(Puzzles) Challenging Mathematical Puzzles (22)
Puzzles : Challenging Mathematical Puzzles
63. To move a Safe, two cylindrical steel bars 7 inches in diameter are used as rollers.
How far will the safe have moved forward when the rollers have made one revolution?
Answer : 
The safe must have moved 22 inches forward.
If the rollers make one revolution, the safe will move the distance equal to the circumference of the roller. Hence, the distance covered by the safe is
= PI * Diameter (or 2 * PI * Radius)
= PI * 7
= 3.14159265 * 7
= 21.99115
= 22 inches approx.
64. If a rook and a bishop of a standard chess set are randomly placed on a chessboard, what is the probability that one is attacking the other?
Note that both are different colored pieces.
Answer : 
The probability of either the Rook or the Bishop attacking the other is 0.3611
A Rook and a Bishop on a standard chessboard can be arranged in 64P2 = 64*63 = 4032 ways
Now, there are 2 cases  Rook attacking Bishop and Bishop attacking Rook. Note that the Rook and the Bishop never attack each other simultaneously. Let's consider both the cases one by one.
Case I  Rook attacking Bishop
The Rook can be placed in any of the given 64 positions and it always attacks 14 positions. Hence, total possible ways of the Rook attacking the Bishop = 64*14 = 896 ways
Case II  Bishop attacking Rook
View the chessboard as a 4 cocentric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.
If the bishop is in one of the outer 28 squares, then it can attack 7 positions. If the bishop is in one of the 20 squares at next innerlevel, then it can attack 9 positions. Similarly if the bishop is in one of the 12 squares at next innerlevel, then it can attack 11 positions. And if the bishop is in one of the 4 squares at next innerlevel (the innermost level), then it can attack 13 positions.
Hence, total possible ways of the Bishop attacking the Rook
= 28*7 + 20*9 + 12*11 + 4*13
= 560 ways
Thus, the required probability is
= (896 + 560) / 4032
= 13/36
= 0.3611
65. In England McDonald's has just launched a new
advertising campaign. The poster shows 8 McDonald's products
and underneath claims there are 40312 combinations of the above items.
Given that the maximum number of items allowed is 8, and you are allowed to have
less than 8 items, and that the order of purchase does not matter (i.e. buying a
burger and fries is the same as buying fries and a burger)
How many possible combinations are there? Are McDonald's correct in claiming
there are 40312 combinations?
Answer : 
Total possible combinations are 12869.
It is given that you can order maximum of 8 items and you are allowed to have
less than 8 items. Also, the order of purchase does not matter. Let's create a
table for ordering total N items using X products.
Items 
Products Used (X) 

1 
2 
3 
4 
5 
6 
7 
8 

1 
1 
 
 
 
 
 
 
 
2 
1 
1 
 
 
 
 
 
 
3 
1 
2 
1 
 
 
 
 
 
4 
1 
3 
3 
1 
 
 
 
 
5 
1 
4 
6 
4 
1 
 
 
 
6 
1 
5 
10 
10 
5 
1 
 
 
7 
1 
6 
15 
20 
15 
6 
1 
 
8 
1 
7 
21 
35 
35 
21 
7 
1 
Total (T) 
8 
28 
56 
70 
56 
28 
8 
1 
Ways
to choose 
8C1 
8C2 
8C3 
8C4 
8C5 
8C6 
8C7 
8C8 
Total
combinations 
64 
784 
3136 
4900 
3136 
784 
64 
1 
Thus, total possible combinations are
= 64 + 784 + 3136 + 4900 + 3136 + 784 + 64 + 1
= 12869