Question

There are \( n \) numbers. When 50 is subtracted from each of these numbers, the sum of the numbers so obtained is -10. When 46 is subtracted from each of the original \( n \) numbers, then the sum of the numbers so obtained is 70. What is the mean of the original \( n \) numbers?

• A. 45.4
• B. 45
• C. 49
• D. 49.5

Hint:

When given a relationship involving sums after adding or subtracting a constant from each number, set up equations based on the sums and solve for the mean.

Answer 1

The Correct Option is D

Let the sum of the original \( n \) numbers be \( S \).
When 50 is subtracted from each number, the new sum is \( S - 50n = -10 \), so: \[ S = 50n - 10 \quad \cdots (1) \] When 46 is subtracted from each number, the new sum is \( S - 46n = 70 \), so: \[ S = 46n + 70 \quad \cdots (2) \] Equating equations (1) and (2): \[ 50n - 10 = 46n + 70 \] Solving for \( n \): \[ 4n = 80 \quad \Rightarrow \quad n = 20 \] Substitute \( n = 20 \) into equation (1): \[ S = 50 \times 20 - 10 = 1000 - 10 = 990 \] Thus, the mean of the numbers is: \[ \frac{S}{n} = \frac{990}{20} = 49.5 \]