NCERT Solutions Ganita Prakash (Part 2) Chapter 5 –106Unchanging Mean! — In-text Questions

Book page 105 Updated on2026-09-05

Q1.

Explore if it is possible to include or remove 2 values such that the mean is unchanged. You may use the following data to experiment with.

Answer

Yes. Reading the dot plot on page 106, the data is

2.5, 5, 6.5, 7, 7.5, 8, 8, 8, 8, 9, 10, 10.5, 11, 12, 12, 13, 15
Number of values = 17, sum = 153, mean = 153 ÷ 17 = 9

To keep the mean at 9 while adding two values, the two must be the same distance below and above 9 — that is, they must add up to 18.

Include 6 and 12 → new sum = 153 + 18 = 171, new count = 19
New mean = 171 ÷ 19 = 9 ✓
Include 4 and 14, or 8 and 10, or 9 and 9 → all give 9 again

Removing works the same way: take out 8 and 10 (they add to 18), leaving a sum of 135 over 15 values, and 135 ÷ 15 = 9.

Why it happens: two new values x and y change the total by x + y and the count by 2. The mean survives only if the extra total is exactly two average-sized shares, i.e. x + y = 2 × 9 = 18. In balance language, one dot pulls left and the other pulls right by an equal amount, so the see-saw does not tip.

Q2.

How about including or removing 3 values without changing the mean? Is it possible?

Answer

Yes — three new values keep the mean at 9 as long as they add up to 3 × 9 = 27.

Include 7, 9 and 11 → sum = 153 + 27 = 180, count = 20
New mean = 180 ÷ 20 = 9 ✓
Include 2, 10 and 15 → 2 + 10 + 15 = 27 → mean still 9 ✓

Why it happens: each extra value must bring exactly its own fair share of 9 on average. They need not each equal 9 — only their total must equal 27, because then the extra total (27) and the extra count (3) are in the same ratio as the old total to the old count.

Tip: the general rule is now clear — including k values leaves the mean unchanged exactly when those k values add up to k × (the mean).

Q3.

Can we include 2 values less than the mean and 1 value greater than the mean, so that the mean remains the same?

Answer

Yes. The two small values pull the mean down; the one large value must pull it up by the same total amount.

Take 5 and 6 (below 9): shortfalls are 9 − 5 = 4 and 9 − 6 = 3, total 7 short
So the third value must be 7 above the mean → 9 + 7 = 16
Check: 5 + 6 + 16 = 27 = 3 × 9 ✓
New mean = (153 + 27) ÷ 20 = 180 ÷ 20 = 9 ✓

Why it happens: what matters is not how many values sit on each side, but the total distance. Two values 4 and 3 below the mean can be balanced by a single value 7 above it — exactly the balance rule from page 104 applied to the newcomers alone.

The book’s own example: the green dots added in the figure are at 7, 7 and 13. The two 7s are each 2 below the mean (total 4) and 13 is 4 above it, so the pulls cancel — and 7 + 7 + 13 = 27 = 3 × 9, exactly as the rule requires.

Q4.

Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same.

Answer

Mirror the previous idea: the two large values must be balanced by one small one.

Take 11 and 13 (above 9): excesses are 2 and 4, total 6 extra
So the third value must be 6 below the mean → 9 − 6 = 3
Check: 11 + 13 + 3 = 27 = 3 × 9 ✓
New mean = (153 + 27) ÷ 20 = 9 ✓

Another set that works: 10 and 12 (excesses 1 and 3, total 4) together with 9 − 4 = 5, since 10 + 12 + 5 = 27.

Check it yourself: pick any two numbers above 9, add up how far above they are, and go that far below 9 for the third. It will work every time.

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