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by gmatmachoman » Thu May 27, 2010 9:09 pm
A man buys shares at a discount of Rs.x. Later he sells all but 10 of the shares he purchased at a
premium of Rs.x. If his investment was Rs.4500 and proceeds from the sale were Rs.6250, how
many shares did he buy originally ? [Assume face value of shares as Rs.100.]

1] 50 2] 40 3] 60 4] 90
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by albatross86 » Thu May 27, 2010 10:11 pm
Man buys N the shares at Rs. (100 - x) per share and invests a total of Rs. 4,500.

This gives us the equation : 100 - x = 4500/N
=> x = 100*(N - 45) / N

He sells all but 10, i.e. (N - 10) shares at a price of (100 + x) per share.

The net proceeds would be: (N - 10) * (100 + x) = 6250

Substituting for x: [N - 10] * (100 + 100*(N-45)/N) = 6250

=> (N-10) * (200N - 4500) = 6250N
=> 200N^2 - 6500N + 45000 = 6250N
=> 200N^2 - 12750N + 45000 = 0
=> 4N^2 - 255N + 900 = 0
=> N = 60 or N = 3.75

Since N must be an integer, the answer is 60.
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by kstv » Thu May 27, 2010 11:44 pm
No of shares bought = 10y

Each share is bought at 100 - x = 4500/10y = 450/y ---(1)

Selling price of each share is 100+x = 6250/10y-10 = 625/(y-1) --------(2)

Adding 1 and 2

200 = 450/y + 625/(y-1)
Answer option are 50,40,60 and 90
so possible values of y are 5, 4,6,9 and correspondingly (y-1) = 4,3,5,9

625 cannot be divided by 4 even no , 3 ( sum of the digits - 6+2+5 = 13 is not a multiple of 3) or 9 ( sum of the digits are not a multiple of 9)
IMO 60
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