Question

A rural child specialist has to find the weight of five children of different ages. He knows from past experience that each of the children weighs less than 30 Kg, and all five weigh different amounts. Unfortunately, the scale available in the village can measure weight only over 30 Kg, so the doctor decides to weigh the children in pairs. His new assistant weighed the children without noting down the names. The ten weights recorded were: 35, 36, 37, 39, 40, 41, 42, 45, 46 and 47 Kg. The weight of the lightest child is:

• A. 15 Kg.
• B. 16 Kg.
• C. 17 Kg.
• D. 18 Kg.

Hint:

Every child's weight appears in exactly four of the ten pair-sums, so the sum of all ten pair-sums equals four times the total weight of all five children.

Answer 1

The Correct Option is B

Step 1: Find the total weight of all five children.
There are 5 children, so the number of distinct pairs is \(\binom{5}{2} = 10\), matching the 10 weights given. Each child is paired with the other four children, so each child's weight is counted in exactly 4 of the 10 pair-sums.
Adding all ten given weights: \(35+36+37+39+40+41+42+45+46+47 = 408\).
Since this total counts each child's weight 4 times, the sum of the five actual weights is \(408 / 4 = 102\) Kg.

Step 2: Identify the smallest and largest pair-sums.
Call the weights \(w_1 < w_2 < w_3 < w_4 < w_5\). The smallest pair-sum must be \(w_1+w_2\), which is 35, and the largest pair-sum must be \(w_4+w_5\), which is 47.

Step 3: Use these to isolate w3.
From \(w_1+w_2 = 35\) and the total \(w_1+w_2+w_3+w_4+w_5 = 102\), we get \(w_3+w_4+w_5 = 102 - 35 = 67\).
From \(w_4+w_5 = 47\), we get \(w_1+w_2+w_3 = 102 - 47 = 55\), so \(w_3 = 55 - 35 = 20\).

Step 4: Find w1 using the second-smallest pair-sum.
The second-smallest pair-sum has to be \(w_1+w_3\), since \(w_1\) paired with the next smallest, \(w_3\), beats pairing \(w_2\) with \(w_3\). That value is 36.
So \(w_1 = 36 - w_3 = 36 - 20 = 16\) Kg.

Step 5: Verify with the remaining weights.
Then \(w_2 = 35 - w_1 = 35 - 16 = 19\). Using the second-largest pair-sum \(w_3+w_5 = 46\) gives \(w_5 = 46 - 20 = 26\), and \(w_4 = 67 - w_3 - w_5 = 67 - 20 - 26 = 21\).
The five weights are 16, 19, 20, 21, 26. Checking all ten pairwise sums gives 35, 36, 37, 42, 39, 40, 45, 41, 46, 47, exactly the given list in a different order.

Final Answer:
The lightest child weighs 16 Kg.
\[ \boxed{16 \text{ Kg}} \]