Rational Numbers

rational numbers class 8 selina

Step by Step solutions of Concise Mathematics ICSE Class-8 Maths chapter 1- Rational Numbers by Selina is provided.

Table of Contents

Exercise: 1-B

Q1: Multiple Choice Type:

i. The sum of two rational numbers is 8, if one of them is \(2\frac{3}{4}\) , the other number is:
Solution:
Let the other number be \(x\)
Given: \[ x + 2\frac{3}{4} = 8 \] Convert mixed number to improper fraction: \[ x + \frac{11}{4} = 8 \] Now subtract \(\frac{11}{4}\) from both sides: \[ x = 8 – \frac{11}{4} \] \[ x = \frac{32}{4} – \frac{11}{4} \] \[ x = \frac{21}{4} \] Convert back to mixed number: \[ \frac{21}{4} = 5\frac{1}{4} \] Correct Answer: (c) \(5\frac{1}{4}\)

ii. For three rational numbers \(\frac{a}{b},\frac{c}{d}\ and\ \frac{e}{f}\), we have:
Solution:
We know the identity: \[ a – (b – c) = a – b + c \] So applying this property: \[ \frac{a}{b}-\left(\frac{c}{d}-\frac{e}{f}\right) = \frac{a}{b} – \frac{c}{d} + \frac{e}{f} \] Correct Answer: (a)

iii. The sum of two rational numbers is -6. If one of them is \(4\frac{1}{2}\) , other number is:
Solution:
Let the other number be \(x\)
We are given: \[ x + 4\frac{1}{2} = -6 \] Convert mixed number: \[ x + \frac{9}{2} = -6 \] Subtract \(\frac{9}{2}\) from both sides: \[ x = -6 – \frac{9}{2} \] \[ x = \frac{-12}{2} – \frac{9}{2} = \frac{-21}{2} \] Convert to mixed number: \[ \frac{-21}{2} = -10\frac{1}{2} \] Correct Answer: (d) \(-10\frac{1}{2}\)

iv. The number subtracted from \(5\frac{2}{3}\) to get \(-1\frac{2}{3}\) is:
Solution:
Let the number be \(x\)
We are given: \[ 5\frac{2}{3} – x = -1\frac{2}{3} \] Convert mixed numbers: \[ \frac{17}{3} – x = \frac{-5}{3} \] Now solve: \[ -x = \frac{-5}{3} – \frac{17}{3} = \frac{-22}{3} \] Multiply both sides by -1: \[ x = \frac{22}{3} = 7\frac{1}{3} \] Correct Answer: (d) \(7\frac{1}{3}\)

v. The number added to \(5\frac{2}{3}\) to get \(-1\frac{2}{3}\) is:
Solution:
Let the number be \(x\)
We are given: \[ 5\frac{2}{3} + x = -1\frac{2}{3} \] Convert mixed numbers: \[ \frac{17}{3} + x = \frac{-5}{3} \] Solve for \(x\): \[ x = \frac{-5}{3} – \frac{17}{3} = \frac{-22}{3} \] Convert to mixed number: \[ \frac{-22}{3} = -7\frac{1}{3} \] Correct Answer: (b) \(-7\frac{1}{3}\)


Q2: Evaluate

i. \(\frac{2}{3}-\frac{4}{5}\)
Step 1: Find LCM of 3 and 5 = 15
Step 2: Convert to like denominators: \[ \frac{2}{3} = \frac{10}{15},\quad \frac{4}{5} = \frac{12}{15} \] Step 3: Subtract: \[ \frac{10}{15} – \frac{12}{15} = \frac{-2}{15} \] Answer: \(\frac{-2}{15}\)

ii. \(\frac{-4}{9}-\frac{2}{-3}\)
Note: \(\frac{2}{-3} = -\frac{2}{3}\) \[ \frac{-4}{9} – \left(-\frac{2}{3}\right) = \frac{-4}{9} + \frac{2}{3} \] Step 1: LCM of 9 and 3 = 9 \[ \frac{-4}{9} + \frac{2}{3} = \frac{-4}{9} + \frac{6}{9} = \frac{2}{9} \] Answer: \(\frac{2}{9}\)

