**Ans : B**

Sol :

Let a, b, c, d, e, and f be the numbers in the set, and let f be the smallest number in the set.

When the smallest number (f) in the set is replaced by 0, the numbers in the set are a, b, c, d, e, and 0.

Quantity A equals the average of these six numbers, which equals

(a + b + c + d + e + 0) / 6 = (a + b + c + d + e) / 6

Instead, if the smallest number in the set is removed, the remaining numbers in the set would be a, b, c, d, and e. Now there are only 5 numbers in the set. Hence, Quantity B, which equals the average of the remaining numbers (five numbers) in the set, equals

(a + b + c + d + e) / 5

Since all the numbers in the set are positive (given), the sum of the five numbers a + b + c + d + e is also positive. Note that dividing a positive number by 5 yields a greater result than dividing it by 6.

Hence,

(a + b + c + d + e) / 5 is greater than (a + b + c + d + e) / 6.

Thus, Quantity B is greater than Quantity A, and the answer is (B).