Achieving Average

A person walks at the rate of 3 mph to his office. How fast does he have to run back home in order to average 6 mph for the round trip? that is for both trips combined, to and fro.

Answer:

He would have to be infinitely fast. This is true no matter what speed he was travelling. Anyone travelling x mph over a road nx long, would take n hours one direction. To average 2x, over 2nx (the total road length), 2nx/n+? = 2x, ? must equal zero, so it took him zero hours, or he went infinitely fast.

Non-empty Delivery

A person having 2 branches of Bananas, each branch having 50 numbers, has to cross a river in 50 boats (because after certain time the first boat man transport the person and the bunch to the second boat, the second one doing the same to the third and so on and so forth until the 50 th boat who reaches the other bank finally) but with the condition that each boatman has to be given a banana per branch he carries. Obeying the conditions stipulated strictly, the poor man cannot make it anything to his home since all 100 bananas would have been exhausted by delivering to the unscrupulous boatmen however he strikes upon a strange idea by which he could accomplish carrying something, if not all.
How did he do it?

Answer:

For the first 25 boats he gives 50 bananas from the same branch so that from the 26th boat onwards, he might have only one branch and therefore is expected to give only one banana, thereby leaving 25 bananas to take it home gleefully.

note: the above problem was communicated to me by a small south indian lad aged 12 by name Ashwin and by another most talented child Bharath, residing in Madras.

Class Strength

In a certain algebra lecture class, Chris and Pat count the students and compare notes. "Hmm, 12/17 of my classmates in here are women," notes Pat. "Funny," recounts Chris, "5/7 of my classmates are women. "They were both right. How many students, men and women, were in the class, and what were the genders of Chris and Pat? Men and women are mutually exclusive sets for the purposes of this problem. Explain your reasoning.

Solution:

Let n be the total number of students in the class, w = the number of women. Now 12 / 17 is less than 5 / 7 , so Pat has to be a woman, as Chris sees more women classmates from his perspective. Then Pat sees that (w-1) / (n-1) = 12 / 17 , and Chris sees w / (n-1) = 5 / 7. The denominator (n-1) can be the product 7*17 = 119, because 12/17 = 84/119 and 5/7 = 85/119 , so since 84 and 85 are 1 apart, we're there. Pat has 84 women classmates, and Chris has 85. Don't forget the 119 doesn't count the observer, so there are 120 students in the class.