Question

In a potato race, 20 potatoes are placed in a line at an interval of 4 metres with the first potato at a distance of 24 metres from the starting point. A contestant is required to bring all the potatoes to the starting point one at a time. How long does he have to run to bring all the potatoes at the starting point?

Answer 1

Hint: First transform the given question into some mathematical form and draw the diagram to visualise the problem. Then think of a concept which can be applied to this question to answer the total length of run needed to bring back all the potatoes. The length of run for each potato forms an arithmetic progression. Then apply the sum of terms of arithmetic progression.

Complete step-by-step answer:

Let us first draw the diagram as shown below. Here A is the starting point and at location B, C, D, E and so on till U 20 potatoes are kept. Distance AB = 24 m. Distance BC = CD = DE = EF = . . . = TU = 4 m.

Contestant starts from A goes to B pick up the potato and come back to A. In this journey the contestant runs for the distance AB and then BA. Let distance covered to pick up the nth potato and transfer it to the starting point be Dn. So total distance covered for picking the first potato is
D1 = AB + BA = 2AB = 24 m + 24 m = 48 m.
Similarly, distance covered to pick the 2nd potato and transfer it to starting point is
D2 = AC + CA = 2AC = 2(24 + 4) = 2 x 28 = 56 m.
And so on writing for every potatoes
D3 = AD + DA = 2AD = 2(24 + 8) = 2 x 32 = 64 m.
D4 = AE + EA = 2EC = 2(24 + 12) = 2 x 36 = 72 m.
In general we get that
$\begin{aligned} & {{D}_{n}}=2\left( 24+4(n-1) \right) \\\ & \Rightarrow {{D}_{n}}=2\left( 20+4n \right) \\\ & \Rightarrow {{D}_{n}}=40+8n \\\ \end{aligned}$
Now we have to add all the Dn from n = 1 to n = 20. Let S denote the sum. Then
$\begin{aligned} & S=\sum\limits_{n=1}^{20}{\left( 40+8n \right)} \\\ & \Rightarrow S=\sum\limits_{n=1}^{20}{40+\sum\limits_{n=1}^{20}{8n}} \\\ & \Rightarrow S=20\times 40+8\sum\limits_{n=1}^{20}{n} \\\ & \Rightarrow S=800+8\sum\limits_{n=1}^{20}{n}\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(i)} \\\ \end{aligned}$
We know that
$\sum\limits_{k=1}^{n}{n=\dfrac{n(n+1)}{2}}$
Applying this into equation first we get
$\begin{aligned} & S=800+8\times \dfrac{20(20+1)}{2} \\\ & \Rightarrow S=800+8\times 210 \\\ & \Rightarrow S=2480 \\\ \end{aligned}$
Hence the total distance travelled by the contestant is 2480 m.
Note: There is one alternate method to add all those Dn . See that all those Dn form an arithmetic progression.
First term is a = 48
Common difference is d = 8
Total number of terms is n = 20
So applying the sum of n terms of AP we get
$\begin{aligned} & S=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\\ & \Rightarrow S=\dfrac{20}{2}\left[ 2(48)+8\left( 20-1 \right) \right] \\\ & \Rightarrow S=\dfrac{20}{2}[96+152] \\\ & \Rightarrow S=2480 \\\ \end{aligned}$
Hence the total distance travelled by contestants is 2480 m.