the number is 3 itself. If you can't use all six packs then use one 9 pack and you can do the rest with six packs.
If the number is not divisible by 3 then use one 20 pack. If the remaining number is divisible by 3 then use the above method for the rest.
If the number still isn't divisible by 3 use a second 20 pack. The remainder must be divisible by 3, in which case use the 6 and 9 packs as above.

The largest impossible number would be such that you would have to subtract 20 twice to get a remainder divisible by 3. However, you can't make 3 itself with 6 and 9 packs. So the largest impossible number is 2*20+3=43.

Milk:
ln(17/15) =~ 0.1252
dc/dt = 100-c
dc = 100-c dt
dc 1/(100-c) = dt
-ln(100-c) = t + K1
t = ln(K2) - ln(100-c) (where K1 = ln(K2)
t = ln(K2/(100-c))
When t=0 c=15 so K2 must be 85
t = ln(85/(100-c))
Solving for t when c=25:
t = ln(85/(100-25)) = ln(85/75) =~ ln(17/15) = 0.1252



Given a coin with probability p of landing on heads after a flip, what is the probability that the number of heads will ever equal the number of tails assuming an infinite number of flips




A store offered triple the GST in savings. A sales clerk calculated the selling price by first reducing the original price by 21% and then adding the 7% GST based on the reduced price. A customer protested, saying that the store should first add the 7% GST and then reduce that total by 21%. They agreed on a compromise: the clerk just reduced the original price by the 14% difference.

How do the three ways compare with one another from the customer's point of view

(a) The clerk's way is the best.
(b) The customer's way is the best.
(c) The compromise is the best.
(d) All three ways are the same.
(e) The compromise is the worst while the other two are the same
(f) Get a life.



If m and n are integers such that 2m - n = 3, then what will m - 2n equal

(a) -3 only
(b) 0 only
(c) Only multiples of 3
(d) Any integer
(e) None of these
(f) Get bent.


BTW Rap, I love your weddin pic.

Biased coin

[size=7]Offhand I'd say never, but it really depends upon the value of p. If p is the probability of heads, then 1-p is the probability of tails. Moreover as the number of flips goes to infinity this is the ratio of heads and tails. However, there is that possibility that heads equals tails, if p is close to 1-p, and the probability of heads is only slightly more likely than tails.

IOW it a skewed game, and, offhand, the probability of heads exceeding tails with a p biased coin would approach 1-p.[/size]

Sales tax

[size=7]Let X be the items marked price

Clerks calc X(1-2(.07))(1+.07)=X(0.79)(1.07)=0.8453X
Customers calc X(1+.07)(1-3(.07))=X(1.07)(0.79)=X(0.8453)
Compromise X(1-.14)=0.86X

Then answer is
(e) The compromise is the worst while the other two are the same [/size]

m and n

[size=7]2m-n=3
so
n=2m-3
and
m-2n=m-2(2m-3)=m-4m+6=-3m+6=-3(m-2)
and -3(m-2)|3 for all integral m

(c) Only multiples of 3 [/size]

Oh, thanks for the compliment on the picture, but that wasn't a wedding picture. That's my college graduation picture. I majored in Polaroid Camera repair.

Rap

[size=8]COIN
How about
SUM(n=1 to infinity) of 2*Cn*p^n*(1-p)^n
where Cn is the nth Catalan number?

STORE
e

M & N
c[/size]

Rap wrote, "I majored in Polaroid Camera repair."

May I ask how your career developed?




Biased coin

Offhand I'd say never, but it really depends upon the value of p. If p is the probability of heads, then 1-p is the probability of tails. Moreover as the number of flips goes to infinity this is the ratio of heads and tails. However, there is that possibility that heads equals tails, if p is close to 1-p, and the probability of heads is only slightly more likely than tails.

IOW it a skewed game, and, offhand, the probability of heads exceeding tails with a p biased coin would approach 1-p.

I like the way you think.



The answer could be; 2*min (p,1-p).

Think of this problem instead as one of a random walk along a number line. Let 0 be the starting point, p be the probability of moving to the right, and q be the probability of moving to the left.
Let B=probability of ever moving one to the left from where you are.
Let A=probability of revisiting current square from the right.

B = q + Aq + A^2q + A^3q + ... = q/(1-A).
A=pB --> B=A/p --> B=A/(1-q).
q/(1-A) = A/(1-q) --> A=q,1-q.

However, A must be less than or equal to both p and q, thus the reasonable solution is A=min(q,1-q).

