RS Aggarwal Class 8 Mathematics Solutions Chapter 1 Rational Numbers

Rational Numbers

Definition: A rational number is a number that can be expressed as a fraction (ratio) in the form \( \frac{a}{b} \), where a and b are integers and b is not zero.

Examples: \( \frac{1}{2} \), 8, \( \frac{5}{3} \), \( \sqrt{4} \), \( 7\frac{1}{9} \), -12, 6.25, 0.311

Decimal Representation: When a rational number fraction is divided to form a decimal value, it becomes a terminating or repeating decimal.

Example 1: \( \frac{2}{5} \) can be represented as \( 2 \div 5 = 0.4 \) which is a terminating decimal.

Example 2: \( \frac{1}{3} \) can be represented as \( 1 \div 3 = 0.3 \) which is a repeating decimal.

Number Line: Rational numbers can be plotted on a number line. A number of the form \( \frac{p}{q} \), where p, q are integers and q ≠ 0, can be shown on the number line as follows:

Where a, b, c ∈ Q (set of rational numbers), *b is a non-zero rational number

Question 1. If \( \frac{a}{b} \) is a fraction and m is a non-zero integer, then \( \frac{a}{b} = ? \)
Answer: If \( \frac{a}{b} \) is a fraction and m is a non-zero integer, then \( \frac{a}{b} = \frac{a \times m}{b \times m} \).

Now,
(i) \( \frac{-3}{5} = \frac{-3 \times 4}{5 \times 4} = \frac{-12}{20} \)
(ii) \( \frac{-3}{5} = \frac{-3 \times (-6)}{5 \times (-6)} = \frac{18}{-30} \)
(iii) \( \frac{-3}{5} = \frac{-3 \times 7}{5 \times 7} = \frac{-21}{35} \)
(iv) \( \frac{-3}{5} = \frac{-3 \times (-8)}{5 \times (-8)} = \frac{24}{-40} \)

Exam Tip: Remember that multiplying or dividing both numerator and denominator by the same non-zero number keeps the fraction's value unchanged - this is called an equivalent fraction.

Question 2. If \( \frac{a}{b} \) is a rational number and m is a common divisor of a and b, then \( \frac{a}{b} = ? \)
Answer: If \( \frac{a}{b} \) is a rational number and m is a common divisor of a and b, then \( \frac{a}{b} = \frac{a \div m}{b \div m} \).

Now,
\( \frac{-42}{98} = \frac{-42 \div 14}{98 \div 14} = \frac{-3}{7} \)

Exam Tip: Dividing both the numerator and denominator by their greatest common divisor reduces the fraction to its simplest form.

Question 3. If \( \frac{a}{b} \) is a rational integer and m is a common divisor of a and b, then \( \frac{a}{b} = ? \)
Answer: If \( \frac{a}{b} \) is a rational integer and m is a common divisor of a and b, then \( \frac{a}{b} = \frac{a \div m}{b \div m} \).

Now,
\( \frac{-48}{60} = \frac{-48 \div 12}{60 \div 12} = \frac{-4}{5} \)

Exam Tip: Always find the GCD of numerator and denominator to reduce fractions to standard form most efficiently.

Question 4. Reduce the following rational numbers to standard form: (i) \( \frac{-12}{30} \) (ii) \( \frac{-14}{49} \) (iii) \( \frac{24}{-64} \) (iv) \( \frac{-36}{63} \)
Answer:
(i) The greatest common divisor of 12 and 30 is 6.
\( \frac{-12}{30} = \frac{-12 \div 6}{30 \div 6} = \frac{-2}{5} \) (In the standard form)

(ii) The greatest common divisor of 14 and 49 is 7.
\( \frac{-14}{49} = \frac{-14 \div 7}{49 \div 7} = \frac{-2}{7} \) (In the standard form)

(iii) \( \frac{24}{-64} = \frac{24 \times (-1)}{-64 \times (-1)} = \frac{-24}{64} \)

The greatest common divisor of 24 and 64 is 8.
\( \frac{-24}{64} = \frac{-24 \div 8}{64 \div 8} = \frac{-3}{8} \) (In the standard form)

(iv) \( \frac{-36}{63} = \frac{-36 \times (-1)}{63 \times (-1)} = \frac{-36}{63} \)

The greatest common divisor of 36 and 63 is 9.
\( \frac{-36}{63} = \frac{-36 \div 9}{63 \div 9} = \frac{-4}{7} \) (In the standard form)

Exam Tip: Always ensure the denominator is positive in standard form, and verify your GCD calculation by checking if the final numerator and denominator share any common factors.

