Discussion

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 5C3 = 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).

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