The worlds first riddle!

[size=8]PIRATES
Abe: 75
Bob: 150
Cal: 300[/size]

Mark wrote, "Can you remind us of the question?"

No can do, but it was about three tanks that were filling up with rainwater and had to be emptied. The rain stopped, started, doubled etc. Do not tell me you have forgotten all about it, while I have been working night and day to make sense of it!


BTW. As I may be away tomorrow, may I congratulate you as you start your third year in the forum. The number of questions you have answered will never be equalled. A truly magnificent record.



In the sixth, seventh, eighth, and ninth basketball games of the season, Mame scored 23, 14, 11, and 20 points, respectively.

Her points-per-game average was higher after nine games than it was after the first five games.

If her average after 10 games was greater than 18, what is the least number of points she could have scored in the 10th game

Thanks. It's been fun.

[size=8]MAME
29 points

She averaged 17 points in games 6-9. Therefore, she could have scored at most 84 points in games 1-5 for a total of 152 in games 1-9. After ten games, she needed a total of at least 181 points. So, she scored at least 29.[/size]

The problem:

Three identical, empty open-top containers (C1, C2, C3) are outside.

- At 0 minutes, a steady rain begins and starts filling the containers at R liters per minute.

- Seven minutes later, Try starts emptying C1 and empties it at rate of T liters per minute.

- As soon as C1 is empty (instantaneously), Try starts emptying C2 and empties it in 42 minutes.

- Then Try starts emptying C3. At that point, the rate of the rainfall doubles to 2R liters per minute, and Try doubles his speed to 2T liters per minute. Try empties C3, but before he does, the rain stops.

- Try goes back to C1, then to C2, emptying them at the original rate of T liters per minute in 44 minutes.

At what minute does Try finish emptying C3?

Try's answer: 132 minutes

I'll get back to you. I have to solve it again.

4 kids: 5, 2, 2, 1

has to be 4 kids.... less than five, and 2 middle kids

youngest is 1, and 2 middle are B and B, and oldest is A years old

AB^2 = 2(A + 2B + 1)

A = (4B+2)/(B^2-2)
then i used a graphing calculator to find integer A when B is an integer and got my answer

RAIN
Try: 132 minutes

How did he do it?


The first column shows the elapsed time with variable (unknown) quantities.
Columns 3-5 show the amount of water in the three containers at the various times.
Column 7 shows the equation that can be derived from the information at that point in time.
From the first two equations, we get x = 14 minutes, T/R = 3/2.
Knowing T/R, from the fourth equation, we get y = 6 minutes.
Knowing x, y, and T/R, from the third equation, we get z = 19 minutes.
The second column shows the elapsed time computed by plugging in the values for x, y, and z.

Thoh:

4 kids: 5, 2, 2, 1

has to be 4 kids.... less than five, and 2 middle kids

youngest is 1, and 2 middle are B and B, and oldest is A years old

AB^2 = 2(A + 2B + 1)

A = (4B+2)/(B^2-2)
then i used a graphing calculator to find integer A when B is an integer and got my answer


Nice and logical.



For those with only pen and paper:
Age possibilities:
3 6 *
4 4 *
1 3 8 **
1 4 5
2 2 4 *
1 1 4 6 **
1 2 2 5
1 2 3 3 **
2 2 2 2 *

* eliminated due to youngest(s) being 1 (wife gave birth a year ago).
** eliminated: if number on the door was equal to oldest, then John would know.
So, it's 1-4-5 or 1-2-2-5; 1-2-2-5 is it, due to "two in the middle"...


RAIN
132

Wow! That was a lucky guess of mine . I wonder how many ways there are to reach the same conclusion. I think I have it as:


R = Rain speed
T = Trys speed
A = hours to empty C1
B = hours of rain in C3
C = hours to empty C3 (Note: AR means A*R)
Try enters C1 at hour 8

Gallons already in C1 : 7R
Rain while Try empties: AR
Trys emptying of C1 : AT
so: AT = 7R + AR ; T / R = (A + 7) / A (1)

Try empties C2 in 42 hours
Gallons already in C2 : R(A + 49)
Rain while Try empties: 42R
Trys emptying of C2 : 42T
so: 42T = R(A + 49) ; T / R = (A + 49) / 42 (2)

(1) (2) : (A + 7) / A = (A + 49) / 42
A^2 + 7A - 294 = 0
(A + 21)(A - 14) = 0 ; A = 14 (3)

