Question

In a shipment of 5 parcels, the average weight of the 4 lightest parcels is 50 kg, while the average weight of the 4 heaviest parcels is 55 kg. What is the difference between the maximum and minimum possible average weight of all 5 parcels?

• A. 1 kg
• B. 2 kg
• C. 3 kg
• D. 4 kg

Answer 1

Answer: (C)

3 kg

Answer 2

Let the weights of the 5 parcels be $w_1, w_2, w_3, w_4, w_5$ in ascending order, so $w_1 \le w_2 \le w_3 \le w_4 \le w_5$.

The average weight of the 4 lightest parcels is 50 kg. The 4 lightest parcels are $w_1, w_2, w_3, w_4$. So, $\frac{w_1 + w_2 + w_3 + w_4}{4} = 50$. The sum of the weights of the 4 lightest parcels is $S_L = w_1 + w_2 + w_3 + w_4 = 4 \times 50 = 200$ kg.

The average weight of the 4 heaviest parcels is 55 kg. The 4 heaviest parcels are $w_2, w_3, w_4, w_5$. So, $\frac{w_2 + w_3 + w_4 + w_5}{4} = 55$. The sum of the weights of the 4 heaviest parcels is $S_H = w_2 + w_3 + w_4 + w_5 = 4 \times 55 = 220$ kg.

Let $S_5$ be the sum of the weights of all 5 parcels: $S_5 = w_1 + w_2 + w_3 + w_4 + w_5$. We can write $S_5 = (w_1 + w_2 + w_3 + w_4) + w_5 = S_L + w_5 = 200 + w_5$. We can also write $S_5 = w_1 + (w_2 + w_3 + w_4 + w_5) = w_1 + S_H = w_1 + 220$.

Equating the two expressions for $S_5$, we get $200 + w_5 = w_1 + 220$. This gives the relationship between the heaviest and lightest parcel: $w_5 - w_1 = 220 - 200 = 20$.

The average weight of all 5 parcels is $A_5 = \frac{S_5}{5} = \frac{200 + w_5}{5} = \frac{w_1 + 220}{5}$. To find the range of $A_5$, we need to find the range of possible values for $w_1$ or $w_5$.

We know the weights are sorted: $w_1 \le w_2 \le w_3 \le w_4 \le w_5$. From $w_1 + w_2 + w_3 + w_4 = 200$, since $w_1 \le w_2 \le w_3 \le w_4$, we have $4w_1 \le w_1 + w_2 + w_3 + w_4 = 200$, which implies $w_1 \le 50$. Also, $w_1 + w_2 + w_3 + w_4 \le w_4 + w_4 + w_4 + w_4 = 4w_4$, so $200 \le 4w_4$, which implies $w_4 \ge 50$.

From $w_2 + w_3 + w_4 + w_5 = 220$, since $w_2 \le w_3 \le w_4 \le w_5$, we have $4w_2 \le w_2 + w_3 + w_4 + w_5 = 220$, which implies $w_2 \le 55$. Also, $w_2 + w_3 + w_4 + w_5 \le w_5 + w_5 + w_5 + w_5 = 4w_5$, so $220 \le 4w_5$, which implies $w_5 \ge 55$.

We have the constraints:

1. $w_1 \le w_2 \le w_3 \le w_4 \le w_5$
2. $w_1 + w_2 + w_3 + w_4 = 200$
3. $w_2 + w_3 + w_4 + w_5 = 220$ From (3) - (2), $w_5 - w_1 = 20$, so $w_5 = w_1 + 20$.

Substitute $w_5 = w_1 + 20$ into (1): $w_1 \le w_2 \le w_3 \le w_4 \le w_1 + 20$.

