Anand Classes brings you a detailed explanation on the multiplication of rational numbers, an important topic for SSC exam preparation including SSC CGL, SSC CHSL, SSC MTS, and SSC GD. In this article, we cover rules, examples, solved problems, and FAQs to strengthen your concepts. Mastering multiplication of fractions and rational numbers will help you solve simplification questions quickly and accurately in competitive exams. Click the print button to download study material and notes.
Multiplication of Rational Numbers
The product of two given fractions is a fraction whose numerator is the product of the numerators of the given fractions and whose denominator is the product of the denominators of the given fractions.
In other words,
$$
\text{Product of two given fractions} = \frac{\text{Product of their numerators}}{\text{Product of their denominators}}
$$
This rule is also true for the product of rational numbers.
$$
\text{Product of two rational numbers} = \frac{\text{Product of their numerators}}{\text{Product of their denominators}}
$$
Thus, if $\frac{a}{b}$ and $\frac{c}{d}$ are any two rational numbers, then
$$
\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
$$
Solved Examples of Multiplication of Rational Numbers for SSC CGL CHSL CPO Exams
Example 1 : Multiply:
(i) $\frac{3}{4} \times \frac{5}{7}$
(ii) $\frac{7}{3} \times (-\frac{5}{4})$
(iii) $\frac{-9}{5} \times 4$
(iv) $\frac{-6}{9} \times (-\frac{29}{8})$
Solution
(i)
$$
\frac{3}{4} \times \frac{5}{7} = \frac{3 \times 5}{4 \times 7} = \frac{15}{28}
$$
(ii)
$$
\frac{7}{3} \times \left(-\frac{5}{4}\right) = \frac{7 \times -5}{3 \times 4} = \frac{-35}{12}
$$
(iii)
$$
\frac{-9}{5} \times 4 = \frac{-9 \times 4}{5 \times 1} = \frac{-36}{5}
$$
(iv)
$$
\frac{-6}{9} \times \left(-\frac{29}{8}\right) = \frac{(-6) \times (-29)}{9 \times 8} = \frac{174}{72} = \frac{29}{12}
$$
Example 2 : Simplify:
(i) $\frac{-8}{7} \times \frac{14}{5}$
(ii) $\frac{13}{6} \times \frac{-18}{91}$
(iii) $\frac{-5}{9} \times \frac{72}{-125}$
(iv) $\frac{-22}{9} \times \frac{-51}{88}$
Solution
(i)
$$
\frac{-8}{7} \times \frac{14}{5} = \frac{-8 \times 14}{7 \times 5} = \frac{-112}{35} = \frac{-16}{5}
$$
(ii)
$$
\frac{13}{6} \times \frac{-18}{91} = \frac{13 \times -18}{6 \times 91} = \frac{-234}{546} = \frac{-3}{7}
$$
(iii)
$$
\frac{-5}{9} \times \frac{72}{-125} = \frac{-5 \times 72}{9 \times -125} = \frac{-360}{-1125} = \frac{8}{25}
$$
(iv)
$$
\frac{-22}{9} \times \frac{-51}{88} = \frac{(-22) \times (-51)}{9 \times 88} = \frac{1122}{792} = \frac{17}{12}
$$
Example 3 : Simplify:
(i) $\frac{-16}{5} \times \frac{20}{8} – \left(\frac{-15}{5} \times \frac{-35}{3}\right)$
(ii) $\left(\frac{-3}{2} \times \frac{4}{5}\right) + \left(\frac{9}{5} \times \frac{-10}{3}\right) – \left(\frac{1}{2} \times \frac{3}{4}\right)$
(iii) $\frac{-7}{18} \times \frac{15}{-7} – \frac{1}{6} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4}$
Solution
(i)
$$
\frac{-16}{5} \times \frac{20}{8} – \left(\frac{-15}{5} \times \frac{-35}{3}\right)
= \frac{-320}{40} – \frac{525}{15}
