Question : Find theÂ probabilityÂ 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 :Â
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 =Â 7/55

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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 :Â
Probability of Red Ball =Â 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) = Â (1/2)[ (44+45) / 99 ] =Â 89/198

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

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Question: A fair coin is tossed 43 times. What is the number of cases whereÂ number ofÂ â€˜Headâ€™ > number of â€˜Tailâ€™?
(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 Â 43C22 + 43C23 + 43C24 .. 43C43
We know that 43C0 + 43C1 + 43C2 ... 43C43 = 2^43
We also know know that our answer has 22 terms which areÂ 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

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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. Â 5.5t = 120
Therefore , Â t = 240/11 = 21 9/11 minutes,
AtÂ 2:21:9/11thÂ min the angle between the hands of a clock would be 60 degrees.

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