Anand Classes provides SSC exam-focused study material on simplification of expressions involving addition and subtraction of rational numbers. This post explains shortcut methods, algorithm-based solutions, and illustrative examples to help candidates quickly solve rational number problems in competitive exams. These SSC maths notes are designed to improve calculation speed, accuracy, and problem-solving skills essential for Quantitative Aptitude sections of SSC CGL, SSC CHSL, SSC CPO, and other government job exams. Click the print button to download study material and notes.
Simplification of Expressions Involving Addition and Subtraction of Rational Numbers
Uptill now, we have been simplifying rational expressions involving addition and subtraction of more than two rational numbers by making groups of pairs of rational numbers having either the same denominator or having some common factor in their denominators, by using commutativity and associativity of addition.
We may simplify expressions involving addition and subtraction more easily by using the following algorithm.
Algorithm : Calculation Involving Addition and Subtraction of Rational Numbers
Step I
Find the LCM of the denominators of all the numbers involved.
Step II
Divide the LCM by the denominator of the first rational number and get a quotient.
Step III
Multiply the first numerator by the quotient obtained in step II and get an integer.
Step IV
Repeat steps II and III for the remaining rational numbers in the sum and obtain integers.
Step V
Retain the given symbols of addition and subtraction between the rationals and get an expression involving integers. Simplify this expression and get an integer.
Step VI
Obtain the required sum equal to the rational number whose numerator is the integer obtained in step V and denominator equal to the LCM obtained in step I. Reduce this number to the lowest form if it is not already so.
Illustration (Explained Using Algorithm)
Simplify:
$$
\frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3}
$$
Step I: Find the LCM of the denominators
The denominators are $5, 3, 15, 5, 3$.
$$
\text{LCM}(5, 3, 15, 5, 3) = 15
$$
Step II: Divide the LCM by each denominator to get quotients
- $15 \div 5 = 3$
- $15 \div 3 = 5$
- $15 \div 15 = 1$
- $15 \div 5 = 3$
- $15 \div 3 = 5$
Step III: Multiply each numerator by its quotient to get integers
- First numerator: $2 \times 3 = 6$
- Second numerator: $8 \times 5 = 40$
- Third numerator: $12 \times 1 = 12$
- Fourth numerator: $4 \times 3 = 12$
- Fifth numerator: $2 \times 5 = 10$
Step IV: Write the new expression with integers, keeping signs
$$
6 + 40 – 12 + 12 – 10
$$
Step V: Simplify the integers
$$
6 + 40 = 46 $$ $$ 46 – 12 = 34 $$ $$ 34 + 12 = 46 $$ $$ 46 – 10 = 36 $$
So, the result of simplification is $36$.
Step VI: Form the rational number with denominator = LCM
$$
\frac{36}{15}
$$
Reduce to lowest terms:
$$
\frac{36}{15} = \frac{12}{5}
$$
Final Answer
$$
\frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3} = \frac{12}{5}
$$
Final Expression in Shorter Form
$$
\frac{2}{5} + \frac{8}{3} – \frac{12}{15} + \frac{4}{5} – \frac{2}{3}
= \frac{(2 \times 3) + (8 \times 5) – (12 \times 1) + (4 \times 3) – (2 \times 5)}{15}
$$
$$
= \frac{6 + 40 – 12 + 12 – 10}{15}
= \frac{58 – 22}{15}
= \frac{36}{15}
= \frac{12}{5}
$$
In order to understand the above procedure properly, let us discuss some more examples.
Solved Examples Involving Addition and Subtraction of Rational Numbers for SSC CGL CHSL CPO Exams
Ex.1 : Simplify: $$
\frac{3}{7} + \frac{-6}{11} + \frac{8}{21} + \frac{-5}{22}
$$
Solution:
LCM of $7, 11, 21, 22$ is:
$$
\text{LCM} = 7 \times 11 \times 3 \times 2 = 462
$$
Now,
$$
\frac{(3 \times 66) + ((-6) \times 42) + (8 \times 22) + ((-5) \times 21)}{462}
$$
$$
= \frac{198 + (-252) + 176 + (-105)}{462}
$$
$$
= \frac{374 – 357}{462} = \frac{17}{462}
$$
Ex.2 : Simplify: $$
\frac{-7}{4} + \frac{5}{3} + \frac{-5}{6} + \frac{1}{3} + \frac{-1}{2}
$$
Solution:
LCM of $4, 3, 6, 3, 2$ is $12$.
