Question: If Michael ended up with Rs. 100 less cash than Chetan at the end of day 5, what was the difference in the number of shares possessed by Michael and Chetan (at the end of day 5)?
Explanation:
Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:
Solving the above two equations simultaneously, we have
The price of the share goes up on 3 days and falls on 2 days.
The three days on which the price rises can be selected in 5 C3 = 10 ways
The following are the 10 cases:
Consider Case 5:
Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.
Change in the number of shares he has = –30 + 20 = –10
Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300
Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.
Change in the number of shares he has = –20
Change in his cash = 10 × (120 + 120) = Rs. 2400
The other cases are evaluated in a similar manner and the data is tabulated as shown above.
Chetan sold on three consecutive days: Cases 1, 2 and 3.
Michael sold only once: Case 3.
Michael ends up with Rs. 100 less cash than Chetan in cases 3, 6 and 8. In each of these cases, both of them hold the same number of shares at the end of day 5.
Hence, option (e).