Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 1 Rational Numbers

Exercise 1(A)

Question 1. Add, each pair of rational numbers, given below, and show that their addition (sum) is also a rational number:
(i) \( \frac{-5}{8} \) and \( \frac{3}{8} \)
(ii) \( \frac{-8}{13} \) and \( \frac{-4}{13} \)
(iii) \( \frac{6}{11} \) and \( \frac{-9}{11} \)
(iv) \( \frac{5}{-26} \) and \( \frac{8}{39} \)
(v) \( \frac{5}{-6} \) and \( \frac{2}{3} \)
(vi) -2 and \( \frac{2}{5} \)
(vii) \( \frac{9}{-4} \) and \( \frac{-3}{8} \)
(viii) \( \frac{7}{-18} \) and \( \frac{8}{27} \)
Answer:
(i) \( \frac{-5}{8} + \frac{3}{8} \)
Since the denominators are identical, the LCM is 8.
\( = \frac{-5 + 3}{8} \)
\( = \frac{-2}{8} = \frac{-1}{4} \)
This value is a rational number.

(ii) \( \frac{-8}{13} + \left( \frac{-4}{13} \right) \)
Since both denominators are 13, the LCM is 13.
\( = \frac{-8 - 4}{13} \)
\( = \frac{-12}{13} \)
This represents a rational number.

(iii) \( \frac{6}{11} + \left( \frac{-9}{11} \right) \)
Because the denominators are the same, the LCM is 11.
\( = \frac{6 - 9}{11} \)
\( = \frac{-3}{11} \)
This result is a rational number.

(iv) \( \frac{5}{-26} + \frac{8}{39} \)
Expressing with positive denominators:
\( = \frac{-5}{26} + \frac{8}{39} \)
Finding the LCM of 26 and 39:
\( 26 = 2 \times 13 \)
\( 39 = 3 \times 13 \)
LCM \( = 2 \times 3 \times 13 = 78 \)
\( = \frac{-5 \times 3}{26 \times 3} + \frac{8 \times 2}{39 \times 2} \)
\( = \frac{-15 + 16}{78} \)
\( = \frac{1}{78} \)
This output is a rational number.

(v) \( \frac{5}{-6} + \frac{2}{3} \)
Expressing with positive denominators:
\( = \frac{-5}{6} + \frac{2}{3} \)
Finding the LCM of 6 and 3:
LCM \( = 6 \)
\( = \frac{-5 \times 1}{6 \times 1} + \frac{2 \times 2}{3 \times 2} \)
\( = \frac{-5 + 4}{6} \)
\( = \frac{-1}{6} \)
This is indeed a rational number.

(vi) \( -2 + \frac{2}{5} \)
\( = \frac{-2}{1} + \frac{2}{5} \)
Since the LCM of 1 and 5 is 5:
\( = \frac{-2 \times 5}{1 \times 5} + \frac{2 \times 1}{5 \times 1} \)
\( = \frac{-10 + 2}{5} \)
\( = \frac{-8}{5} \)
This solution is a rational number.

(vii) \( \frac{9}{-4} + \left(\frac{-3}{8}\right) \)
Expressing with positive denominators:
\( = \frac{-9}{4} + \left(\frac{-3}{8}\right) \)
Finding the LCM of 4 and 8:
LCM \( = 8 \)
\( = \frac{-9 \times 2}{4 \times 2} + \frac{-3 \times 1}{8 \times 1} \)
\( = \frac{-18 - 3}{8} \)
\( = \frac{-21}{8} \)
This is a rational number.

(viii) \( \frac{7}{-18} + \frac{8}{27} \)
Expressing with positive denominators:
\( = \frac{-7}{18} + \frac{8}{27} \)
Finding the LCM of 18 and 27:
LCM \( = 2 \times 3 \times 3 \times 3 = 54 \)
\( = \frac{-7 \times 3}{18 \times 3} + \frac{8 \times 2}{27 \times 2} \)
\( = \frac{-21 + 16}{54} \)
\( = \frac{-5}{54} \)
This final value is a rational number.

In simple words: To add fractions, make their bottom numbers the same by finding their LCM. Once they are equal, add the top numbers. The final answer will always be a fraction, which is a rational number.

Exam Tip: Always make the denominator positive before calculating, as this prevents errors with negative signs in the numerator.

Question 2. Evaluate:
(i) \( \frac{5}{9} + \frac{-7}{6} \)
(ii) \( 4 + \frac{3}{-5} \)
(iii) \( \frac{1}{-15} + \frac{5}{-12} \)
(iv) \( \frac{5}{9} + \frac{3}{-4} \)
(v) \( \frac{-8}{9} + \frac{-5}{12} \)
(vi) \( 0 + \frac{-2}{7} \)
(vii) \( \frac{5}{-11} + 0 \)
(viii) \( 2 + \frac{-3}{5} \)
(ix) \( \frac{4}{-9} + 1 \)
Answer:
(i) \( \frac{5}{9} + \frac{-7}{6} \)
LCM of 9 and 6 is 18:
\( = \frac{5 \times 2}{9 \times 2} - \frac{7 \times 3}{6 \times 3} \)
\( = \frac{10 - 21}{18} \)
\( = \frac{-11}{18} \)

(ii) \( 4 + \frac{3}{-5} \)
\( = \frac{4}{1} - \frac{3}{5} \)
LCM of 1 and 5 is 5:
\( = \frac{4 \times 5}{1 \times 5} - \frac{3 \times 1}{5 \times 1} \)
\( = \frac{20 - 3}{5} \)
\( = \frac{17}{5} = 3\frac{2}{5} \)

(iii) \( \frac{1}{-15} + \frac{5}{-12} \)
\( = \frac{-1}{15} - \frac{5}{12} \)
LCM of 15 and 12 is 60:
\( = \frac{-1 \times 4}{15 \times 4} - \frac{5 \times 5}{12 \times 5} \)
\( = \frac{-4 - 25}{60} \)
\( = \frac{-29}{60} \)

(iv) \( \frac{5}{9} + \frac{3}{-4} \)
\( = \frac{5}{9} - \frac{3}{4} \)
LCM of 9 and 4 is 36:
\( = \frac{5 \times 4}{9 \times 4} - \frac{3 \times 9}{4 \times 9} \)
\( = \frac{20 - 27}{36} \)
\( = \frac{-7}{36} \)

(v) \( \frac{-8}{9} + \frac{-5}{12} \)
LCM of 9 and 12 is 36:
\( = \frac{-8 \times 4}{9 \times 4} - \frac{5 \times 3}{12 \times 3} \)
\( = \frac{-32 - 15}{36} \)
\( = \frac{-47}{36} \)

(vi) \( 0 + \frac{-2}{7} \)
\( = \frac{0}{1} - \frac{2}{7} \)
LCM of 1 and 7 is 7:
\( = \frac{0 \times 7}{1 \times 7} - \frac{2 \times 1}{7 \times 1} \)
\( = \frac{0 - 2}{7} \)
\( = \frac{-2}{7} \)

(vii) \( \frac{5}{-11} + 0 \)
\( = \frac{-5}{11} + \frac{0}{1} \)
LCM of 11 and 1 is 11:
\( = \frac{-5 \times 1}{11 \times 1} + \frac{0 \times 11}{1 \times 11} \)
\( = \frac{-5 + 0}{11} \)
\( = \frac{-5}{11} \)

(viii) \( 2 + \frac{-3}{5} \)
\( = \frac{2}{1} - \frac{3}{5} \)
LCM of 1 and 5 is 5:
\( = \frac{2 \times 5}{1 \times 5} - \frac{3 \times 1}{5 \times 1} \)
\( = \frac{10 - 3}{5} \)
\( = \frac{7}{5} = 1\frac{2}{5} \)

(ix) \( \frac{4}{-9} + 1 \)
\( = \frac{-4}{9} + \frac{1}{1} \)
LCM of 9 and 1 is 9:
\( = \frac{-4 \times 1}{9 \times 1} + \frac{1 \times 9}{1 \times 9} \)
\( = \frac{-4 + 9}{9} \)
\( = \frac{5}{9} \)

In simple words: To find the sum of fractions, first find their common bottom number (LCM). Then, change each fraction to have this common bottom, and finally combine the top numbers.

Exam Tip: Remember that adding zero to any number does not change its value. Similarly, look out for mixed fractions and convert them when needed, keeping the final answers in their simplest terms.

Question 3. Evaluate:
(i) \( \frac{3}{7} + \frac{-4}{9} + \frac{-11}{7} + \frac{7}{9} \)
(ii) \( \frac{2}{3} + \frac{-4}{5} + \frac{1}{3} + \frac{2}{5} \)
(iii) \( \frac{4}{7} + 0 + \frac{-8}{9} + \frac{-13}{7} + \frac{17}{9} \)
(iv) \( \frac{3}{8} + \frac{-5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{-5}{8} + \frac{-2}{7} \)
Answer:
(i) \( \frac{3}{7} + \frac{-4}{9} + \frac{-11}{7} + \frac{7}{9} \)
First, group the fractions with matching denominators:
\( = \left( \frac{3}{7} - \frac{11}{7} \right) + \left( \frac{-4}{9} + \frac{7}{9} \right) \)
\( = \frac{3 - 11}{7} + \frac{-4 + 7}{9} \)
\( = \frac{-8}{7} + \frac{3}{9} \)
Simplifying \( \frac{3}{9} \) gives \( \frac{1}{3} \):
\( = \frac{-8}{7} + \frac{1}{3} \)
Finding the LCM of 7 and 3, which is 21:
\( = \frac{-8 \times 3}{7 \times 3} + \frac{1 \times 7}{3 \times 7} \)
\( = \frac{-24 + 7}{21} = \frac{-17}{21} \)

(ii) \( \frac{2}{3} + \frac{-4}{5} + \frac{1}{3} + \frac{2}{5} \)
Group the fractions together by their denominators:
\( = \left( \frac{2}{3} + \frac{1}{3} \right) + \left( \frac{-4}{5} + \frac{2}{5} \right) \)
\( = \frac{2 + 1}{3} + \frac{-4 + 2}{5} \)
\( = \frac{3}{3} + \left( \frac{-2}{5} \right) \)
Taking the LCM of 3 and 5, which is 15:
\( = \frac{3 \times 5}{3 \times 5} - \frac{2 \times 3}{5 \times 3} \)
\( = \frac{15 - 6}{15} = \frac{9}{15} = \frac{3}{5} \)

(iii) \( \frac{4}{7} + 0 + \frac{-8}{9} + \frac{-13}{7} + \frac{17}{9} \)
Since zero doesn't change the sum, we can skip it and group the remaining terms:
\( = \left( \frac{4}{7} - \frac{13}{7} \right) + \left( \frac{-8}{9} + \frac{17}{9} \right) \)
\( = \frac{4 - 13}{7} + \frac{-8 + 17}{9} \)
\( = \frac{-9}{7} + \frac{9}{9} \)
Simplify the second term to 1:
\( = \frac{-9}{7} + \frac{1}{1} \)
Taking the LCM of 7 and 1, which is 7:
\( = \frac{-9 \times 1}{7 \times 1} + \frac{1 \times 7}{1 \times 7} \)
\( = \frac{-9 + 7}{7} = \frac{-2}{7} \)

(iv) \( \frac{3}{8} + \frac{-5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{-5}{8} + \frac{-2}{7} \)
Regroup terms that share the same denominators:
\( = \left( \frac{3}{8} - \frac{5}{8} \right) + \left( \frac{-5}{12} + \frac{3}{12} \right) + \left( \frac{3}{7} - \frac{2}{7} \right) \)
\( = \frac{3 - 5}{8} + \frac{-5 + 3}{12} + \frac{3 - 2}{7} \)
\( = \frac{-2}{8} + \frac{-2}{12} + \frac{1}{7} \)
Reducing the first two fractions to their simplest terms:
\( = \frac{-1}{4} - \frac{1}{6} + \frac{1}{7} \)
Finding the LCM of 4, 6, and 7, which is 84:
\( = \frac{-1 \times 21}{4 \times 21} - \frac{1 \times 14}{6 \times 14} + \frac{1 \times 12}{7 \times 12} \)
\( = \frac{-21 - 14 + 12}{84} \)
\( = \frac{-35 + 12}{84} = \frac{-23}{84} \)

In simple words: When adding several fractions, group the ones with the same bottom numbers together first. This makes the math much easier because you can combine those parts quickly before dealing with different denominators.

Exam Tip: Simplify individual fractions like \( \frac{3}{9} \) or \( \frac{-2}{8} \) early in the calculation. This keeps the numbers smaller and makes finding the overall LCM much easier.