$$
\frac{((-7)\times 3) + (5 \times 4) + ((-5)\times 2) + (1 \times 4) + ((-1)\times 6)}{12}
$$
$$
= \frac{-21 + 20 – 10 + 4 – 6}{12}
$$
$$
= \frac{-37 + 24}{12} = \frac{-13}{12}
$$
Ex.3 : Simplify : $$
\frac{-12}{5} + \frac{-7}{20} + \frac{3}{14} + \frac{1}{7} + \frac{-1}{10}
$$
Solution:
LCM of $5, 20, 14, 7, 10$ is $140$.
$$
\frac{((-12)\times 28) + ((-7)\times 7) + (3 \times 10) + (1 \times 20) + ((-1)\times 14)}{140}
$$
$$
= \frac{-336 + (-49) + 30 + 20 + (-14)}{140}
$$
$$
= \frac{-399 + 50}{140} = \frac{-349}{140}
$$
Ex.4 : Simplify:
$$
\frac{3}{4} + \frac{-3}{5} + \frac{-2}{3} + \frac{5}{8} + \frac{-4}{15}
$$
Solution:
LCM of $4, 5, 3, 8, 15$ is $120$.
$$
\frac{(3 \times 30) + ((-3)\times 24) + ((-2)\times 40) + (5 \times 15) + ((-4)\times 8)}{120}
$$
$$
= \frac{90 + (-72) + (-80) + 75 + (-32)}{120}
$$
$$
= \frac{-184 + 165}{120} = \frac{-19}{120}
$$
Ex.5 : Simplify:
$$
\frac{6}{7} + \frac{-2}{9} + \frac{-7}{21} + \frac{19}{21}
$$
Solution:
LCM of $7, 9, 21$ is $63$.
$$
\frac{(6 \times 9) + ((-2)\times 63) + ((-7)\times 3) + (19 \times 3)}{63}
$$
$$
= \frac{54 + (-126) + (-21) + 57}{63}
$$
$$
= \frac{111 – 175}{63} = \frac{-64}{63}
$$
Ex.6 : Simplify:
$$
\frac{-2}{3} + \frac{5}{9} + \frac{7}{6}
$$
Solution:
LCM of $3, 9, 6$ is $18$.
$$
\frac{((-2)\times 6) + (5 \times 2) + (7 \times 3)}{18}
$$
$$
= \frac{-12 + 10 + 21}{18} = \frac{19}{18}
$$
Ex.7 : Simplify: $$ \frac{5}{12} + \frac{-5}{18} + \frac{-7}{24} $$
Solution:
LCM of $12, 18, 24$ is $72$.
$$
\frac{(5 \times 6) + ((-5)\times 4) + ((-7)\times 3)}{72}
$$
$$
= \frac{30 + (-20) + (-21)}{72}
$$
$$
= \frac{30 – 41}{72} = \frac{-11}{72}
$$
Frequently Asked Questions (FAQs)
Q1. What is the easiest way to simplify rational number expressions for SSC exams?
The easiest way is to take the LCM of all denominators, convert each rational number to an equivalent fraction with the LCM as the denominator, and then perform addition or subtraction on the numerators.
Q2. Why is simplification of rational numbers important in SSC exams?
Simplification of rational numbers is a key part of the Quantitative Aptitude section in SSC exams. It helps in solving arithmetic problems quickly, reduces calculation errors, and builds confidence for handling complex expressions under time pressure.
Q3. What are some shortcut methods for rational numbers in SSC CGL and SSC CHSL exams?
Shortcut methods include:
- Directly multiplying numerator and denominator to avoid step-by-step fraction conversion.
- Cancelling common factors before taking the LCM.
- Using Vedic maths techniques for quick addition and subtraction.
Q4. Are questions on rational numbers frequently asked in SSC exams?
Yes, rational number problems regularly appear in SSC CGL, SSC CHSL, SSC CPO, and other government competitive exams. They are considered scoring topics because they require simple but accurate calculations.
Q5. Where can I download SSC exam notes on rational numbers?
You can download free SSC exam notes and study material on rational numbers by clicking the print button available at the end of this post.
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