Question

The students of a school decided to beautify the school on an annual day by fixing colorful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored in the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?
A. 830 m, 20 m
B. 728 m, 26 m
C. 619 m, 53 m
D. 513 m, 19 m

Answer 1

Hint: There are 27 flags in total and all the flags are stored at the position of the middle flag which is the 14th flag. So we have 13 flags each on either side of the 14th flag. So consider the 14th flag as the zeroth flag and number the 13 flags on either side from 1 to 13. So from the 0th flag it will take 4 m to and from the 1st flag and 8m to and from the 2nd flag. So this forms an arithmetic progression with 13 terms. Find the sum of the 13 terms and double it because we have to travel on the other side too. And then find the maximum distance Ruchi travelled carrying a flag.
Formula used:
Sum of n terms in an A.P. is $ \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $ , where a is the first term and d is the common difference.

Complete step-by-step answer:
Distance between every two consecutive flags is 2 m.
Let the flags be numbered from 1 to 27.

As we can see the 14th flag is the middle most flag and there are 13 flags on either side.
So rename the 14th flag as the 0th flag and the 13 flags either side from 1 to 13.

Let us first consider the left side of the 0th flag.
Now to travel from 0th to 1st flag, Ruchi takes 2 m and to comeback she takes another 2 m, so totally 4m.
Then to travel to and from the 2nd flag Ruchi takes 8m and then 12m and then 16m.
So the distances travelled will be 4m, 8m, 12m, 16m, …. till 13 terms.
As we can it forms an arithmetic progression with 4 as first term ‘a’ and 4 as common difference ‘d’.
So the total distance travelled by Ruchi on the left side of 0th flag is $ \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $
 $ \Rightarrow \dfrac{{13}}{2}\left( {2 \times 4 + \left( {13 - 1} \right) \times 4} \right) = \dfrac{{13}}{2}\left( {8 + 48} \right) = \dfrac{{13}}{2} \times 56 = 364\;m $
The same distance will be travelled by Ruchi on the right side too.
Therefore total distance travelled by Ruchi to put up all the flags is $ 2 \times 364 = 728\;m $
And the maximum distance travelled by Ruchi carrying a flag is to put the 13th flag on the right side and left side. Therefore the maximum distance travelled carrying a flag is
$\Rightarrow 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 13 \times 2 = 26\;m $
Hence the correct option is Option B; $ 72\; m $ and $ 26\;m $ .
So, the correct answer is “Option B”.

Note: Total distance travelled carrying a flag is different from maximum distance travelled carrying a flag. So do not be confused with this. We considered the series as an A.P. because the difference between every two consecutive distances is the same and the next distance is obtained by adding the common difference to the previous distance.