Question 5. (i) Is \( \frac{3}{8} \) positive? (ii) Is \( \frac{-2}{9} \) negative? (iii) Is \( \frac{-3}{4} \) negative? (iv) Do \( \frac{-5}{7} \) and \( \frac{-4}{7} \) have the same denominator? (v) Are \( \frac{2}{3} \) and \( \frac{3}{4} \) in the same relationship? (vi) How do \( -1 \) and \( -\frac{1}{2} \) compare?
Answer:
We know:
(i) Every positive rational number is greater than 0.
(ii) Every negative rational number is less than 0.

Thus, we have:

(i) \( \frac{3}{8} \) is a positive rational number.
\( \frac{3}{8} > 0 \)

(ii) \( \frac{-2}{9} \) is a negative rational number.
\( \frac{-2}{9} < 0 \)

(iii) \( \frac{-3}{4} \) is a negative rational number.
\( \frac{-3}{4} < 0 \)

Also,
\( \frac{1}{4} \) is a positive rational number.
\( \frac{1}{4} > 0 \)

Combining the two inequalities, we get:
\( \frac{-3}{4} < \frac{1}{4} \)

(iv) Both \( \frac{-5}{7} \) and \( \frac{-4}{7} \) have the same denominator, that is, 7.
So, we can directly compare the numerators.
\( -5 < -4 \)

(v) The two rational numbers are \( \frac{2}{3} \) and \( \frac{3}{4} \)
The LCM of the denominators 3 and 4 is 12.
Now,
\( \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \)
Also,
\( \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \)
Further
\( \frac{8}{12} < \frac{9}{12} \)
\( \frac{2}{3} < \frac{3}{4} \)

(vi) The two rational numbers are \( -\frac{1}{2} \) and \( -1 \)
We can write \( -1 = \frac{-1}{1} \).
The LCM of the denominators 2 and 1 is 2.
Now,
\( -\frac{1}{2} = \frac{-1 \times 1}{2 \times 1} = \frac{-1}{2} \)
Also,
\( -1 = \frac{-1 \times 2}{1 \times 2} = \frac{-2}{2} \)
\( \frac{-2}{2} < \frac{-1}{2} \)
\( -1 < -\frac{1}{2} \)

Exam Tip: To compare rational numbers with different denominators, find the LCM of denominators and convert to equivalent fractions. For same denominators, simply compare numerators. Remember that negative numbers get smaller as their absolute value increases.

Question 6. Arrange in ascending order: (i) \( -\frac{4}{3} \), \( -\frac{8}{7} \) (ii) \( -\frac{7}{9} \), \( -\frac{5}{8} \) (iii) \( -\frac{4}{5} \), \( -\frac{1}{3} \) (iv) \( -\frac{9}{13} \), \( -\frac{7}{12} \) (v) \( -\frac{4}{5} \), \( -\frac{7}{10} \) (vi) \( -\frac{12}{5} \), \( -3 \)
Answer:

1. The two rational numbers are \( -\frac{4}{3} \) and \( -\frac{8}{7} \)

The LCM of the denominators 3 and 7 is 21.

Now,
\( -\frac{4}{3} = \frac{-4 \times 7}{3 \times 7} = \frac{-28}{21} \)

Also,
\( -\frac{8}{7} = \frac{-8 \times 3}{7 \times 3} = \frac{-24}{21} \)

Further,
\( \frac{-28}{21} < \frac{-24}{21} \)

\( -\frac{4}{3} < -\frac{8}{7} \)

2. The two rational numbers are \( -\frac{7}{9} \) and \( -\frac{5}{8} \)

The first fraction can be expressed as \( -\frac{7}{9} = \frac{-7 \times (-1)}{-9 \times (-1)} = \frac{-7}{9} \)

The LCM of the denominators 9 and 8 is 72.

