Mathematics solution NCERT
Class 9 - Chapter 3: The World of Numbers
Question 1. Prove that the following rational numbers are equal:
(i) $$\frac{2}{3}\text{ and }\frac{4}{6}$$
Solution:
Multiply the numerator and denominator of $$\frac{2}{3}$$ by 2.
$$ \begin{aligned} \frac{2}{3} &=\frac{2\times2}{3\times2}\\ &=\frac{4}{6} \end{aligned} $$Since both rational numbers have the same value, they are equal.
Answer:
$$ \boxed{\frac{2}{3}=\frac{4}{6}} $$(ii) $$\frac{5}{4}\text{ and }\frac{10}{8}$$
Solution:
Multiply the numerator and denominator of $$\frac{5}{4}$$ by 2.
$$ \begin{aligned} \frac{5}{4} &=\frac{5\times2}{4\times2}\\ &=\frac{10}{8} \end{aligned} $$Hence, both rational numbers are equal.
Answer:
$$ \boxed{\frac{5}{4}=\frac{10}{8}} $$(iii) $$-\frac{3}{5}\text{ and }-\frac{6}{10}$$
Solution:
Multiply the numerator and denominator of $$-\frac{3}{5}$$ by 2.
$$ \begin{aligned} -\frac{3}{5} &=\frac{-3\times2}{5\times2}\\ &=-\frac{6}{10} \end{aligned} $$Hence, both rational numbers are equal.
Answer:
$$ \boxed{-\frac{3}{5}=-\frac{6}{10}} $$(iv) $$\frac{9}{3}\text{ and }3$$
Solution:
$$ \begin{aligned} \frac{9}{3} &=3 \end{aligned} $$Therefore, both numbers are equal.
Answer:
$$ \boxed{\frac{9}{3}=3} $$Question 2. Find the sum:
(i) $$\frac{2}{5}+\frac{3}{10}$$
Solution:
LCM of 5 and 10 is 10.
$$ \frac{2}{5} = \frac{2\times2}{5\times2} = \frac{4}{10} $$Now, add the fractions.
$$ \begin{aligned} \frac{2}{5}+\frac{3}{10} &=\frac{4}{10}+\frac{3}{10}\\ &=\frac{4+3}{10}\\ &=\frac{7}{10} \end{aligned} $$Answer:
$$ \boxed{\frac{7}{10}} $$(ii) $$\frac{7}{12}+\frac{5}{8}$$
Solution:
LCM of 12 and 8 is 24.
$$ \frac{7}{12} = \frac{7\times2}{12\times2} = \frac{14}{24} $$ $$ \frac{5}{8} = \frac{5\times3}{8\times3} = \frac{15}{24} $$Now, add the fractions.
$$ \begin{aligned} \frac{7}{12}+\frac{5}{8} &=\frac{14}{24}+\frac{15}{24}\\ &=\frac{14+15}{24}\\ &=\frac{29}{24}\\ &=1\frac{5}{24} \end{aligned} $$Answer:
$$ \boxed{\frac{29}{24}} $$or
$$ \boxed{1\frac{5}{24}} $$(iii) $$-\frac{4}{7}+\frac{3}{14}$$
Solution:
LCM of 7 and 14 is 14.
$$ -\frac{4}{7} = -\frac{4\times2}{7\times2} = -\frac{8}{14} $$Now, add the fractions.
$$ \begin{aligned} -\frac{4}{7}+\frac{3}{14} &=-\frac{8}{14}+\frac{3}{14}\\ &=\frac{-8+3}{14}\\ &=-\frac{5}{14} \end{aligned} $$Answer:
$$ \boxed{-\frac{5}{14}} $$Question 3. Find the difference:
(i) $$\frac{5}{6}-\frac{1}{4}$$
Solution:
LCM of 6 and 4 is 12.
$$ \frac{5}{6} = \frac{5\times2}{6\times2} = \frac{10}{12} $$ $$ \frac{1}{4} = \frac{1\times3}{4\times3} = \frac{3}{12} $$Now, subtract the fractions.
$$ \begin{aligned} \frac{5}{6}-\frac{1}{4} &=\frac{10}{12}-\frac{3}{12}\\ &=\frac{10-3}{12}\\ &=\frac{7}{12} \end{aligned} $$Answer:
$$ \boxed{\frac{7}{12}} $$(ii) $$\frac{11}{8}-\frac{3}{4}$$
Solution:
LCM of 8 and 4 is 8.
$$ \frac{3}{4} = \frac{3\times2}{4\times2} = \frac{6}{8} $$Now, subtract the fractions.
$$ \begin{aligned} \frac{11}{8}-\frac{3}{4} &=\frac{11}{8}-\frac{6}{8}\\ &=\frac{11-6}{8}\\ &=\frac{5}{8} \end{aligned} $$Answer:
$$ \boxed{\frac{5}{8}} $$(iii) $$-\frac{7}{9}-\left(-\frac{2}{3}\right)$$
Solution:
Subtracting a negative number is the same as adding its positive value.
$$ -\frac{7}{9}-\left(-\frac{2}{3}\right) = -\frac{7}{9}+\frac{2}{3} $$LCM of 9 and 3 is 9.
$$ \frac{2}{3} = \frac{2\times3}{3\times3} = \frac{6}{9} $$Now, add the fractions.
$$ \begin{aligned} -\frac{7}{9}+\frac{6}{9} &=\frac{-7+6}{9}\\ &=-\frac{1}{9} \end{aligned} $$Answer:
$$ \boxed{-\frac{1}{9}} $$Question 4. Find the product:
(i) $$\frac{2}{3}\times\frac{3}{10}$$
Solution:
Multiply the numerators and the denominators.
$$ \begin{aligned} \frac{2}{3}\times\frac{3}{10} &=\frac{2\times3}{3\times10}\\ &=\frac{6}{30} \end{aligned} $$Simplify the fraction by dividing the numerator and denominator by 6.
$$ \begin{aligned} \frac{6}{30} &=\frac{1}{5} \end{aligned} $$Answer:
$$ \boxed{\frac{1}{5}} $$(ii) $$\frac{7}{11}\times\frac{5}{8}$$
Solution:
Multiply the numerators and the denominators.
$$ \begin{aligned} \frac{7}{11}\times\frac{5}{8} &=\frac{7\times5}{11\times8}\\ &=\frac{35}{88} \end{aligned} $$Since 35 and 88 have no common factor other than 1, the fraction is already in its simplest form.
