February 11th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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

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February 9th, 2013 by Ravi Handa

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.

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January 25th, 2013 by Ravisankar Vemuri

Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?
(A) ≥ 105 (B) ≤ 123 (C) ≥ 125 (D) ≥ 100 and ≤ 125 (E) ≥ 105 and ≤ 123
Solution:
The key point:
“Each of Sara’s 5 friends has twenty five friends” => Sara is one among these 25 friends.

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Posted in Quant Funda, Solved Examples, XAT