Question

Q1.In a potato race a bucket is placed at the starting point, which is 5 m from the first potato and the other potatoes are placed 3m apart in a straight line . There are ten potato in the lineA competitor starts from the bucket, picks up the nearest potato runs back with it drops in the bucket runs back to pick up the next potato, runs to the bucket to drop it in and continues
in the same way until all the potatoes are in the bucket. What is the
Total distance (in metres) the competitor has to run
[Hint: To pick up the first potato and the second potato the total distance (in metres) run by acompetitor is 2 x 5 + 2 multiply (5 + 3)





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Answer 1

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Answer 2

Step-by-step explanation:

Answer: 370

Step-by-step explanation:

Given that we have 10 potatoes.

First one is 5 meter from the starting point.

Second is 3 m from the first potato, that implies its 5+3=8 metre from the starting point.

Third one is 8+3 = 11 m from Starting point

Fourth one is 11+3 = 14 m from the starting point.

And so on,

Hence,

Distance of each potato from the starting point is respectively as follows,

5 , 8 , 11 , 14 , 17 , 20 , 23 ,. .. ... ..upto 10 terms

Let us assume that there is a potato placed at x m from the starting point.

Competitor goes to x m and pick up the potato and come back upto x m to staring point.

Hence to pick up and put down in bucket any potato, competitor needs to travel 2x m.

So,

To get first potato; needs to run = 2×5 = 10

To get second potato; needs to run = 2×8 = 16

And so on upto 10 potaoes

Hence we have a series of distance,

10 , 16 , 22 . .. ... . upto 10 terms.

Total distance = 10 + 16 + 22 + . .... ..

This is the AP, having

First term a = 10

Common Difference (d) = 6

Number of terms (n) = 10

Total Distance (S) = ?

We know that,

$\begin{gathered}S=\frac{n}{2}[2a+(n-1)d]\\\;\\S=\frac{10}{2}[2\times10+(10-1)6]\\\;\\S=5[20+(9\times6)]\\\;\\S=5[20+54]\\\;\\S=5\times74\\\;\\S=370\;\;\text{m}\end{gathered}$

Hence, Total distance the competitor has to run is 370 m.