Question

What is the serial number of a boy in the middle of the row of 45 boys?

Answer 1

Hint: First of all, we shall understand how the boys are arranged. Then we will decide which way to rank the boys. After deciding which side to be the start and which side to be the end of the row, we can opt one of two approaches. Either we can develop a formula for any number of items or boys in our case, in the row or we can physically count. For our question, we will try to derive the formula. Once the formula is derived, we can put the value for the number of boys and get the rank or serial number of the middle position.

Complete step-by-step answer:
It is given to us that there are 45 boys in a row. From the word “row”, we can deduce that they are standing in a single linear arrangement. Thus, in a linear arrangement, we have a starting point and an ending point. For our question, we will take the left end of the row as the starting point and the right end as the ending point.
Therefore, the boy at the leftmost end will have serial number 1, the second will have serial number 2 and so on up to serial number 45 at the right most end.
1 2 3 … 45
Now, to generalise, let the number of items in a row be 2n + 1. We can clearly see that there are an odd number of items.
Now, the term in the middle means it has an equal item to its right as well as left. So, the maximum equal number in two arrangements from 2n + 1 items will have n items each. This will leave 1 item, which will have n items to the left and n items to the right. The position this middle term will be (n+1).
But (n + 1) is half of (2n + 2), that is (2n + 1 + 1).
Hence, the position of middle term if we have 2n + 1 items in a row is $\dfrac{2n+1+1}{2}$.
We know that for our question, 2n + 1 = 45.
Therefore, the position and hence serial number of the middle term will be $\dfrac{45+1}{2}=23$.

Note: For the physical counting process, we will start counting from the left end and the right end and look for the position which will be common from both ends and have the same rank or serial number will be the middle position. This is possible for a small number of items. If the number gets bigger, formula method is preferred.