NCERT Solutions for Class 9th Maths Chapter 3 .4 Filling the Spaces: Fractions and Rational Numbers — Exercise Set 3.3

Book page 493 Updated on2026-09-08

Q1.

Prove that the following rational numbers are equal: (i) 2/3 and 4/6 (ii) 5/4 and 10/8 (iii) −3/5 and −6/10 (iv) 9/3 and 3

Answer

The chapter's test is: a/b = c/d exactly when ad = bc. Apply it to each pair.

Part	Cross products ad and bc	Equal?
(i) 2/3, 4/6	2 × 6 = 12 and 3 × 4 = 12	yes
(ii) 5/4, 10/8	5 × 8 = 40 and 4 × 10 = 40	yes
(iii) −3/5, −6/10	(−3) × 10 = −30 and 5 × (−6) = −30	yes
(iv) 9/3, 3/1	9 × 1 = 9 and 3 × 3 = 9	yes

The same four facts seen through equivalent fractions:

4/6 = (2 × 2)/(2 × 3) = 2/3
10/8 = (2 × 5)/(2 × 4) = 5/4
−6/10 = (2 × −3)/(2 × 5) = −3/5
9/3 = (3 × 3)/(3 × 1) = 3

Why it happens: ad = bc is just the cleared-denominator form of a/b = c/d — multiply both sides by bd and the fractions disappear. That is why the test needs no division and works even when the fractions are negative.

Tip: in (iv), a whole number is written as 3/1 before the test is applied. Every integer is a rational number with denominator 1.

Q2.

Find the sum: (i) 2/5 + 3/10 (ii) 7/12 + 5/8 (iii) −4/7 + 3/14

Answer

(i) LCM(5, 10) = 10
2/5 + 3/10 = 4/10 + 3/10 = 7/10
7 and 10 are co-prime → 7/10

(ii) 12 = 2² × 3, 8 = 2³ → LCM = 24
7/12 + 5/8 = 14/24 + 15/24 = 29/24 (= 1 5/24)

(iii) LCM(7, 14) = 14
−4/7 + 3/14 = −8/14 + 3/14 = −5/14
5 and 14 are co-prime → −5/14

Why it happens: the rule a/b + c/b = (a + c)/b only works when the parts are of equal size, so every sum begins by rewriting both fractions over a common denominator. After that, only the numerators are added — the denominator names the size of the part and does not change.

Check it yourself: in (iii) the answer must be negative, since 4/7 = 8/14 is larger than 3/14.

Q3.

Find the difference: (i) 5/6 − 1/4 (ii) 11/8 − 3/4 (iii) −7/9 − (−2/3)

Answer

(i) LCM(6, 4) = 12
5/6 − 1/4 = 10/12 − 3/12 = 7/12

(ii) LCM(8, 4) = 8
11/8 − 3/4 = 11/8 − 6/8 = 5/8

(iii) subtracting a negative is adding its opposite
−7/9 − (−2/3) = −7/9 + 2/3
= −7/9 + 6/9 = −1/9

Why it happens: part (iii) uses the integer rule of Section 3.3 inside a fraction. Since 2/3 = 6/9 is smaller than 7/9, adding it does not quite reach 0, so the answer stays negative — a useful check before you compute anything.

Q4.

Find the product: (i) 2/3 × 3/10 (ii) 7/11 × 5/8 (iii) −4/7 × 5/14

Answer

The rule is a/b × c/d = ac/bd. Cancelling common factors before multiplying keeps the numbers small.

(i) 2/3 × 3/10 = (2 × 3)/(3 × 10)
cancel 3, cancel 2 → 1/5
= 1/5

(ii) 7/11 × 5/8 = 35/88
88 = 2³ × 11, 35 = 5 × 7 — no common factor
= 35/88

(iii) −4/7 × 5/14 = −20/98
divide numerator and denominator by 2
= −10/49

Why it happens: a product of a positive and a negative rational is negative — the same debt-and-fortune rule as for integers, since the sign travels with the numerator.

Q5.

Find the quotient: (i) 2/3 ÷ 3/10 (ii) 7/11 ÷ 5/8 (iii) −4/7 ÷ 5/14

Answer

Dividing by c/d means multiplying by its reciprocal d/c, which needs c ≠ 0.

(i) 2/3 ÷ 3/10 = 2/3 × 10/3 = 20/9

(ii) 7/11 ÷ 5/8 = 7/11 × 8/5 = 56/55

(iii) −4/7 ÷ 5/14 = −4/7 × 14/5
= −56/35, divide both by 7
= −8/5

Why it happens: the reciprocal of c/d is the number that multiplies it to 1, since (c/d) × (d/c) = cd/dc = 1. So dividing by c/d and multiplying by d/c must give the same result. Note in (i) and (ii) the quotient is larger than the first number — dividing by a fraction less than 1 makes things bigger.

Q6.

Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.

Answer

Left side
1/2 + 3/4 = 2/4 + 3/4 = 5/4
(5/4) × (8/3) = 40/12 = 10/3

Right side
1/2 × 8/3 = 8/6 = 4/3
3/4 × 8/3 = 24/12 = 2 = 6/3
4/3 + 6/3 = 10/3

Both sides equal 10/3, so the statement is proved.

Why it happens: this is the distributive law (p + q)r = pr + qr with p = 1/2, q = 3/4, r = 8/3. It is not a coincidence about these particular fractions — the law holds for every triple of rational numbers, and the equality would still be true if 8/3 were replaced by any rational number at all.

Q7.

Simplify the following using the distributive property: 7/9 (6/7 − 3/4).

Answer

Distribute 7/9 across the bracket:

7/9 × 6/7 − 7/9 × 3/4
= 42/63 − 21/36
= 2/3 − 7/12
= 8/12 − 7/12
= 1/12

Doing the bracket first must give the same answer — a useful check:

6/7 − 3/4 = 24/28 − 21/28 = 3/28
7/9 × 3/28 = 21/252 = 1/12

Why it happens: distributivity is stated in the chapter for addition, p(q + r) = pq + pr; it extends to subtraction because q − r means q + (−r). Here the first route is lighter, since 7/9 × 6/7 cancels the 7 immediately.

Q8.

Find the rational number x such that: 5/6 (x + 3/5) = 5/6 x + 1/2.

Answer

Expand the left side using the distributive law:

5/6 (x + 3/5) = 5/6 · x + 5/6 · 3/5
5/6 × 3/5 = 15/30 = 1/2
so the left side is 5x/6 + 1/2

5x/6 + 1/2 = 5x/6 + 1/2
subtract 5x/6 from both sides → 1/2 = 1/2

The equation reduces to a true statement with no x left in it. So it is an identity: every rational number x satisfies it.

Why it happens: the equation is nothing but the distributive law itself, written out for p = 5/6, q = x, r = 3/5, with the product 5/6 × 3/5 already simplified to 1/2. A law that holds for all rationals cannot single out one value of x. Had the right side been 5x/6 + 1/3, say, we would have reached 1/2 = 1/3, a false statement, and then no x would work.

Check it yourself: try x = 2. Left side: 5/6 (2 + 3/5) = 5/6 × 13/5 = 13/6. Right side: 5/6 × 2 + 1/2 = 5/3 + 1/2 = 10/6 + 3/6 = 13/6. Equal — as it must be for any x.

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