= -8 – 35 = -43
$$
(ii)
$$
\left(\frac{-3}{2} \times \frac{4}{5}\right) + \left(\frac{9}{5} \times \frac{-10}{3}\right) – \left(\frac{1}{2} \times \frac{3}{4}\right)
= \frac{-12}{10} + \frac{-90}{15} – \frac{3}{8}
= \frac{-6}{5} – 6 – \frac{3}{8}
$$
Taking LCM of 40:
$$
= \frac{-48}{40} – \frac{240}{40} – \frac{15}{40} = \frac{-303}{40}
$$
(iii)
$$
\frac{-7}{18} \times \frac{15}{-7} – \frac{1}{6} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4}
= \frac{-105}{-126} – \frac{1}{24} + \frac{1}{8}
= \frac{5}{6} – \frac{1}{24} + \frac{1}{8}
$$
Taking LCM of 24:
$$
= \frac{20}{24} – \frac{1}{24} + \frac{3}{24} = \frac{22}{24} = \frac{11}{12}
$$
Important Questions Answers of Multiplication of Rational Numbers for SSC CGL CHSL CPO Exams
Question 1. Multiply the following Rational Numbers (Level-1)
(i) $\frac{7}{11} \times \frac{5}{4}$
Solution:
$$
\frac{7}{11} \times \frac{5}{4} = \frac{7 \times 5}{11 \times 4} = \frac{35}{44}
$$
(ii) $\frac{5}{7} \times \frac{-3}{4}$
Solution:
$$
\frac{5}{7} \times \frac{-3}{4} = \frac{5 \times -3}{7 \times 4} = \frac{-15}{28}
$$
(iii) $\frac{-2}{9} \times \frac{5}{11}$
Solution:
$$
\frac{-2}{9} \times \frac{5}{11} = \frac{-2 \times 5}{9 \times 11} = \frac{-10}{99}
$$
(iv) $\frac{-3}{17} \times \frac{5}{4}$
Solution:
$$
\frac{-3}{17} \times \frac{5}{4} = \frac{-3 \times 5}{17 \times 4} = \frac{-15}{68}
$$
(v) $\frac{9}{-7} \times \frac{36}{-11}$
Solution:
$$
\frac{9}{-7} \times \frac{36}{-11} = \frac{-9 \times -36}{7 \times 11} = \frac{324}{77}
$$
(vi) $\frac{-11}{13} \times \frac{-21}{7}$
Solution:
$$
\frac{-11}{13} \times \frac{-21}{7} = \frac{231}{91}
$$
(vii) $\frac{-3}{5} \times \frac{-4}{7}$
Solution:
$$
\frac{-3}{5} \times \frac{-4}{7} = \frac{12}{35}
$$
(viii) $\frac{-15}{11} \times 7$
Solution:
$$
\frac{-15}{11} \times \frac{7}{1} = \frac{-105}{11}
$$
Question 2. Multiply the following Rational Numbers (Level-2)
(i) $\frac{-5}{17} \times \frac{51}{-60}$
Solution:
$$
\frac{-5}{17} \times \frac{51}{-60} = \frac{-5 \times 51}{17 \times -60} = \frac{255}{1020} = \frac{1}{4}
$$
(ii) $\frac{-6}{11} \times \frac{-55}{36}$
Solution:
$$
\frac{-6}{11} \times \frac{-55}{36} = \frac{330}{396} = \frac{5}{6}
$$
(iii) $\frac{-8}{25} \times \frac{-5}{16}$
Solution:
$$
\frac{-8}{25} \times \frac{-5}{16} = \frac{40}{400} = \frac{1}{10}
$$
(iv) $\frac{6}{7} \times \frac{-49}{36}$
Solution:
$$
\frac{6}{7} \times \frac{-49}{36} = \frac{-294}{252} = \frac{-7}{6}
$$
(v) $\frac{8}{-9} \times \frac{-7}{-16}$
Solution:
$$
\frac{-8}{9} \times \frac{7}{16} = \frac{-56}{144} = \frac{-7}{18}
$$
(vi) $\frac{-8}{9} \times \frac{3}{64}$
Solution:
$$
\frac{-8}{9} \times \frac{3}{64} = \frac{-24}{576} = \frac{-1}{24}
$$
Question 3. Simplify each of the following and express the result as a rational number in
standard form. (Level-3)
(i) $\frac{-16}{21} \times \frac{14}{5}$
Solution:
$$
\frac{-16}{21} \times \frac{14}{5} = \frac{-224}{105} = \frac{-32}{15}
$$
(ii) $\frac{7}{6} \times \frac{-3}{28}$
Solution:
$$
\frac{7}{6} \times \frac{-3}{28} = \frac{-21}{168} = \frac{-1}{8}
