Five algebra problems

YES or NO Logic Problem
In order to join the math club potential members must guess a secret whole number from 1 to 50, by asking yes-or-no questions only.

An excellent logician, wanted to join, and asked the following four questions:

Is the number greater than 25?
Is the number evenly divisible by 2?
Is the number evenly divisible by 3?
Is the number evenly divisible by 5?

After s/he was told the answers, s/he did some figuring and said, "I still don't have enough information. Is the number a perfect square?" When s/he got the reply "NO," The logician knew what the number was.

Can you determine what the secret number was

Somebody at the Pentagon probably can.

That's a good problem, very tricky. The real clue is that the logician knew the number from the clues. I think the answer is 5. If the logician knows the answer after the fifth question, but not after the fourth, then the answers supplied after four questions must leave a couple of options available, but after the fifth leave only one acceptable answer. If the answers are NO, NO, NO, YES to the first four, only 5 and 25 meet the requirements. Since 25 is subsequently weeded out, 5 must be the answer.

Congratulations Markr, (cryptic), and Engineer for the excellent reasoning. 5 it is.

I didn't have any takers for this in the Riddles forum:

A man set out at noon to walk from Appleminster to Boneyham, and a friend of his started at 1:20 P.M. on the same day to walk from Boneyham to Appleminster. They met on the road at 2:24 P.M., and each man reached his destination at exactly the same time. Can you say at what time they both arrived?

Riddles, are where the mysteries of life are decoded.
I distinctly remember answering this exact question one Sunday pm. :wink:

Try: Sorry about that. I looked, but couldn't find any answers. What was your answer? The only response I could find was something about blue.

markr


4:00 PM

Yep - here's another:

Two trains, A and B, leave Pickleminster for Quickville at the same time as two trains, C and D, leave Quickville for Pickleminster. A passes C 288 miles from Pickleminster and D 315 miles from Pickleminster. B passes C 280 miles from Quickville and D half way between Pickleminster and Quickville. Now, what is the distance from Pickleminster to Quickville? Every train runs uniformly at an ordinary rate.

Not very clean
OK, I got a distance of 504 miles. My approach...

from the first clue:
Va * T1 = 288
Vc * T1 = D-288
so (1) Va/Vc = 288 / (D-288)

similarly from the next two clues..
(2) Va/Vd = 315/(D-315)
(3) Vb/Vc = (D-280)/280 - Note the change in reference to Quickville

And from the last clue, Vb must equal Vd for them to meet half way, so Va/Vd = Va/Vb

So equation (1)/(3) = (2) and you can solve for D.

BINGO!

We're all familiar with this problem:
Smith takes the train daily from his work in the city back to the suburb where he lives and is met at the station by his wife, who drives him home. One day Smith finishes his work earlier than usual and arrives in his suburb on hour earlier than usual. He then walks home from the station, and meets his wife on the way. The wife stops and takes Smith home; this brings Smith home 10 minutes earlier than usual. How long did Smith walk? The time for stopping and picking up is to be disregarded, and it is also assumed that the wife arrives at the station as the same moment as the train.

Here's the twist:
The road rises from the station to the house, so that the car travels more slowly from the station to the house than from the house to the station. We assume that the road rises equally sharply everywhere, so that the car travels equally fast everywhere from the house to the station, and similarly on its way back, but then at a lower speed.

Once again, Smith arrives in his suburb one hour earlier than usual, walks home from the station, and meets his wife. The latter, however, does not notice Smith, and drives on. "What a pity," Smith says to himself. "If my wife had seen me, I would have been home 18 minutes earlier than usual. I'll keep on walking; my wife will catch up with me in 21 minutes, and I'll be home at the usual time." How long had Smith been walking when his wife drove past him on her way to the station? The time the wife needs to ascertain that Smith is not on the train is to be disregarded; so it is assumed that after her arrival at the station his wife immediately returns to join Smith.