To find the minimum possible average weight $A_{5, min}$, we need to minimize $w_1$. Consider the sum $w_1 + w_2 + w_3 + w_4 = 200$. Since $w_2 \le w_3 \le w_4 \le w_1 + 20$, we have $w_2 + w_3 + w_4 \le 3(w_1 + 20)$. $w_1 + (w_2 + w_3 + w_4) = 200$. $w_1 + w_2 + w_3 + w_4 = 200$. Since $w_1 \le w_2 \le w_3 \le w_4$, we have $w_4 \ge w_1$. Also, $w_4 \le w_1 + 20$. From $w_1 + w_2 + w_3 + w_4 = 200$, we have $w_1 = 200 - (w_2 + w_3 + w_4)$. To minimize $w_1$, we need to maximize $w_2 + w_3 + w_4$. We know $w_2 \le w_3 \le w_4$ and $w_4 \le w_1 + 20$. The maximum value of $w_4$ is $w_1 + 20$. The maximum values of $w_2, w_3$ are limited by $w_4$. Consider the case where $w_2 = w_3 = w_4 = w_1 + 20$. Then $w_1 + 3(w_1 + 20) = 200$. $w_1 + 3w_1 + 60 = 200$. $4w_1 = 140$. $w_1 = 35$. In this case, $w_1 = 35$, $w_2 = 35+20 = 55$, $w_3 = 55$, $w_4 = 55$, $w_5 = w_1 + 20 = 35 + 20 = 55$. The weights are 35, 55, 55, 55, 55. Let's check the conditions: Sorted: 35 $\le$ 55 $\le$ 55 $\le$ 55 $\le$ 55. Yes. Average of 4 lightest (35, 55, 55, 55): $\frac{35+55+55+55}{4} = \frac{200}{4} = 50$. Yes. Average of 4 heaviest (55, 55, 55, 55): $\frac{55+55+55+55}{4} = \frac{220}{4} = 55$. Yes. This is a valid set of weights. The sum is $35 + 55 + 55 + 55 + 55 = 255$. The average is $\frac{255}{5} = 51$. This corresponds to the minimum possible value of $w_1$, which is 35. Minimum sum $S_{5, min} = 220 + w_{1, min} = 220 + 35 = 255$. Minimum average $A_{5, min} = \frac{255}{5} = 51$.

To find the maximum possible average weight $A_{5, max}$, we need to maximize $w_1$. Consider the sum $w_1 + w_2 + w_3 + w_4 = 200$. Since $w_1 \le w_2 \le w_3 \le w_4$, to maximize $w_1$, we need to make $w_2, w_3, w_4$ as close to $w_1$ as possible. Consider the case where $w_1 = w_2 = w_3 = w_4$. Then $w_1 + w_1 + w_1 + w_1 = 200$, so $4w_1 = 200$, $w_1 = 50$. In this case, $w_1 = 50$, $w_2 = 50$, $w_3 = 50$, $w_4 = 50$. Then $w_5 = w_1 + 20 = 50 + 20 = 70$. The weights are 50, 50, 50, 50, 70. Let's check the conditions: Sorted: 50 $\le$ 50 $\le$ 50 $\le$ 50 $\le$ 70. Yes. Average of 4 lightest (50, 50, 50, 50): $\frac{50+50+50+50}{4} = \frac{200}{4} = 50$. Yes. Average of 4 heaviest (50, 50, 50, 70): $\frac{50+50+50+70}{4} = \frac{220}{4} = 55$. Yes. This is a valid set of weights. The sum is $50 + 50 + 50 + 50 + 70 = 270$. The average is $\frac{270}{5} = 54$. This corresponds to the maximum possible value of $w_1$, which is 50. Maximum sum $S_{5, max} = 220 + w_{1, max} = 220 + 50 = 270$. Maximum average $A_{5, max} = \frac{270}{5} = 54$.

The range of possible average weights is from 51 kg to 54 kg. The difference between the maximum and minimum possible average weight is $A_{5, max} - A_{5, min} = 54 - 51 = 3$ kg.

The final answer is $\boxed{3 kg}$.