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Answer: Option 3

Let X be the rupees he initially had. He gave for the cancer fund one rupee more than half of what he had. i.e.,[1 +(X/2)]. Remaining money = X-(1+X/2) = [(X/2) - 1. he gave for poor people's, rupee 2 more than half what he remain with, = [2+{1/2*(X/2-1)}] = [2+{(X-2)/4}] = (6+X)/4 Now, remaining money = ((X/2)-1) - ((6+X)/4) = (X-10)/4. Again he gave 3 rupees more than half of what he had for orphanage, [3+(1/2*((X-10)/4))] = 3+[(X-10)/8] = (14+X)/8 now left money,[{(X-10)/4]-[(14+X)/8]} = [(2X-X-20-14)/8] = (X-34)/8 As given, finally he had one rupee remaining so (X-34)/8 = 1 So,X-34 = 8 X = 8+34 = 42 Hence, Rohit had Rs. 42 initially in his pocket.

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