iii. \(-1-\frac{4}{9}\)
Convert \(-1\) to fractional form: \[ -1 = \frac{-9}{9} \] \[ \frac{-9}{9} – \frac{4}{9} = \frac{-13}{9} \] Answer: \(\frac{-13}{9}\)

iv. \(\frac{-2}{7}-\frac{3}{14}\)
LCM of 7 and 14 = 14 \[ \frac{-2}{7} = \frac{-4}{14} \] \[ \frac{-4}{14} – \frac{3}{-14} = \frac{-4}{14} + \frac{3}{14} = \frac{-1}{14} \] Answer: \(\frac{-1}{14}\)

v. \(\frac{-5}{18}-\frac{-2}{9}\)
Note: \(\frac{-2}{9} = -\frac{4}{18}\) \[ \frac{-5}{18} – (-\frac{4}{18}) = \frac{-5}{18} + \frac{4}{18} = \frac{-1}{18} \] Answer: \(\frac{-1}{18}\)

vi. \(\frac{5}{21}-\frac{-13}{42}\)
Note: Subtracting negative = adding: \[ \frac{5}{21} + \frac{13}{42} \] LCM of 21 and 42 = 42 \[ \frac{5}{21} = \frac{10}{42} \] \[ \frac{10}{42} + \frac{13}{42} = \frac{23}{42} \] Answer: \(\frac{23}{42}\)


Q3: Subtract

i. \(\frac{5}{8}\ from\ \frac{-3}{8})\)
This means: \(\frac{-3}{8} – \frac{5}{8} = \frac{-8}{8} = -1\)
Answer: \(-1\)

ii. \(\frac{-8}{11}\ from\ \frac{4}{11}\)
\(\frac{4}{11} – \frac{-8}{11} = \frac{4}{11} + \frac{8}{11} = \frac{12}{11}\)
Answer: \(\frac{12}{11}\)

iii. \(\frac{4}{9}\ from\ \frac{-5}{9}\)
\(\frac{-5}{9} – \frac{4}{9} = \frac{-9}{9} = -1\)
Answer: \(-1\)

iv. \(\frac{1}{4}\ from\ \frac{-3}{8}\)
LCM of 4 and 8 = 8 \[ \frac{1}{4} = \frac{2}{8}, \quad \frac{-3}{8} – \frac{2}{8} = \frac{-5}{8} \] Answer: \(\frac{-5}{8}\)

v. \(\frac{-5}{8}\ from\ \frac{-13}{16}\)
Convert to like denominators: LCM = 16 \[ \frac{-5}{8} = \frac{-10}{16} \] \[ \frac{-13}{16} – \frac{-10}{16} = \frac{-13}{16} + \frac{10}{16} = \frac{-3}{16} \] Answer: \(\frac{-3}{16}\)

vi. \(\frac{-9}{22}\ from\ \frac{5}{33}\)
LCM of 22 and 33 = 66 \[ \frac{-9}{22} = \frac{-27}{66}, \quad \frac{5}{33} = \frac{10}{66} \] \[ \frac{10}{66} – \frac{-27}{66} = \frac{10}{66} + \frac{27}{66} = \frac{37}{66} \] Answer: \(\frac{37}{66}\)


Q4: The sum of two rational numbers is \(\frac{9}{20}\). If one of them is \(\frac{2}{5}\), find the other.

Solution:
Let the other number be \(x\).
We are given: \[ x + \frac{2}{5} = \frac{9}{20} \] Subtract \(\frac{2}{5}\) from both sides: \[ x = \frac{9}{20} – \frac{2}{5} \]Convert \(\frac{2}{5}\) to have a denominator of 20: \[ \frac{2}{5} = \frac{8}{20} \]Now subtract: \[ x = \frac{9}{20} – \frac{8}{20} = \frac{1}{20} \]Answer: \(\frac{1}{20}\)


Q5: The sum of two rational numbers is \(\frac{-2}{3}\). If one of them is \(\frac{-8}{15}\), find the other.

