Rational Numbers

Q1: Multiple Choice Type

i. \(\frac{0}{5}\) is a rational number. \(\frac{0}{8}\) is a rational number, then \(\frac{0}{5}\div\frac{0}{8}\) is :
We know: \[ \frac{0}{5} = 0 \quad \text{and} \quad \frac{0}{8} = 0 \] So, \[ \frac{0}{5} \div \frac{0}{8} = 0 \div 0 \] But dividing zero by zero is undefined.
Answer: d. undefined

ii. a and bare two rational numbers such that a + b = 0; then:
\[ a + b = 0 \Rightarrow b = -a \] So \(a\) and \(b\) are equal in magnitude but opposite in sign.
Answer: c. a and b are numerically equal but opposite in sign

iii. The product of rational number \(\frac{3}{8}\) and its additive inverse is:
Additive inverse of \(\frac{3}{8}\) is \(-\frac{3}{8}\)
\[ \frac{3}{8} \times \left(-\frac{3}{8}\right) = -\frac{9}{64} \]
Answer: d. \(-\frac{9}{64}\) iv. The sum of rational number \(\frac{2}{3}\) and its reciprocal is:
Reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\)
\[ \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6} = 2\frac{1}{6} \]
Answer: b. \(2\frac{1}{6}\)

v. The product of two rational numbers is -1, if one of them is \(\frac{2}{5}\), then other is:
Let the other number be \(x\), then:
\[ \frac{2}{5} \times x = -1 \Rightarrow x = -1 \div \frac{2}{5} = -1 \times \frac{5}{2} = -\frac{5}{2} \]
Answer: c. \(-\frac{5}{2}\)

vi. Statement 1: For a rational number \(\frac{7}{9}, \frac{7}{9}-0=\frac{7}{9}\) but \(0-\frac{7}{9}=-\frac{7}{9}\). Hence subtraction has only right identity.
Statement 2: Subtraction has no identity.
Which of following options is correct?
From Statement 1: \[ \frac{7}{9} – 0 = \frac{7}{9} \quad \text{(True)} \\ 0 – \frac{7}{9} = -\frac{7}{9} \quad \text{(Also true)} \] So subtraction has no two-sided identity.
✅ Statement 1 is true
❌ Statement 2 is false
Answer: c. Statement 1 is true, and statement 2 is false.

vii. Assertion (A): Additive inverse of \(\frac{2}{5}\ is-\frac{5}{2}\).
Reason (R): For every non-zero rational number ‘a’, ‘-a’ such that a + (-a) = 0
Check Assertion:
Additive inverse of \(\frac{2}{5}\) is \(-\frac{2}{5}\), not \(-\frac{5}{2}\)
✅ Assertion is false
Check Reason:
Yes, for every rational number \(a\), \[ a + (-a) = 0 \] ✅ Reason is true
Answer: d. A is false, but R is true

viii. Assertion (A): Multiplicative inverse of \(-\frac{7}{5}\ is\ -\frac{5}{7}\).
Reason (R): For every non-zero rational number ‘a’, there is a rational number \(\frac{1}{a}\) such that \(a\times\left(\frac{1}{a}\right)=1\).
Check Assertion:
Multiplicative inverse of \(-\frac{7}{5}\) is indeed \(-\frac{5}{7}\), because: \[ -\frac{7}{5} \times -\frac{5}{7} = \frac{35}{35} = 1 \] ✅ Assertion is true
Check Reason:
This is the definition of multiplicative inverse: \[ a \times \frac{1}{a} = 1 \] ✅ Reason is true
✅ R correctly explains A.
Answer: a. Both A and R are correct, and R is the correct explanation for A