Question 4. For each pair of rational numbers, verify commutative property of addition of rational numbers:
(i) \( \frac{-8}{7} \) and \( \frac{5}{14} \)
(ii) \( \frac{5}{9} \) and \( \frac{5}{-12} \)
(iii) \( \frac{-4}{5} \) and \( \frac{-13}{-15} \)
(iv) \( \frac{2}{-5} \) and \( \frac{11}{-15} \)
(v) 3 and \( \frac{-2}{7} \)
(vi) -2 and \( \frac{3}{-5} \)
Answer:
(i) We need to show that: \( \frac{-8}{7} + \frac{5}{14} = \frac{5}{14} + \frac{-8}{7} \)
Left hand side (LHS):
\( \frac{-8}{7} + \frac{5}{14} \)
The lowest common multiple (LCM) of 7 and 14 is 14.
\( \implies \frac{-8 \times 2}{7 \times 2} + \frac{5 \times 1}{14 \times 1} \)
\( \implies \frac{-16}{14} + \frac{5}{14} \)
\( \implies \frac{-16 + 5}{14} \)
\( \implies \frac{-11}{14} \)
Right hand side (RHS):
\( \frac{5}{14} + \frac{-8}{7} \)
\( \implies \frac{5 \times 1}{14 \times 1} + \frac{-8 \times 2}{7 \times 2} \)
\( \implies \frac{5 - 16}{14} \)
\( \implies \frac{-11}{14} \)
Since both LHS and RHS match, the property is verified.

(ii) We need to show that: \( \frac{5}{9} + \frac{5}{-12} = \frac{5}{-12} + \frac{5}{9} \)
We can write \( \frac{5}{-12} \) as \( \frac{-5}{12} \).
LHS:
\( \frac{5}{9} + \frac{-5}{12} \)
The LCM of 9 and 12 is 36.
\( \implies \frac{5 \times 4}{9 \times 4} + \frac{-5 \times 3}{12 \times 3} \)
\( \implies \frac{20 - 15}{36} \)
\( \implies \frac{5}{36} \)
RHS:
\( \frac{-5}{12} + \frac{5}{9} \)
\( \implies \frac{-5 \times 3}{12 \times 3} + \frac{5 \times 4}{9 \times 4} \)
\( \implies \frac{-15 + 20}{36} \)
\( \implies \frac{5}{36} \)
Since both calculations give the same result, the commutative property is correct.

(iii) We need to show that: \( \frac{-4}{5} + \frac{-13}{-15} = \frac{-13}{-15} + \frac{-4}{5} \)
We can simplify \( \frac{-13}{-15} \) to \( \frac{13}{15} \).
LHS:
\( \frac{-4}{5} + \frac{13}{15} \)
The LCM of 5 and 15 is 15.
\( \implies \frac{-4 \times 3}{5 \times 3} + \frac{13 \times 1}{15 \times 1} \)
\( \implies \frac{-12 + 13}{15} \)
\( \implies \frac{1}{15} \)
RHS:
\( \frac{13}{15} + \frac{-4}{5} \)
\( \implies \frac{13 \times 1}{15 \times 1} + \frac{-4 \times 3}{5 \times 3} \)
\( \implies \frac{13 - 12}{15} \)
\( \implies \frac{1}{15} \)
The property is verified as both sides are equal.

(iv) We need to show that: \( \frac{2}{-5} + \frac{11}{-15} = \frac{11}{-15} + \frac{2}{-5} \)
We can write the fractions as \( \frac{-2}{5} \) and \( \frac{-11}{15} \).
LHS:
\( \frac{-2}{5} + \frac{-11}{15} \)
The LCM of 5 and 15 is 15.
\( \implies \frac{-2 \times 3}{5 \times 3} + \frac{-11 \times 1}{15 \times 1} \)
\( \implies \frac{-6 - 11}{15} \)
\( \implies \frac{-17}{15} \)
RHS:
\( \frac{-11}{15} + \frac{-2}{5} \)
\( \implies \frac{-11 \times 1}{15 \times 1} + \frac{-2 \times 3}{5 \times 3} \)
\( \implies \frac{-11 - 6}{15} \)
\( \implies \frac{-17}{15} \)
Both sides match, so the property holds true.

(v) We need to show that: \( 3 + \frac{-2}{7} = \frac{-2}{7} + 3 \)
We can write 3 as \( \frac{3}{1} \).
LHS:
\( \frac{3}{1} + \frac{-2}{7} \)
The LCM of 1 and 7 is 7.
\( \implies \frac{3 \times 7}{1 \times 7} + \frac{-2 \times 1}{7 \times 1} \)
\( \implies \frac{21 - 2}{7} \)
\( \implies \frac{19}{7} \)
RHS:
\( \frac{-2}{7} + \frac{3}{1} \)
\( \implies \frac{-2 \times 1}{7 \times 1} + \frac{3 \times 7}{1 \times 7} \)
\( \implies \frac{-2 + 21}{7} \)
\( \implies \frac{19}{7} \)
Both equations yield the same answer, so the property is correct.

(vi) We need to show that: \( -2 + \frac{3}{-5} = \frac{3}{-5} + (-2) \)
We can write the numbers as \( \frac{-2}{1} \) and \( \frac{-3}{5} \).
LHS:
\( \frac{-2}{1} + \frac{-3}{5} \)
The LCM of 1 and 5 is 5.
\( \implies \frac{-2 \times 5}{1 \times 5} + \frac{-3 \times 1}{5 \times 1} \)
\( \implies \frac{-10 - 3}{5} \)
\( \implies \frac{-13}{5} \)
RHS:
\( \frac{-3}{5} + \frac{-2}{1} \)
\( \implies \frac{-3 \times 1}{5 \times 1} + \frac{-2 \times 5}{1 \times 5} \)
\( \implies \frac{-3 - 10}{5} \)
\( \implies \frac{-13}{5} \)
Since both LHS and RHS are equal, the property is verified.
In simple words: The commutative property means that when you add two numbers, the order does not matter. You will get the same final answer whether you add the first number to the second, or the second to the first.

Exam Tip: When proving commutative property, always solve LHS and RHS separately. Always write negative signs in the numerator before finding the common denominator.

Question 5. For each set of rational numbers, given below, verify the associative property of addition of rational numbers:
(i) \( \frac { 1 }{ 2 } \), \( \frac { 2 }{ 3 } \) and \( \frac { -1 }{ 6 } \)
(ii) \( \frac { -2 }{ 5 } \), \( \frac { 4 }{ 15 } \) and \( \frac { -7 }{ 10 } \)
(iii) \( \frac { -7 }{ 9 } \), \( \frac { 2 }{ -3 } \) and \( \frac { -5 }{ 18 } \)
(iv) -1, \( \frac { 5 }{ 6 } \) and \( \frac { -2 }{ 3 } \)
Answer:
(i) We need to show that: \( \frac{1}{2} + \left(\frac{2}{3} + \frac{-1}{6}\right) = \left(\frac{1}{2} + \frac{2}{3}\right) + \frac{-1}{6} \)
LHS:
\( \frac{1}{2} + \left(\frac{2}{3} + \frac{-1}{6}\right) \)
The LCM of 3 and 6 is 6.
\( \implies \frac{1}{2} + \left(\frac{2 \times 2}{3 \times 2} + \frac{-1 \times 1}{6 \times 1}\right) \)
\( \implies \frac{1}{2} + \left(\frac{4 - 1}{6}\right) \)
\( \implies \frac{1}{2} + \frac{3}{6} \)
The LCM of 2 and 6 is 6.
\( \implies \frac{1 \times 3}{2 \times 3} + \frac{3 \times 1}{6 \times 1} \)
\( \implies \frac{3 + 3}{6} \)
\( \implies \frac{6}{6} = 1 \)
RHS:
\( \left(\frac{1}{2} + \frac{2}{3}\right) + \frac{-1}{6} \)
The LCM of 2 and 3 is 6.
\( \implies \left(\frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2}\right) + \frac{-1}{6} \)
\( \implies \left(\frac{3 + 4}{6}\right) + \frac{-1}{6} \)
\( \implies \frac{7}{6} + \frac{-1}{6} \)
\( \implies \frac{7 - 1}{6} \)
\( \implies \frac{6}{6} = 1 \)
LHS and RHS are equal to 1, so the associative property is verified.

(ii) We need to show that: \( \frac{-2}{5} + \left(\frac{4}{15} + \frac{-7}{10}\right) = \left(\frac{-2}{5} + \frac{4}{15}\right) + \frac{-7}{10} \)
LHS:
\( \frac{-2}{5} + \left(\frac{4}{15} + \frac{-7}{10}\right) \)
The LCM of 15 and 10 is 30.
\( \implies \frac{-2}{5} + \left(\frac{4 \times 2}{15 \times 2} + \frac{-7 \times 3}{10 \times 3}\right) \)
\( \implies \frac{-2}{5} + \left(\frac{8 - 21}{30}\right) \)
\( \implies \frac{-2}{5} + \frac{-13}{30} \)
The LCM of 5 and 30 is 30.
\( \implies \frac{-2 \times 6}{5 \times 6} + \frac{-13 \times 1}{30 \times 1} \)
\( \implies \frac{-12 - 13}{30} \)
\( \implies \frac{-25}{30} = \frac{-5}{6} \)
RHS:
\( \left(\frac{-2}{5} + \frac{4}{15}\right) + \frac{-7}{10} \)
The LCM of 5 and 15 is 15.
\( \implies \left(\frac{-2 \times 3}{5 \times 3} + \frac{4 \times 1}{15 \times 1}\right) + \frac{-7}{10} \)
\( \implies \left(\frac{-6 + 4}{15}\right) + \frac{-7}{10} \)
\( \implies \frac{-2}{15} + \frac{-7}{10} \)
The LCM of 15 and 10 is 30.
\( \implies \frac{-2 \times 2}{15 \times 2} + \frac{-7 \times 3}{10 \times 3} \)
\( \implies \frac{-4 - 21}{30} \)
\( \implies \frac{-25}{30} = \frac{-5}{6} \)
LHS and RHS are both \( \frac{-5}{6} \), so they are equal.

(iii) We need to show that: \( \frac{-7}{9} + \left(\frac{2}{-3} + \frac{-5}{18}\right) = \left(\frac{-7}{9} + \frac{2}{-3}\right) + \frac{-5}{18} \)
We can write \( \frac{2}{-3} \) as \( \frac{-2}{3} \).
LHS:
\( \frac{-7}{9} + \left(\frac{-2}{3} + \frac{-5}{18}\right) \)
The LCM of 3 and 18 is 18.
\( \implies \frac{-7}{9} + \left(\frac{-2 \times 6}{3 \times 6} + \frac{-5 \times 1}{18 \times 1}\right) \)
\( \implies \frac{-7}{9} + \left(\frac{-12 - 5}{18}\right) \)
\( \implies \frac{-7}{9} + \frac{-17}{18} \)
The LCM of 9 and 18 is 18.
\( \implies \frac{-7 \times 2}{9 \times 2} + \frac{-17 \times 1}{18 \times 1} \)
\( \implies \frac{-14 - 17}{18} \)
\( \implies \frac{-31}{18} \)
RHS:
\( \left(\frac{-7}{9} + \frac{-2}{3}\right) + \frac{-5}{18} \)
The LCM of 9 and 3 is 9.
\( \implies \left(\frac{-7 \times 1}{9 \times 1} + \frac{-2 \times 3}{3 \times 3}\right) + \frac{-5}{18} \)
\( \implies \left(\frac{-7 - 6}{9}\right) + \frac{-5}{18} \)
\( \implies \frac{-13}{9} + \frac{-5}{18} \)
The LCM of 9 and 18 is 18.
\( \implies \frac{-13 \times 2}{9 \times 2} + \frac{-5 \times 1}{18 \times 1} \)
\( \implies \frac{-26 - 5}{18} \)
\( \implies \frac{-31}{18} \)
The property is verified because LHS = RHS.

(iv) We need to show that: \( -1 + \left(\frac{5}{6} + \frac{-2}{3}\right) = \left(-1 + \frac{5}{6}\right) + \frac{-2}{3} \)
We can write -1 as \( \frac{-1}{1} \).
LHS:
\( \frac{-1}{1} + \left(\frac{5}{6} + \frac{-2}{3}\right) \)
The LCM of 6 and 3 is 6.
\( \implies \frac{-1}{1} + \left(\frac{5 \times 1}{6 \times 1} + \frac{-2 \times 2}{3 \times 2}\right) \)
\( \implies \frac{-1}{1} + \left(\frac{5 - 4}{6}\right) \)
\( \implies \frac{-1}{1} + \frac{1}{6} \)
The LCM of 1 and 6 is 6.
\( \implies \frac{-1 \times 6}{1 \times 6} + \frac{1 \times 1}{6 \times 1} \)
\( \implies \frac{-6 + 1}{6} \)
\( \implies \frac{-5}{6} \)
RHS:
\( \left(\frac{-1}{1} + \frac{5}{6}\right) + \frac{-2}{3} \)
The LCM of 1 and 6 is 6.
\( \implies \left(\frac{-1 \times 6}{1 \times 6} + \frac{5 \times 1}{6 \times 1}\right) + \frac{-2}{3} \)
\( \implies \left(\frac{-6 + 5}{6}\right) + \frac{-2}{3} \)
\( \implies \frac{-1}{6} + \frac{-2}{3} \)
The LCM of 6 and 3 is 6.
\( \implies \frac{-1 \times 1}{6 \times 1} + \frac{-2 \times 2}{3 \times 2} \)
\( \implies \frac{-1 - 4}{6} \)
\( \implies \frac{-5}{6} \)
Since both sides match, the associative property is verified.
In simple words: The associative property means that when adding three numbers, you can group them in any way you like. You will get the same final answer whether you add the first two first or the last two first.