Redo the above only reverse the words left and right and A will still equal min(q,1-q). One ramification of this is that the probability of revisiting 0 is the same from both the left as the right. This stands to reason since any path has a mirror image on the other side of equal probability. So the answer is 2*min(q,1-q). Where I am a little uncomfortable is dismissing the other solution of A. I believe I can do so but can not put into words why.

Below are the results of computer trials which have verified the answer above. For each probability p 100,000 trials were conducted, in which a trial ended when either the number of heads equalled the number of tails, or the difference was more than 25. It seems reasonable to assume that for p<=.4 if the difference ever equals 25 it is very unlikely that the gap ever be closed. A 'win' means the number of heads was ever equal to the number of tails. The number of losses would be 100,000 minus the number of wins.




Sales tax

Let X be the items marked price

Clerks calc X(1-2(.07))(1+.07)=X(0.79)(1.07)=0.8453X
Customers calc X(1+.07)(1-3(.07))=X(1.07)(0.79)=X(0.8453)
Compromise X(1-.14)=0.86X

Then answer is
(e) The compromise is the worst while the other two are the same


Since 1.07 times 0.79 is 0.8103, both the customer's way and the clerk's way yield a discount approximately 19%.



m and n

2m-n=3
so
n=2m-3
and
m-2n=m-2(2m-3)=m-4m+6=-3m+6=-3(m-2)
and -3(m-2)|3 for all integral m

(c) Only multiples of 3


We have m - 2n = m - 2n + 2m - n - 3 = 3(m - n - 1).


Mark:

COIN
How about
SUM(n=1 to infinity) of 2*Cn*p^n*(1-p)^n
where Cn is the nth Catalan number?

You may be onto something with that.


STORE
e

M & N
c





If x is x% of y, and y is y% of z, where x, y and z are positive real numbers, what is z

(a) 100
(b) 200
(c) 10000
(d) Does not exist
(e) Cannot be determined
(f) Is this a joke?



He! He! You will like this - NOT!
What is the value of $1, invested for one year at 100% interest, compounded infinitely



Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers


Do NOT try this at home!

Zzzzzzzzzzzz's (I'm awake---I/m awake)

[size=7]Ignoring the first relation (it's unnecessary) then
y=yz/100
so
1=z/100
and z=100 and the answer is (a)[/size]

$1 compounded infinite times

[size=7]Compounding function
The value of $1 invested for 1year at interest rate i compounded n times per year is $(1+(i/n))^n.
In this case the value of i is 100% so interest is (1) as n goes to infinity
So the value is the lim(1+(1/n))^n as n->infinity
This is one definition of e
So the value is $e or about $2.72[/size]

Russian roulette

[size=7]Probability that the first player will win/lose, assuming it a 6 shot revolver
On the first pass=1/6
On the second pass=(5/6*5/6)1/6
On the third pass=(5/6*5/6)(5/6*5/6)1/6
On the nth pass=(5/6)^2n*1/6=(25/36)^n*1/6
So the probability of winning on an infinite number of passes is the sum of all passes
Or
Win/loss=(1/6)*Sum(25/36)^n as n ->infinity
And
(25/36)^n as n ->infinity=1/(1-(25/36))=1/(11/36)=36/11
So the probability of winning/losing is 1/6*36/11=6/11
Note:
If the game was played with a Rugar or a S&W small frame revolver (5 shot) the odds of the first player winning/losing increases to
1/5(1/(1-(16/25))=1/5(25/9)=5/9
And with an H&R 22 cal 9 shot revolver the win/loss probability decreases to
1/9(1/(1-64/81)0=1/9(81/17)=9/17
If the players used a 1911 45 ACP the first player's win/loss will be assured[/size]

The Polaroid repair business is bad---damn digital cameras

Rap

[size=8]XYZ
y = 100, z = 100

INTEREST
$e

ROULETTE
Let x = P(first wins)
x = 5/6 * (1-x)
x = 5/11
Therefore, P(first loses) = 6/11[/size]

Listen up guy's; Your answers far exceed the standard of the question. Take it down a notch.

10% of the people in a certain population use an illegal drug. A drug test yields the correct result 90% of the time, whether the person uses drugs or not. A random person is forced to take the drug test and the result is positive.

What is the probability he uses drugs


Rap, call me old fashioned, but they will never replace the camera obscura.

Drug using probability

[size=7]Use the second form of Bayes Theorem

PE(H)=P(H)Ph(E)/[P(H)Ph(E)=P(~H)P~H(E)]
Where PE(H) is the expectation of positive test accuracy

Where
P(H)is the probability that the testee is a drug user P(E)=0.1
PH(E) is the expected test accuracy (sensitivity?) PH(E)=0.9
P(~H) is the probability that the testee is not a drug user -- this case it is yes/no so P~(H)=1-P(H)=1-0.1=0.9
P~H(E) is the probability that the test is in error (False Positive)---another yes/no situation P~H(E)=1-PH(E)=1-0.9=0.1

Putting in the values into
PE(H)=P(H)Ph(E)/[P(H)PH(E)=P(~H)P~H(E)]=0.1*0.9/(0.1*0.9+0.9*0.1)=0.09/0.18=0.5

So if the odds are 50:50 that the test is accurate.

