NCERT Solutions Ganita Prakash (Part 1) Chapter 5 Breaking Even — In-text Questions

Book page 115 Updated on2026-09-05

Q1.
Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?
Answer

No — it holds for any count of numbers. Nothing in the argument used the number four.

For n numbers a₁ ± a₂ ± … ± aₙ
there are 2ⁿ⁻¹ expressions,
and switching one sign changes the value by 2aᵢ — still even

So all 2n–1 expressions carry the parity of a₁ + a₂ + … + an.

NumbersHow many expressionsShared parity
3, 4, 52² = 4even (3 + 4 + 5 = 12)
3, 4, 5, 62³ = 8even (18)
3, 4, 5, 6, 72⁴ = 16odd (25)
Check it yourself: 3 – 4 + 5 = 4, 3 + 4 – 5 = 2, 3 – 4 – 5 = – 6 — all even, matching 3 + 4 + 5 = 12.
Q2.
We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even. 43 + 37, 672 – 348, 4 × 347 × 3, 708 – 477, 809 + 214, 119 × 303, 543 – 479, 513³
Answer

Four of them are even: 43 + 37, 672 – 348, 4 × 347 × 3 and 543 – 479.

ExpressionParity reasoningEven?
43 + 37odd + odd = evenYes
672 – 348even – even = evenYes
4 × 347 × 3a product with the factor 4 in it must be evenYes
708 – 477even – odd = oddNo
809 + 214odd + even = oddNo
119 × 303odd × odd = oddNo
543 – 479odd – odd = evenYes
513³odd × odd × odd = oddNo
Check it yourself (afterwards!): 80, 324, 4164, 231, 1023, 36057, 64 and 135005697. The parity rules got every one right without any of that work.
Q3.
Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers. 2a + 2b, 3g + 5h, 4m + 2n, 2u – 4v, 13k – 5k, 6m – 3n, x² + 2, b² + 1, 4k × 3j
Answer

Five of the nine are always even: 2a + 2b, 4m + 2n, 2u – 4v, 13k – 5k and 4k × 3j. Each of them has 2 as a factor of the whole expression.

ExpressionRewrittenAlways even?
2a + 2b2(a + b)Yes
3g + 5hNo
4m + 2n2(2m + n)Yes
2u – 4v2(u – 2v)Yes
13k – 5k8k = 2(4k)Yes
6m – 3n3(2m – n)No
x² + 2No
b² + 1No
4k × 3j12kj = 2(6kj)Yes
Why it happens: once an expression is written as 2 × (something whole), 2 is a factor of it no matter what the letters stand for. The four that fail have no such factor — a single counterexample kills each one: 3g + 5h = 3 when g = 1, h = 0; 6m – 3n = 3 when m = n = 1; x² + 2 = 11 when x = 3; b² + 1 = 5 when b = 2.
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