Exam Tip: Be very careful when simplifying fractions inside parentheses. Always perform operations inside the brackets first according to the BODMAS rule.

Question 6. Write the additive inverse (negative) of:
(i) \( \frac {-3}{ 8 } \)
(ii) \( \frac { 4 }{ -9 } \)
(iii) \( \frac { -7 }{ 5 } \)
(iv) \( \frac { -4 }{ -13 } \)
(v) 0
(vi) -2
(vii) 1
(viii) \( -\frac { 1 }{ 3 } \)
(ix) \( \frac { -3 }{ -1 } \)
Answer:
(i) The additive inverse of \( \frac{-3}{8} \) is \( \frac{3}{8} \).
(ii) Since \( \frac{4}{-9} = \frac{-4}{9} \), its additive inverse is \( \frac{4}{9} \).
(iii) The additive inverse of \( \frac{-7}{5} \) is \( \frac{7}{5} \).
(iv) Since \( \frac{-4}{-13} = \frac{4}{13} \), its additive inverse is \( \frac{-4}{13} \).
(v) The additive inverse of 0 is 0.
(vi) The additive inverse of -2 is 2.
(vii) The additive inverse of 1 is -1.
(viii) The additive inverse of \( -\frac{1}{3} \) is \( \frac{1}{3} \).
(ix) Since \( \frac{-3}{-1} = 3 \), its additive inverse is -3.
In simple words: The additive inverse is just the opposite of a number. When you add a number and its additive inverse together, the answer is always zero.

Exam Tip: Simplify the signs of the rational number first. For example, if a fraction has two minus signs like \( \frac{-a}{-b} \), it is a positive number, so its additive inverse must be negative.

Question 7. Fill in the blanks:
(i) Additive inverse of \( \frac{-5}{-12} \) = .......... .
(ii) \( \frac{-5}{-12} \) + its additive inverse = .......... .
(iii) If \( \frac{a}{b} \) is additive inverse of \( \frac{-c}{d} \), then \( \frac{-c}{d} \) is additive inverse of .......... .
Also so \( \frac{a}{b} + \frac{-c}{d} - \left( \frac{-c}{d} + \frac{a}{b} \right) \) = .......... .
Answer:
(i) \( -\frac{5}{12} \)
(ii) \( 0 \)
(iii) \( \frac{a}{b} \)
Also so \( \frac{a}{b} + \frac{-c}{d} - \left( \frac{-c}{d} + \frac{a}{b} \right) = 0 \)
In simple words: The additive inverse is the number you add to get zero. If you add any fraction to its opposite, the final sum is always zero.

Exam Tip: Remember that \( \frac{-a}{-b} \) is a positive fraction, so its additive inverse must be negative. Keep this sign change in mind during tests.

Question 8. State, true or false:
(i) \( \frac{7}{9} = \frac{7+5}{9+5} \)
(ii) \( \frac{7}{9} = \frac{7-5}{9-5} \)
(iii) \( \frac{7}{9} = \frac{7 \times 5}{9 \times 5} \)
(iv) \( \frac{7}{9} = \frac{7 \div 5}{9 \div 5} \)
(v) \( \frac{-5}{-12} \) is a negative rational number
(vi) \( \frac{-13}{25} \) is smaller than \( \frac{-25}{13} \)
Answer:
(i) False
(ii) False
(iii) True
(iv) True
(v) False
(vi) False
In simple words: Adding or subtracting the same number to the top and bottom changes a fraction. Only multiplying or dividing keeps it equal. Also, negative fractions with two minus signs are actually positive.

Exam Tip: Remember that dividing or multiplying both terms of a fraction by a non-zero number creates an equivalent fraction, while adding or subtracting does not.

Exercise 1(B)

Question 1. Evaluate:
(i) \( \frac{2}{3} - \frac{4}{5} \)
(ii) \( \frac{-4}{9} - \frac{2}{-3} \)
(iii) \( -1 - \frac{4}{9} \)
(iv) \( \frac{-2}{7} - \frac{3}{-14} \)
(v) \( \frac{-5}{18} - \frac{-2}{9} \)
(vi) \( \frac{5}{21} - \frac{-13}{42} \)
Answer:
(i) To calculate \( \frac{2}{3} - \frac{4}{5} \), find the LCM of 3 and 5, which is 15.
\( \implies \frac{2 \times 5}{3 \times 5} - \frac{4 \times 3}{5 \times 3} \)
\( \implies \frac{10}{15} - \frac{12}{15} \)
\( \implies \frac{10 - 12}{15} = \frac{-2}{15} \)

(ii) For \( \frac{-4}{9} - \frac{2}{-3} \), write the second fraction with a positive denominator:
\( \implies \frac{-4}{9} - \frac{-2}{3} \)
\( \implies \frac{-4}{9} + \frac{2}{3} \)
The LCM of 9 and 3 is 9.
\( \implies \frac{-4}{9} + \frac{2 \times 3}{3 \times 3} \)
\( \implies \frac{-4}{9} + \frac{6}{9} \)
\( \implies \frac{-4 + 6}{9} = \frac{2}{9} \)

(iii) For \( -1 - \frac{4}{9} \), rewrite the integer -1 as a fraction:
\( \implies \frac{-1}{1} - \frac{4}{9} \)
Using the common denominator 9:
\( \implies \frac{-1 \times 9}{1 \times 9} - \frac{4}{9} \)
\( \implies \frac{-9 - 4}{9} = \frac{-13}{9} \)

(iv) To evaluate \( \frac{-2}{7} - \frac{3}{-14} \), rewrite the denominators first:
\( \implies \frac{-2}{7} - \frac{-3}{14} \)
\( \implies \frac{-2}{7} + \frac{3}{14} \)
The LCM of 7 and 14 is 14.
\( \implies \frac{-2 \times 2}{7 \times 2} + \frac{3}{14} \)
\( \implies \frac{-4 + 3}{14} = \frac{-1}{14} \)

(v) Simplify \( \frac{-5}{18} - \frac{-2}{9} \):
\( \implies \frac{-5}{18} + \frac{2}{9} \)
The LCM of 18 and 9 is 18.
\( \implies \frac{-5}{18} + \frac{2 \times 2}{9 \times 2} \)
\( \implies \frac{-5 + 4}{18} = \frac{-1}{18} \)

(vi) Solve \( \frac{5}{21} - \frac{-13}{42} \):
\( \implies \frac{5}{21} + \frac{13}{42} \)
The LCM of 21 and 42 is 42.
\( \implies \frac{5 \times 2}{21 \times 2} + \frac{13}{42} \)
\( \implies \frac{10 + 13}{42} = \frac{23}{42} \)
In simple words: When subtracting fractions, make the bottom numbers match by finding their LCM. Remember that subtracting a negative number is the same as adding.

Exam Tip: Always check if your final answer can be simplified to lower terms. Keeping the signs clear at each step will prevent simple arithmetic errors.

Question 2. Subtract:
(i) \( \frac{5}{8} \) from \( \frac{-3}{8} \)
(ii) \( \frac{-8}{11} \) from \( \frac{4}{11} \)
(iii) \( \frac{4}{9} \) from \( \frac{-5}{9} \)
(iv) \( \frac{1}{4} \) from \( \frac{-3}{8} \)
(v) \( \frac{-5}{8} \) from \( \frac{-13}{16} \)
(vi) \( \frac{-9}{22} \) from \( \frac{5}{33} \)
Answer:
(i) Subtract \( \frac{5}{8} \) from \( \frac{-3}{8} \):
\( \implies \frac{-3}{8} - \frac{5}{8} \)
\( \implies \frac{-3 - 5}{8} = \frac{-8}{8} = -1 \)

(ii) Subtract \( \frac{-8}{11} \) from \( \frac{4}{11} \):
\( \implies \frac{4}{11} - \left( \frac{-8}{11} \right) \)
\( \implies \frac{4 + 8}{11} = \frac{12}{11} = 1\frac{1}{11} \)

(iii) Subtract \( \frac{4}{9} \) from \( \frac{-5}{9} \):
\( \implies \frac{-5}{9} - \frac{4}{9} \)
\( \implies \frac{-5 - 4}{9} = \frac{-9}{9} = -1 \)

(iv) Subtract \( \frac{1}{4} \) from \( \frac{-3}{8} \):
\( \implies \frac{-3}{8} - \frac{1}{4} \)
The LCM of 8 and 4 is 8.
\( \implies \frac{-3}{8} - \frac{1 \times 2}{4 \times 2} \)
\( \implies \frac{-3 - 2}{8} = \frac{-5}{8} \)

(v) Subtract \( \frac{-5}{8} \) from \( \frac{-13}{16} \):
\( \implies \frac{-13}{16} - \left( \frac{-5}{8} \right) \)
\( \implies \frac{-13}{16} + \frac{5}{8} \)
The LCM of 16 and 8 is 16.
\( \implies \frac{-13}{16} + \frac{5 \times 2}{8 \times 2} \)
\( \implies \frac{-13 + 10}{16} = \frac{-3}{16} \)

(vi) Subtract \( \frac{-9}{22} \) from \( \frac{5}{33} \):
\( \implies \frac{5}{33} - \left( \frac{-9}{22} \right) \)
\( \implies \frac{5}{33} + \frac{9}{22} \)
The LCM of 33 and 22 is 66.
\( \implies \frac{5 \times 2}{33 \times 2} + \frac{9 \times 3}{22 \times 3} \)
\( \implies \frac{10 + 27}{66} = \frac{37}{66} \)
In simple words: When the problem says "subtract A from B", it means you write B minus A. If A has a minus sign, it turns into addition.

Exam Tip: Be very careful with the phrase "subtract A from B". This always means \( B - A \). Writing \( A - B \) is a common mistake that leads to the incorrect sign.

Question 3. The sum of two rational numbers is \( \frac{9}{20} \). If one of them is \( \frac{2}{5} \), find the other.
Answer: Let the second rational number be \( x \).
The sum of both rational numbers is \( \frac{9}{20} \).
\( \implies x + \frac{2}{5} = \frac{9}{20} \)
To solve for \( x \), subtract \( \frac{2}{5} \) from \( \frac{9}{20} \):
\( \implies x = \frac{9}{20} - \frac{2}{5} \)
The LCM of 20 and 5 is 20.
\( \implies x = \frac{9}{20} - \frac{2 \times 4}{5 \times 4} \)
\( \implies x = \frac{9 - 8}{20} \)
\( \implies x = \frac{1}{20} \)
So, the other rational number is \( \frac{1}{20} \).
In simple words: If you know the total sum and one of the parts, you can easily find the other part by subtracting the known part from the total.

Exam Tip: Clearly write down the variable you are solving for, like \(x\). This makes your steps easy for the examiner to read and grade.

Question 4. The sum of two rational numbers is \( \frac{-2}{3} \). If one of them is \( \frac{-8}{15} \), find the other.
Answer: Let the other rational number be \( x \).
The sum of the two numbers is given as \( \frac{-2}{3} \).
\( \implies x + \left( \frac{-8}{15} \right) = \frac{-2}{3} \)
To find \( x \), subtract \( \frac{-8}{15} \) from both sides:
\( \implies x = \frac{-2}{3} - \left( \frac{-8}{15} \right) \)
\( \implies x = \frac{-2}{3} + \frac{8}{15} \)
The LCM of 3 and 15 is 15.
\( \implies x = \frac{-2 \times 5}{3 \times 5} + \frac{8}{15} \)
\( \implies x = \frac{-10 + 8}{15} \)
\( \implies x = \frac{-2}{15} \)
Thus, the other rational number is \( \frac{-2}{15} \).
In simple words: Subtract the known number from the total sum. Subtracting a negative number means you add it to the sum instead.

Exam Tip: Keep an eye on negative signs. When you subtract a negative number, like \( -(-\frac{8}{15}) \), it becomes positive.

Question 5. The sum of the two rational numbers is -6. If one of them is \( \frac{-8}{5} \), find the other.
Answer: Let the other rational number be \( x \).
According to the question, their sum is -6.
\( \implies x + \left( \frac{-8}{5} \right) = -6 \)
\( \implies x = -6 - \left( \frac{-8}{5} \right) \)
\( \implies x = \frac{-6}{1} + \frac{8}{5} \)
The LCM of 1 and 5 is 5.
\( \implies x = \frac{-6 \times 5}{1 \times 5} + \frac{8}{5} \)
\( \implies x = \frac{-30 + 8}{5} \)
\( \implies x = \frac{-22}{5} = -4\frac{2}{5} \)
Therefore, the other rational number is \( \frac{-22}{5} \) (or \( -4\frac{2}{5} \)).
In simple words: Write the integer -6 as \(\frac{-6}{1}\) to easily subtract the other fraction from it by finding a common denominator.

