Question

The average weight of three men A, B and C is 84 kg. D joins them and the average weight of the four becomes 80 kg. If E whose weight is 3 kg more than that of D replaces A, the average weight of B, C, D and E becomes 79 kg. The weight of A is.

• A. 65 kg
• B. 70 kg
• C. 75 kg
• D. 80 kg

Hint:

When dealing with averages and sums, break the problem into smaller steps by calculating the total weight of the individuals involved.

Answer 1

The Correct Option is C

The average weight of A, B, and C is 84 kg. So, the total weight of A, B, and C is: \[ \frac{\text{Weight of A + B + C}}{3} = 84 \quad \Rightarrow \quad \text{Weight of A + B + C} = 84 \times 3 = 252 \, \text{kg}. \] After D joins, the average weight of A, B, C, and D is 80 kg, so the total weight of these four is: \[ \frac{\text{Weight of A + B + C + D}}{4} = 80 \quad \Rightarrow \quad \text{Weight of A + B + C + D} = 80 \times 4 = 320 \, \text{kg}. \] Thus, the weight of D is: \[ \text{Weight of D} = 320 - 252 = 68 \, \text{kg}. \] When E replaces A, the average weight of B, C, D, and E becomes 79 kg. So, the total weight of B, C, D, and E is: \[ \frac{\text{Weight of B + C + D + E}}{4} = 79 \quad \Rightarrow \quad \text{Weight of B + C + D + E} = 79 \times 4 = 316 \, \text{kg}. \] Thus, the weight of E is: \[ \text{Weight of E} = 316 - (B + C + D) = 316 - (252 + 68) = 316 - 320 = -4 \, \text{kg}. \] Therefore, the weight of A is \( \boxed{75} \, \text{kg}. \)