ix. Assertion (A): \(\frac{1}{2}+2=\frac{5}{2}\), which is a rational number.
Reason (R): If \(\frac{p}{q}\ and\ \frac{r}{s}\) are any two rational numbers then \(\frac{p}{q}+\frac{r}{s}=\frac{r}{s}+\frac{p}{q}\)
Check Assertion: \[ \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \] ✅ This is a rational number. So, assertion is true.
Check Reason: \[ \frac{p}{q} + \frac{r}{s} = \frac{r}{s} + \frac{p}{q} \] ✅ Commutativity of addition holds. Reason is true
But ❌ R is not a direct explanation of why \(\frac{5}{2}\) is rational — it’s just a property.
Answer: b. Both A and R are correct, and R is not the correct explanation for A

x. Assertion (A): 0 and \(\frac{11}{12}\) are two rational numbers and \(\frac{11}{12}\neq0\) then \(0\div\frac{11}{12}=0\), a rational number.
Reason (R): If a rational number is divided by some non-zero rational number, the result is always a rational number.
Check Assertion: \[ 0 \div \frac{11}{12} = 0 \times \frac{12}{11} = 0 \] ✅ 0 is a rational number. So, assertion is true
Check Reason:
Yes, dividing any rational number by non-zero rational number gives rational result.
✅ Reason is true
✅ And here, R explains A correctly.
Answer: a. Both A and R are correct, and R is the correct explanation for A

Q2: Write :

i. The rational number that does not have a reciprocal.
Any non-zero rational number has a reciprocal. But,
0 is the only rational number which does not have a reciprocal
Because:
Reciprocal of a rational number \( \frac{a}{b} \) is \( \frac{b}{a} \).
But, if \( a = 0 \), then \( \frac{b}{a} = \frac{b}{0} \), which is undefined.
So,
\( \frac{0}{1} \) has no reciprocal.

ii. The rational numbers that are equal to their reciprocal.
Let a number \( x \) be equal to its reciprocal: \[ x = \frac{1}{x} \] Multiplying both sides by \( x \): \[ x \times x = 1 \Rightarrow x^2 = 1 \] \[ x = \pm 1 \]So, the rational numbers which are equal to their reciprocals are \( 1 \text{ and } -1 \)

iii. The reciprocal of \(-\frac{8}{17}+\left(\frac{-8}{17}\right)\).
First simplify the expression: \[ -\frac{8}{17} + \left( \frac{-8}{17} \right) = -\frac{8}{17} – \frac{8}{17} = -\frac{16}{17} \]Now find the reciprocal: \[ \text{Reciprocal of } -\frac{16}{17} = -\frac{17}{16} \]Answer: \( -\frac{17}{16} \)

Q3: Write five rational numbers between \(-\frac{3}{2}\ and\ \frac{5}{3}\).

Step 1: Find LCM of denominators 2 and 3 \[ \text{LCM}(2, 3) = 6 \]Step 2: Convert to like denominators: \[ -\frac{3}{2} = \frac{-9}{6},\quad \frac{5}{3} = \frac{10}{6} \]Step 3: Multiply numerator and denominator by \(5 + 1 = 6\) to create space for 5 numbers \[ \frac{-9 \times 6}{6 \times 6} = \frac{-54}{36},\quad \frac{10 \times 6}{6 \times 6} = \frac{60}{36} \]Step 4: Find five numbers between -54 and 60 with denominator 36:
Pick numerators between -54 and 60: \[ \frac{-30}{36},\ \frac{-15}{36},\ \frac{0}{36},\ \frac{18}{36},\ \frac{45}{36} \]You may simplify them too: \[ \frac{-5}{6},\ \frac{-5}{12},\ 0,\ \frac{1}{2},\ \frac{5}{4} \]∴ Required rational numbers: \( \frac{-5}{6},\ \frac{-5}{12},\ 0,\ \frac{1}{2},\ \frac{5}{4} \)

Q4: Write five rational numbers greater than -4.

Step 1: Understand the condition
We are asked to find rational numbers that are greater than -4.
That means any rational number which lies to the right of -4 on the number line.
Step 2: Write five rational numbers just greater than -4
Examples of rational numbers greater than -4:
Pick any 5 of them. For example:

• -3
• -2
• -1
• 0
• 1 and so on

These are all valid rational numbers greater than -4.
Answer: -3, -2, -1, 0, 1 and so on

Q5: What should be added to \(-2\frac{1}{2}\) to get \(-3\frac{1}{3}\)?