Exam Tip: Writing integers like -6 as a fraction over 1 is a useful trick to prevent errors when finding a common denominator.

Question 6. Which rational number should be added to \( \frac{-7}{8} \) to get \( \frac{5}{9} \)?
Answer: Let the required rational number be \( x \).
This gives the equation:
\( \implies \frac{-7}{8} + x = \frac{5}{9} \)
Solve for \( x \) by isolating it on one side:
\( \implies x = \frac{5}{9} - \left( \frac{-7}{8} \right) \)
\( \implies x = \frac{5}{9} + \frac{7}{8} \)
The LCM of 9 and 8 is 72.
\( \implies x = \frac{5 \times 8}{9 \times 8} + \frac{7 \times 9}{8 \times 9} \)
\( \implies x = \frac{40 + 63}{72} \)
\( \implies x = \frac{103}{72} = 1\frac{31}{72} \)
So, the number to be added is \( \frac{103}{72} \) (or \( 1\frac{31}{72} \)).
In simple words: To find out what to add to a fraction to reach your target, subtract the starting fraction from that target.

Exam Tip: Write your final answer as a mixed number if the fraction is improper, as this demonstrates a complete and thorough solution.

Question 7. Which rational number should be added to \( \frac{-5}{9} \) to get \( \frac{-2}{3} \)?
Answer: Let the missing rational number be \( x \).
This can be written as:
\( \implies \frac{-5}{9} + x = \frac{-2}{3} \)
Solve for \( x \) by subtracting \( \frac{-5}{9} \) from \( \frac{-2}{3} \):
\( \implies x = \frac{-2}{3} - \left( \frac{-5}{9} \right) \)
\( \implies x = \frac{-2}{3} + \frac{5}{9} \)
The LCM of 3 and 9 is 9.
\( \implies x = \frac{-2 \times 3}{3 \times 3} + \frac{5}{9} \)
\( \implies x = \frac{-6 + 5}{9} \)
\( \implies x = \frac{-1}{9} \)
So, the required rational number is \( \frac{-1}{9} \).
In simple words: Subtract the number you have from your goal. Keep the denominators matching to finish the subtraction.

Exam Tip: Pay attention to the negative sign in the final answer. Missed signs are the most common source of lost marks on algebraic questions.

Question 8. Which rational number should be subtracted from \( \frac{-5}{6} \) to get \( \frac{4}{9} \)?
Answer: Let the rational number to be subtracted be \( x \).
This gives the expression:
\( \implies \frac{-5}{6} - x = \frac{4}{9} \)
Rearrange to solve for \( x \):
\( \implies x = \frac{-5}{6} - \frac{4}{9} \)
The LCM of 6 and 9 is 18.
\( \implies x = \frac{-5 \times 3}{6 \times 3} - \frac{4 \times 2}{9 \times 2} \)
\( \implies x = \frac{-15 - 8}{18} \)
\( \implies x = \frac{-23}{18} = -1\frac{5}{18} \)
Thus, the number to be subtracted is \( \frac{-23}{18} \) (or \( -1\frac{5}{18} \)).
In simple words: To find what needs to be taken away from a value to get a target, subtract that target from the starting value.

Exam Tip: Be careful with the algebraic setup. For "subtracted from \(A\)", use the equation \(A - x = B\), which simplifies directly to \(x = A - B\).

Question 9.
(i) What should be subtracted from -2 to get \( \frac{3}{8} \)
(ii) What should be added to -2 to get \( \frac{3}{8} \)
Answer:
(i) Let the required number to be subtracted be \( x \).
\( \implies -2 - x = \frac{3}{8} \)
\( \implies -x = \frac{3}{8} + 2 \)
Convert 2 to a fraction with a denominator of 8:
\( \implies -x = \frac{3 + 16}{8} \)
\( \implies -x = \frac{19}{8} \)
\( \implies x = \frac{-19}{8} = -2\frac{3}{8} \)
So, the number to be subtracted is \( \frac{-19}{8} \).

(ii) Let the required number to be added be \( x \).
\( \implies -2 + x = \frac{3}{8} \)
\( \implies x = \frac{3}{8} + 2 \)
Using a common denominator of 8:
\( \implies x = \frac{3 + 16}{8} \)
\( \implies x = \frac{19}{8} = 2\frac{3}{8} \)
So, the number to be added is \( \frac{19}{8} \).
In simple words: The first part solves for a number being subtracted, which gives a negative result. The second part solves for a number being added, which gives a positive result.

Exam Tip: Check that your signs are correct during transposition. Moving a term to the opposite side of the equals sign reverses its sign.

Question 10. Evaluate:
(i) \( \frac{3}{7} + \frac{-4}{9} - \frac{-11}{7} - \frac{7}{9} \)
(ii) \( \frac{2}{3} + \frac{-4}{5} - \frac{1}{3} - \frac{2}{5} \)
(iii) \( \frac{4}{7} - \frac{-8}{9} - \frac{-13}{7} + \frac{17}{9} \)
Answer:
(i) Group the fractions with the same denominators:
\( \implies \left( \frac{3}{7} - \frac{-11}{7} \right) + \left( \frac{-4}{9} - \frac{7}{9} \right) \)
\( \implies \left( \frac{3 + 11}{7} \right) + \left( \frac{-4 - 7}{9} \right) \)
\( \implies \frac{14}{7} + \left( \frac{-11}{9} \right) \)
Since \( \frac{14}{7} = 2 \), we get:
\( \implies 2 - \frac{11}{9} \)
\( \implies \frac{2 \times 9 - 11}{9} \)
\( \implies \frac{18 - 11}{9} = \frac{7}{9} \)

(ii) Group terms by their denominators 3 and 5:
\( \implies \left( \frac{2}{3} - \frac{1}{3} \right) + \left( \frac{-4}{5} - \frac{2}{5} \right) \)
\( \implies \frac{1}{3} + \left( \frac{-4 - 2}{5} \right) \)
\( \implies \frac{1}{3} - \frac{6}{5} \)
The LCM of 3 and 5 is 15.
\( \implies \frac{1 \times 5 - 6 \times 3}{15} \)
\( \implies \frac{5 - 18}{15} = \frac{-13}{15} \)

(iii) Group terms by their denominators 7 and 9:
\( \implies \left( \frac{4}{7} - \frac{-13}{7} \right) - \left( \frac{-8}{9} - \frac{17}{9} \right) \)
\( \implies \left( \frac{4 + 13}{7} \right) - \left( \frac{-8 - 17}{9} \right) \)
\( \implies \frac{17}{7} - \left( \frac{-25}{9} \right) \)
\( \implies \frac{17}{7} + \frac{25}{9} \)
The LCM of 7 and 9 is 63.
\( \implies \frac{17 \times 9 + 25 \times 7}{63} \)
\( \implies \frac{153 + 175}{63} \)
\( \implies \frac{328}{63} = 5\frac{13}{63} \)
In simple words: When you have to add or subtract many fractions, group the ones with the same bottom numbers first. This makes the arithmetic much simpler.

Exam Tip: Grouping terms with common denominators using the associative property is an excellent way to save time and prevent mistakes on exams.

Exercise 1(C)

Question 1. Evaluate:
(i) \( \frac{-14}{5} \times \frac{-6}{7} \)
(ii) \( \frac{7}{6} \times \frac{-18}{91} \)
(iii) \( \frac{-125}{72} \times \frac{9}{-5} \)
(iv) \( \frac{-11}{9} \times \frac{-51}{-44} \)
(v) \( -\frac{16}{5} \times \frac{20}{8} \)
Answer:
(i) \( \frac{-14}{5} \times \frac{-6}{7} \)
First, we multiply the numerators together and the denominators together:
\( = \frac{(-14) \times (-6)}{5 \times 7} \)
Simplify by dividing \( -14 \) and \( 7 \) by \( 7 \):
\( = \frac{(-2) \times (-6)}{5 \times 1} \)
\( = \frac{12}{5} = 2\frac{2}{5} \)

(ii) \( \frac{7}{6} \times \frac{-18}{91} \)
Multiply the numerators and the denominators:
\( = \frac{7 \times (-18)}{6 \times 91} \)
We can simplify this by dividing \( 7 \) and \( 91 \) by \( 7 \), and dividing \( -18 \) and \( 6 \) by \( 6 \):
\( = \frac{1 \times (-3)}{1 \times 13} = \frac{-3}{13} \)

(iii) \( \frac{-125}{72} \times \frac{9}{-5} \)
Combine into a single fraction:
\( = \frac{(-125) \times 9}{72 \times (-5)} \)
We divide \( -125 \) and \( -5 \) by \( -5 \) to get \( 25 \) and \( 1 \). We also divide \( 9 \) and \( 72 \) by \( 9 \) to get \( 1 \) and \( 8 \):
\( = \frac{25 \times 1}{8 \times 1} \)
\( = \frac{25}{8} = 3\frac{1}{8} \)

(iv) \( \frac{-11}{9} \times \frac{-51}{-44} \)
Write as a single fraction:
\( = \frac{(-11) \times (-51)}{9 \times (-44)} \)
Simplify by dividing both \( -11 \) and \( -44 \) by \( -11 \) to get \( 1 \) and \( 4 \):
\( = \frac{1 \times (-51)}{9 \times 4} \)
\( = \frac{-51}{36} \)
Divide the numerator and denominator by \( 3 \) to reduce the fraction:
\( = \frac{-17}{12} \)

(v) \( -\frac{16}{5} \times \frac{20}{8} \)
Multiply across:
\( = \frac{(-16) \times 20}{5 \times 8} \)
Reduce by dividing \( -16 \) and \( 8 \) by \( 8 \) to get \( -2 \) and \( 1 \), and dividing \( 20 \) and \( 5 \) by \( 5 \) to get \( 4 \) and \( 1 \):
\( = \frac{(-2) \times 4}{1 \times 1} \)
\( = -8 \)
In simple words: To multiply fractions, you multiply the top numbers together and the bottom numbers together. Before getting the final answer, you can make the numbers smaller by dividing the top and bottom by the same numbers.

Exam Tip: Always look for common factors in the numerators and denominators to simplify the fractions before multiplying. This prevents calculation errors with larger numbers.

Question 2. Multiply:
(i) \( \frac{5}{6} \) and \( \frac{8}{9} \)
(ii) \( \frac{2}{7} \) and \( \frac{-14}{9} \)
(iii) \( \frac{-7}{8} \) and \( 4 \)
(iv) \( \frac{36}{-7} \) and \( \frac{-9}{28} \)
(v) \( \frac{-7}{10} \) and \( \frac{-8}{15} \)
(vi) \( \frac{3}{-2} \) and \( \frac{-7}{3} \)
Answer:
(i) \( \frac{5}{6} \times \frac{8}{9} \)
Combine the numerators and denominators:
\( = \frac{5 \times 8}{6 \times 9} \)
Simplify by dividing \( 8 \) and \( 6 \) by \( 2 \):
\( = \frac{5 \times 4}{3 \times 9} = \frac{20}{27} \)

(ii) \( \frac{2}{7} \times \frac{-14}{9} \)
Multiply the numerators and denominators:
\( = \frac{2 \times (-14)}{7 \times 9} \)
Divide \( -14 \) and \( 7 \) by \( 7 \) to simplify:
\( = \frac{2 \times (-2)}{1 \times 9} = \frac{-4}{9} \)

(iii) \( \frac{-7}{8} \times 4 \)
Express the whole number \( 4 \) as \( \frac{4}{1} \) and multiply:
\( = \frac{(-7) \times 4}{8 \times 1} \)
Simplify by dividing \( 4 \) and \( 8 \) by \( 4 \):
\( = \frac{(-7) \times 1}{2 \times 1} = \frac{-7}{2} = -3\frac{1}{2} \)

(iv) \( \frac{36}{-7} \times \frac{-9}{28} \)
Multiply across:
\( = \frac{36 \times (-9)}{(-7) \times 28} \)
Simplify by dividing \( 36 \) and \( 28 \) by \( 4 \):
\( = \frac{9 \times (-9)}{(-7) \times 7} = \frac{-81}{-49} = \frac{81}{49} = 1\frac{32}{49} \)

(v) \( \frac{-7}{10} \times \frac{-8}{15} \)
Combine:
\( = \frac{(-7) \times (-8)}{10 \times 15} \)
Simplify by dividing \( -8 \) and \( 10 \) by \( 2 \):
\( = \frac{(-7) \times (-4)}{5 \times 15} = \frac{28}{75} \)

(vi) \( \frac{3}{-2} \times \frac{-7}{3} \)
Multiply across:
\( = \frac{3 \times (-7)}{(-2) \times 3} \)
Simplify by canceling the common factor \( 3 \):
\( = \frac{1 \times (-7)}{(-2) \times 1} = \frac{-7}{-2} = \frac{7}{2} = 3\frac{1}{2} \)
In simple words: When you multiply a fraction by a whole number, think of the whole number as having a 1 on the bottom. Multiply the tops and bottoms, then simplify. Remember that multiplying two negative numbers gives a positive result.

