Question

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Answer 1

The distances of potatoes are as follows.

5, 8, 11, 14...

It can be observed that these distances are in A.P.

a = 5

d = 8 − 5

d = 3

`S_n = n/2 [2a+(n-1)d]`

`S_10=10/2[2(5)+(10-1)3]`

= 5[10 + 9 × 3]

= 5(10 + 27)

= 5(37)

= 185

Every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be twice as much.

Therefore, total distance that the competitor will run = 2 × 185

= 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14,...

distance run by the competitor for collecting these potatoes is two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

10, 16, 22, 28, 34,...

a = 10

d = 16 − 10

d = 6

S₁₀ = ?

`S_10 = 10/2 [2 xx 10 + (10 - 1)6]`

= 5[20 + 54]

= 5(74)

= 370

Therefore, the competitor will run a total distance of 370 m.

Answer 2

Distance covered by the competitor to pick and drop the first potato = 2 × 5 m = 10 m

Distance covered by the competitor to pick and drop off the second potato

= 2 × (5 + 3) m

= 2 × 8 m

= 16 m

Distance covered by the competitor to pick and drop the third potato

= 2 × (5 + 3 + 3) m

= 2 × 11 m

= 22 m and so on.

∴ Total distance covered by the competitor = 10 m + 16 m + 22 m + ... up to 10 terms

This is an arithmetic series.

Here, a = 10, d = 16 – 10 = 6 and n = 10

Using the formula, `S_n = n/2 [2a + (n - 1)d]`, we get

`S_10 = 10/2 [2 xx 10 + (10 - 1) xx 6]`

= 5 × (20 + 54)

= 5 × 74

= 370

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