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 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).
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)].
Solution:
Distance of 10 potatoes from the bucket is such that, the first potato is 5m away from the bucket and other potatoes are placed 3m apart in a straight line.
To find the total distance the competitor has to run, we will find the sum of the AP if they are in an arithmetic Progression
The sum of the first n terms of an Arithmetic Progression is given by Sₙ = n/2 [2a + (n - 1) d].
Here, a is the first term, d is a common difference and n is the number of terms.
The distances of potatoes are as follows. 5, 8, 11, 14, ...
It can be observed that these distances are in A.P.
- First term, a = 5
- Common difference, d = 8 - 5 = 3
- Number of terms, n = 10
We know that the sum of n terms is given by the formula Sₙ = n/2 [2a + (n - 1) d]
S₁₀= 10/2 [2 × 5 + (10 - 1) × 3]
= 5 [10 + 27]
= 5 × 37
S₁₀ = 185
Like every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times the total distance.
Therefore, the total distance that the competitor will run = 2 × 185 m = 370 m
Alternatively,
The distances of potatoes from the bucket are 5, 8, 11, 14...
The distance run by the competitor for collecting these potatoes is two times the distance at which the potatoes have been kept.
Therefore, distances to be run are
First potato, a = 2 × 5m =10m
Second potato, a₂ = 2 × (5 + 3) m = 16m
Third potato, a₃ = 2 × (5 + 2 × 3) m = 22m
Number of potatoes, n = 10
10, 16, 22, 28, 34,....
a = 10
d = 16 - 10 = 6
Sₙ = n/2 [2a + (n - 1) d]
= 10/2 [2 × 10 + (10 - 1)6]
= 5[20 + 54]
= 5 × 74
= 370
Therefore, the competitor will run a total distance of 370 m.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 5
Video Solution:
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 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).
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?
Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 20
Summary:
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. The competitor will run a total distance of 370 m.
☛ Related Questions:
- Find the sums given below (i) 7 + 10 ½ + 14 +....+ 84 (ii) 34 + 32 + 30 +....+ 10 (iii) - 5 + (- 8) + (- 11) +....+ (- 230).
- In an AP (i) Given a = 5, d = 3, an = 50 , find n and Sn. (ii) Given a = 7, a13 = 35 , find d and S13. (iii) Given a12 = 37, d = 3, find a and S12. (iv) Given a3 = 15, S10 = 125, find d and a10. (v) Given d = 5, S9 = 75 , find a and a9. (vi) Given a = 2, d = 8, Sn = 90 , find n and an. (vii) Given a = 8, an = 62, Sn = 210 , find n and d. (viii) Given an = 4, d = 2, Sn = - 14 , find n and a. (ix) Given a = 3, n = 8, S = 192 , find d. (x) Given l = 28, S = 144 and there are total 9 terms. Find a.
- How many terms of the AP. 9, 17, 25 ... must be taken to give a sum of 636?
- The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
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