Exam Tip: Be extra careful with signs. Two minus signs in a product cancel out to become positive, whereas an odd number of minus signs keeps the product negative.

Question 3. Evaluate:
(i) \( \left(\frac{2}{-3} \times \frac{5}{4}\right) + \left(\frac{5}{9} \times \frac{3}{-10}\right) \)
(ii) \( \left(2 \times \frac{1}{4}\right) - \left(\frac{-18}{7} \times \frac{-7}{15}\right) \)
(iii) \( \left(-5 \times \frac{2}{15}\right) - \left(-6 \times \frac{2}{9}\right) \)
(iv) \( \left(\frac{8}{5} \times \frac{-3}{2}\right) + \left(\frac{-3}{10} \times \frac{9}{16}\right) \)
Answer:
(i) \( \left(\frac{2}{-3} \times \frac{5}{4}\right) + \left(\frac{5}{9} \times \frac{3}{-10}\right) \)
Solve the multiplication inside each bracket first:
\( = \left(\frac{2 \times 5}{(-3) \times 4}\right) + \left(\frac{5 \times 3}{9 \times (-10)}\right) \)
Simplify each term by reducing common factors:
\( = \left(\frac{1 \times 5}{(-3) \times 2}\right) + \left(\frac{1 \times 1}{3 \times (-2)}\right) \)
\( = \frac{-5}{6} + \frac{-1}{6} \)
Now, add the two simplified fractions together:
\( = \frac{-5 - 1}{6} = \frac{-6}{6} = -1 \)

(ii) \( \left(2 \times \frac{1}{4}\right) - \left(\frac{-18}{7} \times \frac{-7}{15}\right) \)
First, calculate the product inside both sets of brackets:
\( = \left(\frac{2 \times 1}{1 \times 4}\right) - \left(\frac{(-18) \times (-7)}{7 \times 15}\right) \)
Simplify by reducing common factors in both terms:
\( = \left(\frac{1 \times 1}{1 \times 2}\right) - \left(\frac{(-18) \times (-1)}{1 \times 15}\right) \)
\( = \frac{1}{2} - \frac{18}{15} \)
Find the LCM of the denominators \( 2 \) and \( 15 \), which is \( 30 \):
\( = \frac{1 \times 15}{2 \times 15} - \frac{18 \times 2}{15 \times 2} \)
\( = \frac{15 - 36}{30} \)
\( = \frac{-21}{30} \)
Divide the numerator and denominator by \( 3 \) to simplify:
\( = \frac{-7}{10} \)

(iii) \( \left(-5 \times \frac{2}{15}\right) - \left(-6 \times \frac{2}{9}\right) \)
Multiply the numbers in both brackets:
\( = \left(\frac{(-5) \times 2}{1 \times 15}\right) - \left(\frac{(-6) \times 2}{1 \times 9}\right) \)
Simplify the fractions by dividing common terms:
\( = \left(\frac{(-1) \times 2}{1 \times 3}\right) - \left(\frac{(-2) \times 2}{1 \times 3}\right) \)
\( = \frac{-2}{3} - \left(\frac{-4}{3}\right) \)
Subtracting a negative fraction is the same as adding it:
\( = \frac{-2 + 4}{3} = \frac{2}{3} \)

(iv) \( \left(\frac{8}{5} \times \frac{-3}{2}\right) + \left(\frac{-3}{10} \times \frac{9}{16}\right) \)
Solve the multiplications first:
\( = \left(\frac{8 \times (-3)}{5 \times 2}\right) + \left(\frac{(-3) \times 9}{10 \times 16}\right) \)
Simplify the first bracket by dividing \( 8 \) and \( 2 \) by \( 2 \):
\( = \left(\frac{4 \times (-3)}{5 \times 1}\right) + \left(\frac{(-3) \times 9}{10 \times 16}\right) \)
\( = \frac{-12}{5} + \left(\frac{-27}{160}\right) \)
The LCM of \( 5 \) and \( 160 \) is \( 160 \). Convert the first fraction:
\( = \frac{(-12) \times 32}{5 \times 32} + \frac{(-27) \times 1}{160 \times 1} \)
\( = \frac{-384 - 27}{160} = \frac{-411}{160} \)
In simple words: First solve the multiplication inside the brackets. After simplifying the fractions, find a common bottom number (LCM) to add or subtract them.

Exam Tip: Always follow BODMAS - PEMDAS rules. Do the multiplication inside the brackets first before performing addition or subtraction between the terms.

Question 4. Multiply each rational number, given below, by one (1):
(i) \( \frac{7}{-5} \)
(ii) \( \frac{-3}{-4} \)
(iii) \( 0 \)
(iv) \( \frac{-8}{13} \)
(v) \( \frac{-6}{-7} \)
Answer:
(i) \( \frac{7}{-5} \times 1 = 1 \times \left(\frac{7}{-5}\right) = \frac{7}{-5} \)
(ii) \( \frac{-3}{-4} \times 1 = 1 \times \left(\frac{-3}{-4}\right) = \frac{3}{4} \)
(iii) \( 0 \times 1 = 1 \times 0 = 0 \)
(iv) \( \frac{-8}{13} \times 1 = 1 \times \left(\frac{-8}{13}\right) = \frac{-8}{13} \)
(v) \( \frac{-6}{-7} \times 1 = 1 \times \left(\frac{-6}{-7}\right) = \frac{6}{7} \)
In simple words: Multiplying any number by 1 does not change its value. Remember to also simplify double negative signs to positive.

Exam Tip: Remember that 1 is the multiplicative identity. Any rational number multiplied by 1 remains the same number.

Question 5. For each pair of rational numbers, given below, verify that the multiplication is commutative:
(i) \( \frac{-1}{5} \) and \( \frac{2}{9} \)
(ii) \( \frac{5}{-3} \) and \( \frac{13}{-11} \)
(iii) \( 3 \) and \( \frac{-8}{9} \)
(iv) \( 0 \) and \( \frac{-12}{17} \)
Answer:
(i) To verify the commutative property, we multiply the numbers in both orders:
First order:
\( \frac{-1}{5} \times \frac{2}{9} = \frac{(-1) \times 2}{5 \times 9} = \frac{-2}{45} \)
Reverse order:
\( \frac{2}{9} \times \left(\frac{-1}{5}\right) = \frac{2 \times (-1)}{9 \times 5} = \frac{-2}{45} \)
Both results are the same. Hence:
\( \therefore \frac{-1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{-1}{5} \)

(ii) Let's multiply the rational numbers both ways:
First order:
\( \frac{5}{-3} \times \frac{13}{-11} = \frac{5 \times 13}{(-3) \times (-11)} = \frac{65}{33} \)
Reverse order:
\( \frac{13}{-11} \times \frac{5}{-3} = \frac{13 \times 5}{(-11) \times (-3)} = \frac{65}{33} \)
Since the products match:
\( \therefore \frac{5}{-3} \times \frac{13}{-11} = \frac{13}{-11} \times \frac{5}{-3} \)

(iii) Multiply in both directions:
First order:
\( 3 \times \frac{-8}{9} = \frac{3}{1} \times \frac{-8}{9} = \frac{1 \times (-8)}{1 \times 3} = \frac{-8}{3} \)
Reverse order:
\( \frac{-8}{9} \times 3 = \frac{-8}{9} \times \frac{3}{1} = \frac{(-8) \times 1}{3 \times 1} = \frac{-8}{3} \)
Because both values are identical:
\( \therefore 3 \times \frac{-8}{9} = \frac{-8}{9} \times 3 \)

(iv) Multiply in both directions:
First order:
\( 0 \times \frac{-12}{17} = \frac{0 \times (-12)}{1 \times 17} = 0 \)
Reverse order:
\( \frac{-12}{17} \times 0 = \frac{(-12) \times 0}{17 \times 1} = 0 \)
Both calculations give zero:
\( \therefore 0 \times \frac{-12}{17} = \frac{-12}{17} \times 0 \)
In simple words: The commutative property means that when you multiply two numbers, the order does not matter. Multiplying A by B gives the same answer as multiplying B by A.

Exam Tip: To show verification, always write down both the left-hand side (LHS) and right-hand side (RHS) calculations separately and show they are equal.

Question 6. Write the reciprocal (multiplicative inverse) of each rational number, given below:
(i) \( 5 \)
(ii) \( -3 \)
(iii) \( \frac{5}{11} \)
(iv) \( \frac{-7}{-8} \)
(v) \( \frac{-7}{-8} \)
(vi) \( \frac{15}{-17} \)
Answer:
(i) The reciprocal of \( 5 \) is \( \frac{1}{5} \).
(ii) The reciprocal of \( -3 \) is \( \frac{1}{-3} = -\frac{1}{3} \).
(iii) The reciprocal of \( \frac{5}{11} \) is \( \frac{11}{5} = 2\frac{1}{5} \).
(iv) The reciprocal of \( \frac{-7}{-8} = \frac{7}{8} \) is \( \frac{8}{7} = 1\frac{1}{7} \).
(v) The reciprocal of \( \frac{-7}{-8} = \frac{7}{8} \) is \( \frac{8}{7} = 1\frac{1}{7} \).
(vi) The reciprocal of \( \frac{15}{-17} \) is \( \frac{-17}{15} = -1\frac{2}{15} \).
In simple words: To find the reciprocal of a fraction, just turn it upside down. The top number goes to the bottom, and the bottom number goes to the top.

Exam Tip: When finding a reciprocal, the sign of the number does not change. A positive number has a positive reciprocal, and a negative number has a negative reciprocal.

Question 7. Find the reciprocal (multiplicative inverse) of:
(i) \( \frac{3}{5} \times \frac{2}{3} \)
(ii) \( \frac{-8}{3} \times \frac{13}{-7} \)
(iii) \( \frac{-3}{5} \times \frac{-1}{13} \)
Answer:
(i) First, calculate the product:
\( \frac{3}{5} \times \frac{2}{3} = \frac{3 \times 2}{5 \times 3} = \frac{1 \times 2}{5 \times 1} = \frac{2}{5} \)
The reciprocal of \( \frac{2}{5} \) is \( \frac{5}{2} \).

(ii) First, find the product:
\( \frac{-8}{3} \times \frac{13}{-7} = \frac{(-8) \times 13}{3 \times (-7)} = \frac{-104}{-21} = \frac{104}{21} \)
The reciprocal of \( \frac{104}{21} \) is \( \frac{21}{104} \).

(iii) Multiply first:
\( \frac{-3}{5} \times \frac{-1}{13} = \frac{(-3) \times (-1)}{5 \times 13} = \frac{3}{65} \)
The reciprocal of \( \frac{3}{65} \) is \( \frac{65}{3} = 21\frac{2}{3} \).
In simple words: First multiply the fractions to get a single fraction, then swap the top and bottom numbers of the result to get its reciprocal.

Exam Tip: Do not just invert the starting fractions and multiply them. Always simplify first to get a single fraction, then write its reciprocal clearly in the final step.

Question 8. Verify that \( (x + y) \times z = x \times z + y \times z \), if
(i) \( x = \frac{4}{5}, y = \frac{-2}{3} \) and \( z = -4 \)
(ii) \( x = 2, y = \frac{4}{5} \) and \( z = \frac{3}{-10} \)
Answer:
(i) Given \( x = \frac{4}{5}, y = \frac{-2}{3}, z = -4 \)
Let's evaluate both sides of the equation:
Left-Hand Side (L.H.S.):
\( (x + y) \times z = \left(\frac{4}{5} + \frac{-2}{3}\right) \times (-4) \)
Using a common denominator of \( 15 \) for addition:
\( = \left(\frac{4 \times 3}{5 \times 3} + \frac{(-2) \times 5}{3 \times 5}\right) \times (-4) \)
\( = \left(\frac{12 - 10}{15}\right) \times (-4) \)
\( = \frac{2}{15} \times (-4) = \frac{-8}{15} \)
Right-Hand Side (R.H.S.):
\( x \times z + y \times z = \frac{4}{5} \times (-4) + \frac{-2}{3} \times (-4) \)
\( = \frac{-16}{5} + \frac{8}{3} \)
We find the LCM of \( 5 \) and \( 3 \) is \( 15 \):
\( = \frac{-48 + 40}{15} = \frac{-8}{15} \)
Here, L.H.S. = R.H.S., so the statement is verified.

