Anand Classes is the leading SSC coaching center in Jalandhar, providing high-quality study material, detailed SSC math notes, and step-by-step solutions for rational number problems to help aspirants crack exams like SSC CGL, SSC CHSL, SSC MTS, and other government competitive exams. With expert faculty guidance and simplified explanations, students can strengthen their concepts in quantitative aptitude, practice previous year questions, and improve exam speed and accuracy. Click the print button to download study material and notes.
Problem 1. Simplify each of the following and write as a rational number in the form of p/q:
(i) $ \frac{3}{4} + \frac{5}{6} + \frac{-7}{8} $
Solution:
$2 = 2 \times 2$
$6 = 2 \times 3$
$8 = 2 \times 2 \times 2$
LCM is $2 \times 2 \times 2 \times 3 = 24$
$$
= \frac{3 \times 6 + 5 \times 4 + \left(-7 \times 3\right)}{24}
= \frac{18 + 20 – 21}{24}
= \frac{38 – 21}{24}
= \frac{17}{24}
$$
(ii) $ \frac{2}{3} + \frac{-5}{6} + \frac{-7}{9} $
Solution:
LCM of $3, 6$ and $9$ is $18$
$$
= \frac{2 \times 6 + \left(-5 \times 3\right) + \left(-7 \times 2\right)}{18}
= \frac{12 – 15 – 14}{18}
= \frac{12 – 29}{18}
= \frac{-17}{18}
$$
(iii) $ \frac{-11}{2} + \frac{7}{6} + \frac{-5}{8} $
Solution:
$2 = 2 \times 1$
$6 = 2 \times 3$
$8 = 2 \times 2 \times 2$
LCM is $2 \times 2 \times 2 \times 3 = 24$
$$
= \frac{\left(-11 \times 12\right) + 7 \times 4 + \left(-5 \times 3\right)}{24}$$
$$= \frac{-132 + 28 – 15}{24}
= \frac{-147 + 28}{24}
= \frac{-119}{24}
$$
(iv) $ \frac{-4}{5} + \frac{-7}{10} + \frac{-8}{15} $
Solution:
$10 = 5 \times 2$
$15 = 3 \times 5$
LCM is $5 \times 2 \times 3 = 30$
$$
= \frac{\left(-4 \times 6\right) + \left(-7 \times 3\right) + \left(-8 \times 2\right)}{30}
= \frac{-24 – 21 – 16}{30}
= \frac{-61}{30}
$$
(v) $ \frac{-9}{10} + \frac{22}{15} + \frac{13}{-20} $
Solution:
This can be written as $ \frac{-9}{10} + \frac{22}{15} + \frac{-13}{20} $
$10 = 2 \times 5$
$15 = 3 \times 5$
$20 = 2 \times 2 \times 5$
LCM is $2 \times 2 \times 3 \times 5 = 60$
$$ = \frac{\left(-9 \times 6\right) + 22 \times 4 + \left(-13 \times 3\right)}{60} $$
$$ = \frac{-54 + 88 – 39}{60}
= \frac{-93 + 88}{60}
= \frac{-5}{60}
= \frac{-1}{12}$$
(vi) $ \frac{5}{3} + \frac{3}{-2} + \frac{-7}{3} + 3 $
Solution:
This can be written as $ \frac{5}{3} + \frac{-3}{2} + \frac{-7}{3} + \frac{3}{1} $
LCM is $6$
$$
= \frac{5 \times 2 + \left(-3 \times 3\right) + \left(-7 \times 2\right) + 3 \times 6}{6}
= \frac{10 – 9 – 14 + 18}{6}
= \frac{28 – 23}{6}
= \frac{5}{6}
$$
Problem 2. Express each of the following as a rational number of the form p/q:
(i) $ \frac{-8}{3} + \frac{-1}{4} + \frac{-11}{6} + \frac{3}{8} + (-3) $
Solution:
$4 = 2 \times 2$
$6 = 2 \times 3$
$8 = 2 \times 2 \times 2$
LCM is $2 \times 2 \times 2 \times 3 = 24$
$$ = \frac{\left(-8 \times 8\right) + \left(-1 \times 6\right) + \left(-11 \times 4\right) + 3 \times 3 + \left(-3 \times 24\right)}{24}$$
$$= \frac{-64 – 6 – 44 + 9 – 72}{24}
= \frac{-186 + 9}{24}
= \frac{-177}{24}
= \frac{-59}{8}
$$
(ii) $ \frac{6}{7} + 1 + \frac{-7}{9} + \frac{19}{21} + \frac{-12}{7} $
Solution:
$ \left(\frac{6}{7} + \frac{-12}{7}\right) + \left(\frac{-7}{9}\right) + \frac{19}{21} + 1 $ (Taking numbers with same denominators together)
$$
= \frac{6 – 12}{7} + \frac{-7}{9} + \frac{19}{21} + \frac{1}{1}
= \frac{-6}{7} + \frac{-7}{9} + \frac{19}{21} + \frac{1}{1}
$$
$9 = 3 \times 3$
$21 = 3 \times 7$
LCM of $7, 1, 9$ and $21$ is $63$
