NCERT Solutions Class 9 Maths Exercise 3.3 — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers, Exercise Set 3.3. All 8 questions solved.

With zero and the negatives in place, the number system is closed under subtraction but still not under division — $2 \div 3$ is no integer. The rational numbers repair that: every number of the form $\frac{p}{q}$ with $p, q$ integers and $q \neq 0$.

This exercise is the arithmetic drill for them, and four rules cover all of it:

$$\frac{a}{b} = \frac{c}{d} \iff ad = bc \qquad\qquad \frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$$

$$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \qquad\qquad \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$$

Key insight. Multiplication and division need no common denominator; addition and subtraction do. That single distinction explains why Questions 4 and 5 are so much shorter than Questions 2 and 3 — and why the commonest error in the whole exercise is putting the denominators over a common value before multiplying, where it is simply wasted work.

Question 1

Prove that the following rational numbers are equal:

Solution. Two fractions are equal exactly when their cross products agree: $\frac{a}{b} = \frac{c}{d}$ precisely when $ad = bc$. This avoids any division and works whatever the signs.

(i) $\dfrac{2}{3}$ and $\dfrac{4}{6}$

$$2 \times 6 = 12 \qquad 3 \times 4 = 12$$

Equal. (Equivalently, $\frac{4}{6} = \frac{2 \times 2}{2 \times 3} = \frac{2}{3}$ after cancelling the common factor $2$.)

(ii) $\dfrac{5}{4}$ and $\dfrac{10}{8}$

$$5 \times 8 = 40 \qquad 4 \times 10 = 40$$

Equal.

(iii) $-\dfrac{3}{5}$ and $-\dfrac{6}{10}$

$$(-3) \times 10 = -30 \qquad 5 \times (-6) = -30$$

Equal. The negative sign belongs to the whole fraction and can sit with the numerator throughout.

(iv) $\dfrac{9}{3}$ and $3$

Write the integer as a fraction, $3 = \frac{3}{1}$, and cross-multiply:

$$9 \times 1 = 9 \qquad 3 \times 3 = 9$$

Equal. This is the fact that every integer is also a rational number.

Question 2

Find the sum:

Solution. Addition needs a common denominator. Use the lowest common multiple rather than the product of the denominators — it keeps the numbers small and often leaves the answer already in lowest terms.

(i) $\dfrac{2}{5} + \dfrac{3}{10}$

The LCM of $5$ and $10$ is $10$:

$$\frac{2}{5} = \frac{4}{10} \quad\Longrightarrow\quad \frac{4}{10} + \frac{3}{10} = \frac{7}{10}$$

(ii) $\dfrac{7}{12} + \dfrac{5}{8}$

The LCM of $12$ and $8$ is $24$ (not $96$):

$$\frac{7}{12} = \frac{14}{24}, \qquad \frac{5}{8} = \frac{15}{24}$$ $$\frac{14}{24} + \frac{15}{24} = \frac{29}{24}$$

$29$ is prime and does not divide $24$, so this is already in lowest terms. As a mixed number it is $1\frac{5}{24}$.

(iii) $-\dfrac{4}{7} + \dfrac{3}{14}$

The LCM of $7$ and $14$ is $14$:

$$-\frac{4}{7} = -\frac{8}{14} \quad\Longrightarrow\quad -\frac{8}{14} + \frac{3}{14} = \frac{-8 + 3}{14} = -\frac{5}{14}$$

The negative wins because $\frac{8}{14}$ is the larger of the two in size.

Question 3

Find the difference:

Solution. Same method as addition, with the sign of the second fraction reversed.