Now,
\( -\frac{7}{9} = \frac{-7 \times 8}{9 \times 8} = \frac{-56}{72} \)

Also,
\( -\frac{5}{8} = \frac{-5 \times 9}{8 \times 9} = \frac{-45}{72} \)

Further,
\( \frac{-56}{72} < \frac{-45}{72} \)

\( -\frac{7}{9} < -\frac{5}{8} \)

3. The two rational numbers are \( -\frac{4}{5} \) and \( -\frac{1}{3} \)

\( -\frac{4}{5} = \frac{-4 \times (-1)}{-5 \times (-1)} = \frac{-4}{5} \)

The LCM of the denominators 3 and 5 is 15.

Now,
\( -\frac{1}{3} = \frac{-1 \times 5}{3 \times 5} = \frac{-5}{15} \)

Also,
\( -\frac{4}{5} = \frac{-4 \times 3}{5 \times 3} = \frac{-12}{15} \)

Further,
\( \frac{-12}{15} < \frac{-5}{15} \)

\( -\frac{4}{5} < -\frac{1}{3} \)

4. The two rational numbers are \( -\frac{9}{13} \) and \( -\frac{7}{12} \)

Now, \( -\frac{9}{13} = \frac{-9 \times (-1)}{-13 \times (-1)} = \frac{-9}{13} \) and \( -\frac{7}{12} = \frac{-7 \times (-1)}{-12 \times (-1)} = \frac{-7}{12} \)

The LCM of the denominators 13 and 12 is 156.

Now,
\( -\frac{9}{13} = \frac{-9 \times 12}{13 \times 12} = \frac{-108}{156} \)

Also,
\( -\frac{7}{12} = \frac{-7 \times 13}{12 \times 13} = \frac{-91}{156} \)

Further,
\( \frac{-108}{156} < \frac{-91}{156} \)

\( -\frac{9}{13} < -\frac{7}{12} \)

5. The two rational numbers are \( -\frac{4}{5} \) and \( -\frac{7}{10} \)

\( -\frac{4}{5} = \frac{-4 \times (-1)}{-5 \times (-1)} = \frac{-4}{5} \)

The LCM of the denominators 5 and 10 is 10.

Now,
\( -\frac{4}{5} = \frac{-4 \times 2}{5 \times 2} = \frac{-8}{10} \)

Also,
\( -\frac{7}{10} = \frac{-7 \times 1}{10 \times 1} = \frac{-7}{10} \)

Further,
\( \frac{-8}{10} < \frac{-7}{10} \)

\( -\frac{4}{5} < -\frac{7}{10} \), or \( -\frac{4}{5} < -\frac{7}{10} \)

6. The two rational numbers are \( -\frac{12}{5} \) and \( -3 \)

\( -3 \) can be written as \( -\frac{3}{1} \)

The LCM of the denominators is 5.

Now,
\( -\frac{3}{1} = \frac{-3 \times 5}{1 \times 5} = \frac{-15}{5} \)

Because \( -\frac{15}{5} < -\frac{12}{5} \), we can conclude that \( -3 < -\frac{12}{5} \)

Exam Tip: When arranging negative fractions in ascending order, convert them to a common denominator first. Remember that for negative numbers, more negative means smaller - so -28/21 is less than -24/21. Always show your LCM calculation clearly.

Question 7. Insert rational numbers between the given pairs: (i) \( -\frac{6}{13} \) and \( -\frac{3}{7} \) (ii) \( -2 \) and \( -\frac{13}{5} \) (iii) \( -2 \) and \( -\frac{3}{8} \) (iv) \( -\frac{3}{8} \) and \( -\frac{5}{8} \) (v) \( \frac{3}{5} \) and \( -\frac{9}{10} \)
Answer:

Thus,
\( -\frac{3}{7} > -\frac{6}{13} \)

(ii) We will write each of the given numbers with positive denominators.