Step 1: Convert mixed fractions into improper fractions\[ -2\frac{1}{2} = -\left(\frac{2 \times 2 + 1}{2}\right) = -\frac{5}{2} \] \[ -3\frac{1}{3} = -\left(\frac{3 \times 3 + 1}{3}\right) = -\frac{10}{3} \]Step 2: Let the number to be added be \(x\)
We know: \[ – \frac{5}{2} + x = -\frac{10}{3} \]Step 3: Solve the equation
To find \(x\), subtract \(-\frac{5}{2}\) from both sides: \[ x = -\frac{10}{3} + \frac{5}{2} \]Step 4: Take LCM and simplify
LCM of 3 and 2 is 6: \[ x = \left( -\frac{10 \times 2}{6} + \frac{5 \times 3}{6} \right) = \left( -\frac{20}{6} + \frac{15}{6} \right) \] \[ x = \frac{-20 + 15}{6} = \frac{-5}{6} \]Answer: \(-\frac{5}{6}\)

Q6: What should be subtracted from \(2\frac{1}{2}\) to get \(-3\frac{1}{3}\)?

Step 1: Convert mixed fractions into improper fractions
\[ 2\frac{1}{2} = \frac{5}{2}, \quad -3\frac{1}{3} = -\frac{10}{3} \]Step 2: Let the number to be subtracted be \(x\)
We know: \[ \frac{5}{2} – x = -\frac{10}{3} \]Step 3: Solve for \(x\)
Subtract \(\frac{5}{2}\) from both sides: \[ -x = -\frac{10}{3} – \frac{5}{2} \]Take LCM of 3 and 2 = 6: \[ -x = \left( \frac{-10 \times 2}{6} – \frac{5 \times 3}{6} \right) = \left( \frac{-20 – 15}{6} \right) = \frac{-35}{6} \]Now multiply both sides by -1: \[ x = \frac{35}{6} \]Answer: \(\frac{35}{6}\)

Q7: If \(m=-\frac{7}{9}\ and\ n=\frac{5}{6}\), verify that:

i. \(m-n\neq\ n-m\)
We are given:
\(m = -\frac{7}{9}\), \(n = \frac{5}{6}\)
Now calculate \(m – n\): \[ m – n = -\frac{7}{9} – \frac{5}{6} \]Take LCM of 9 and 6 = 18: \[ = \left( \frac{-14}{18} – \frac{15}{18} \right) = \frac{-29}{18} \]Now calculate \(n – m\): \[ n – m = \frac{5}{6} – (-\frac{7}{9}) = \frac{5}{6} + \frac{7}{9} \]Take LCM of 6 and 9 = 18: \[ = \left( \frac{15}{18} + \frac{14}{18} \right) = \frac{29}{18} \]So we have: \[ m – n = \frac{-29}{18}, \quad n – m = \frac{29}{18} \]Hence, \(m – n \ne n – m\)

ii. \(-(m+n)=(-m)+(-n)\)
Calculate \(m + n\): \[ m + n = -\frac{7}{9} + \frac{5}{6} \]Take LCM of 9 and 6 = 18: \[ = \left( \frac{-14}{18} + \frac{15}{18} \right) = \frac{1}{18} \]Now calculate \(-(m+n)\): \[ -(m+n) = -\left( \frac{1}{18} \right) = \frac{-1}{18} \]Now calculate \(-m\) and \(-n\): \[ -m = \frac{7}{9}, \quad -n = -\frac{5}{6} \]Now calculate \((-m) + (-n)\): \[ = \frac{7}{9} – \frac{5}{6} \]Take LCM of 9 and 6 = 18: \[ = \left( \frac{14}{18} – \frac{15}{18} \right) = \frac{-1}{18} \]Hence, \(-(m+n) = (-m) + (-n)\)

Q8: Represent rational numbers \(-\frac{7}{3}\ and\ \frac{7}{4}\) on same number line.