(i) $\dfrac{5}{6} – \dfrac{1}{4}$

LCM of $6$ and $4$ is $12$:

$$\frac{5}{6} = \frac{10}{12}, \qquad \frac{1}{4} = \frac{3}{12}$$ $$\frac{10}{12} – \frac{3}{12} = \frac{7}{12}$$

(ii) $\dfrac{11}{8} – \dfrac{3}{4}$

LCM of $8$ and $4$ is $8$:

$$\frac{3}{4} = \frac{6}{8} \quad\Longrightarrow\quad \frac{11}{8} – \frac{6}{8} = \frac{5}{8}$$

(iii) $-\dfrac{7}{9} – \left(-\dfrac{2}{3}\right)$

Deal with the double sign first — subtracting a negative is adding a positive, the rule from Exercise Set 3.2:

$$-\frac{7}{9} – \left(-\frac{2}{3}\right) = -\frac{7}{9} + \frac{2}{3}$$

LCM of $9$ and $3$ is $9$:

$$-\frac{7}{9} + \frac{6}{9} = \frac{-7 + 6}{9} = -\frac{1}{9}$$

Question 4

Find the product:

Solution. Multiply numerators together and denominators together. No common denominator is needed. Cancelling common factors before multiplying keeps the numbers small.

(i) $\dfrac{2}{3} \times \dfrac{3}{10}$

The $3$ in the first denominator cancels the $3$ in the second numerator:

$$\frac{2}{\cancel{3}} \times \frac{\cancel{3}}{10} = \frac{2}{10} = \frac{1}{5}$$

(ii) $\dfrac{7}{11} \times \dfrac{5}{8}$

Nothing cancels — $7$, $11$, $5$ and $8$ share no factors:

$$\frac{7 \times 5}{11 \times 8} = \frac{35}{88}$$

(iii) $-\dfrac{4}{7} \times \dfrac{5}{14}$

A negative times a positive is negative:

$$-\frac{4 \times 5}{7 \times 14} = -\frac{20}{98}$$

Both parts are even, so divide by $2$:

$$-\frac{20}{98} = -\frac{10}{49}$$

Question 5

Find the quotient:

Solution. Dividing by a fraction is multiplying by its reciprocal — turn the second fraction upside down and multiply.

(i) $\dfrac{2}{3} \div \dfrac{3}{10}$

$$\frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$$

(ii) $\dfrac{7}{11} \div \dfrac{5}{8}$

$$\frac{7}{11} \times \frac{8}{5} = \frac{56}{55}$$

(iii) $-\dfrac{4}{7} \div \dfrac{5}{14}$

$$-\frac{4}{7} \times \frac{14}{5}$$

The $7$ cancels into the $14$, leaving $2$:

$$-\frac{4 \times 2}{5} = -\frac{8}{5}$$

Notice that dividing by $\frac{5}{14}$, a number smaller than $1$, made the answer larger in size than $\frac{4}{7}$ — which is what dividing by a fraction less than one always does.

Question 6

Show that: $\left( \dfrac{1}{2} + \dfrac{3}{4} \right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$.

Solution. This is the distributive law for rational numbers. “Show that” means work out both sides independently and compare — do not assume the law and use it.

Left-hand side. Add inside the bracket first:

$$\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$$ $$\frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}$$

Right-hand side. Two products, then a sum:

$$\frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3}, \qquad \frac{3}{4} \times \frac{8}{3} = \frac{24}{12} = 2$$ $$\frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$$

Both sides equal $\frac{10}{3}$, so the statement holds.

Question 7

Simplify the following using the distributive property: $\dfrac{7}{9}\left( \dfrac{6}{7} – \dfrac{3}{4} \right)$.

Solution. The question names the method, so distribute the $\frac{7}{9}$ across the bracket:

$$\frac{7}{9} \times \frac{6}{7} \;-\; \frac{7}{9} \times \frac{3}{4}$$

First product. The $7$s cancel:

$$\frac{\cancel{7}}{9} \times \frac{6}{\cancel{7}} = \frac{6}{9} = \frac{2}{3}$$

Second product.

$$\frac{7}{9} \times \frac{3}{4} = \frac{21}{36} = \frac{7}{12}$$

Subtract, over the LCM of $3$ and $12$, which is $12$:

$$\frac{2}{3} – \frac{7}{12} = \frac{8}{12} – \frac{7}{12} = \frac{1}{12}$$

Check by the direct route — bracket first:

$$\frac{6}{7} – \frac{3}{4} = \frac{24}{28} – \frac{21}{28} = \frac{3}{28}, \qquad \frac{7}{9} \times \frac{3}{28} = \frac{21}{252} = \frac{1}{12} \; \checkmark$$

Question 8

Find the rational number $x$ such that: $\dfrac{5}{6}\left( x + \dfrac{3}{5} \right) = \dfrac{5}{6}x + \dfrac{1}{2}$.

