Question: If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once during the five days, what was the price of MCS 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:
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.
∴ The price of the share at the end of day 3 = Rs. 110
Hence, option (c).