Question: If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the share at the end of day 3?
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:
x + y = 5 and 10x – 10y = 10
Solving the above two equations simultaneously, we have
x = 3 and y = 2
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.
This information corresponds to case 2 from the table. The price at the end of day 3 was Rs. 90.
Hence, option (a).