Solution:
Let the other number be \(x\).
We are given: \[ x + \left(\frac{-8}{15}\right) = \frac{-2}{3} \]To isolate \(x\), add \(\frac{8}{15}\) to both sides: \[ x = \frac{-2}{3} + \frac{8}{15} \]Convert \(\frac{-2}{3}\) to denominator 15: \[ \frac{-2}{3} = \frac{-10}{15} \]Now add: \[ x = \frac{-10}{15} + \frac{8}{15} = \frac{-2}{15} \]Answer: \(\frac{-2}{15}\)


Q6: The sum of the two rational numbers is -6. If one of them is \(\frac{-8}{5}\) , find the other.

Solution:
Let the other number be \(x\).
We are given: \[ x + \left(\frac{-8}{5}\right) = -6 \]
To isolate \(x\), add \(\frac{8}{5}\) to both sides: \[ x = -6 + \frac{8}{5} \]Convert \(-6\) to a fraction with denominator 5: \[ -6 = \frac{-30}{5} \]Now add: \[ x = \frac{-30}{5} + \frac{8}{5} = \frac{-22}{5} \]Convert the improper fraction to a mixed number: \[ \frac{-22}{5} = -4\frac{2}{5} \]Answer: \(-4\frac{2}{5}\)


Q7: Which rational number should be added to \(\frac{-7}{8}\) to get \(\frac{5}{9}\)?

Solution:
Let the required number be \(x\).
We are given: \[ x + \left(\frac{-7}{8}\right) = \frac{5}{9} \]To isolate \(x\), add \(\frac{7}{8}\) to both sides: \[ x = \frac{5}{9} + \frac{7}{8} \]Now find the LCM of 9 and 8:
LCM(9, 8) = 72
Convert both fractions to have denominator 72: \[ \frac{5}{9} = \frac{40}{72}, \quad \frac{7}{8} = \frac{63}{72} \]Add the two fractions: \[ x = \frac{40}{72} + \frac{63}{72} = \frac{103}{72} \]Convert the improper fraction to a mixed number: \[ \frac{103}{72} = 1\frac{31}{72} \]Answer: \(1\frac{31}{72}\)


Q8: Which rational number should be added to \(\frac{-5}{9}\) to get \(\frac{-2}{3}\)?

Solution:
Let the required number be \(x\).
We are given: \[ x + \left(\frac{-5}{9}\right) = \frac{-2}{3} \]To isolate \(x\), add \(\frac{5}{9}\) to both sides: \[ x = \frac{-2}{3} + \frac{5}{9} \] Now find the LCM of 3 and 9:
LCM(3, 9) = 9
Convert both fractions to have denominator 9: \[ \frac{-2}{3} = \frac{-6}{9} \]Now add: \[ x = \frac{-6}{9} + \frac{5}{9} = \frac{-1}{9} \]Answer: \(\frac{-1}{9}\)


Q9: Which rational number should be subtracted from \(\frac{-5}{6}\) to get \(\frac{4}{9}\)?

Solution:
Let the required number be \(x\).
According to the question: \[ \frac{-5}{6} – x = \frac{4}{9} \]To find \(x\), subtract \(\frac{-5}{6}\) from both sides: \[ -x = \frac{4}{9} – \left(\frac{-5}{6}\right) \]Use the rule: subtracting a negative is like adding: \[ -x = \frac{4}{9} + \frac{5}{6} \]Find the LCM of 9 and 6:
LCM(9, 6) = 18
Convert both to have denominator 18: \[ \frac{4}{9} = \frac{8}{18}, \quad \frac{5}{6} = \frac{15}{18} \]Now add: \[ -x = \frac{8}{18} + \frac{15}{18} = \frac{23}{18} \]Multiply both sides by -1: \[ x = -\frac{23}{18} \]Answer: \(\frac{-23}{18}\)


