1) What is the average speed of the car ?

a) The car covered 25% of distance at 45 Km per Hour.

b) The car covered the remaining 75% of distance at 55 Km /

Hour.

Let the total distance be 100X

Time taken for first 25x = 25X/45

Time taken to cover remaining 75X = 75X/55

Total time taken = 25X/45 + 75X/55 =

LCM of 45 and 55 = 495

= 275X+675X/ 495 = 950X/495

So average speed = 100X/ ( 950X/495) = 49500/950 Km per

hour.

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2)What is the remainder when factorial 11 divided by 17 ?

11*10*9*8*7*6*5*4*3*2 Mod 17

11*10 Mod 17 = 8

9*8 Mod 17= 4

7*6 Mod 17= 8

5*4*3*2 Mod 17 = 1

Now the remaining balances are 8*4*8*1

64 mod 17 = 13

13* 4 Mod 17= 1

Hence the Answer is 1——————————————————————————

3)What is the remainder when difference of the 5th power

and the 4th power of 59 is divided by 11? ( 59^5-59^4)

59^5-59^4 mod 11

59^4( 59-1)

= 58*59^4

58 mod 11= 3

59 mod 11= 4

so we are left with 3*4^4 which is equal to 3*(4^2)^2

4^2 mod 11= 5

so we are left with 3*5*5 =75

75 mod 11= 9———————————————————–

4)At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?

Answer

4:21:49.5

Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.

For every minute, minute hand travels 6 degrees.

Hence, for X minutes it will travel 6 * X degrees.For every minute, hour hand

travels 1/2 degrees.

Hence, for X minutes it will travel X/2 degrees.At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,

6 * X = 120 + X/2

12 * X = 240 + X

11 * X = 240

X = 21.8182

X = 21 minutes 49.5 secondsHence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.

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5)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.

SubmAnswerThe 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 waysCase 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 waysThus, the required probability is

= (896 + 560) / 4032

= 13/36

= 0.3611

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