Dittee has 10 purses and each of which having an average of 25 lipsticks in each purse. Further, it is known that each purse has a minimum of 8 lipsticks and none of the purse has identical number of lipsticks.

Determine that what can be the maximum number of lipsticks that Dittee shall have in any of her purse?

142

The correct answer is Option (1) → 142

Given:

• Number of purses = 10
• Average lipsticks per purse = 25 ⇒ Total lipsticks = 10 × 25 = 250
• Each purse has at least 8 lipsticks
• No two purses have the same number of lipsticks

Objective: Maximize the number of lipsticks in one purse.

Assume the lipstick counts in purses are distinct integers: $x_1, x_2, ..., x_{10}$, with $x_i \geq 8$ and all $x_i$ different.

To maximize the largest value, minimize the sum of the other 9 values using the smallest distinct integers ≥ 8:

Smallest 9 values ≥ 8 and distinct: 8, 9, 10, 11, 12, 13, 14, 15, 16

Sum = $8 + 9 + ... + 16 = \frac{9}{2}(8 + 16) = \frac{9 \times 24}{2} = 108$

Thus, the maximum possible value of the 10th purse = $250 - 108 = 142$