NCERT Solutions · Class 9 Mathematics The World of Numbers

Cross-multiplication test.Two rational numbers \(\displaystyle \frac{a}{b}\) and \(\displaystyle \frac{c}{d}\) (with \(\displaystyle b, d \neq 0\)) are equal exactly when\[a \times d = b \times c. \]Why this works: multiplying both fractions by \(\displaystyle bd\) turns \(\displaystyle \frac{a}{b}\) into \(\displaystyle ad\) and \(\displaystyle \frac{c}{d}\) into \(\displaystyle bc\), and multiplying both sides of an equality by the same non-zero number cannot change whether it is true. Equivalently, multiplying the numerator and denominator of a fraction by the same non-zero number does not change its value.(i) \(\displaystyle \frac{2}{3}\) and \(\displaystyle \frac{4}{6}\). \[2 \times 6 = 12, \qquad 3 \times 4 = 12. \] Equal. Directly: \(\displaystyle \frac{4}{6} = \frac{2 \times 2}{2 \times 3} = \frac{2}{3}\).(ii) \(\displaystyle \frac{5}{4}\) and \(\displaystyle \frac{10}{8}\). \[5 \times 8 = 40, \qquad 4 \times 10 = 40. \] Equal. Directly: \(\displaystyle \frac{10}{8} = \frac{2 \times 5}{2 \times 4} = \frac{5}{4}\).(iii) \(\displaystyle -\frac{3}{5}\) and \(\displaystyle -\frac{6}{10}\). \[(-3) \times 10 = -30, \qquad 5 \times (-6) = -30. \] Equal. Directly: \(\displaystyle -\frac{6}{10} = -\frac{2 \times 3}{2 \times 5} = -\frac{3}{5}\).(iv) \(\displaystyle \frac{9}{3}\) and \(\displaystyle 3\). Write the whole number as a fraction, \(\displaystyle 3 = \frac{3}{1}\). Then \[9 \times 1 = 9, \qquad 3 \times 3 = 9. \] Equal. Directly: \(\displaystyle \frac{9}{3} = \frac{3 \times 3}{3 \times 1} = \frac{3}{1} = 3\).In all four cases the cross products match, so each pair names the same rational number.

Common denominator, then add the numerators.Fractions can only be added once the pieces are the same size, so first rewrite each pair with the LCM of the denominators.(i) \(\displaystyle \frac{2}{5} + \frac{3}{10}\). LCM of \(\displaystyle 5\) and \(\displaystyle 10\) is \(\displaystyle 10\). Multiply the first fraction above and below by \(\displaystyle 2\): \[\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}. \] \[\frac{4}{10} + \frac{3}{10} = \frac{4 + 3}{10} = \frac{7}{10}. \] \(\displaystyle 7\) and \(\displaystyle 10\) share no factor, so this is in lowest terms.(ii) \(\displaystyle \frac{7}{12} + \frac{5}{8}\). \(\displaystyle 12 = 2^2 \times 3\) and \(\displaystyle 8 = 2^3\), so the LCM is \(\displaystyle 2^3 \times 3 = 24\). \[\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}, \qquad \frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}. \] \[\frac{14}{24} + \frac{15}{24} = \frac{29}{24}. \] \(\displaystyle 29\) is prime, so no cancelling is possible. (As a mixed number, \(\displaystyle 1\frac{5}{24}\) — sensible, since \(\displaystyle \frac{7}{12}\) and \(\displaystyle \frac{5}{8}\) are each a little over a half.)(iii) \(\displaystyle -\frac{4}{7} + \frac{3}{14}\). LCM of \(\displaystyle 7\) and \(\displaystyle 14\) is \(\displaystyle 14\). \[-\frac{4}{7} = -\frac{4 \times 2}{7 \times 2} = -\frac{8}{14}. \] \[-\frac{8}{14} + \frac{3}{14} = \frac{-8 + 3}{14} = \frac{-5}{14} = -\frac{5}{14}. \] The answer is negative because the negative part is larger in size, which matches the estimate \(\displaystyle -0.571 + 0.214 \approx -0.357\).(i) \(\displaystyle \dfrac{7}{10}\) (ii) \(\displaystyle \dfrac{29}{24}\) (iii) \(\displaystyle -\dfrac{5}{14}\).