Step 1: Convert rational numbers to mixed fractions
\[ -\frac{7}{3} = -2\frac{1}{3}, \quad \frac{7}{4} = 1\frac{3}{4} \]So, we need a number line from at least -3 to +2.
Step 2: Divide intervals accordingly
👉 For \(-\frac{7}{3}\), divide each whole number into 3 equal parts (thirds).
👉 For \(\frac{7}{4}\), divide each whole number into 4 equal parts (fourths).
Step 3: Mark the points
– To mark \(-\frac{7}{3}\), move **2 wholes and 1 third to the left** of 0.
– To mark \(\frac{7}{4}\), move **1 whole and 3 fourths to the right** of 0.
🧮 Here’s a sketch-style representation idea (draw this on paper or digitally):

←──●───|───Φ───●───|───|───●───|───|───●──|──|──|──●──|──|──Φ──●───→
  -3          -2          -1           0           1           2   
           ↑                                                ↑    
   -7/3 marked                                        7/4 marked
   

Hence, \(-\frac{7}{3}\) lies between -3 and -2, and \(\frac{7}{4}\) lies between 1 and 2 on the number line.

Q9:

i. Add: \(-\frac{5}{6}\ and\ \frac{3}{8}\)
Step 1: Find LCM of 6 and 8: \[ \text{LCM}(6, 8) = 24 \]Step 2: Convert to like denominators: \[ -\frac{5}{6} = -\frac{20}{24}, \quad \frac{3}{8} = \frac{9}{24} \]Step 3: Add the numerators: \[ -\frac{20}{24} + \frac{9}{24} = \frac{-11}{24} \]Answer: \(-\frac{11}{24}\)

ii. Subtract: \(\frac{3}{8}\ from\ -\frac{5}{6}\)
Step 1: Convert to like denominators: \[ \frac{3}{8} = \frac{9}{24}, \quad -\frac{5}{6} = -\frac{20}{24} \]Step 2: Subtract: \[ \frac{-20}{24} – (\frac{9}{24}) = -\frac{9}{24} – \frac{20}{24} = -\frac{29}{24} \]Answer: \(-\frac{29}{24}\)

iii. Multiply: \(-\frac{5}{6}\ and\ \frac{3}{8}\)
\[ -\frac{5}{6} \times \frac{3}{8} = -\frac{15}{48} = -\frac{5}{16} \]Answer: \(-\frac{5}{16}\)

iv. Divide: \(\frac{3}{8}\ by\ -\frac{5}{6}\)
Step 1: Multiply by reciprocal: \[ \frac{3}{8} \div -\frac{5}{6} = \frac{3}{8} \times -\frac{6}{5} = -\frac{18}{40} = -\frac{9}{20} \]Answer: \(-\frac{9}{20}\)

Q10:

i. By what number should \(\frac{12}{17}\) be multiplied to get \(-\frac{4}{7}\)?
Let the required number be \(x\).
Then, \[ \frac{12}{17} \times x = -\frac{4}{7} \]To isolate \(x\), divide both sides by \(\frac{12}{17}\): \[ x = \frac{-\frac{4}{7}}{\frac{12}{17}} = -\frac{4}{7} \times \frac{17}{12} \]Now multiply: \[ x = -\frac{4 \times 17}{7 \times 12} = -\frac{68}{84} \]Simplify: \[ x = -\frac{17}{21} \]Answer: \(-\frac{17}{21}\)

ii. By what number should \(\frac{12}{17}\) be divided to get \(-\frac{4}{7}\)?
Let the required number be \(y\).
Then, \[ \frac{12}{17} \div y = -\frac{4}{7} \]This implies: \[ y = \frac{\frac{12}{17}}{-\frac{4}{7}} = \frac{12}{17} \times -\frac{7}{4} \]Now multiply: \[ y = -\frac{12 \times 7}{17 \times 4} = -\frac{84}{68} \]Simplify: \[ y = -\frac{21}{17} \]Answer: \(-\frac{21}{17}\)