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NCERT Solutions for Class 7th Maths Chapter 5 Figure it Out — Representative Values

Book page 101 Updated on2026-09-19

Q1.
Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.
Answer

5 bounces per attempt.

Total bounces = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5
= 8 + 9 + 5 + 4 + 6 + 3 + 5
= 40
Number of attempts = 8
Average = 40 ÷ 8 = 5 bounces
Why it happens: Shreyas's best attempt was 9 and his worst was 2. The mean of 5 sits between them and balances them out — the 4 extra bounces above 5 in his best attempt pay for the 3 missing bounces in his worst.
Tip: the median here is also 5. Sorted, the data is 2, 3, 4, 5, 5, 6, 6, 9, and the two middle values are 5 and 5. Mean and median agree because there is no outlier.
Q2.
Try the activity above on your own. Collect data for 7 or more attempts and find the average.
Answer

This is an activity to do yourself. Here is the method, and then one worked sample.

  • Bounce the ball on the bat and count until it drops. That is one attempt.
  • Write the count down immediately, before the next attempt.
  • Do at least 7 attempts. Do not throw away a bad attempt — a 1 or a 0 is real data.
  • Add all the counts, then divide by the number of attempts.

Sample answer: suppose your 8 attempts give 3, 5, 1, 7, 4, 6, 4, 2.

Total = 3 + 5 + 1 + 7 + 4 + 6 + 4 + 2 = 32
Average = 32 ÷ 8 = 4 bounces per attempt
Why it happens: the more attempts you record, the steadier the average becomes. With only 2 attempts one lucky try changes everything; with 10 attempts one lucky try moves the average very little.
Try This: record your attempts again after a week of practice. If your average has gone up, you now have evidence that you improved — not just a feeling.
Q3.
Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?
Answer

This is a data-collection activity. Here is how to do it well, and a sample.

  • Pick one plant and stay with it — a hibiscus, a jasmine, a marigold.
  • Count at the same time each day, say every morning. Counting at different times of day would mix two different things.
  • Record all 7 days, including days when nothing bloomed. A 0 is a real value here and must be counted.

Sample answer: a hibiscus plant gives 4, 6, 0, 3, 5, 7, 3 flowers over one week.

Total = 4 + 6 + 0 + 3 + 5 + 7 + 3 = 28
Average = 28 ÷ 7 = 4 flowers per day
Why it happens: blooming is uneven — a warm day may give 7 flowers and a cloudy one none. The average smooths this into a single number you can use to compare one week with the next, or one plant with another.
Try This: also write down the highest and the lowest count. "About 4 a day, between 0 and 7" says far more than "about 4 a day".
Q4.
Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil: 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?
Answer

Neither. Their averages are exactly the same — about 17.43 seconds each.

Nikhil's total = 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122 s
Nikhil's mean = 122 ÷ 7 ≈ 17.43 s

Sunil's total = 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122 s
Sunil's mean = 122 ÷ 7 ≈ 17.43 s
Why it happens: the two lists are made of almost the same numbers in a different order, so their totals are identical. Since both ran the same number of times, identical totals must give identical means.

The means tie, so look at the spread — and there the two runners are not the same.

FastestSlowestRangeMedian
Nikhil16 s19 s3 s17 s
Sunil16 s20 s4 s17 s
The strongest thing you can say: on average they are equally quick, and their medians are equal too. But Nikhil is more consistent — his times stay inside a 3-second band, while Sunil's stretch over 4 seconds. Sunil's slowest run (20 s) is worse than anything Nikhil produced. For a one-off race that hardly matters; for reliable performance, Nikhil looks the safer pick.
Q5.
The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.
Answer

Mean enrolment = 1859 students.

Total = 1555 + 1670 + 1750 + 2013 + 2040 + 2126
1555 + 1670 = 3225
3225 + 1750 = 4975
4975 + 2013 = 6988
6988 + 2040 = 9028
9028 + 2126 = 11154

Mean = 11154 ÷ 6 = 1859 students
Why it happens: 1859 lies between the smallest year (1555) and the largest (2126), as every mean must. It is the number of students the school would have had each year if the same number had come every year.
Check it yourself: 1859 × 6 = 11154 — the same total we started with. A quick sanity check: the six numbers all lie between 1500 and 2200, so the mean had to land in that band.
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