$$
(iii) $\frac{-19}{36} \times 16$
Solution:
$$
\frac{-19}{36} \times \frac{16}{1} = \frac{-304}{36} = \frac{-76}{9}
$$
(iv) $\frac{-13}{9} \times \frac{27}{-26}$
Solution:
$$
\frac{-13}{9} \times \frac{27}{-26} = \frac{351}{234} = \frac{3}{2}
$$
(v) $\frac{-9}{16} \times \frac{-64}{-27}$
Solution:
$$
\frac{-9}{16} \times \frac{-64}{-27} = \frac{576}{-432} = \frac{-4}{3}
$$
(vi) $\frac{-50}{7} \times \frac{14}{3}$
Solution:
$$
\frac{-50}{7} \times \frac{14}{3} = \frac{-700}{21} = \frac{-100}{3}
$$
(vii) $\frac{-11}{9} \times \frac{-81}{-88}$
Solution:
$$
\frac{-11}{9} \times \frac{-81}{-88} = \frac{891}{-792} = \frac{-9}{8}
$$
(viii) $\frac{-5}{9} \times \frac{72}{-25}$
Solution:
$$
\frac{-5}{9} \times \frac{72}{-25} = \frac{360}{-225} = \frac{8}{5}
$$
Question 4. Simplify each of the following and express the result as a rational number in
standard form. (Level-4)
(i) $\left(\frac{25}{8} \times \frac{2}{5}\right) – \left(\frac{3}{5} \times \frac{-10}{9}\right)$
Solution:
$$
\frac{50}{40} – \frac{-30}{45} = \frac{5}{4} + \frac{2}{3} = \frac{23}{12}
$$
(ii) $\left(\frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{1}{2} \times 6\right)$
Solution:
$$
\frac{1}{8} + \frac{3}{1} = \frac{25}{8}
$$
(iii) $(-5 \times \frac{2}{15}) – (-6 \times \frac{2}{9})$
Solution:
$$
\frac{-10}{15} – \frac{-12}{9} = \frac{-2}{5} + \frac{4}{3} = \frac{2}{3}
$$
(iv) $\left(\frac{-9}{4} \times \frac{5}{3}\right) + \left(\frac{13}{2} \times \frac{5}{6}\right)$
Solution:
$$
\frac{-45}{12} + \frac{65}{12} = \frac{20}{12} = \frac{5}{3}
$$
(v) $\left(\frac{-4}{3} \times \frac{12}{-5}\right) + \left(\frac{3}{7} \times \frac{21}{15}\right)$
Solution:
$$
\frac{48}{15} + \frac{3}{5} = \frac{57}{15} = \frac{19}{5}
$$
(vi) $\left(\frac{13}{5} \times \frac{8}{3}\right) – \left(\frac{-5}{2} \times \frac{11}{3}\right)$
Solution:
$$
\frac{104}{15} – \frac{-55}{6} = \frac{483}{30}
$$
(vii) $\left(\frac{13}{7} \times \frac{11}{26}\right) – \left(\frac{-4}{3} \times \frac{5}{6}\right)$
Solution:
$$
\frac{11}{14} + \frac{10}{9} = \frac{239}{126}
$$
Question 5. Simplify each of the following and express the result as a rational number in
standard form. (Level-5)
(i) $\left(\frac{3}{2} \times \frac{1}{6}\right) + \left(\frac{5}{3} \times \frac{7}{2}\right) – \left(\frac{13}{8} \times \frac{4}{3}\right)$
Solution:
$$
\frac{1}{4} + \frac{35}{6} – \frac{13}{6} = \frac{47}{12}
$$
(ii) $\left(\frac{1}{4} \times \frac{2}{7}\right) – \left(\frac{5}{14} \times \frac{-2}{3}\right) + \left(\frac{3}{7} \times \frac{9}{2}\right)$
Solution:
$$
\frac{1}{14} + \frac{5}{21} + \frac{27}{14} = \frac{94}{42}
$$
(iii) $\left(\frac{13}{9} \times \frac{-15}{2}\right) + \left(\frac{7}{3} \times \frac{8}{5}\right) + \left(\frac{3}{5} \times \frac{1}{2}\right)$
Solution:
$$
\frac{-65}{6} + \frac{56}{15} + \frac{3}{10} = \frac{-204}{30}
$$
(iv) $\left(\frac{3}{11} \times \frac{5}{6}\right) – \left(\frac{9}{12} \times \frac{4}{3}\right) + \left(\frac{5}{13} \times \frac{6}{15}\right)$
Solution:
$$
\frac{5}{22} – 1 + \frac{2}{13} = \frac{-177}{286}
$$
