Question 431 markSection E

  1. (ii) What is the distance covered to pick up the second potato and drop it in bucket ?

Arithmetic Progressions
Terms of an AP - distance for second potato

Official answer

16 m — the competitor runs 8 m to the second potato (at 5+3 m5 + 3\text{ m} from the bucket) and 8 m back, so the total distance = 2×8=162 \times 8 = 16 m.

Answer keywords:16 msecond potatoto and fro5 + 3 = 8 mdouble the distancearithmetic progression5 m3 m apart

Marking scheme

  1. 1.1 mark: correct final distance 16 m obtained as 2×(5+3)2 \times (5 + 3).
  2. 2.Full credit if the student writes 2×8=162 \times 8 = 16 m; award the mark only if the return trip (the factor 2) is included.

Hint

The second potato is 5+3=85 + 3 = 8 m away — remember the competitor also runs back, so double it.

Quick oral answer

The second potato is 5+3=85 + 3 = 8 m from the bucket, and since the competitor runs there and back, the distance covered is 2×8=162 \times 8 = 16 m.

Common mistakes

  1. 1.Answering 8 m by forgetting that the competitor must run back to the bucket to drop the potato — the distance must be doubled.
  2. 2.Taking the second potato at 3 m or 6 m instead of 5+3=85 + 3 = 8 m by ignoring the 5 m gap of the first potato.
  3. 3.Using 2×5=102 \times 5 = 10 m (the first potato's distance) instead of the second potato's distance.

Analysis and explanation

This part checks whether the to-and-fro nature of the race is modelled correctly before any AP formula is used.


Concept

  • The competitor must return to the bucket after every potato, so the running distance for a potato = 2 × (its distance from the bucket).
  • Second potato is at 5+3=85 + 3 = 8 m from the bucket, so the run = 2×8=162 \times 8 = 16 m.

Key points

  • The to-and-fro distances 10, 16, 22, … form an AP with a=10,d=6a = 10, d = 6 — this exact AP is reused in the later parts.

Common mistakes (परीक्षा में सावधानी)

  • Forgetting the return trip and writing only the one-way distance, 8 m, instead of 16 m.

Real-world

  • The same to-and-fro modelling is used for shuttle runs and relay-style events in sports.

Interesting facts

The potato-race problem is a direct application of the sum of an AP that appears as a worked example in the NCERT Class 10 Chapter 5 (Arithmetic Progressions).

Potato races were popular field-day events in 19th-century schools; the to-and-fro rule makes them a perfect natural model for a linearly increasing sequence.

Because each return trip adds a fixed 6 m (2×3 m2 \times 3\text{ m}) to the previous potato's distance, the situation is a physical demonstration of a constant common difference.

Question details

Q43 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section E
Marks
1
Chapter
Arithmetic Progressions
Topic
Terms of an AP - distance for second potato
Bloom's level
apply
Syllabus topic
Unit 2: Algebra — Arithmetic Progressions: nth term and to-and-fro distance modelling (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why do we multiply the distance by 2?

Because the competitor must carry each potato back to the bucket. So for every potato the person runs the distance to it once and the same distance back, making the total twice the one-way distance.

How far is the second potato from the bucket?

The first potato is 5 m from the bucket and the potatoes are 3 m apart, so the second potato is 5+3=85 + 3 = 8 m from the bucket.

Do the distances for each potato form an AP?

Yes. The to-and-fro distances are 10 m, 16 m, 22 m, … which form an AP with first term 10 and common difference 6.