Brain Teaser

patcooke · Jul 17, 2011

Logical analytical approach for this one:
if 100 horses eat 100 bales in 100 days then:

dividing the number of horses by 10: ten horses will eat ten bales of day in one hundred days

then:

dividing the number of days by ten: ten horses will eat one bale of hay in ten days

Elmer · Jul 17, 2011

Is the correct answer. As one bloke who insisted it would be ten bales, once we convinced him it would be only one reposted with "They'll be f****n skinny horses then!"

Over to you!

patcooke · Jul 17, 2011

Right - the circumference of the Earth around the equator is approximately 25,000 miles. Assuming it were possible, it would take 25,000 miles of cable to lay a cable completely round the equator flat to the Earth. If it was then required to raise the cable to run completely round at the top of 10 foot high poles all the way round the equator approximately how much more cable would be required?

beeaitch · Jul 17, 2011

Ok, I'll give this a shot.

Circumference = pi * diameter so if the circumference is 25,000 miles then -
25000 = pi * d so
25000/pi = d therefore
d = 7,958 miles

Then if you increase the diameter by 20 feet... that basically does nothing... damn... I guess 26,000ish miles?

I'm probably going about this all wrong.

Elmer · Jul 17, 2011

My guess is an extra 62' 8"

patcooke · Jul 18, 2011

Yep - it's actually 62.8' rather than 62'8" but we're not gonna quibble over that are we - for those who didn't (or who still don't) get it:

The circumference is given by the formula pi x D where pi = 3.14 to 2 dp and D = diameter

Poles ten feet high increase the diameter by 20 feet so the two circumferences differ by 3.14 x 20 = 62.8 feet.

Must admit I was surprised by how small the answer was the first time I saw this puzzle and thought I must have made a silly mistake. Well done Elmer - looking forwards to your next one . . .

Elmer · Jul 18, 2011

I'm the same as you Pat. Thought I'd hit some keys wrong as 628 miles "felt" more like it.

Pat I'm leaving it to you to set the next puzzle. I just can't think of one!

patcooke · Jul 18, 2011

Straight forward calculating problem:

A car and a coach set off from the same point, at the same time, to travel the same 140 mile journey. The car travels at 48 mph and the coach travels at 40 mph. If the car arrives at noon, what time will the coach arrive?

alfred01 · Jul 21, 2011

If the coach is going to run at 40 mph for 140 miles, he'd probably be dead the first day! :ahaha:

patcooke · Jul 22, 2011

alfred01 said:
If the coach is going to run at 40 mph for 140 miles, he'd probably be dead the first day!

your coach has gender?

alfred01 · Jul 23, 2011

patcooke said:
your coach has gender?

The last coach I had was in high school (35 years ago) and yes he had gender. Don't all coaches? lol :wave:

Just joking around.

patcooke · Jul 24, 2011

alfred01 said:
The last coach I had was in high school (35 years ago) and yes he had gender. Don't all coaches? lol Just joking around.

I know - so was I! Don't seem to be getting far with the most recent puzzle though . . .

alfred01 · Jul 24, 2011

OK. I'm going with 12:36 pm.

Elmer · Jul 24, 2011

My coach went downhill and used gravitational foce to assist it. I'll say 12.25.

patcooke · Jul 25, 2011

Looks like 12.36 is about as close as we might get - not sure why you didn't get it spot on but the explanation is:

To arrive at noon the car must have travelled 140/48 x 60 minutes = 175 minutes
So the start time was 2hrs 55mins before noon or 9.05am

The bus would take 140/40 x 60 minutes to travel the journey = 210 minutes
So the bus arrives 3hrs 30mins after setting off
9.05am + 3hrs 30mins = 12.35pm

alfred01 · Jul 25, 2011

It was my calculators fault. OK. here we go.

I know a word of letters three, add two and fewer there will be. What word am I?

patcooke · Jul 26, 2011

Must be "few" - add "er" and it becomes "fewer"!

alfred01 · Jul 26, 2011

You got it Pat. Was it too easy or are you that smart?

patcooke · Jul 26, 2011

alfred01 said:
You got it Pat. Was it too easy or are you that smart?

I reckon it's like a lot of questions of this type - you either spot it straight away or sit scratching yer head for an hour or two.

patcooke · Jul 26, 2011

How can you throw a ball as hard as you can and have it come back to you. It doesn't bounce off anything, there is nothing attached to it, and no one else catches or throws it back to you.

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