$$ = \frac{\left(-6 \times 9\right) + \left(-7 \times 7\right) + 19 \times 3 + 1 \times 63}{63}$$
$$ = \frac{-54 – 49 + 57 + 63}{63}
= \frac{-103 + 120}{63}
= \frac{17}{63}
$$
(iii) $ \frac{15}{2} + \frac{9}{8} + \frac{-11}{3} + 6 + \frac{-7}{6} $
Solution:
$ \frac{15}{2} + \frac{9}{8} + \left(\frac{-11}{3}\right) + \frac{6}{1} + \left(\frac{-7}{6}\right) $
LCM of $2, 8, 3, 1$ and $6$ is $24$
$$
= \frac{15 \times 12 + 9 \times 3 + \left(-11 \times 8\right) + 6 \times 24 + \left(-7 \times 4\right)}{24}$$
$$= \frac{180 + 27 – 88 + 144 – 28}{24}
= \frac{351 – 116}{24}
= \frac{235}{24}
$$
(iv) $ \frac{-7}{4} + 0 + \frac{-9}{5} + \frac{19}{10} + \frac{11}{14} $
Solution:
$4 = 2 \times 2$
$5 = 5 \times 1$
$10 = 2 \times 5$
$14 = 2 \times 7$
LCM is $2 \times 2 \times 5 \times 7 = 140$
$$
= \frac{\left(-7 \times 35\right) + \left(-9 \times 28\right) + 19 \times 14 + 11 \times 10}{140}$$
$$= \frac{-245 – 252 + 266 + 110}{140}
= \frac{-497 + 376}{140}
= \frac{-121}{140}
$$
(v) $ \frac{-7}{4} + \frac{5}{3} + \frac{-1}{2} + \frac{-5}{6} + 2 $
Solution:
LCM of $4, 3, 2$ and $6$ is $12$
$$
= \frac{\left(-7 \times 3\right) + 5 \times 4 + \left(-1 \times 6\right) + \left(-5 \times 2\right) + 2 \times 12}{12}$$
$$= \frac{-21 + 20 – 6 – 10 + 24}{12}
= \frac{-37 + 44}{12}
= \frac{7}{12}
$$
Problem 3. Simplify:
(i) $ \frac{-3}{2} + \frac{5}{4} + \frac{-7}{4} $
Solution:
Taking numbers with the same denominators together
$$
= \frac{-3}{2} + \frac{5 – 7}{4}
= \frac{-3}{2} – \frac{2}{4}
$$
LCM of $2$ and $4$ is $4$
$$
= \frac{-3 \times 2 – 2 \times 1}{4}
= \frac{-6 – 2}{4}
= \frac{-8}{4}
= -2
$$
(ii) $ \frac{5}{3} + \frac{-7}{6} + \frac{-2}{3} $
Solution:
Taking numbers with same denominators together
$$
\left(\frac{5}{3} + \frac{-2}{3}\right) + \frac{-7}{6}
= \frac{5 – 2}{3} + \frac{-7}{6}
= \frac{3}{3} + \frac{-7}{6}
$$
LCM of $3$ and $6$ is $6$
$$
= \frac{3 \times 2 + \left(-7 \times 1\right)}{6}
= \frac{6 – 7}{6}
= \frac{-1}{6}
$$
(iii) $ \frac{5}{4} – \frac{7}{6} – \left(\frac{-2}{3}\right) $
Solution:
This can be written as $ \frac{5}{4} – \frac{7}{6} + \frac{2}{3} $
LCM of $4, 6$ and $3$ is $12$
$$
= \frac{5 \times 3 – 7 \times 2 + 2 \times 4}{12}
= \frac{15 – 14 + 8}{12}
= \frac{23 – 14}{12}
= \frac{9}{12}
= \frac{3}{4}
$$
(iv) $ \frac{-2}{5} – \left(\frac{-3}{10}\right) – \left(\frac{-4}{7}\right) $
Solution:
This can be written as: $ \frac{-2}{5} + \frac{3}{10} + \frac{4}{7} $
LCM of $5, 10$ and $7$ is $70$
$$
= \frac{\left(-2 \times 14\right) + 3 \times 7 + 4 \times 10}{70}
= \frac{-28 + 21 + 40}{70}
= \frac{-28 + 61}{70}
= \frac{33}{70}
$$
(v) $ \frac{5}{6} + \frac{-2}{5} – \left(\frac{-2}{15}\right) $
Solution:
This can be written as $ \frac{5}{6} + \frac{-2}{5} + \frac{2}{15} $
$6 = 2 \times 3$
$5 = 5 \times 1$
$15 = 3 \times 5$
LCM is $2 \times 3 \times 5 = 30$
$$
= \frac{5 \times 5 + \left(-2 \times 6\right) + 2 \times 2}{30}
= \frac{25 – 12 + 4}{30}
= \frac{29 – 12}{30}
= \frac{17}{30}
$$
(vi) $ \frac{3}{8} – \left(\frac{-2}{9}\right) + \left(\frac{-5}{36}\right) $
Solution:
This can be written as $ \frac{3}{8} + \frac{2}{9} – \frac{5}{36} $
$8 = 2 \times 2 \times 2$
$9 = 3 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
LCM is $2 \times 2 \times 2 \times 3 \times 3 = 72$
$$
= \frac{3 \times 9 + 2 \times 8 – 5 \times 2}{72}
= \frac{27 + 16 – 10}{72}
= \frac{43 – 10}{72}
= \frac{33}{72}
= \frac{11}{24}
$$
Frequently Asked Questions (FAQs) For SSC Exam
Q1. What is the easiest way to simplify rational number expressions for SSC exams?