(ii) Given \( x = 2, y = \frac{4}{5}, z = \frac{3}{-10} \)
Let's verify the expression:
Left-Hand Side (L.H.S.):
\( (x + y) \times z = \left(\frac{2}{1} + \frac{4}{5}\right) \times \frac{3}{-10} \)
\( = \left(\frac{2 \times 5 + 4 \times 1}{1 \times 5}\right) \times \frac{3}{-10} \)
\( = \frac{14}{5} \times \frac{3}{-10} = \frac{7 \times 3}{5 \times (-5)} = \frac{-21}{25} \)
Right-Hand Side (R.H.S.):
\( x \times z + y \times z = 2 \times \frac{3}{-10} + \frac{4}{5} \times \frac{3}{-10} \)
\( = \frac{3}{-5} + \frac{6}{-25} \)
Express with a common denominator of \( 25 \):
\( = \frac{-3 \times 5}{5 \times 5} + \frac{-6 \times 1}{25 \times 1} \)
\( = \frac{-15 - 6}{25} = \frac{-21}{25} \)
Since L.H.S. = R.H.S., the identity holds true.
In simple words: This shows the distributive law. Adding first and then multiplying gives the same result as multiplying each part separately and then adding them.

Exam Tip: When simplifying terms with negative denominators like \( \frac{3}{-10} \), always shift the minus sign to the numerator to avoid sign confusion during addition.

Question 9. Verify that \( x \times (y - z) = x \times y - x \times z \), if
(i) \( x = \frac{4}{5}, y = -\frac{7}{4} \) and \( z = 3 \)
(ii) \( x = \frac{3}{4}, y = \frac{8}{9} \) and \( z = -5 \)
Answer:
(i) Given \( x = \frac{4}{5}, y = -\frac{7}{4}, z = 3 \)
Let's calculate both sides:
Left-Hand Side (L.H.S.):
\( x \times (y - z) = \frac{4}{5} \times \left(\frac{-7}{4} - 3\right) \)
Write \( 3 \) as \( \frac{12}{4} \) to subtract:
\( = \frac{4}{5} \times \left(\frac{-7 - 12}{4}\right) \)
\( = \frac{4}{5} \times \left(\frac{-19}{4}\right) = \frac{-19}{5} \)
Right-Hand Side (R.H.S.):
\( x \times y - x \times z = \frac{4}{5} \times \left(-\frac{7}{4}\right) - \frac{4}{5} \times 3 \)
\( = \frac{-7}{5} - \frac{12}{5} \)
\( = \frac{-7 - 12}{5} = \frac{-19}{5} \)
Since LHS and RHS match, this is verified.

(ii) Given \( x = \frac{3}{4}, y = \frac{8}{9}, z = -5 \)
Let's check both sides:
Left-Hand Side (L.H.S.):
\( x \times (y - z) = \frac{3}{4} \times \left(\frac{8}{9} - (-5)\right) \)
Subtracting a negative number is the same as adding:
\( = \frac{3}{4} \times \left(\frac{8}{9} + 5\right) \)
\( = \frac{3}{4} \times \left(\frac{8 + 45}{9}\right) \)
\( = \frac{3}{4} \times \frac{53}{9} = \frac{1 \times 53}{4 \times 3} = \frac{53}{12} \)
Right-Hand Side (R.H.S.):
\( x \times y - x \times z = \frac{3}{4} \times \frac{8}{9} - \frac{3}{4} \times (-5) \)
\( = \frac{2}{3} - \left(\frac{-15}{4}\right) \)
\( = \frac{2}{3} + \frac{15}{4} \)
Find the common denominator of \( 12 \):
\( = \frac{2 \times 4 + 15 \times 3}{12} = \frac{8 + 45}{12} = \frac{53}{12} \)
Because L.H.S. = R.H.S., the statement is verified.
In simple words: This is the distributive law over subtraction. Multiplying a number by the difference of two other numbers gives the same answer as multiplying them separately first and then subtracting.

Exam Tip: Be very careful with signs like \( - (-5) \), which simplifies to \( +5 \). A minor sign error is the most common cause of verification failure.

Question 10. Name the multiplication property of rational numbers shown below :
(i) \( \frac{3}{5} \times \frac{-8}{9} = \frac{-8}{9} \times \frac{3}{5} \)
(ii) \( \frac{-3}{4} \times \left( \frac{5}{7} \times \frac{-8}{15} \right) = \left( \frac{-3}{4} \times \frac{5}{7} \right) \times \frac{-8}{15} \)
(iii) \( \frac{4}{5} \times \left( \frac{3}{-8} + \frac{-4}{7} \right) = \frac{4}{5} \times \frac{3}{-8} + \frac{4}{5} \times \frac{-4}{7} \)
(iv) \( \frac{-7}{5} \times \frac{5}{-7} = 1 \)
(v) \( \frac{8}{-9} \times 1 = 1 \times \frac{8}{-9} = \frac{8}{-9} \)
(vi) \( \frac{-3}{4} \times 0 = 0 \)
Answer:
(i) Commutative property of multiplication.
(ii) Associative property of multiplication.
(iii) Distributive property of multiplication over addition.
(iv) Existence of multiplicative inverse.
(v) Existence of multiplicative identity.
(vi) Property of zero under multiplication.
In simple words: These rules show how rational numbers behave when we multiply them. They tell us that changing the order or grouping of numbers does not change the final result.

Exam Tip: To identify the property, look at what changes from one side to the other. If only the grouping of numbers changes, it is associative, but if the order of numbers changes, it is commutative.

Question 11. Fill in the blanks:
(i) The product of two positive rational numbers is always ……………
(ii) The product of two negative rational numbers is always ……………
(iii) If two rational numbers have opposite signs then their product is always …………..
(iv) The reciprocal of a positive rational number is ………. and the reciprocal of a negative rational number is ……………
(v) Rational number 0 has ………….. reciprocal.
(vi) The product of a rational number and its reciprocal is ………..
(vii) The numbers ……….. and ……….. are their own reciprocals.
(viii) If m is reciprocal of n, then the reciprocal of n is ………….
Answer:
(i) positive.
(ii) positive.
(iii) negative.
(iv) positive, negative.
(v) no.
(vi) 1.
(vii) 1 and -1.
(viii) m.
In simple words: These are basic rules of multiplication. Multiplying identical signs gives a positive answer, while multiplying opposite signs gives a negative answer. Finding a reciprocal simply means flipping a fraction upside down.

Exam Tip: Remember that zero is the only rational number that has no reciprocal because division by zero is not allowed.

Exercise 1(D)

Question 1. Evaluate:
(i) \( 1 \div \frac{1}{3} \)
(ii) \( 3 \div \frac{3}{5} \)
(iii) \( -\frac{5}{12} \div \frac{1}{16} \)
(iv) \( -\frac{21}{16} \div \left( \frac{-7}{8} \right) \)
(v) \( 0 \div \left( -\frac{4}{7} \right) \)
(vi) \( \frac{8}{-5} \div \frac{24}{25} \)
(vii) \( -\frac{3}{4} \div (-9) \)
(viii) \( \frac{3}{4} \div \left( -\frac{5}{12} \right) \)
(ix) \( -5 \div \left( -\frac{10}{11} \right) \)
(x) \( \frac{-7}{11} \div \left( \frac{-3}{44} \right) \)
Answer:
(i) \( 1 \div \frac{1}{3} = 1 \times \frac{3}{1} = 3 \)
(ii) \( 3 \div \frac{3}{5} = 3 \times \frac{5}{3} = \frac{3 \times 5}{3} = 5 \)
(iii) \( -\frac{5}{12} \div \frac{1}{16} = -\frac{5}{12} \times \frac{16}{1} = \frac{-5 \times 16}{12} = \frac{-5 \times 4}{3} = \frac{-20}{3} = -6\frac{2}{3} \)
(iv) \( -\frac{21}{16} \div \left( \frac{-7}{8} \right) = -\frac{21}{16} \times \frac{8}{-7} = \frac{21 \times 8}{16 \times 7} = \frac{3 \times 1}{2 \times 1} = \frac{3}{2} = 1\frac{1}{2} \)
(v) \( 0 \div \left( -\frac{4}{7} \right) = 0 \times \left( -\frac{7}{4} \right) = 0 \)
(vi) \( \frac{8}{-5} \div \frac{24}{25} = \frac{8}{-5} \times \frac{25}{24} = \frac{2 \times 5}{(-1) \times 6} = \frac{1 \times 5}{(-1) \times 3} = \frac{-5}{3} = -1\frac{2}{3} \)
(vii) \( -\frac{3}{4} \div (-9) = -\frac{3}{4} \times \frac{1}{-9} = \frac{(-1) \times 1}{4 \times (-3)} = \frac{1}{12} \)
(viii) \( \frac{3}{4} \div \left( -\frac{5}{12} \right) = \frac{3}{4} \times \left( -\frac{12}{5} \right) = \frac{3 \times (-3)}{1 \times 5} = -\frac{9}{5} = -1\frac{4}{5} \)
(ix) \( -5 \div \left( -\frac{10}{11} \right) = -5 \times \frac{11}{-10} = \frac{1 \times 11}{1 \times 2} = \frac{11}{2} = 5\frac{1}{2} \)
(x) \( \frac{-7}{11} \div \left( \frac{-3}{44} \right) = \frac{-7}{11} \times \left( \frac{44}{-3} \right) = \frac{(-7) \times 4}{1 \times (-3)} = \frac{28}{3} = 9\frac{1}{3} \)
In simple words: To divide one fraction by another, multiply the first fraction by the flipped version of the second fraction. Simplify by dividing common numbers from the top and bottom.

Exam Tip: Be very careful with negative signs. If you have an odd number of minus signs, the final answer is negative. If you have an even number, it is positive.

Question 2. Divide:
(i) 3 by \( \frac{1}{3} \)
(ii) -2 by \( \left( -\frac{1}{2} \right) \)
(iii) 0 by \( \frac{7}{-9} \)
(iv) \( \frac{-5}{8} \) by \( \frac{1}{4} \)
(v) \( -\frac{3}{4} \) by \( \frac{9}{16} \)
Answer:
(i) \( 3 \div \frac{1}{3} = 3 \times \frac{3}{1} = 9 \)
(ii) \( -2 \div \left( -\frac{1}{2} \right) = -2 \times \frac{2}{-1} = 4 \)
(iii) \( 0 \div \frac{7}{-9} = 0 \times \frac{-9}{7} = 0 \)
(iv) \( \frac{-5}{8} \div \frac{1}{4} = \frac{-5}{8} \times \frac{4}{1} = \frac{-5 \times 1}{2 \times 1} = \frac{-5}{2} = -2\frac{1}{2} \)
(v) \( -\frac{3}{4} \div \frac{9}{16} = -\frac{3}{4} \times \frac{16}{9} = \frac{(-1) \times 4}{1 \times 3} = \frac{-4}{3} = -1\frac{1}{3} \)
In simple words: Replace the word "by" with a division sign. Flip the second term upside down and multiply it by the first term.

Exam Tip: If you divide zero by any rational number, the result will always stay zero.

Question 3. The product of two rational numbers is -2. If one of them is \( \frac{4}{7} \), find the other.
Answer: Let us assume the unknown rational number is \( x \). We are given that when we multiply these two numbers, their total product is \( -2 \). One of these numbers is \( \frac{4}{7} \).
This gives us the relationship:
\( x \times \frac{4}{7} = -2 \)
To find our missing value, we can divide the total product by the given number:
\( x = -2 \div \frac{4}{7} \)
\( \implies x = -2 \times \frac{7}{4} \)
\( \implies x = \frac{-1 \times 7}{1 \times 2} \)
\( \implies x = -\frac{7}{2} = -3\frac{1}{2} \)
Therefore, the other rational number is \( -3\frac{1}{2} \).
In simple words: If you have the final product and one of the factors, you can get the other factor by dividing the product by the known factor.

Exam Tip: Always convert improper fractions like \( -\frac{7}{2} \) to mixed fractions like \( -3\frac{1}{2} \) to give a neat final answer.

Question 4. The product of two numbers is \( \frac{-4}{9} \). If one of them is \( \frac{-2}{27} \), find the other.
Answer: We are told that the product of two rational numbers is \( -\frac{4}{9} \). One of these numbers is \( -\frac{2}{27} \).
Let the other rational number be \( y \).
We can write this as:
\( y \times \left( -\frac{2}{27} \right) = -\frac{4}{9} \)
To find \( y \), we divide the product by the known rational number:
\( y = -\frac{4}{9} \div \left( -\frac{2}{27} \right) \)
\( \implies y = -\frac{4}{9} \times \frac{27}{-2} \)
\( \implies y = \frac{(-4) \times 27}{9 \times (-2)} \)
\( \implies y = \frac{2 \times 3}{1 \times 1} = 6 \)
So, the other rational number is \( 6 \).
In simple words: Divide the final product by the first fraction to find the second number. Dividing two negative fractions results in a positive whole number here.