If you lost your job, or were refused employment strictly on the basis of this single test---I know several attorneys who'd be more than willing to take your case for a third of the potential settlement.

If I were an employer, I'd be looking for a better follow up drug test.[/size]

Camera Obscura. Got one of those at home. Made it out of a 1 pound coffee can

Rap

[size=7]DRUGS
.5[/size]

Rap:

Ignoring the first relation (it's unnecessary) then
y=yz/100
so
1=z/100
and z=100 and the answer is (a)


$1 compounded infinite times

Compounding function
The value of $1 invested for 1year at interest rate i compounded n times per year is $(1+(i/n))^n.
In this case the value of i is 100% so interest is (1) as n goes to infinity
So the value is the lim(1+(1/n))^n as n->infinity
This is one definition of e
So the value is $e or about $2.72 .


The value of $1 invested for n years at interest rate i, compounded x times per year is (1+(i/x))nx.
In this case f(x) = the value of $1 invested at 100% interest compounded n times per year = (1+(1/x))x, where x approaches infinity.
The following table shows values of f(x) for various values of x:



As x approaches infinity it can be seen that f(x) approaches e = ~ 2.71828182846.




Russian roulette

Probability that the first player will win/lose, assuming it a 6 shot revolver
On the first pass=1/6
On the second pass=(5/6*5/6)1/6
On the third pass=(5/6*5/6)(5/6*5/6)1/6
On the nth pass=(5/6)^2n*1/6=(25/36)^n*1/6
So the probability of winning on an infinite number of passes is the sum of all passes
Or
Win/loss=(1/6)*Sum(25/36)^n as n ->infinity
And
(25/36)^n as n ->infinity=1/(1-(25/36))=1/(11/36)=36/11
So the probability of winning/losing is 1/6*36/11=6/11 /
Note:
If the game was played with a Rugar or a S&W small frame revolver (5 shot) the odds of the first player winning/losing increases to
1/5(1/(1-(16/25))=1/5(25/9)=5/9

And with an H&R 22 cal 9 shot revolver the win/loss probability decreases to
1/9(1/(1-64/81)0=1/9(81/17)=9/17

If the players used a 1911 45 ACP the first player's win/loss will be assured.


Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?
The answer is 6/11 = ~ 54.5%

p = 1/6 + (5/6)*(1-p)
6p = 1 + 5 - 5p
11p = 6
p = 6/11



What is the probability he uses drugs?

Use the second form of Bayes Theorem

PE(H)=P(H)Ph(E)/[P(H)Ph(E)=P(~H)P~H(E)]
Where PE(H) is the expectation of positive test accuracy

Where
P(H)is the probability that the testee is a drug user P(E)=0.1
PH(E) is the expected test accuracy (sensitivity?) PH(E)=0.9
P(~H) is the probability that the testee is not a drug user -- this case it is yes/no so P~(H)=1-P(H)=1-0.1=0.9
P~H(E) is the probability that the test is in error (False Positive)---another yes/no situation P~H(E)=1-PH(E)=1-0.9=0.1

Putting in the values into
PE(H)=P(H)Ph(E)/[P(H)PH(E)=P(~H)P~H(E)]=0.1*0.9/(0.1*0.9+0.9*0.1)=0.09/0.18=0.5

So if the odds are 50:50 that the test is accurate. :

If you lost your job, or were refused employment strictly on the basis of this single test---I know several attorneys who'd be more than willing to take your case for a third of the potential settlement.
If I were an employer, I'd be looking for a better follow up drug test.



I give you the long version:

The probability of event A given event B is Pr(A and B)/Pr(B).
In this specific case this is [(.1)(.9)]/[(.1)(.9)+(.9)(.1)] = 1/2.



Mark:

XYZ
y = 100, z = 100

INTEREST
$e

ROULETTE
Let x = P(first wins)
x = 5/6 * (1-x)
x = 5/11
Therefore, P(first loses) = 6/11

DRUGS
.5


Rap writes, "Camera Obscura. Got one of those at home. Made it out of a 1 pound coffee can"

I doubt many can make the same claim!