Common denominator, then subtract the numerators.(i) \(\displaystyle \frac{5}{6} - \frac{1}{4}\). \(\displaystyle 6 = 2 \times 3\) and \(\displaystyle 4 = 2^2\), so the LCM is \(\displaystyle 2^2 \times 3 = 12\). \[\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}, \qquad \frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}. \] \[\frac{10}{12} - \frac{3}{12} = \frac{7}{12}. \] Check by adding back: \(\displaystyle \frac{7}{12} + \frac{3}{12} = \frac{10}{12} = \frac{5}{6}\). Correct.(ii) \(\displaystyle \frac{11}{8} - \frac{3}{4}\). LCM of \(\displaystyle 8\) and \(\displaystyle 4\) is \(\displaystyle 8\). \[\frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}. \] \[\frac{11}{8} - \frac{6}{8} = \frac{5}{8}. \](iii) \(\displaystyle -\frac{7}{9} - \left(-\frac{2}{3}\right)\). Subtracting a negative is adding the positive, so \[-\frac{7}{9} - \left(-\frac{2}{3}\right) = -\frac{7}{9} + \frac{2}{3}. \] LCM of \(\displaystyle 9\) and \(\displaystyle 3\) is \(\displaystyle 9\), and \(\displaystyle \frac{2}{3} = \frac{6}{9}\): \[-\frac{7}{9} + \frac{6}{9} = \frac{-7 + 6}{9} = -\frac{1}{9}. \] Sense check: \(\displaystyle -\frac{7}{9} \approx -0.778\) and \(\displaystyle -\frac{2}{3} \approx -0.667\); the first is the smaller (further left), so the difference should be a small negative number, and \(\displaystyle -\frac{1}{9} \approx -0.111\) fits.(i) \(\displaystyle \dfrac{7}{12}\) (ii) \(\displaystyle \dfrac{5}{8}\) (iii) \(\displaystyle -\dfrac{1}{9}\).

Multiply numerators together and denominators together, cancelling common factors first.For rational numbers, \(\displaystyle \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}\). No common denominator is needed — that is only for addition.(i) \(\displaystyle \frac{2}{3} \times \frac{3}{10}\). Cancel the \(\displaystyle 3\) top and bottom, and cancel a factor \(\displaystyle 2\) between \(\displaystyle 2\) and \(\displaystyle 10\): \[\frac{2}{3} \times \frac{3}{10} = \frac{2 \times 3}{3 \times 10} = \frac{6}{30} = \frac{1}{5}. \](ii) \(\displaystyle \frac{7}{11} \times \frac{5}{8}\). \[\frac{7 \times 5}{11 \times 8} = \frac{35}{88}. \] \(\displaystyle 35 = 5 \times 7\) and \(\displaystyle 88 = 2^3 \times 11\) share no factor, so this is already in lowest terms.(iii) \(\displaystyle -\frac{4}{7} \times \frac{5}{14}\). A negative times a positive is negative. Multiply the sizes: \[\frac{4 \times 5}{7 \times 14} = \frac{20}{98}, \] and \(\displaystyle 20\) and \(\displaystyle 98\) are both even, so divide top and bottom by \(\displaystyle 2\): \[\frac{20}{98} = \frac{10}{49}. \] Restoring the sign: \[-\frac{4}{7} \times \frac{5}{14} = -\frac{10}{49}. \](i) \(\displaystyle \dfrac{1}{5}\) (ii) \(\displaystyle \dfrac{35}{88}\) (iii) \(\displaystyle -\dfrac{10}{49}\).

Dividing by a fraction means multiplying by its reciprocal.\[\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} \quad (c \neq 0). \]Why: \(\displaystyle \frac{c}{d} \times \frac{d}{c} = 1\), so multiplying by \(\displaystyle \frac{d}{c}\) undoes multiplication by \(\displaystyle \frac{c}{d}\) — exactly what dividing is supposed to do.(i) \(\displaystyle \frac{2}{3} \div \frac{3}{10}\). \[\frac{2}{3} \times \frac{10}{3} = \frac{20}{9}. \] Check: \(\displaystyle \frac{20}{9} \times \frac{3}{10} = \frac{60}{90} = \frac{2}{3}\). Correct.(ii) \(\displaystyle \frac{7}{11} \div \frac{5}{8}\). \[\frac{7}{11} \times \frac{8}{5} = \frac{56}{55}. \] \(\displaystyle 56 = 2^3 \times 7\) and \(\displaystyle 55 = 5 \times 11\) share no factor. (The answer is just over \(\displaystyle 1\), which is right: \(\displaystyle \frac{7}{11} \approx 0.636\) is slightly bigger than \(\displaystyle \frac{5}{8} = 0.625\).)(iii) \(\displaystyle -\frac{4}{7} \div \frac{5}{14}\). \[-\frac{4}{7} \times \frac{14}{5} = -\frac{4 \times 14}{7 \times 5} = -\frac{56}{35}. \] Both \(\displaystyle 56\) and \(\displaystyle 35\) are divisible by \(\displaystyle 7\): \[-\frac{56}{35} = -\frac{8}{5}. \] Check: \(\displaystyle -\frac{8}{5} \times \frac{5}{14} = -\frac{40}{70} = -\frac{4}{7}\). Correct.(i) \(\displaystyle \dfrac{20}{9}\) (ii) \(\displaystyle \dfrac{56}{55}\) (iii) \(\displaystyle -\dfrac{8}{5}\).