Answer:
$$ \boxed{\frac{35}{88}} $$(iii) $$-\frac{4}{7}\times\frac{5}{14}$$
Solution:
Multiply the numerators and the denominators.
$$ \begin{aligned} -\frac{4}{7}\times\frac{5}{14} &=\frac{-4\times5}{7\times14}\\ &=\frac{-20}{98} \end{aligned} $$Simplify the fraction by dividing the numerator and denominator by 2.
$$ \begin{aligned} \frac{-20}{98} &=-\frac{10}{49} \end{aligned} $$Answer:
$$ \boxed{-\frac{10}{49}} $$Question 5. Find the quotient:
(i) $$\frac{2}{3}\div\frac{3}{10}$$
Solution:
To divide two rational numbers, multiply the first rational number by the reciprocal of the second rational number.
$$ \begin{aligned} \frac{2}{3}\div\frac{3}{10} &=\frac{2}{3}\times\frac{10}{3}\\ &=\frac{2\times10}{3\times3}\\ &=\frac{20}{9} \end{aligned} $$Answer:
$$ \boxed{\frac{20}{9}} $$(ii) $$\frac{7}{11}\div\frac{5}{8}$$
Solution:
Multiply the first rational number by the reciprocal of the second rational number.
$$ \begin{aligned} \frac{7}{11}\div\frac{5}{8} &=\frac{7}{11}\times\frac{8}{5}\\ &=\frac{7\times8}{11\times5}\\ &=\frac{56}{55} \end{aligned} $$Answer:
$$ \boxed{\frac{56}{55}} $$(iii) $$-\frac{4}{7}\div\frac{5}{14}$$
Solution:
Multiply the first rational number by the reciprocal of the second rational number.
$$ \begin{aligned} -\frac{4}{7}\div\frac{5}{14} &=-\frac{4}{7}\times\frac{14}{5}\\ &=\frac{-4\times14}{7\times5}\\ &=\frac{-56}{35}\\ &=-\frac{8}{5} \end{aligned} $$Answer:
$$ \boxed{-\frac{8}{5}} $$Question 6. Show that:
$$ \left(\frac{1}{2}+\frac{3}{4}\right)\times\frac{8}{3} = \frac{1}{2}\times\frac{8}{3} + \frac{3}{4}\times\frac{8}{3} $$Solution:
Left-hand side (LHS):
$$ \begin{aligned} \left(\frac{1}{2}+\frac{3}{4}\right)\times\frac{8}{3} &=\left(\frac{2}{4}+\frac{3}{4}\right)\times\frac{8}{3}\\ &=\frac{5}{4}\times\frac{8}{3}\\ &=\frac{40}{12}\\ &=\frac{10}{3} \end{aligned} $$Right-hand side (RHS):
$$ \begin{aligned} \frac{1}{2}\times\frac{8}{3} + \frac{3}{4}\times\frac{8}{3} &=\frac{8}{6}+\frac{24}{12}\\ &=\frac{4}{3}+2\\ &=\frac{4}{3}+\frac{6}{3}\\ &=\frac{10}{3} \end{aligned} $$Since,
$$ \boxed{\text{LHS}=\text{RHS}=\frac{10}{3}} $$Hence, proved.
Question 7. Simplify the following using the distributive property:
$$\frac{7}{9}\left(\frac{6}{7}-\frac{3}{4}\right)$$
Solution:
Using the distributive property,
$$ \begin{aligned} \frac{7}{9}\left(\frac{6}{7}-\frac{3}{4}\right) &=\frac{7}{9}\times\frac{6}{7} - \frac{7}{9}\times\frac{3}{4} \end{aligned} $$ $$ \begin{aligned} &=\frac{42}{63}-\frac{21}{36}\\ &=\frac{2}{3}-\frac{7}{12} \end{aligned} $$Taking LCM of 3 and 12, we get 12.
$$ \begin{aligned} &=\frac{8}{12}-\frac{7}{12}\\ &=\frac{1}{12} \end{aligned} $$Answer:
$$ \boxed{\frac{1}{12}} $$Question 8. Find the rational number $$x$$ such that:
$$ \frac{5}{6}\left(x+\frac{3}{5}\right)=\frac{5}{6}x+\frac{1}{2} $$Solution:
Expand the left-hand side using the distributive property.
$$ \begin{aligned} \frac{5}{6}x+\frac{5}{6}\times\frac{3}{5} = \frac{5}{6}x+\frac{1}{2} \end{aligned} $$ $$ \begin{aligned} \frac{5}{6}x+\frac{15}{30} = \frac{5}{6}x+\frac{1}{2} \end{aligned} $$ $$ \begin{aligned} \frac{5}{6}x+\frac{1}{2} = \frac{5}{6}x+\frac{1}{2} \end{aligned} $$Both sides are identical for every value of $$x$$.
Answer:
$$ \boxed{\text{Any rational number }x\text{ satisfies the given equation.}} $$