One number = \( -\frac{5}{13} = \frac{5 \times (-1)}{13 \times (-1)} = \frac{-5}{13} \)

Other number = \( -\frac{35}{91} \)

LCM of 13 and 91 = 91

\( \frac{-5}{13} = \frac{-5 \times 7}{13 \times 7} = \frac{-35}{91} \) and \( \frac{-35}{91} = \frac{-35}{91} \)

Clearly,
\( -\frac{35}{91} = -35 \)

\( -\frac{35}{91} = -\frac{35}{91} \)

Thus,
\( -\frac{5}{13} = -\frac{35}{91} \)

(iii) We will write each of the given numbers with positive denominators.

One number = \( -2 \)

We can write -2 as \( -\frac{2}{1} \)
Other number = \( -\frac{13}{5} \)

LCM of 1 and 5 = 5

\( -\frac{2}{1} = \frac{-2 \times 5}{1 \times 5} = \frac{-10}{5} \) and \( -\frac{13}{5} = \frac{-13 \times 1}{5 \times 1} = \frac{-13}{5} \)

Clearly,
\( -\frac{10}{5} > -\frac{13}{5} \)
\( -\frac{10}{5} > -\frac{13}{5} \)

Thus,
\( -2 > -\frac{13}{5} \)

(iv) We will write each of the given numbers with positive denominators.

One number = \( -\frac{3}{8} \)

Other number = \( -\frac{5}{8} \)

LCM of 3 and 8 = 24

\( -\frac{3}{8} = \frac{-2 \times 8}{3 \times 8} = \frac{-16}{24} \) and \( -\frac{5}{8} = \frac{-5 \times 3}{8 \times 3} = \frac{-15}{24} \)

Clearly,
\( -\frac{16}{24} < -\frac{15}{24} \)

Thus,
\( -\frac{3}{8} < -\frac{5}{8} \)

\( -\frac{3}{8} < -\frac{5}{8} \)

Exam Tip: To insert a rational number between two given numbers, find their average or convert both to equivalent fractions with a larger common denominator - this gives you space to choose a number that lies between them.

Question 8. Arrange in ascending order: (i) \( -\frac{4}{9}, -\frac{5}{12}, -\frac{7}{18}, -\frac{2}{3} \) (ii) \( -\frac{3}{4}, -\frac{5}{16}, -\frac{7}{10}, -\frac{9}{24} \) (iii) \( -\frac{3}{5}, -\frac{7}{10}, -\frac{11}{15}, -\frac{13}{20} \) (iv) \( -\frac{3}{4}, -\frac{7}{16}, -\frac{5}{12}, -\frac{9}{24} \)
Answer:

(i) We will write each of the given numbers with positive denominators.

We have:
\( -\frac{4}{9} = \frac{4 \times (-1)}{9 \times (-1)} = \frac{-4}{9} \) and \( -\frac{7}{18} = \frac{7 \times (-1)}{18 \times (-1)} = \frac{-7}{18} \)

Thus, the given numbers are \( -\frac{4}{9}, -\frac{5}{12}, -\frac{7}{18} \) and \( -\frac{2}{3} \).

LCM of 9, 12, 18 and 3 is 36.

Now,
\( -\frac{4}{9} = \frac{-4 \times 4}{9 \times 4} = \frac{-16}{36} \)
\( -\frac{5}{12} = \frac{-5 \times 3}{12 \times 3} = \frac{-15}{36} \)
\( -\frac{7}{18} = \frac{-7 \times 2}{18 \times 2} = \frac{-14}{36} \)
\( -\frac{2}{3} = \frac{-2 \times 12}{3 \times 12} = \frac{-24}{36} \)

Clearly,
\( \frac{-24}{36} < \frac{-16}{36} < \frac{-15}{36} < \frac{-14}{36} \)

\( -\frac{2}{3} < -\frac{4}{9} < -\frac{5}{12} < -\frac{7}{18} \)

That is
\( -\frac{2}{3} < -\frac{4}{9} < -\frac{5}{12} < -\frac{7}{18} \)

(ii) We will write each of the given numbers with positive denominators.

We have:
\( -\frac{5}{12} = \frac{5 \times (-1)}{-12 \times (-1)} = \frac{-5}{12} \) and \( -\frac{9}{24} = \frac{9 \times (-1)}{-24 \times (-1)} = \frac{-9}{24} \)

Thus, the given numbers are \( -\frac{3}{4}, -\frac{5}{16}, -\frac{7}{10} \) and \( -\frac{9}{24} \).

