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NCERT Solutions Ganita Prakash (Part 1) Chapter 6 Parity of Expressions — In-text Questions

Book page 132 Updated on2026-09-05

Q1.
Find the parity of the number of small squares in these grids: (a) 27 × 13 (b) 42 × 78 (c) 135 × 654
Answer
GridParity of the two sidesParity of the countActual product
(a)27 × 13odd, oddodd351
(b)42 × 78even, eveneven3276
(c)135 × 654odd, eveneven88290
Why it happens: a product is odd only when every factor is odd. A single even factor supplies a 2, and that makes the whole product even.
Check it yourself: the parity was found without multiplying — the products are shown only to confirm the answers.
Q2.
Come up with an expression that always has even parity. Some examples are: 100p and 48w – 2. Try to find more.
Answer

Any expression in which every term is a multiple of 2 always gives an even value.

2n  •  6k  •  10m  •  4a + 8  •  12t – 6  •  100p  •  48w – 2

Test one of them:

4a + 8  →  a = 1: 12,   a = 2: 16,   a = 3: 20  — all even
Why it happens: write the expression as 2 × (something). For example 4a + 8 = 2(2a + 4). A number with 2 as a factor can always be arranged in pairs, so it is even.
Q3.
Come up with expressions that always have odd parity.
Answer

Take an always-even expression and add or subtract 1 (or any odd number).

2n + 1  •  2m – 1  •  4n + 3  •  6k + 5  •  8a + 3  •  10p – 7

Test one of them:

6k + 5  →  k = 1: 11,   k = 2: 17,   k = 3: 23  — all odd
Why it happens: 6k is a collection of complete pairs; adding 5 leaves one dot without a partner (since 5 itself is odd). So the value can never be even.
Q4.
Come up with other expressions, like 3n + 4, which could have either odd or even parity.
Answer

Use a term whose coefficient is odd, so that the parity flips as n changes.

3n + 4  •  5n + 2  •  7n + 1  •  9n – 4  •  n + 3
n5n + 2Parity
17odd
212even
317odd
422even
Why it happens: in 5n + 2 the part 5n is odd when n is odd and even when n is even. Adding the fixed even number 2 does not change that, so the answer alternates between odd and even.
Q5.
The expression 6k + 2 evaluates to 8, 14, 20,... (for k = 1, 2, 3,...) — many even numbers are missing. Are there expressions using which we can list all the even numbers? Hint: All even numbers have a factor 2.
Answer

Yes — the expression is 2n.

n = 1, 2, 3, 4, 5, …
2n = 2, 4, 6, 8, 10, …  — every even number appears, none is skipped.

Compare with 6k + 2:

ExpressionValuesWhat is missed
6k + 28, 14, 20, 26, …2, 4, 6, 10, 12, 16, …
2n2, 4, 6, 8, 10, …nothing
Why it happens: 6k + 2 = 2(3k + 1), so it only produces even numbers whose half is of the form 3k + 1 — it jumps in steps of 6. The expression 2n moves in steps of 2 and therefore catches every even number.
Q6.
Are there expressions using which we can list all odd numbers?
Answer

Yes — 2n – 1 (with n = 1, 2, 3, …).

n = 1 → 2(1) – 1 = 1
n = 2 → 2(2) – 1 = 3
n = 3 → 2(3) – 1 = 5
n = 4 → 2(4) – 1 = 7  … and so on

The expression 2n + 1 also gives only odd numbers, but starting from n = 1 it gives 3, 5, 7, … and misses 1.

Why it happens: every odd number is one less than an even number. Since 2n runs through all even numbers, 2n – 1 runs through all odd numbers.
Q7.
What would be the nth term for multiples of 2? Or, what is the nth even number?
Answer

The nth even number is 2n.

1st even number = 2 × 1 = 2
2nd even number = 2 × 2 = 4
5th even number = 2 × 5 = 10
23rd even number = 2 × 23 = 46
Why it happens: the even numbers are exactly the multiples of 2, and the nth multiple of 2 is 2 × n.
Q8.
What is the 100th odd number?
Answer

The 100th odd number is 199.

100th even number = 2 × 100 = 200
At every position the odd number is one less than the even number:
Even: 2, 4, 6, 8, 10, …
Odd:  1, 3, 5, 7,  9, …
100th odd number = 200 – 1 = 199
Why it happens: the two lists march together, position by position, and the odd list is always exactly 1 behind — so 2n and 2n – 1 give the nth even and odd numbers.
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