The easiest method is to calculate the LCM of the denominators, then express each fraction with the same denominator, and finally add or subtract the numerators. For example, in $\frac{3}{4} + \frac{5}{6} – \frac{7}{8}$, the LCM of 4, 6, and 8 is 24. Converting all fractions to this denominator makes the calculation simple. This technique is extremely useful in SSC math preparation, where speed and accuracy play a big role. Practicing with SSC study material pdf helps in mastering this method.
Q2. Why is simplification of rational numbers important in SSC exams?
Simplification is one of the most scoring topics in SSC CGL, SSC CHSL, and SSC CPO exams. Since questions are generally straightforward, candidates who practice well can save valuable time. Rational numbers also form the base for other topics like algebra and percentages, so mastering them ensures better performance in the quantitative aptitude section. Regular practice from SSC math notes and SSC exam study material will give you a strong edge.
Q3. What are some shortcut methods for rational numbers in SSC exams?
Some useful shortcuts include cancelling common factors before taking the LCM, grouping fractions with the same denominator, and applying quick tricks from Vedic Maths. These methods reduce calculation steps and improve accuracy. For aspirants targeting SSC CGL Tier 1 and Tier 2, such tricks can save up to a minute per question. Referring to SSC quantitative aptitude notes and solving mock tests ensures you apply these shortcuts effectively in the exam.
Q4. Are questions on rational numbers frequently asked in SSC exams?
Yes, questions on rational numbers appear regularly in the SSC quantitative aptitude section. Generally, 2–3 problems are seen in exams like SSC CGL and SSC CHSL. While they are considered easy, even a small mistake can lead to a wrong answer. Practicing previous year SSC math papers and revising from SSC study material pdf helps you understand the exact level of difficulty expected.
Q5. Where can I download SSC exam notes on rational numbers?
You can download SSC notes and SSC study material pdf on rational numbers from reputed coaching centers like Anand Classes or from online learning platforms. These notes usually include solved examples, short tricks, and practice problems designed for SSC exams. Keeping a compiled set of SSC math notes will make revision much faster before the actual test.
Q6. What are the best books to prepare rational numbers for SSC exams?
Some of the most recommended books are Quantitative Aptitude by R.S. Aggarwal, Fast Track Objective Arithmetic by Rajesh Verma, and NCERT basics for Class 7 to 10. These books cover rational numbers and related concepts in detail. Along with these, candidates should practice from SSC exam practice sets and SSC math study material pdf for targeted preparation. Using these resources ensures a strong foundation as well as exam-oriented practice.
Q7. How much time should I devote daily for SSC math preparation?
Aspirants preparing for SSC exams should ideally spend at least 2–3 hours daily on quantitative aptitude, with 20–30 minutes dedicated to rational numbers and simplification. Regular timed practice helps improve speed, which is crucial in exams like SSC CGL, SSC CHSL, and SSC MTS. Solving topic-wise worksheets, revising from SSC math notes, and attempting SSC study material pdfs can boost confidence and accuracy.
Q8. What is the best strategy to prepare rational numbers for SSC exams?
The best strategy is to first revise the basic rules of fractions and LCM/HCF, then move on to practice a variety of problems. Start with solved examples, then attempt previous year SSC questions and mock tests. Consistency is key — revising short tricks regularly and solving daily practice problems ensures you don’t make mistakes under exam pressure. Having access to downloadable SSC notes and organized SSC preparation material also makes revision much faster and more effective.
To learn NDA Polity Notes, Click Here (Borrowed Features of Indian Constitution)
📚 Buy Study Material & Join Our Coaching
For premium study materials specially designed for SSC CGL CHSL Classes, visit our official study material portal:
👉 https://publishers.anandclasses.co.in/
For NDA Study Material, Click Here
For JEE/NEET/CBSE Study Material, Click Here
To enroll in our offline or online coaching programs, visit our coaching center website:
👉 https://anandclasses.co.in/
📞 Call us directly at: +91-94631-38669
💬 WhatsApp Us Instantly
Need quick assistance or want to inquire about classes and materials?
📲 Click below to chat instantly on WhatsApp:
👉 Chat on WhatsApp
🎥 Watch Video Lectures
Get access to high-quality video lessons, concept explainers, and revision tips by subscribing to our official YouTube channel:
👉 Neeraj Anand Classes – YouTube Channel
To learn JEE Chemistry Periodic Table Chaper, click here