NCERT Solutions Ganita Prakash (Part 2) Chapter 5 Finding the Unknown — In-text Questions

Book page 109 Updated on2026-09-05

Q1.

Coach Balwan noted down the weights of the kushti players (wrestlers) and the mean as shown. But one value that was written down got smudged. Can you find out the missing value?

Answer

The recorded weights are 42, 40, 39, 33, 48, 38, 42, 35, 32 and one smudged value w, and the mean of all 10 is 39.2 kg.

Sum of the nine known weights
= 42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 = 349

(349 + w) ÷ 10 = 39.2
349 + w = 39.2 × 10 = 392
w = 392 − 349 = 43

The missing weight is 43 kg.

Why it works: the mean tells you the total in disguise. If ten players average 39.2 kg, they must weigh 392 kg between them. Subtract the nine weights you can read and only the smudged one is left. Working backwards from the mean to the total is the key move in every ‘find the unknown’ problem of this kind.

Q2.

Venkayya keeps track of the coconut harvest in his farm. He calculates the average harvest per tree as 25.6. His son verifies the counts and finds that one tree’s harvest count is incorrectly noted as 3 more than the actual number. Can you find the correct average if the number of trees is 15?

Answer

The correct average is 25.4 coconuts per tree.

Wrong total = average × number of trees = 25.6 × 15 = 384
One tree was counted 3 too many, so the correct total = 384 − 3 = 381
Correct average = 381 ÷ 15 = 25.4

Why it works: you never need the individual tree counts. The average multiplied by the number of trees recovers the total, and an error of 3 in one tree is an error of 3 in the total. Spreading that mistake of 3 over 15 trees changes the average by 3 ÷ 15 = 0.2, which is exactly the drop from 25.6 to 25.4.

Tip: this is the ‘adding a value’ formula in another guise — an error of e in the total shifts the mean by e ÷ n. The more trees, the less one miscount matters.

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