Q1. [4]
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes, as shown in the figure :
A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket.
Based on the above information, answer the following questions :
- (i) What is the distance covered to pick up the first potato and drop it in bucket ? [1]
- (ii) What is the distance covered to pick up the second potato and drop it in bucket ? [1]
- (iii) What is the total distance the competitor has to run ? [2]
Previously asked in CBSE board exam
2026 30/1/1 Q36
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding stimulus
Model Answer
(i) The first potato is 5 m from the bucket. The competitor runs 5 m to pick it up and 5 m to return.
Distance = 5 + 5 = 10 m
(ii) The second potato is 5 + 3 = 8 m from the bucket. The competitor runs 8 m and returns 8 m.
Distance = 8 + 8 = 16 m
(iii) Distance covered for the $n$-th potato = $2[5 + (n-1) \times 3]$ m
So the distances form an AP: 10, 16, 22, … with $a = 10$, $d = 6$, $n = 10$.
Total distance $S_{10} = \dfrac{10}{2}[2(10) + (10-1)(6)]$
$= 5[20 + 54] = 5 \times 74 = \mathbf{370 \text{ m}}$
Source: Arithmetic Progressions, Application/Case Study
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Explanation
- Each trip (to pick up potato $n$ and return) covers $2 \times$ (distance of that potato from bucket).
- Distances of potatoes from bucket: 5, 8, 11, … (AP with $a=5$, $d=3$), so trips form AP 10, 16, 22, … ($a=10$, $d=6$).
- Use $S_n = \frac{n}{2}[2a + (n-1)d]$ for the total. Examiners award 1 mark each for parts (i) and (ii), and for part (iii) they look for correct AP identification + correct formula application + final answer (2 marks).
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