Q10:

i. What should be subtracted from \(-2\) to get \(\frac{3}{8}\)?Solution:
Let the required number be \(x\).
According to the question: \[ -2 – x = \frac{3}{8} \]Now, isolate \(x\): \[ -x = \frac{3}{8} + 2 \]Convert 2 into a fraction with denominator 8: \[ 2 = \frac{16}{8} \Rightarrow -x = \frac{3}{8} + \frac{16}{8} = \frac{19}{8} \]Multiply both sides by -1: \[ x = -\frac{19}{8} \]Answer: \(\frac{-19}{8}\)

ii. What should be added to -2 to get \(\frac{3}{8}\)?Solution:

Let the required number be \(x\).

According to the question: \[ -2 + x = \frac{3}{8} \]Now, isolate \(x\): \[ x = \frac{3}{8} + 2 \]Convert 2 to a fraction with denominator 8: \[ 2 = \frac{16}{8} \Rightarrow x = \frac{3}{8} + \frac{16}{8} = \frac{19}{8} \]Answer: \(\frac{19}{8}\)


Q11: Evaluate:

i. \(\frac{3}{7}+\frac{-4}{9}-\frac{-11}{7}-\frac{7}{9}\)
Solution:
Group terms with same denominators: \[ \left(\frac{3}{7} – \frac{-11}{7}\right) + \left(\frac{-4}{9} – \frac{7}{9}\right) \]Simplify each group: \[ = \left(\frac{3}{7} + \frac{11}{7}\right) + \left(\frac{-4 – 7}{9}\right) = \left(\frac{14}{7}\right) + \left(\frac{-11}{9}\right) \]Now simplify: \[ = 2 – \frac{11}{9} = \frac{18}{9} – \frac{11}{9} = \frac{7}{9} \]Answer: \(\frac{7}{9}\)

ii. \(\frac{2}{3}+\frac{-4}{5}-\frac{1}{3}-\frac{2}{5}\)
Solution:

Group terms with same denominators: \[ \left(\frac{2}{3} – \frac{1}{3}\right) + \left(\frac{-4}{5} – \frac{2}{5}\right) \] Simplify: \[ = \left(\frac{1}{3}\right) + \left(\frac{-6}{5}\right) \]Find LCM of 3 and 5 = 15: \[ \frac{1}{3} = \frac{5}{15}, \quad \frac{-6}{5} = \frac{-18}{15} \]Now subtract: \[ \frac{5}{15} – \frac{18}{15} = \frac{-13}{15} \]Answer: \(\frac{-13}{15}\)

iii. \(\frac{4}{7}-\frac{-8}{9}-\frac{-13}{7}+\frac{17}{9}\)
Solution:
Group like denominators: \[ \left(\frac{4}{7} + \frac{13}{7}\right) + \left(\frac{8}{9} + \frac{17}{9}\right) \]Simplify: \[ = \frac{17}{7} + \frac{25}{9} \]Find LCM of 7 and 9 = 63: \[ \frac{17}{7} = \frac{153}{63}, \quad \frac{25}{9} = \frac{175}{63} \]Now add: \[ \frac{153 + 175}{63} = \frac{328}{63} \]Answer: \(\frac{328}{63}\)


previous
next

Share the Post:

Leave a Comment

Your email address will not be published. Required fields are marked *

Related Posts​

  • Identities
    Step by Step solutions of Test Yourself Concise Mathematics ICSE Class-8 Maths chapter 12- Identities by Selina is provided.
  • Identities
    Step by Step solutions of Exercise- 12B Concise Mathematics ICSE Class-8 Maths chapter 12- Identities by Selina is provided.

Join Our Newsletter

Name
Email
The form has been submitted successfully!
There has been some error while submitting the form. Please verify all form fields again.

Scroll to Top