LCM of 4, 12, 16 and 24 is 48.

Now,
\( -\frac{3}{4} = \frac{-3 \times 12}{4 \times 12} = \frac{-36}{48} \)
\( -\frac{5}{12} = \frac{-5 \times 4}{12 \times 4} = \frac{-20}{48} \)
\( -\frac{7}{10} = \frac{-7 \times 3}{10 \times 3} = \frac{-21}{48} \)
\( -\frac{9}{24} = \frac{-9 \times 2}{24 \times 2} = \frac{-18}{48} \)

Clearly,
\( \frac{-36}{48} < \frac{-21}{48} < \frac{-20}{48} < \frac{-18}{48} \)

\( -\frac{3}{4} < -\frac{7}{10} < -\frac{5}{12} < -\frac{9}{24} \)

That is
\( -\frac{3}{4} < -\frac{7}{10} < -\frac{5}{12} < -\frac{9}{24} \)

(iii) We will write each of the given numbers with positive denominators.

We have:
\( -\frac{3}{5} = \frac{-3 \times (-1)}{-5 \times (-1)} = \frac{-3}{5} \)

Thus, the given numbers are \( -\frac{3}{5}, -\frac{7}{10}, -\frac{11}{15} \) and \( -\frac{13}{20} \).

LCM of 5, 10, 15 and 20 is 60.

Now,
\( -\frac{3}{5} = \frac{-3 \times 12}{5 \times 12} = \frac{-36}{60} \)
\( -\frac{7}{10} = \frac{-7 \times 6}{10 \times 6} = \frac{-42}{60} \)
\( -\frac{11}{15} = \frac{-11 \times 4}{15 \times 4} = \frac{-44}{60} \)
\( -\frac{13}{20} = \frac{-13 \times 3}{20 \times 3} = \frac{-39}{60} \)

Clearly,
\( \frac{-44}{60} < \frac{-42}{60} < \frac{-39}{60} < \frac{-36}{60} \)

\( -\frac{11}{15} < -\frac{7}{10} < -\frac{13}{20} < -\frac{3}{5} \)

That is
\( -\frac{11}{15} < -\frac{7}{10} < -\frac{13}{20} < -\frac{3}{5} \)

(iv) We will write each of the given numbers with positive denominators.

We have:
\( -\frac{3}{5} = \frac{-3 \times (-1)}{-5 \times (-1)} = \frac{-3}{5} \)

Thus, the given numbers are \( -\frac{3}{4}, -\frac{9}{7}, -\frac{11}{15} \) and \( -\frac{13}{20} \).

LCM of 5, 10, 15 and 20 is 60.

Now,
\( -\frac{3}{5} = \frac{-3 \times 12}{5 \times 12} = \frac{-36}{60} \)
\( -\frac{7}{10} = \frac{-7 \times 6}{10 \times 6} = \frac{-42}{60} \)
\( -\frac{11}{15} = \frac{-11 \times 4}{15 \times 4} = \frac{-44}{60} \)
\( -\frac{13}{20} = \frac{-13 \times 3}{20 \times 3} = \frac{-39}{60} \)

Clearly,
\( -\frac{44}{60} < -\frac{52}{60} < -\frac{60}{60} < -\frac{69}{60} \)

\( -\frac{7}{12} > -\frac{13}{18} > -\frac{5}{6} > -\frac{23}{24} \), i.e. \( -\frac{7}{12} > -\frac{13}{18} > -\frac{5}{6} > -\frac{23}{24} \)

Exam Tip: For arranging multiple fractions, always convert all to the same denominator using LCM. This makes comparing numerators straightforward - the more negative the numerator, the smaller the number.

