Fun math problem that stumps everyone

ERD50 said:
edit/add: Another weak solution: You didn't say the pipe was fixed in the ground, just "sticking out". Just pull it out, turn it upside down, and the ball falls out!


-ERD50
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Or jump up and push up on the top of the box to dislodge the pipe from the hole in the ground, then push high on the lower side of the box to tip the box over. Use the match to light your victory cigar.

You also might be able to suck on the pipe and get enough suction to get the ball out.
 
NW-Bound said:
A man did not have a watch, only a good clock. However, he let the clock's battery run down, and now he did not know the time to set it after replacing the battery. He did not have a phone to call his friend a few blocks away to ask for the time.

He walked over to visit his friend and spent an evening. He then walked back home and set his clock accurately.

How did our man do this, as he would need to know how long the walk took?
Click to expand...

Are we assuming his walk took the same time each way?

He set the clock (to an arbitrary time) before leaving. Upon arrival at his friend's, he noted the time difference between his clock (at departure) and his friend's clock at arrival. (Diff1)

The next day, he also notes the difference between the departure time (on friend's clock) and arrival time at home. (Diff2)

The absolute difference (or offset) between his clock and his friend's clock = 0.5*(Diff1+Diff2).
 
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Since the pipe is 1 foot diameter and 1 foot long and has a 1 foot less 1 mm diameter ball in it, the ball is right at the top of the pipe. Just grab it with your hand and pull it out.
 
I commend Out-to-Lunch for solving the last two problems, but the solutions include unstated assumptions. If the 12" tube is buried in ground such as sand that will absorb any liquid, OtL's method fails. In the clock quiz OtL does make the solution dependent on the walks taking the same time, but any number of things, like a steep hill, could make that untrue.
 
Sure there are ASSUMPTIONS in this puzzle....

Robbie it's stated its a ping pong ball...

Also thinking outside the box would have not thought of a sand filled basement ....

In the bottom of the pipe is a ping pong ball with a diameter that is one millimeter smaller than the inner diameter of the pipe.
 
f35phixer said:
Originally Posted by Out-to-Lunch View Post
In addition to the match, string, etc., do you also have a full bladder?
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BINGO Lunch !:LOL:
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OK, I *did* laugh at that one! With a good dose of groaning as well.

I was wondering about the combination of inches and millimeters though. And the unspecified 8x8x8 units. The wording was weird, a ping pong ball with a diameter 1 mm less than the pipe? Wouldn't a ping-pong ball be a standard diameter, so the pipe would be 1 mm larger? No difference, just the more common way to use the standard as the reference.

-ERD50
 
Out-to-Lunch said:
Are we assuming his walk took the same time each way?

He set the clock (to an arbitrary time) before leaving. Upon arrival at his friend's, he noted the time difference between his clock (at departure) and his friend's clock at arrival. (Diff1)

The next day, he also notes the difference between the departure time (on friend's clock) and arrival time at home. (Diff2)

The absolute difference (or offset) between his clock and his friend's clock = 0.5*(Diff1+Diff2).
Click to expand...


You've got it, although I have an explanation that is easier to grasp.

Yes, we have to assume that his stride was the same in both directions, so that the travel times were the same.

Our man noted the time on his clock before departure (T1). Upon arrival at his friend's, he noted the time on the good clock (T2). When he left his friend's home, he noted the time again (T3). And when he got home, he looked at his clock (T4).

He computed the elapsed time on his clock. He knew how long he visited his friend. The difference between the two elapsed times was the time he walked to/fro. Dividing the above by 2 and he got the time he walked one way. He added that time to his friend's clock time as he started walking home, and that's the correct time he arrived back at home.

Travel time = 1/2 [ (T4-T1) - (T3-T2) ]

Time upon home arrival = T3 + Travel time


A bit of algebraic manipulation will show that the above is the same as the calculation described by Out-to-Lunch.
 
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Here is an old one:

A mother is 21 years older than her son. In exactly 6 years, the son will be one-fifth his mother’s age. The question is: what is the father doing right now?
 
After solving for the son's age, I have to conclude that the parents are busy conceiving him. :)
 
ERD50 said:
In what way is that misusing the technology? We have speed limits, this is a technological solution to enforcement. Sounds good to me. Catch those speeders, make them pay! All without a cop having to risk his/her life on a traffic stop.

Seems all good to me.

-ERD50
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I will agree to disagree about this one if you will. Not worth a food fight though YMMV.
 
You would need to move infinitely fast. Trick question.


Sk08 said:
I stumbled upon this simple looking math problem last week and showed it to several people and have not found anyone that got it right, including my daughters 6th grade math teacher. It took Albert Einstein an hour to figure it out.

If you travel at a speed of 30mph in the 1st mile, how fast do you have to go in the second mile so you have an average speed of 60mph over the 2 mile stretch?
Click to expand...
 
All actual math problems have answers. The answer to this one is "infinite speed". 30mph for 2 hours + infinite = 60mph in 2 hours... Math isn't bounded by observable physics. That's why we have infinite and imaginary (sq root of a negative) numbers to show us on paper what we can't do on earth.
 
I believe the young lady named Bright could do it. As in:

"There was a young lady named Bright,
Who could travel much faster than light,
She set out one day, in a relative way,
And returned on the previous night!"
 
Averaging

Here is my problem with supposedly unanswerable math problems. The speeds listed are not stated to be absolute and as such can be considered to be estimates. So that being said if you travel at 30 mph for two minutes then travel at the speed of light for the next mile, the time involved is 200000th of a second for the second mile. So in essence you have traveled 2 miles in 120.000005 seconds. This equals 60.0000025 miles per hour.
 
Wotfan2 said:
Here is my problem with supposedly unanswerable math problems. The speeds listed are not stated to be absolute and as such can be considered to be estimates. So that being said if you travel at 30 mph for two minutes then travel at the speed of light for the next mile, the time involved is 200000th of a second for the second mile. So in essence you have traveled 2 miles in 120.000005 seconds. This equals 60.0000025 miles per hour.
Click to expand...

Nope. I refer you to my earlier post https://www.early-retirement.org/fo...hat-stumps-everyone-113785-3.html#post2764616

The speed is actually (120/120.000005368) x 60mph = 59.999997316 mph. (using your approximation, it would be 59.9999975 mph)

In sum, if it takes ANY amount of time longer than 2 minutes to travel two miles, the speed is less than 60 mph.
 
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Another math problem

Years ago, when things were less expensive: 3 salesmen go to a hotel and decide to stay in the same room to save money. The price was $30, so each man paid $10. After they were in the room the desk clerk remembered they were having a special rate reduction this week and the price should have only been $25. So, he gives the bell hop 5 one dollar bills to take back to the men. On the way up the bell hop puts $2 in his pocket and gives the men $1 each back. So now each man has paid $9 each, that is $27, the bell hop stole $2. what happened to the other dollar since they paid $30 total to begin with?
 
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