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NCERT Solutions Ganita Prakash (Part 2) Chapter 5 In-text Questions — The Balancing Act

Book page 104 Updated on2026-09-05

Q1.
Can you explain how the mean is the centre of each collection?
Answer

Measure how far each value is from the mean, and add the distances up on each side. The two totals come out equal.

CollectionMeanDistances on the leftDistances on the right
6, 7, 871 (from 6)1 (from 8)
3, 6, 963 (from 3)3 (from 9)
2, 4, 953 + 1 = 44 (from 9)
4, 11, 15106 (from 4)1 + 5 = 6
Why it happens: think of the number line as a see-saw with a dot of equal weight at each value. The mean is the point where the see-saw balances, because the pull of everything on the left exactly cancels the pull of everything on the right. In the third collection, 2 and 4 are close in but there are two of them (3 + 1 = 4), while the single value 9 is far out (4) — small distances in bulk balance one large distance.
Q2.
Mark the mean for the collections below.
Answer

Reading the four dot plots and computing:

Dot plotValuesSumMean
Top left (orange)11, 13, 17, 196060 ÷ 4 = 15
Top right (orange)5, 6, 15, 164242 ÷ 4 = 10.5
Bottom left (green)10, 10, 11, 174848 ÷ 4 = 12
Bottom right (green)3, 5, 10, 123030 ÷ 4 = 7.5

Mark a cross at 15, 10.5, 12 and 7.5. Notice that in two of these the mean falls between the marked numbers, not on one of them.

Q3.
Can you explain how the mean is the centre of each collection?
Answer

Again, add up the distances on either side of the mean:

11, 13, 17, 19 with mean 15 → LHS = 4 + 2 = 6, RHS = 2 + 4 = 6 ✓
5, 6, 15, 16 with mean 10.5 → LHS = 5.5 + 4.5 = 10, RHS = 4.5 + 5.5 = 10 ✓
10, 10, 11, 17 with mean 12 → LHS = 2 + 2 + 1 = 5, RHS = 5 ✓
3, 5, 10, 12 with mean 7.5 → LHS = 4.5 + 2.5 = 7, RHS = 2.5 + 4.5 = 7 ✓
Why it happens: the third collection is the interesting one. Its mean, 12, is nowhere near the midpoint of the extremes 10 and 17 (that would be 13.5). It sits low because three of the four values are down near 10 and only one is out at 17. The balance is 1 + 2 + 2 on the left against 5 on the right — three short pulls against one long one.
Q4.
Verify that this holds for all the collections of data shown earlier.
Answer

It holds for every one of the eight collections on pages 103 and 104:

CollectionMeanLHS totalRHS total
6, 7, 8711
3, 6, 9633
2, 4, 953 + 1 = 44
4, 11, 151061 + 5 = 6
11, 13, 17, 19154 + 2 = 62 + 4 = 6
5, 6, 15, 1610.55.5 + 4.5 = 104.5 + 5.5 = 10
10, 10, 11, 17122 + 2 + 1 = 55
3, 5, 10, 127.54.5 + 2.5 = 72.5 + 4.5 = 7
Why it always works: let the mean be a. Then x1 + x2 + … + xn = na, so (x1 − a) + (x2 − a) + … + (xn − a) = na − na = 0. The values above a give positive terms, the values below a give negative terms, and since the whole lot adds to zero the two groups must be equal in size. That is precisely ‘LHS total = RHS total’.
Q5.
Can there be more than one such ‘centre’? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal?
Answer

No — there is exactly one such point, and it is the mean.

Why it happens: suppose you slide the balance point from the mean a to a + d, with d > 0. Every value below the new point is now d further away, and every value above it is d closer. So the left total goes up and the right total goes down — they can no longer be equal. Sliding to a − d does the opposite. Since moving either way spoils the balance, only one point can balance the data.
Tip: in algebra: (x1 − c) + (x2 − c) + … + (xn − c) = na − nc, which is zero only when c = a.
Q6.
In the case of the collection 10, 10, 11, and 17 whose mean is 12, suppose there is a different centre larger than 12.
Answer

Take the trial centre to be 13 instead of 12 and measure again.

At 12 → LHS = 2 + 2 + 1 = 5, RHS = 5 ✓ balanced
At 13 → LHS = 3 + 3 + 2 = 8, RHS = 4 ✗ not balanced
At 11 → LHS = 1 + 1 + 0 = 2, RHS = 6 ✗ not balanced

Moving up by 1 made all three left-hand distances grow by 1 each (a gain of 3) and the single right-hand distance shrink by 1. The totals part company at once.

Why it happens: when you move the trial centre up by d, the left total grows by (number of values on the left) × d and the right total falls by (number of values on the right) × d. They can only stay equal if d = 0. So 12 is the only centre.
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