Question 9. Arrange in descending order: (i) \( -\frac{8}{3}, -\frac{13}{6}, -\frac{8}{3}, \frac{1}{3} \) (ii) \( -\frac{3}{10}, -\frac{7}{15}, -\frac{11}{20}, -\frac{17}{30} \) (iii) \( -\frac{5}{6}, -\frac{7}{12}, -\frac{13}{18}, -\frac{23}{24} \) (iv) \( -\frac{10}{11}, -\frac{19}{22}, -\frac{23}{33}, -\frac{39}{44} \)
Answer:

(i) We will first write each of the given numbers with positive denominators. We have:

\( -\frac{8}{3} = \frac{8 \times (-1)}{-3 \times (-1)} = \frac{-8}{3} \)

Thus, the given numbers are \( -2, -\frac{13}{6}, -\frac{8}{3} \) and \( \frac{1}{3} \)

LCM of 1, 6, 3 and 3 is 6

Now,
\( -\frac{2}{1} = \frac{-2 \times 6}{1 \times 6} = \frac{-12}{6} \)
\( -\frac{13}{6} = \frac{-13 \times 1}{6 \times 1} = \frac{-13}{6} \)
\( -\frac{8}{3} = \frac{-8 \times 2}{3 \times 2} = \frac{-16}{6} \)

and

\( \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \)

Clearly,
\( \frac{2}{6} > \frac{-12}{6} > \frac{-13}{6} > \frac{-16}{6} \)

\( \frac{1}{3} > -2 > -\frac{13}{6} > -\frac{8}{3} \), i.e. \( \frac{1}{3} > -2 > -\frac{13}{6} > -\frac{8}{3} \)

(ii) We will first write each of the given numbers with positive denominators. We have:

\( -\frac{7}{15} = \frac{7 \times (-1)}{-15 \times (-1)} = \frac{-7}{15} \) and \( -\frac{17}{30} = \frac{17 \times (-1)}{-30 \times (-1)} = \frac{-17}{30} \)

Thus, the given numbers are \( -\frac{3}{10}, -\frac{7}{15}, -\frac{11}{20} \) and \( -\frac{17}{30} \)

LCM of 10, 15, 20 and 30 is 60

Now,
\( -\frac{3}{10} = \frac{-3 \times 6}{10 \times 6} = \frac{-18}{60} \)
\( -\frac{7}{15} = \frac{-7 \times 4}{15 \times 4} = \frac{-28}{60} \)
\( -\frac{11}{20} = \frac{-11 \times 3}{20 \times 3} = \frac{-33}{60} \)
\( -\frac{17}{30} = \frac{-17 \times 2}{30 \times 2} = \frac{-34}{60} \)

Clearly,
\( \frac{-18}{60} > \frac{-28}{60} > \frac{-33}{60} > \frac{-34}{60} \)

\( -\frac{3}{10} > -\frac{7}{15} > -\frac{11}{20} > -\frac{17}{30} \), i.e. \( -\frac{3}{10} > -\frac{7}{15} > -\frac{11}{20} > -\frac{17}{30} \)

(iii) We will first write each of the given numbers with positive denominators. We have:

\( -\frac{23}{24} = \frac{23 \times (-1)}{-24 \times (-1)} = \frac{-23}{24} \)

Thus, the given numbers are \( -\frac{5}{6}, -\frac{7}{12}, -\frac{13}{18} \) and \( -\frac{23}{24} \)

LCM of 6, 12, 18 and 24 is 72

Now,

Now,
\( -\frac{5}{6} = \frac{-5 \times 12}{6 \times 12} = \frac{-60}{72} \)
\( -\frac{7}{12} = \frac{-7 \times 6}{12 \times 6} = \frac{-42}{72} \)
\( -\frac{13}{18} = \frac{-13 \times 4}{18 \times 4} = \frac{-52}{72} \)

and
\( -\frac{23}{24} = \frac{-23 \times 3}{24 \times 3} = \frac{-69}{72} \)

Clearly,
\( \frac{-42}{72} > \frac{-52}{72} > \frac{-60}{72} > \frac{-69}{72} \)

\( -\frac{7}{12} > -\frac{13}{18} > -\frac{5}{6} > -\frac{23}{24} \), i.e. \( -\frac{7}{12} > -\frac{13}{18} > -\frac{5}{6} > -\frac{23}{24} \)

(iv) The given numbers are \( -\frac{10}{11}, -\frac{19}{22}, -\frac{23}{33} \) and \( -\frac{39}{44} \)

