Answer & Solution
Answer: Option C
Solution:
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.