Solution. Begin by expanding the left-hand side, as with any equation of this shape:

$$\frac{5}{6}\left( x + \frac{3}{5} \right) = \frac{5}{6}x + \frac{5}{6} \times \frac{3}{5}$$

Work out that last product. The $5$s cancel:

$$\frac{5}{6} \times \frac{3}{5} = \frac{3}{6} = \frac{1}{2}$$

So the left-hand side is

$$\frac{5}{6}x + \frac{1}{2}$$

which is exactly the right-hand side. Subtracting one from the other:

$$\left( \frac{5}{6}x + \frac{1}{2} \right) – \left( \frac{5}{6}x + \frac{1}{2} \right) = 0$$

The $x$ terms cancel and so do the constants, leaving $0 = 0$ — a statement that is true whatever $x$ happens to be.

So this is an identity, not an equation with one solution. Every rational number satisfies it. Try $x = 0$: both sides give $\frac{1}{2}$. Try $x = 4$: the left is $\frac{5}{6}\left(\frac{23}{5}\right) = \frac{23}{6}$ and the right is $\frac{20}{6} + \frac{1}{2} = \frac{23}{6}$. Try $x = -\frac{3}{5}$: both sides give $0$.

This is not a trick or a misprint. The equation is simply the distributive law of Question 6 written with a letter in it, and a law holds for every value — which is what makes it a law. The question’s phrasing invites you to expect a single number, and the honest answer is that no single number is picked out.

Common mistakes

• Question 1, dividing instead of cross-multiplying. $\frac{2}{3} = 0.666\ldots$ and $\frac{4}{6} = 0.666\ldots$ only looks like a proof, because the decimals never end and have to be cut off. Cross-multiplication is exact.
• Question 2(ii), using $96$ as the common denominator. Multiplying the denominators always works but gives $\frac{116}{96}$, which then needs cancelling back to $\frac{29}{24}$. The LCM saves that step.
• Question 2, adding the denominators. $\frac{2}{5} + \frac{3}{10} = \frac{5}{15}$ is the classic slip. Denominators are never added; they are made equal first.
• Question 3(iii), dropping one minus sign. $-\frac{7}{9} – \left(-\frac{2}{3}\right)$ becomes $-\frac{7}{9} + \frac{2}{3}$, not $-\frac{7}{9} – \frac{2}{3}$ — which would give $-\frac{13}{9}$.
• Question 4, finding a common denominator first. Multiplication does not need one. Doing it anyway is not wrong, just three times the work, and it hides the cancellation that makes part (i) a one-line answer.
• Question 5, inverting the wrong fraction. $\frac{2}{3} \div \frac{3}{10}$ is $\frac{2}{3} \times \frac{10}{3}$, not $\frac{3}{2} \times \frac{3}{10}$. Only the divisor is turned over.
• Question 5, expecting division to make things smaller. Dividing by a fraction less than $1$ makes the result bigger, as part (iii) shows.
• Question 6, assuming the distributive law to prove it. “Show that” requires both sides to be evaluated separately and then compared; using the law to rewrite one side into the other proves nothing.
• Question 8, forcing a single answer. Reaching $0 = 0$ and then writing “$x = 0$” is the natural mistake here. $0 = 0$ means every value works, not that the value is zero — and the difference matters, since $x = 0$ is only one of infinitely many solutions.

Practise next

• Exercise Set 3.4 — the same rational numbers placed on the number line, where the arithmetic here turns into position and distance.
• The End-of-Chapter Exercises, Questions 6, 7 and 8 — more rational arithmetic, with Question 8 a genuine equation to solve in contrast to the identity in Question 8 here.