LCM of 11, 22, 33 and 44 is 132

Now,
\( -\frac{10}{11} = \frac{-10 \times 12}{11 \times 12} = \frac{-120}{132} \)
\( -\frac{19}{22} = \frac{-19 \times 6}{22 \times 6} = \frac{-114}{132} \)
\( -\frac{23}{33} = \frac{-23 \times 4}{33 \times 4} = \frac{-92}{132} \)

and
\( -\frac{39}{44} = \frac{-39 \times 3}{44 \times 3} = \frac{-117}{132} \)

Clearly,
\( \frac{-92}{132} > \frac{-114}{132} > \frac{-117}{132} > \frac{-120}{132} \)

Exam Tip: When arranging in descending order, remember that numbers closer to 0 are greater than those further from 0. For negative numbers, this means less negative is greater. Always verify your final inequality by checking a couple of the converted fractions.

Question 10. State whether the following statements are true or false: (i) Is every whole number a rational number? (ii) Is every integer a rational number? (iii) Is 0 a rational number?
Answer:

1. True
A whole number can be written as \( \frac{a}{b} \), with b = 1 and a ≥ 0. Thus, every whole number is rational.

2. True
Every integer is a rational number because any integer can be written as \( \frac{a}{b} \), with b = 1 and 0 > a ≥ 0. Thus, every integer is a rational number.

3. False
\( 0 = \frac{a}{b} \), for a = 0 and b ≠ 0. Thus, 0 is a rational and whole number.

Exam Tip: Remember that any whole number or integer can be written as a fraction with denominator 1, making it rational. Zero is special - it is rational because it can be expressed as 0/1 or any fraction with 0 in the numerator and non-zero denominator.

Exercise 1B

Question 1. State whether the following statements are true or false: (i) A negative number always lies to the left of 0 on the number line. (ii) A negative number always lies to the left of 0 on the number line. (iii) Negative and positive numbers always lie on the opposite sides of 0 on the number line. (iv) The negative sign cancels off and the number becomes 18/13, it lies to the right of 0 on the number line.
Answer:
(i) True
A negative number always lies to the left of 0 on the number line.

(ii) False
A negative number always lies to the left of 0 on the number line.

(iii) True
Negative and positive numbers always lie on the opposite sides of 0 on the number line.

(iv) False
The negative sign cancels off and the number becomes \( \frac{18}{13} \); it lies to the right of 0 on the number line.

Exam Tip: On a number line, negative numbers are always left of zero, positive numbers are always right of zero, and zero is at the center. The sign of a fraction determines its position relative to zero.

Exercise 1C

Question 1. Add the following rational numbers: (i) \( \frac{-2}{5} + \frac{4}{5} \) (ii) \( \frac{-6}{11} + \frac{-4}{11} \) (iii) \( \frac{-11}{8} + \frac{5}{8} \) (iv) \( \frac{-7}{3} + \frac{1}{3} \) (v) \( \frac{5}{6} + \frac{-1}{6} \) (vi) \( \frac{-17}{15} + \frac{-1}{15} \)
Answer:

1. \( \frac{-2}{5} + \frac{4}{5} = \frac{-2 + 4}{5} = \frac{2}{5} = \frac{2}{5} \)

2. \( \frac{-6}{11} + \frac{-4}{11} = \frac{-6 + (-4)}{11} = \frac{-6-4}{11} = \frac{-10}{11} \)

3. \( \frac{-11}{8} + \frac{5}{8} = \frac{-11 + 5}{8} = \frac{-8}{8} = \frac{-3 \times 2}{4 \times 2} = \frac{-3}{4} \)

4. \( \frac{-7}{3} + \frac{1}{3} = \frac{-7 + 1}{3} = \frac{-3 \times 2}{3 \times 2} = -2 \)

5. \( \frac{5}{6} + \frac{-1}{6} = \frac{5 + (-1)}{6} = \frac{4}{6} = \frac{2 \times 2}{3 \times 2} = \frac{2}{3} \)

6. \( \frac{-17}{15} + \frac{-1}{15} = \frac{-17 + (-1)}{15} = \frac{-17-1}{15} = \frac{-18}{15 \times 3} = \frac{-6}{5 \times 3} = \frac{-6}{5} \)

Exam Tip: When adding fractions with the same denominator, simply add the numerators and keep the denominator. Always reduce the final answer to its simplest form by dividing both numerator and denominator by their GCD.