Exam Tip: When dividing two negative values, remember that they cancel out to give a positive result. Keep track of this to avoid sign errors.

Question 5. m and n are two rational numbers such that \( m \times n = -\frac{25}{9} \).
(i) if \( m = \frac{5}{3} \), find n,
(ii) if \( n = -\frac{10}{9} \), find m.
Answer: We know that multiplying \( m \) and \( n \) gives \( -\frac{25}{9} \).
(i) If we substitute \( m = \frac{5}{3} \) into the equation:
\( \frac{5}{3} \times n = -\frac{25}{9} \)
\( \implies n = -\frac{25}{9} \div \frac{5}{3} \)
\( \implies n = -\frac{25}{9} \times \frac{3}{5} \)
\( \implies n = \frac{-25 \times 3}{9 \times 5} \)
\( \implies n = \frac{-5 \times 1}{3 \times 1} = -\frac{5}{3} \)

(ii) If we substitute \( n = -\frac{10}{9} \) into the equation:
\( m \times \left( -\frac{10}{9} \right) = -\frac{25}{9} \)
\( \implies m = -\frac{25}{9} \div \left( -\frac{10}{9} \right) \)
\( \implies m = -\frac{25}{9} \times \frac{9}{-10} \)
\( \implies m = \frac{(-25) \times 9}{9 \times (-10)} \)
\( \implies m = \frac{5 \times 1}{1 \times 2} = \frac{5}{2} = 2\frac{1}{2} \)
In simple words: When you know the product of two variables, you can find one variable by dividing the product by the other known variable.

Exam Tip: Be methodical when cross-canceling numbers in fractions. Crossing out common factors beforehand makes the calculations much easier.

Question 6. By what number must \( \frac{-3}{4} \) be multiplied so that the product is \( \frac{-9}{16} \)?
Answer: Let the required multiplier be \( a \).
According to the given condition:
\( a \times \left( -\frac{3}{4} \right) = -\frac{9}{16} \)
We can solve for \( a \) by dividing our target product by the starting fraction:
\( a = -\frac{9}{16} \div \left( -\frac{3}{4} \right) \)
\( \implies a = -\frac{9}{16} \times \left( -\frac{4}{3} \right) \)
\( \implies a = \frac{(-9) \times (-4)}{16 \times 3} \)
\( \implies a = \frac{3 \times 1}{4 \times 1} = \frac{3}{4} \)
Therefore, the fraction should be multiplied by \( \frac{3}{4} \).
In simple words: To find the missing number, divide the final product by the fraction you started with.

Exam Tip: You can quickly check your answer by multiplying your result with the original number to see if it yields the target product.

Question 7. By what number should \( \frac{-8}{13} \) be multiplied to get 16?
Answer: Let the unknown multiplier be \( x \).
The question states that multiplying \( \frac{-8}{13} \) by this number yields 16:
\( x \times \left( \frac{-8}{13} \right) = 16 \)
To find \( x \), we divide our target value 16 by the fraction:
\( x = 16 \div \left( \frac{-8}{13} \right) \)
\( \implies x = 16 \times \left( \frac{13}{-8} \right) \)
\( \implies x = \frac{16 \times 13}{-8} \)
\( \implies x = (-2) \times 13 = -26 \)
Thus, the required number is \( -26 \).
In simple words: Find the missing factor by dividing the target number (16) by the fraction. Since the target is positive and the fraction is negative, the final answer must be negative.

Exam Tip: Do not lose track of the negative sign when simplifying whole numbers with fraction denominators.

Question 8. If \( 3\frac{1}{2} \) litres of milk costs Rs. 49, find the cost of one litre of milk?
Answer: We are given that \( 3\frac{1}{2} \) litres of milk costs a total of Rs. 49.
First, let us turn the mixed fraction into an improper fraction:
\( 3\frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{7}{2} \) litres.
To find the cost of just one litre, we divide the total cost by the total quantity of milk in litres:
Price of one litre = \( 49 \div \frac{7}{2} \)
\( \implies \text{Price} = 49 \times \frac{2}{7} \)
\( \implies \text{Price} = \frac{49 \times 2}{7} \)
\( \implies \text{Price} = 7 \times 2 = \text{Rs. } 14 \)
Therefore, one litre of milk costs Rs. 14.
In simple words: Find the price of one litre by dividing the total amount of money spent by the total amount of milk purchased.

Exam Tip: Remember to write the final currency unit "Rs." alongside your numerical answer to get full credit.

Question 9. Cost of \( 3\frac{2}{5} \) metre of cloth is Rs. \( 88\frac{1}{2} \). What is the cost of 1 metre of cloth?
Answer: The total price for \( 3\frac{2}{5} \) metres of cloth is Rs. \( 88\frac{1}{2} \).
First, we convert both mixed fractions into improper fractions:
Length of cloth = \( 3\frac{2}{5} = \frac{17}{5} \) metres.
Total cost = Rs. \( 88\frac{1}{2} = \text{Rs. } \frac{177}{2} \).
To find the price of one metre of cloth, we divide the total cost by the length of the cloth:
Price of one metre = \( \frac{177}{2} \div \frac{17}{5} \)
\( \implies \text{Price} = \frac{177}{2} \times \frac{5}{17} \)
\( \implies \text{Price} = \text{Rs. } \frac{885}{34} \)
Converting this improper fraction back into a mixed fraction:
\( \text{Price} = \text{Rs. } 26\frac{1}{34} \)
So, the cost of one metre of cloth is Rs. \( 26\frac{1}{34} \).
In simple words: Change both mixed fractions into normal fractions. Then, divide the price by the length to find out the rate for a single metre.

Exam Tip: If your final improper fraction cannot be simplified, divide the numerator by the denominator to express your answer as a mixed fraction.

Question 10. Divide the sum of \( \frac{3}{7} \) and \( \frac{-5}{14} \) by \( \frac{-1}{2} \).
Answer: We need to first find the sum of \( \frac{3}{7} \) and \( \frac{-5}{14} \), and then divide that result by \( \frac{-1}{2} \).
Step 1: Calculate the sum of the two fractions:
\( \text{Sum} = \frac{3}{7} + \left( \frac{-5}{14} \right) \)
The least common multiple (LCM) of 7 and 14 is 14. We rewrite the first fraction:
\( \text{Sum} = \frac{3 \times 2}{7 \times 2} - \frac{5}{14} \)
\( \implies \text{Sum} = \frac{6}{14} - \frac{5}{14} \)
\( \implies \text{Sum} = \frac{6 - 5}{14} = \frac{1}{14} \)

Step 2: Divide this sum by \( \frac{-1}{2} \):
\( \text{Result} = \frac{1}{14} \div \left( \frac{-1}{2} \right) \)
\( \implies \text{Result} = \frac{1}{14} \times \frac{2}{-1} \)
\( \implies \text{Result} = \frac{1 \times (-1)}{7 \times 1} = -\frac{1}{7} \)
So, the final value is \( -\frac{1}{7} \).
In simple words: Add the first two fractions together using a common denominator. Then, multiply that sum by the reciprocal of the third fraction.

Exam Tip: When dividing by \( \frac{-1}{2} \), remember that multiplying by \( \frac{2}{-1} \) is mathematically equivalent to multiplying by \( -2 \).

Question 11. Find \( (m + n) \div (m - n) \), if :
(i) \( m = \frac{2}{3} \) and \( n = \frac{3}{2} \)
(ii) \( m = \frac{3}{4} \) and \( n = \frac{4}{3} \)
(iii) \( m = \frac{4}{5} \) and \( n = -\frac{3}{10} \)
Answer:
(i) Given: \( m = \frac{2}{3} \) and \( n = \frac{3}{2} \)
Substitute these into the expression \( (m + n) \div (m - n) \):
\( m + n = \frac{2}{3} + \frac{3}{2} \)
Find the LCM of denominators 3 and 2, which is 6:
\( m + n = \frac{2 \times 2 + 3 \times 3}{6} = \frac{4 + 9}{6} = \frac{13}{6} \)
Similarly, compute the difference:
\( m - n = \frac{2}{3} - \frac{3}{2} = \frac{2 \times 2 - 3 \times 3}{6} = \frac{4 - 9}{6} = -\frac{5}{6} \)
Now, divide the first result by the second result:
\( (m + n) \div (m - n) = \frac{13}{6} \div \left(-\frac{5}{6}\right) \)
\( = \frac{13}{6} \times \left(-\frac{6}{5}\right) = -\frac{13}{5} \)

(ii) Given: \( m = \frac{3}{4} \) and \( n = \frac{4}{3} \)
Substitute these values into the formula:
\( m + n = \frac{3}{4} + \frac{4}{3} \)
The LCM of 4 and 3 is 12:
\( m + n = \frac{3 \times 3 + 4 \times 4}{12} = \frac{9 + 16}{12} = \frac{25}{12} \)
Next, calculate the difference:
\( m - n = \frac{3}{4} - \frac{4}{3} = \frac{3 \times 3 - 4 \times 4}{12} = \frac{9 - 16}{12} = -\frac{7}{12} \)
Now perform the division:
\( (m + n) \div (m - n) = \frac{25}{12} \div \left(-\frac{7}{12}\right) \)
\( = \frac{25}{12} \times \left(-\frac{12}{7}\right) = -\frac{25}{7} \)

(iii) Given: \( m = \frac{4}{5} \) and \( n = -\frac{3}{10} \)
Substitute the values to get:
\( m + n = \frac{4}{5} + \left(-\frac{3}{10}\right) = \frac{4}{5} - \frac{3}{10} \)
With LCM of 5 and 10 being 10:
\( m + n = \frac{4 \times 2 - 3 \times 1}{10} = \frac{8 - 3}{10} = \frac{5}{10} = \frac{1}{2} \)
Next, calculate the difference:
\( m - n = \frac{4}{5} - \left(-\frac{3}{10}\right) = \frac{4}{5} + \frac{3}{10} \)
\( m - n = \frac{4 \times 2 + 3 \times 1}{10} = \frac{8 + 3}{10} = \frac{11}{10} \)
Divide the sum by the difference:
\( (m + n) \div (m - n) = \frac{1}{2} \div \frac{11}{10} \)
\( = \frac{1}{2} \times \frac{10}{11} = \frac{5}{11} \)
In simple words: To solve these, first add the two fractions, and then subtract the second fraction from the first. Finally, divide the sum you got by the difference.

Exam Tip: When dividing fractions, remember to invert the second fraction (the divisor) and change the operation to multiplication, keeping careful track of negative signs.

Question 12. The product of two rational numbers is -5. If one of these numbers is \( \frac{-7}{15} \), find the other.
Answer:
Let the unknown rational number be represented by \( x \).
We are given that one of the numbers is \( \frac{-7}{15} \).
Their product is \( -5 \).
We can write this as an equation:
\( \frac{-7}{15} \times x = -5 \)
Multiply both sides by 15 to clear the fraction:
\( -7x = -5 \times 15 \)
\( -7x = -75 \)
Solve for \( x \) by dividing both sides by -7:
\( x = \frac{-75}{-7} = \frac{75}{7} \)
So, the other rational number is \( \frac{75}{7} \).
In simple words: If you multiply a known number by an unknown number \( x \) to get a result, you can find \( x \) by dividing the result by the known number. Dividing \( -5 \) by \( \frac{-7}{15} \) gives us \( \frac{75}{7} \).

Exam Tip: Pay close attention to signs. Dividing a negative number by another negative number always results in a positive value.

Question 13. Divide the sum of \( \frac{5}{8} \) and \( \frac{-11}{12} \) by the difference of \( \frac{3}{7} \) and \( \frac{5}{14} \).
Answer:
First, find the sum of \( \frac{5}{8} \) and \( \frac{-11}{12} \):
\( \text{Sum} = \frac{5}{8} + \left( \frac{-11}{12} \right) = \frac{5}{8} - \frac{11}{12} \)
The LCM of 8 and 12 is 24. Thus:
\( \text{Sum} = \frac{5 \times 3 - 11 \times 2}{24} = \frac{15 - 22}{24} = -\frac{7}{24} \)

Next, calculate the difference between \( \frac{3}{7} \) and \( \frac{5}{14} \). Since the order of subtraction is not specified, we can compute this in two ways:
Case A: \( \frac{3}{7} - \frac{5}{14} \)
The LCM of 7 and 14 is 14:
\( \text{Difference} = \frac{3 \times 2 - 5 \times 1}{14} = \frac{6 - 5}{14} = \frac{1}{14} \)

Case B: \( \frac{5}{14} - \frac{3}{7} \)
\( \text{Difference} = \frac{5 \times 1 - 3 \times 2}{14} = \frac{5 - 6}{14} = -\frac{1}{14} \)

Now, divide the sum by the difference obtained in each case:
For Case A:
\( -\frac{7}{24} \div \frac{1}{14} = -\frac{7}{24} \times \frac{14}{1} = -\frac{7 \times 7}{12 \times 1} = -\frac{49}{12} = -4\frac{1}{12} \)

For Case B:
\( -\frac{7}{24} \div \left( -\frac{1}{14} \right) = -\frac{7}{24} \times \left( -\frac{14}{1} \right) = \frac{7 \times 7}{12 \times 1} = \frac{49}{12} = 4\frac{1}{12} \)
In simple words: First, add the first two fractions to get \( -\frac{7}{24} \). Then find the difference between the next two fractions, which gives either \( \frac{1}{14} \) or \( -\frac{1}{14} \). Dividing the sum by this difference gives the final result of \( -4\frac{1}{12} \) or \( 4\frac{1}{12} \).

