Tickle Your Grey Cells
Puzzle corner is a dedicated section for honing analytical, mathematical and reasoning skills. In this section every week you will find five Puzzles. Theses mind twisters have been carefully selected after a thorough research and provide right inputs for tickling your grey cells. Solving puzzles is considered very good exercise so as to remain mentally active. Puzzle solving enhances memory and processing speed.
Puzzle is one of the most popular category in competitive examinations. All the major Indian MBA entrance exams like CAT, MAT, XAT and SNAP have direct questions or questions based on similar logics in written test. Besides MBA exams, competitive exams like GRE, GMAT, Bank PO, SSC & other government exams include puzzle questions. Most of the companies while recruiting through campus includes Puzzles in the written tests or at interview stage. Hence those who are preparing for campus placement tests this section is a must.

John goes to a furniture auction. All purchases must be paid for in cash. He goes to the ATM and draws out Rs.40,000 (Currency notes are of the denomination of Re.1, Rs.2, Rs.5, Rs.10, Rs.20, Rs.50 and Rs.100). Since John does not want to be seen carrying that much money, he places it in 16 envelopes numbered 1 through 16. Each envelope contains the least number of currency notes possible (i.e. no two tens in place of a twenty).
At the auction he makes a successful bid of Rs.8456 for a table. He hands the auctioneer envelopes number(s) 4, 9, and 14. After opening the envelopes the auctioneer finds exactly the right amount. How many Rs.2 notes did the auctioneer find in the envelopes?


Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of currency notes possible" is only to misguide you. This is always possible for any amount!!! One more thing to notice here is that John must have placed money in envelopes in such a way that if he bids for any amount less than Rs.40000, he should be able to pick them in terms of envelopes.
First envelope contains, 2^{0 }= Re.1
Second envelope contains, 2^{1} = Rs.2
Third envelope contains, 2^{2} = Rs.4
Fourth envelope contains, 2^{3} = Rs.8 and so on...
Hence the amount in envelopes is Re.1, Rs.2, Rs.4, Rs.8, Rs.16, Rs.32, Rs.64, Rs.128, Rs.256, Rs.512, Rs.1024, Rs.2048, Rs.4096, Rs.8192, Rs. 16384, Rs. 7233.
Last envelope (No. 16) contains only Rs.7233 as total amount is only Rs.40000.
Now as he bids for Rs.8456 and gives envelope number 4, 9 and 14 which contains Rs.8, Rs.256 and Rs.8192 respectively.
Envelope No 4 contains one Rs 5, one Rs 2, one Re 1.
Envelope No 9 contains two Rs.100 note, one Rs.50 note, one Rs.5 note and one Re.1 note.
Envelope No 14 contains eightyone Rs.100 note, one Rs.50 note, four Rs.10 note and one Rs.2 note.
Hence the auctioneer will find only two Rs.2 notes in the envelopes.

Everyday in his business Prakash had to weigh amounts from 1 kg to 364 kgs, to the nearest kg. What is the minimum number of different weights required and how heavy should they be?


The minimum number is 6 and they should weigh 1, 3, 9, 27, 81 and 243 kgs.

If Kumbkaran eats 60 pounds of rice every day EXCEPT every 7th day on which he only eats 50 pounds of rice. If he continues this, how many pounds of rice will he eat in 150 days?


It is given that on every 7th day he eats 50 pounds of rice i.e. on day number 7, 14, 21, 28, .... 140, 147 kumbkaran eats 50 pounds of rice.
Total number of 7th days = 150/7 = 21 ( he eats 50 pounds)
Hence, the normal days are = 150  21 = 129 ( he eats 50 pounds)
Thus, in 150 days, he will eat
= (129) * (60) + (21) * (50)
= 7740 + 1050
= 8790 pounds

There are 3 points labeled X, Y and Z. Then there are other 3 points labeled 5, 6 and 7. The aim of the puzzle is to connect point X with points 5, 6 and 7, point Y with points 5, 6 and 7 and point Z with points 5, 6 and 7. While connecting the points you have to follow one rule  the lines cannot cross over each other.
X Y Z
5 6 7
You can arrange the points in order as long as the lines DO NOT cross over each other.


There is no solution to it, if you consider 2 dimensions. It is impossible to join each of points X, Y and Z with points 5, 6 and 7 without lines crossing each other.
There is solution, if you consider 3 dimensions. Consider a circular base and a line perpendicular to it passing from the center. Now take any 3 points along the perimeter of the circular base as points 5, 6 and 7. Similarly take any 3 points along the perpendicular line as points X, Y and Z.
Now it is quite simple to join each of points X, Y and Z with points 5, 6 and 7 without any of the lines crossing each other.
The other possible 3D structure is Pyramid. Take points 5, 6 and 7 as vertices of the triangular base and points X, Y and Z along the height of the Pyramid which is perpendicular to the triangular base and passing through the apex.

An eagle is flying between two trains, each travelling towards each other on the same track at 80 km/h. The eagle reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time. The eagle is flying at 50 km/h. If the eagle flies 150 km before the trains meet, how far apart were the trains initially?


The eagle is flying at the speed of 50 km/h and covers 150 km. Hence, the eagle flies for 3 hours after trains started. It's obvious that trains met 3 hours after they started travelling towards each other. Also, trains were travelling at the speed of 80 km/h. So, each train traveled 240 km before they met. Hence, the trains were 480 km apart initially.