Distributive property — evaluate each side separately and compare.Left-hand side. Do the bracket first. LCM of \(\displaystyle 2\) and \(\displaystyle 4\) is \(\displaystyle 4\): \[\frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}. \] Then \[\frac{5}{4} \times \frac{8}{3} = \frac{5 \times 8}{4 \times 3} = \frac{40}{12} = \frac{10}{3}. \]Right-hand side. Work out the two products, then add. \[\frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3}, \qquad \frac{3}{4} \times \frac{8}{3} = \frac{24}{12} = 2. \] \[\frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}. \]Compare. \[\text{LHS} = \frac{10}{3} = \text{RHS}. \]What this shows. Multiplication distributes over addition for rational numbers, just as it does for whole numbers: \(\displaystyle a \times (b + c) = a \times b + a \times c\). Here \(\displaystyle a = \frac{8}{3}\), \(\displaystyle b = \frac{1}{2}\), \(\displaystyle c = \frac{3}{4}\), and the two sides came out equal — you may either add first and multiply once, or multiply twice and add. (In decimals both sides are \(\displaystyle 3.\overline{3}\), another quick confirmation.)Hence \(\displaystyle \left(\frac{1}{2}+\frac{3}{4}\right) \times \frac{8}{3}=\frac{1}{2} \times \frac{8}{3}+\frac{3}{4} \times \frac{8}{3} = \dfrac{10}{3}\), as required.

Distributive property: spread the \(\displaystyle \frac{7}{9}\) over both terms.\[\frac{7}{9}\left(\frac{6}{7}-\frac{3}{4}\right) = \frac{7}{9} \times \frac{6}{7} \;-\; \frac{7}{9} \times \frac{3}{4} \]First product. The \(\displaystyle 7\)s cancel: \[\frac{7}{9} \times \frac{6}{7} = \frac{6}{9} = \frac{2}{3}. \]Second product. \[\frac{7}{9} \times \frac{3}{4} = \frac{21}{36} = \frac{7}{12} \] (dividing top and bottom by \(\displaystyle 3\)).Subtract. LCM of \(\displaystyle 3\) and \(\displaystyle 12\) is \(\displaystyle 12\), and \(\displaystyle \frac{2}{3} = \frac{8}{12}\): \[\frac{8}{12} - \frac{7}{12} = \frac{1}{12}. \]Check the other way round (bracket first). \[\frac{6}{7} - \frac{3}{4} = \frac{24}{28} - \frac{21}{28} = \frac{3}{28}, \qquad \frac{7}{9} \times \frac{3}{28} = \frac{21}{252} = \frac{1}{12}. \] Both routes give the same value, which is the distributive property in action.\(\displaystyle \dfrac{1}{12}\).

Expand the bracket first, then compare the two sides.Apply the distributive property to the left-hand side: \[\frac{5}{6}\left(x+\frac{3}{5}\right) = \frac{5}{6}x + \frac{5}{6} \times \frac{3}{5}. \]Work out that constant term — the \(\displaystyle 5\)s cancel: \[\frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2}. \]So the left-hand side is \[\frac{5}{6}x + \frac{1}{2}, \] which is exactly the right-hand side, whatever \(\displaystyle x\) is.Subtracting \(\displaystyle \frac{5}{6}x + \frac{1}{2}\) from both sides of the given equation leaves \[0 = 0, \] a statement that is true for every value of \(\displaystyle x\) and contains no information about \(\displaystyle x\).Verification with a few values.

\(\displaystyle x\)	LHS \(\displaystyle =\frac{5}{6}\left(x+\frac{3}{5}\right)\)	RHS \(\displaystyle =\frac{5}{6}x+\frac{1}{2}\)
\(\displaystyle 0\)	\(\displaystyle \frac{5}{6}\times\frac{3}{5}=\frac{1}{2}\)	\(\displaystyle 0+\frac{1}{2}=\frac{1}{2}\)
\(\displaystyle 2\)	\(\displaystyle \frac{5}{6}\times\frac{13}{5}=\frac{13}{6}\)	\(\displaystyle \frac{10}{6}+\frac{3}{6}=\frac{13}{6}\)
\(\displaystyle -\frac{3}{5}\)	\(\displaystyle \frac{5}{6}\times 0 = 0\)	\(\displaystyle -\frac{1}{2}+\frac{1}{2}=0\)

Every value works.So there is no single value to "find". The equation is simply the distributive property written out for \(\displaystyle \frac{5}{6}\), \(\displaystyle x\) and \(\displaystyle \frac{3}{5}\), and it holds only because the constant works out exactly right: \(\displaystyle \frac{5}{6} \times \frac{3}{5} = \frac{1}{2}\). (Had the right-hand side read, say, \(\displaystyle \frac{5}{6}x + \frac{1}{3}\), the equation would have reduced to \(\displaystyle \frac{1}{2} = \frac{1}{3}\) and had no solution at all.)The equation is an identity, not a condition on \(\displaystyle x\): every rational number \(\displaystyle x\) satisfies it.