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Q20:

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 $2 \times 5 + 2 \times (5 + 3)$]

Solution :

Given:

  • Distance of the first potato from the bucket = $5$ m.
  • Distance between consecutive potatoes = $3$ m.
  • Total number of potatoes ($n$) = $10$.
  • The competitor must run from the bucket to the potato and back to the bucket for each potato.

To Find:

The total distance the competitor has to run to collect all $10$ potatoes.

Bucket P1 P2 3m apart

Step 1: Determine the distance for each potato run.

The competitor runs to the potato and returns to the bucket. Thus, the distance for each potato is $2 \times (\text{distance from bucket to potato})$.

  • Distance for 1st potato: $d_1 = 2 \times 5 = 10$ m.
  • Distance for 2nd potato: $d_2 = 2 \times (5 + 3) = 2 \times 8 = 16$ m.
  • Distance for 3rd potato: $d_3 = 2 \times (5 + 3 + 3) = 2 \times 11 = 22$ m.

Step 2: Identify the Arithmetic Progression (AP).

The sequence of distances is $10, 16, 22, \dots$

  • First term ($a$) = $10$.
  • Common difference ($d$) = $16 - 10 = 6$.
  • Number of terms ($n$) = $10$.

Step 3: Apply the sum formula for an AP.

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

Step 4: Substitute the values into the formula.

$S_{10} = \frac{10}{2} [2(10) + (10 - 1)6]$

$S_{10} = 5 [20 + (9 \times 6)]$

$S_{10} = 5 [20 + 54]$

$S_{10} = 5 [74]$

$S_{10} = 370$

Justification:

Since the distance increases by a constant amount ($2 \times 3 = 6$ m) for each subsequent potato, the sequence of distances forms an arithmetic progression. The sum of these distances represents the total path covered by the competitor.

Final Answer: The total distance the competitor has to run is 370 metres.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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