(1) (3) : 2T = 3R

Try empties C3; both speeds double
Gallons already in C3 : R(A + 49) = 63R
Rain while Try empties : 2RB
Trys emptying of C3 : 3RC
so: 3RC = 2RB + 63R ; C = (2B + 63) / 3 (4)

Try empties C1 and C2 in 44 hours, normal speed
Gallons now in C1 and C2: R(42 + 4B)
Trys emptying of C1/C2: 44(3R / 2)
so: R(42 + 4B) = 44(3R / 2) ; B = 6 (5)

(4) (5) C = 25

Hour at which Try enters C3 : 7 + 14 [A] + 42…………………= 63
Hour at which rain stopped : 7 + 14 + 42 + 6 ………….= 69
Hour at which Try is finished: 7 + 14 + 42 + 25 [C] + 44 = 132



It is starting to rain on Marks parade:

-open-top empty container C is outside
-at 0 hours, a steady heavy rain starts
-rain enters C at rate of R gallons per hour (gph)
-Mark jumps in C as soon as the rain starts
-Mark takes water out at 1/6 R gph for 9 hours
-at this point, rain slows to 1/3 R gph, and Mark triples his speed
-13 hours later, the rain stops
-at this point, Mark reduces his current speed by 8 gph,
and empties C in 10 hours

OK; if Mark handled a total of 778 gallons in this 32 hour period, what is the size of C in gallons, and the rain speed at start

[size=8]RAIN 2
492 gallons
R=66[/size]

Whilst we agree on the R=66. There appears to be a difference of 99 in the gallons, this is interesting. Time will tell.

I got the same R and C as Try....maybe the mistake is in assuming Markr is volumeless

I see the error of my ways.
[size=8]393[/size]gallons

Thoh wrote, "I got the same R and C as Try"

This has never happened before; I have added you to my Christmas card list.


In any event, we all agree that:


R = Rain original speed
Easy to see that Mark's 778 gallons are this way:
9(R / 6) + 13(R / 2) + 10(R / 2 - 8) = 778
Therefore: R = 66

Total rained gallons = 9(66) + 13(22) = 880

1- C overflowed during start 9 hour period (102 gallons)
2- 13hour period begins with a full C
3- Mark emptied C plus 286 gallons (the 13 hour rainfall), in last 2 sessions, or over 23 hours:
C + 13(R / 3) = 13(R / 2) + 10(R / 2 - 8)
Since R=66, then C = 393 gallons.





For the purposes of this puzzle, we now have a new weight:
Similar to the pound/ounce or Kilo/whatever. We now have the JIZZ/KIXX.

(I took the opportunity to give the lonely j/k/x/z some work!).
Mark orders a bag of Ontario potatoes, asking for j JIZZES and k KIXXES.
The potato man goofs and prepares a bag weighing k JIZZES and j KIXXES.
The total KIXXES Mark ordered are less than 500.

YOU: good; see ya later
ME : the clerk's error caused Mark's order to be reduced by P%, P being an integral in the range 1 to 99.
YOU: gotta go
ME : ah c'mon, how many KIXXES in a JIZZ?
YOU: oh boy (leave, come back with printout); hmmm...83 possibilities; gimme a clue
ME: P is equal to this number here
YOU: that sure helps, but need another clue
ME : Mark's order was over 100 KIXXES ...and now you know...

How many KIXXES in a JIZZ

haven't started on the jizzes, kixxes problem yet

[size=8]1,000,000,000,000,000,000,000,000,000 (octillion)

zero[/size]

[size=8]KIXXES IN A JIZZ
I found 243 different valid combinations (84 different total KIXXES values). I'm not sure why the problem says 83.

A 35% reduction is the only case where knowing that the order was greater than 100 KIXXES reduces the possibilities to one.
8 KIXXES/JIZZ yields a total order of 60 KIXXES.
43 KIXXES/JIZZ yields a total order of 480 KIXXES.

There are 43 KIXXES/JIZZ.[/size]

Now, here's one for you:

Same problem but different last statement.
Replace
"Mark's order was over 100 KIXXES."
with
"Mark's order was a multiple of five KIXXES."

I hope you kept your data from the original problem.

7:28PM

4 x 182 = 728

Thoh, you make a convincing case. I will go check my watch. In the meantime:






Using only pennies, nickels, dimes, quarters, half dollars, $1 bills, $5 bills, and $10 bills, what is the most money that you can have WITHOUT being able to change a $20 bill

[size=8]$20.19[/size]