# Solved Example #8

February 11th, 2013 by

Question : Without stoppage, a train travels at an average speed of 75km/h. With stoppages it covers the same distance at an average speed of 60 km/h. How many minutes per hour does the train stop?     Answer :  This seems like a pretty straightforward question but can confuse because of the way it is asked. The best way to understand it, in my opinion, is to assume a distance. Let us say that the distance that the train covers is 300 km. (I am taking 300 km because it is divisible by both 75 & 60)   Case 1: Time taken without stoppages = 300 / 75 = 4

# Solved Example #7

February 9th, 2013 by

Question : Find the number of ternary sequences of length 4 where 0 is not followed by 1 and 1 is not followed by 2. (Ternary sequence of length n is a sequence having n terms and each term is either 0,1 or 2) a. 47 b.37 c. 72 d.54 e.44   Answer :   Total ternary sequences possible = 3^4 = 81 {There are 4 positions and 3 choices for each position}   Invalid sequences: Sequences which contain 01. The location for 01 can be selected in 3C2 or 3 ways. {01xx, x01x, xx01} The other two digits can be selected in 3*3 or 9 ways. {Filling up

# Solved Example #6

February 9th, 2013 by

Question : The sum of the reciprocals of the positive divisors of 120 ? a.2 b.3 c.4 d.6     Answer :  Sum of its factors = (1 + 2 + 4 + 8)(1 + 3)(1 + 5) = 360 Sum of the reciprocals of its factors = 360 / 120 = 3

# Solved Example #5

February 9th, 2013 by

Question : Find the probability that a rectangle selected at random from a 10*10 square is a square? a.17/55 b.7/55 c.9/55 d.16/55   Answer :  Total number of rectangles = 11C2 * 11C2 = 55*55 Total number of squares = 1^2 + 2^2 ....... 10^2 = 10*11*21/6 = 5*11*7 = 55*7 Required probability = 55*7 / 55*55 = 7/55

# Solved Example #4

February 9th, 2013 by

Question : There are 4 red & 5 green balls in bag A and 5 red & 6 green balls in bag B. If a bag is selected at random and a ball is selected from that, what is the probability that it is red?   Answer :  Probability of Red Ball = Bag A is selected & Red Ball is selected + Bag B is selected & Red Ball is selected ie. (1/2)*(4/9) + (1/2)*(5/11) =  (1/2)[ (44+45) / 99 ] = 89/198

# Solved Example #3

February 9th, 2013 by

Question : Find the last two digits of (24)^37298    Answer Last two digits of 24^Odd = 24 Last two digits of 24^Even = 76

# Solved Example #2

February 9th, 2013 by

Question: A fair coin is tossed 43 times. What is the number of cases where number of ‘Head’ > number of ‘Tail’? (A) 2^43 (B) (2^43)-43 (C) 2^42 (D) None of the above.   Answer: No. of heads > No. of tails => No. of heads can be 22, 23, 24... 43 Therefore  43C22 + 43C23 + 43C24 .. 43C43 We know that 43C0 + 43C1 + 43C2 ... 43C43 = 2^43 We also know know that our answer has 22 terms which are 43C22 + 43C23 + 43C24 .. 43C43   These 22 terms are equal to = 43C0 + 43C1 + 43C2 .. 43C21 Both of them are equal and add up to 2^43

# Solved Example #1

February 9th, 2013 by

Question: At what time between 2 and 3 the angle between the hands of clock is 60 degrees? Answer: At 2 o'clock the angle is 60 degree. After t minutes past 2: Minute hand from the vertical = 6t Hour hand from the vertical = 60 + 0.5t Required angle is : 6t - (60 + 0.5t) = 60 ie.  5.5t = 120 Therefore ,  t = 240/11 = 21 9/11 minutes, At 2:21:9/11th min the angle between the hands of a clock would be 60 degrees.