Exam Tip: Unless specified otherwise, "difference" can mean subtracting either number from the other. Computing both cases ensures you do not lose marks for ambiguity.

Exercise 1(E)

Question 1. Draw a number line and mark \( \frac{3}{4} \), \( \frac{7}{4} \), \( \frac{-3}{4} \) and \( \frac{-7}{4} \) on it.
Answer:
To represent these rational numbers on a number line, we first observe that all of them have a denominator of 4. This means each unit interval (for example, between 0 and 1, or between 0 and -1) can be split into 4 equal subdivisions, where each mark represents \( \frac{1}{4} \).
- Moving 3 divisions to the right of 0 gives \( \frac{3}{4} \).
- Moving 7 divisions to the right of 0 gives \( \frac{7}{4} \).
- Moving 3 divisions to the left of 0 gives \( \frac{-3}{4} \).
- Moving 7 divisions to the left of 0 gives \( \frac{-7}{4} \).
The plotted points are shown in red on the number line below:

In simple words: Since the bottom number of all these fractions is 4, split the space between each whole number on your line into 4 equal sections. Then, count that many steps from 0 to find and mark each point.

Exam Tip: Be sure to keep the distance between integers uniform on your number line, and clearly label the direction of positive and negative numbers.

Question 2. On a number line mark the points \( \frac{2}{3} \), \( \frac{-8}{3} \), \( \frac{7}{3} \), \( \frac{-2}{3} \) and -2.
Answer:
Since the fractions have a denominator of 3, we divide each whole unit on the number line into 3 equal parts. Each subdivision corresponds to \( \frac{1}{3} \).
- \( \frac{2}{3} \) is located 2 units to the right of 0.
- \( \frac{-8}{3} \) is located 8 units to the left of 0.
- \( \frac{7}{3} \) is located 7 units to the right of 0.
- \( \frac{-2}{3} \) is located 2 units to the left of 0.
- \( -2 \) is located at the integer -2 (which is the same as \( \frac{-6}{3} \)).
The required points are marked in red below:

In simple words: Since we are working with thirds, divide each integer gap into 3 equal parts. Then count out the steps from 0: left for negative values, right for positive values, and mark each point.

Exam Tip: Remember that any integer like -2 can be written as a fraction like \( \frac{-6}{3} \), which helps you locate it easily among other thirds.

Question 3. Insert one rational number between:
(i) 7 and 8
(ii) 3.5 and 5
(iii) 2 and 3.2
(iv) 4.2 and 3.6
(v) \( \frac{1}{2} \) and 2
Answer:
To find a rational number between any two numbers \( a \) and \( b \), we can use the formula \( \frac{a + b}{2} \).

(i) Between 7 and 8:
\( \frac{7 + 8}{2} = \frac{15}{2} = 7.5 \)

(ii) Between 3.5 and 5:
\( \frac{3.5 + 5}{2} = \frac{8.5}{2} = 4.25 \)

(iii) Between 2 and 3.2:
\( \frac{2 + 3.2}{2} = \frac{5.2}{2} = 2.6 \)

(iv) Between 4.2 and 3.6:
\( \frac{4.2 + 3.6}{2} = \frac{7.8}{2} = 3.9 \)

(v) Between \( \frac{1}{2} \) and 2:
\( \frac{\frac{1}{2} + 2}{2} = \frac{\frac{1 + 4}{2}}{2} = \frac{5}{4} = 1.25 \)
In simple words: To find a number exactly in the middle of two other numbers, just add them together and divide the sum by 2.

Exam Tip: When finding a rational number between a decimal and an integer, you can either convert both to decimals or both to fractions before calculating the average.

Question 4. Insert two rational numbers between :
(i) 6 and 7
(ii) 4.8 and 6
(iii) 2.7 and 6.3
Answer:
To find multiple rational numbers between two given numbers, we can repeatedly find the average of the boundaries or the newly found numbers.

(i) Between 6 and 7:
First, find the midpoint of 6 and 7:
\( \text{First Number} = \frac{6 + 7}{2} = \frac{13}{2} = 6.5 \)
Now, find a second number by taking the average of 6 and 6.5:
\( \text{Second Number} = \frac{6 + 6.5}{2} = \frac{12.5}{2} = 6.25 \)
Thus, the two rational numbers between 6 and 7 are 6.25 and 6.5.

(ii) Between 4.8 and 6:
First, find the midpoint of 4.8 and 6:
\( \text{First Number} = \frac{4.8 + 6}{2} = \frac{10.8}{2} = 5.4 \)
Next, find a second number between 4.8 and 5.4 by taking their average:
\( \text{Second Number} = \frac{4.8 + 5.4}{2} = \frac{10.2}{2} = 5.1 \)
Thus, the two rational numbers between 4.8 and 6 are 5.1 and 5.4.

(iii) Between 2.7 and 6.3:
First, find the midpoint of 2.7 and 6.3:
\( \text{First Number} = \frac{2.7 + 6.3}{2} = \frac{9}{2} = 4.5 \)
Next, find a second number between 4.5 and 6.3 by taking their average:
\( \text{Second Number} = \frac{4.5 + 6.3}{2} = \frac{10.8}{2} = 5.4 \)
Thus, the two rational numbers between 2.7 and 6.3 are 4.5 and 5.4.
In simple words: To find two numbers in between any two given numbers, find the exact middle of the original two first. Then, find the middle point between that new number and one of your original numbers.

Exam Tip: There are infinitely many rational numbers between any two given numbers. Repeatedly using the midpoint formula \( \frac{a+b}{2} \) is a systematic way to find them.

Question 5. Insert three rational numbers between :
(i) 3 and 4
(ii) 10 and 12
Answer:
To find three rational numbers between two numbers, we can first find the middle number, and then find the midpoints of the two resulting intervals.

(i) Between 3 and 4:
First, find the midpoint of 3 and 4:
\( \text{Midpoint 1} = \frac{3 + 4}{2} = 3.5 \)
Now, find the midpoint between 3 and 3.5:
\( \text{Midpoint 2} = \frac{3 + 3.5}{2} = 3.25 \)
Next, find the midpoint between 3.5 and 4:
\( \text{Midpoint 3} = \frac{3.5 + 4}{2} = 3.75 \)
Thus, the three rational numbers are 3.25, 3.5, and 3.75.

(ii) Between 10 and 12:
First, find the midpoint of 10 and 12:
\( \text{Midpoint 1} = \frac{10 + 12}{2} = 11 \)
Now, find the midpoint between 10 and 11:
\( \text{Midpoint 2} = \frac{10 + 11}{2} = 10.5 \)
Next, find the midpoint between 11 and 12:
\( \text{Midpoint 3} = \frac{11 + 12}{2} = 11.5 \)
Thus, the three rational numbers are 10.5, 11, and 11.5.
In simple words: To find three numbers in between, first find the exact center of the two outer numbers. Then, find the center points of the left and right halves to get your three numbers.

Exam Tip: This method of finding successive averages works best when the numbers are close integers. It creates equally spaced values between the two boundaries.

Question 6. Insert five rational numbers between \( \frac{3}{5} \) and \( \frac{2}{3} \).
Answer:
First, find a common denominator for the two fractions. The LCM of the denominators 5 and 3 is 15.
Convert both fractions to have this denominator:
\( \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \)
\( \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \)

Since we need to insert 5 rational numbers, we multiply the numerator and denominator of both fractions by \( 5 + 1 = 6 \):
\( \frac{9}{15} = \frac{9 \times 6}{15 \times 6} = \frac{54}{90} \)
\( \frac{10}{15} = \frac{10 \times 6}{15 \times 6} = \frac{60}{90} \)

Now, we can choose the five integers between 54 and 60 as our numerators:
The required rational numbers are:
\( \frac{55}{90}, \frac{56}{90}, \frac{57}{90}, \frac{58}{90}, \frac{59}{90} \)

Simplifying these fractions to their lowest terms:
- \( \frac{55}{90} = \frac{11}{18} \)
- \( \frac{56}{90} = \frac{28}{45} \)
- \( \frac{57}{90} = \frac{19}{30} \)
- \( \frac{58}{90} = \frac{29}{45} \)
- \( \frac{59}{90} \)

So, the five rational numbers are \( \frac{11}{18} \), \( \frac{28}{45} \), \( \frac{19}{30} \), \( \frac{29}{45} \), and \( \frac{59}{90} \).
In simple words: To find five fractions between two different fractions, first make their bottom numbers the same. Since there are no whole numbers between 9 and 10, multiply both by 6. This gives you a gap of five numbers in the middle.

Exam Tip: When you need to insert \( n \) rational numbers, always convert the fractions to like denominators, and then multiply the numerator and denominator by \( n + 1 \) to create enough spacing.

Question 7. Insert six rational numbers between \( \frac{5}{6} \) and \( \frac{8}{9} \).
Answer:
First, find a common denominator for the two fractions. The LCM of the denominators 6 and 9 is 18.
Convert both fractions to have this common denominator:
\( \frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18} \)
\( \frac{8}{9} = \frac{8 \times 2}{9 \times 2} = \frac{16}{18} \)

Since we need to insert 6 rational numbers, we multiply the numerator and denominator of both fractions by \( 6 + 1 = 7 \):
\( \frac{15}{18} = \frac{15 \times 7}{18 \times 7} = \frac{105}{126} \)
\( \frac{16}{18} = \frac{16 \times 7}{18 \times 7} = \frac{112}{126} \)

The six integers between 105 and 112 give us the numerators of the intermediate fractions:
\( \frac{106}{126}, \frac{107}{126}, \frac{108}{126}, \frac{109}{126}, \frac{110}{126}, \frac{111}{126} \)

Simplifying each fraction to its simplest form:
- \( \frac{106}{126} = \frac{53}{63} \)
- \( \frac{107}{126} \)
- \( \frac{108}{126} = \frac{6}{7} \)
- \( \frac{109}{126} \)
- \( \frac{110}{126} = \frac{55}{63} \)
- \( \frac{111}{126} = \frac{37}{42} \)

Thus, the six rational numbers are \( \frac{53}{63} \), \( \frac{107}{126} \), \( \frac{6}{7} \), \( \frac{109}{126} \), \( \frac{55}{63} \), and \( \frac{37}{42} \).
In simple words: First, rewrite both fractions so they share a bottom number of 18. Since there is no gap between 15/18 and 16/18, multiply both top and bottom by 7. This creates a clear space of six steps in between.

Exam Tip: Be sure to reduce each final fraction to its simplest form, as examiners often award full marks only for fully simplified answers.

Question 8. Insert seven rational numbers between 2 and 3.
Answer:
To find 7 rational numbers between 2 and 3, we can convert both integers into fractions with a common denominator. Since we want 7 numbers, we can multiply and divide by \( 7 + 1 = 8 \):
\( 2 = 2 \times \frac{8}{8} = \frac{16}{8} \)
\( 3 = 3 \times \frac{8}{8} = \frac{24}{8} \)

Now, choose the seven integers between 16 and 24 to be our numerators over the denominator of 8:
\( \frac{17}{8}, \frac{18}{8}, \frac{19}{8}, \frac{20}{8}, \frac{21}{8}, \frac{22}{8}, \frac{23}{8} \)

Simplify these fractions where possible:
- \( \frac{17}{8} = 2\frac{1}{8} \)
- \( \frac{18}{8} = \frac{9}{4} = 2\frac{1}{4} \)
- \( \frac{19}{8} = 2\frac{3}{8} \)
- \( \frac{20}{8} = \frac{5}{2} = 2\frac{1}{2} \)
- \( \frac{21}{8} = 2\frac{5}{8} \)
- \( \frac{22}{8} = \frac{11}{4} = 2\frac{3}{4} \)
- \( \frac{23}{8} = 2\frac{7}{8} \)

Thus, the seven rational numbers are \( 2\frac{1}{8} \), \( 2\frac{1}{4} \), \( 2\frac{3}{8} \), \( 2\frac{1}{2} \), \( 2\frac{5}{8} \), \( 2\frac{3}{4} \), and \( 2\frac{7}{8} \).
In simple words: To find seven numbers between 2 and 3, turn them into fractions with a bottom number of 8. 2 becomes 16/8 and 3 becomes 24/8. Then write down the seven fractions that sit between them.

Exam Tip: Expressing final fractions in mixed-number form makes it easy to verify that they indeed lie between 2 and 3.