You (yes you) have ten light bulbs. Five have an average life of 100 hours, and the other five have a average life of 200 hours. These light bulbs have a memory less property in that their current age (measured in how long they have already been on) has no bearing on their future life expectancy. Assuming they are all already on what is the expected number of hours before the first one burns out

Hint: The density function for this kind of light bulb with average life of n hours is f(x)=1/n * e-x/n.



About how many lines can one rotate a regular hexagon through some angle x, 0° < x < 360°, so that the hexagon again occupies its original position

(a) 1
(b) 3
(c) 4
(d) 6
(e) 7
(f) 360



AB is a diameter of a circle of radius 1 unit. CD is a chord perpendicular to AB that cuts AB at E. If the arc CAD is 2/3 of the circumference of the circle, what is the length of the segment AE

(a) 2/3
(b) 3/2
(c) 3/3
(d) Get a job
(e) None of these
(f) Are you serious

[size=8]HEXAGON
Does the line have to be in the same plane as the hexagon? During rotation, does the hexagon have to remain in the plane it is in?

If there are no restrictions, then I'd say 7:
one line through the center that is perpendicular to the plane (like an axle)
three lines through the centers of opposite sides (rotation outside the plane)
three lines through opposite vertices (rotation outside the plane)

CIRCLE & CHORD
(b) 3/2

LIGHT BULBS
I have no idea.[/size]

LIGHT BULBS

When you see the answer, you will know why!


You are in a race in which the starting line is at a certain point on a straight beach. The finish line is in the water. One way to arrive at the finish line is to run 4 kilometers down the beach, make a 90 degree turn and swim 1 kilometer. However, you may cut into the water at any point. You speed on land is 6 k.p.h and you speed in water is 2 k.p.h..

At what point, measured from the starting line, should you cut into the water

[size=8]BEACH RACE
4 - sqrt(2)/4 km[/size]

Light Bulb

[size=7]I had no idea what a density function was, but I figured it had something to do with mean time to failure (MTTF-shades of a past QA life) so I wikipedia-ed MTTF and lo and behold was a link to exponential density functions of the form lmbda*exp(-lmbda*t)

Ok- so now the long lived light bulbs (LLLB) have a lmbda of 1/200 hrs and the short lived light bulbs (SLLB) have a lmbda of 1/100 hrs. and the density function is the probability of a light bulb operating after t hrs of operation (the wikipedia link also explained the definition of memory less, or more properly memoryless)

So now I had an understanding of the problem, and hopefully the solution.

Probability of a single light bulb failing is
Pf(E)=1-exp(-lmbda*t) where lmbda is the MTTF (200 hrs for a LLLB and 1000 hrs for a SLLB)

This problem asks for the expected time for a failure of a single light bulb out of a sample of 5 LLLBs and 5 SLLBs.
Since the probability of failure is one minus the probability of surviving so many hours, then what I want is to figure out what the probability of simultaneous survival of the 10 bulbs, which is (for simplicities sake I'll call the lmbda of LLLBs k0 and the lmbda of SLLBs k1)

Ps(E)=(exp(-k0*t))^5(exp(-k1*t))^5=exp(-5(k0+k1)t)
And
Pf(E)=1-Ps(E)
Moreover the MTTF (lmbda) of the combined set of bulbs is 5(k0+k1)
Putting in k0=1/200 & k1=1/100 then
5(1/200+1/100)=5*3/200=3/40 (hrs^-1)
So I would expect a single light bulb to fail at or about 40/3 hrs.[/size]

Hexagon

I would guess that a hexagon can rotate 60o. But that isn't on the answer menu. So I'm not convinced I understand the question

Circle Arc

[size=7]1+sin(30)=3/2[/size]

Race

[size=7]I saw a similar problem in an article on high school mathematics titled "My Dog can do Calculus." The set up was a lakeshore game of fetch by the author and his Labrador.

The shore leg is 4-x km, the sea leg is (x^2+1)^(1/2) km
The time for the race is
t=(4-x)/6+(x^2+1)^(1/2)/2 hr
optimize by setting dt/dx=0
-1/6+2x/(2*2(x^2+1)^(1/2))=0
-2(x^2+1)^(1/2)+3(2x)=0
(x^2+1)^(1/2)=3x
x^2+1=9x^2
8x^2=1
x=(1/8)^(1/2) km
so run down the beach 4-(1/8)^(1/2) before entering the water for a best time of 1 hour 8 min 17 sec (squaring the corners would take 1 hr 10 min)[/size]

Rap

Coffee Can Obscura

I made it a few years back to show my son a partial solar eclipse, Put a pinhole in the bottom of the can and focused the image of the sun on a flat white rock. Playing with it later we found it worked well as a camera obscura to make cameo silhouettes.

Rap