(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 chess-board 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 chess-board as a 4 co-centric 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 inner-level, then it can attack 9 positions. Similarly if the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions. And if the bishop is in one of the 4 squares at next inner-level (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
Ordered
(N)

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
X products from
8 products (W)

8C1

8C2

8C3

8C4

8C5

8C6

8C7

8C8

Total combinations
(T*W)

64

784

3136

4900

3136

784

64

1


Thus, total possible combinations are
= 64 + 784 + 3136 + 4900 + 3136 + 784 + 64 + 1
= 12869

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