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Below are three problems dealing with speed and time calculations. Question 1 Two buses leaving from two stations 20 km away from each other travel with constant speed of 45km/hr towards each other.Find the time taken to cross each other if the length of each bus is 100m. a)10sec b)4sec c)15sec d)20sec Answer : b)4sec Solution : The time taken to cross each other = (a + b) / (u + v) sec. Here, a = b = length of the two buses = 100m = 1/10 km. and u = v = speed of the two buses = 45km/hr. Time taken to cross each other = (a + b) / (u + v) sec = (1/10 + 1/10) / (45 + 45) hours 1 / (10 x 90) = 1/900 hours. since 1 hour = 3600 sec, 1/900 hour = 3600/900 = 4 sec Hence the answer is 4 sec. Question 2 A bus starts from A at 7 a.m and reaches the destination B at 7.30 a.m while a cyclist starts from B at 7 a.m and reaches A at 8.30 a.m. At what time the bus and the cyclist will cross each other? a)7.34 a.m b)7.49 a.m c)7.23 a.m d)8.01 a.m Answer : c)7.23a.m Soluti

Below are three model problems dealing with the formulae (x+y) 2 = x 2 + y 2 + 2xy and (x-y) 2 = x 2 + y 2 - 2xy Question 1 Find X when X - Y = 3 and X 2 + Y 2 = 89 where X and Y are integers. a)10 b)-5 c)-10 d)-3 Answer : b)-5. Solution : We know that (x - y) 2 = x 2 + y 2 - 2xy Sub. the given values, 3 2 = 89 - 2XY 9 - 89 = -2XY 80 = 2XY XY = 40 Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows: (1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8) X - Y is a positive integer(3), so X will be greater than Y. By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8. i.e., X is either 8 or -5. From the given options we can conclude that X = -5. Question 2 If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is: a)2 b)1 c)12 d)4 Answer : a)2 Solution : Let A and B be two integers. Then A + B = 24 and AB = 143. We know t