FAQs on Multiplication of Rational Numbers
Q1. What is the rule for multiplying two fractions?
When multiplying two fractions, the product is obtained by multiplying the numerators together and denominators together.
For example:
$$
\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
$$
This same rule applies in all competitive exams including SSC CGL, SSC CHSL, and SSC GD. Practicing this rule helps in solving simplification questions quickly.
If you want to revise more, you can download SSC notes, SSC study material pdf, and SSC math notes preparation content.
Q2. Is the multiplication rule same for rational numbers?
Yes. Rational numbers can be expressed in the form $\frac{p}{q}$ where $q \neq 0$. The multiplication of two rational numbers follows the same rule as fractions:
$$
\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
$$
In SSC exam preparation, questions involving negative rational numbers are frequently asked to test accuracy with signs.
For more practice, download SSC notes or SSC math study material pdf for preparation.
Q3. How do we simplify rational number multiplication problems?
The key steps are:
- Multiply the numerators.
- Multiply the denominators.
- Simplify the fraction by dividing with common factors.
- Pay attention to negative signs.
Example:
$$
\frac{-8}{9} \times \frac{3}{64} = \frac{-24}{576} = \frac{-1}{24}
$$
This approach is useful in SSC exams like SSC CGL and SSC MTS where time-saving tricks are important.
For quick revision, use SSC math notes preparation and download SSC study material pdf.
Q4. Why is multiplication of rational numbers important in SSC exams?
Simplification, number system, and algebra in SSC exams often include multiplication of fractions and rational numbers. A strong grip on this saves time in both Tier-1 and Tier-2 exams. Many SSC aspirants revise rational numbers before practicing advanced topics.
For practice, download SSC notes and SSC math study material pdf to strengthen your basics.
Q5. What mistakes should students avoid in rational number multiplication?
- Ignoring negative signs.
- Forgetting to reduce the final fraction into simplest form.
- Mixing multiplication with addition or subtraction of fractions.
Example of mistake:
$$
\frac{2}{3} \times \frac{3}{4} \neq \frac{5}{7}
$$
Correct:
$$
\frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}
$$
Avoiding these mistakes will help in SSC exam preparation and speed tests.
Aspirants can download SSC math notes, SSC study material pdf, and other SSC exam notes for regular practice.
Q6. Where can I find practice problems on multiplication of rational numbers for SSC preparation?
You can find practice problems in SSC preparation books, coaching class materials, and online PDFs. Practicing directly from solved examples like the ones above is the best way. Repeated practice ensures accuracy in the exam hall.
For convenience, download SSC notes, SSC math study material pdf, and SSC exam notes preparation guides.
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