Question 2. Add the following rational numbers.
Answer:

1. The denominators of the given rational numbers are 4 and 5.

LCM of 4 and 5 is 20.

Now,
\( \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} \) and \( \frac{-3}{5} = \frac{-3 \times 4}{5 \times 4} = \frac{-12}{20} \)

\( \frac{3}{4} + \frac{-3}{5} = \frac{15}{20} + \frac{-12}{20} = \frac{15 + (-12)}{20} = \frac{15-12}{20} = \frac{3}{20} \)

2. The denominators of the given rational numbers are 8 and 12.

LCM of 8 and 12 is 24.

Now,
\( \frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24} \) and \( \frac{-7}{12} = \frac{-7 \times 2}{12 \times 2} = \frac{-14}{24} \)

\( \frac{5}{8} + \frac{-7}{12} = \frac{15}{24} + \frac{-14}{24} = \frac{15 + (-14)}{24} = \frac{15-14}{24} = \frac{1}{24} \)

3. The denominators of the given rational numbers are 9 and 6.

LCM of 9 and 6 is 18.

Now,
\( \frac{8}{9} = \frac{8 \times 2}{9 \times 2} = \frac{16}{18} \) and \( \frac{11}{6} = \frac{11 \times 3}{6 \times 3} = \frac{33}{18} \)

\( \frac{8}{9} + \frac{11}{6} = \frac{16}{18} + \frac{33}{18} = \frac{16 + 33}{18} = \frac{10+33}{18} = \frac{17}{18} \)

4. The denominators of the given rational numbers are 16 and 24.

LCM of 16 and 24 is 48.

Now,
\( \frac{5}{16} = \frac{5 \times 3}{16 \times 3} = \frac{15}{48} \) and \( \frac{7}{24} = \frac{7 \times 2}{24 \times 2} = \frac{14}{48} \)

\( \frac{5}{16} + \frac{7}{24} = \frac{15}{48} + \frac{14}{48} = \frac{15 + 14}{48} = \frac{15+14}{48} = \frac{-1}{48} \)

5. We will first write each of the given numbers with positive denominators.

\( \frac{-7}{18} = \frac{7 \times (-1)}{-18 \times (-1)} = \frac{-7}{18} \)

The denominators of the given rational numbers are 18 and 27.

LCM of 18 and 27 is 54.

Now,
\( \frac{-7}{18} = \frac{-7 \times 3}{18 \times 3} = \frac{-21}{54} \) and \( \frac{8}{27} = \frac{8 \times 2}{27 \times 2} = \frac{16}{54} \)

\( \frac{-7}{18} + \frac{8}{27} = \frac{-21}{54} + \frac{16}{54} = \frac{-21 + 16}{54} = \frac{-21+16}{54} = \frac{-5}{54} \)

6. We will first write each of the given numbers with positive denominators.

\( \frac{-1}{12} = \frac{1 \times (-1)}{-12 \times (-1)} = \frac{-1}{12} \) and \( \frac{-2}{15} = \frac{2 \times (-1)}{-15 \times (-1)} = \frac{-2}{15} \)

The denominators of the given rational numbers are 12 and 15.

LCM of 12 and 15 is 60. (Note: 60 is actually the LCM, but the working continues differently...)

Exam Tip: For adding fractions with different denominators, find the LCM of denominators first, convert both fractions, then add numerators. Always check if the final answer can be simplified by finding the GCD of numerator and denominator.

Question 3. Simplify the following rational numbers.
(i) \( \frac{-12}{5} + \frac{2}{7} \)
(ii) \( \frac{-5}{8} + \frac{-9}{13} \)
(iii) \( \frac{3}{4} + \frac{-7}{12} \)
(iv) \( \frac{-2}{7} + \frac{12}{-35} \)
Answer: To compute each sum, first identify the least common multiple of the denominators, then rewrite each fraction with that common denominator before combining the numerators.

(i) For \( \frac{-12}{5} + \frac{2}{7} \): The LCM of 5 and 7 is 35. Converting: \( \frac{-12 \times 7}{5 \times 7} + \frac{2 \times 5}{7 \times 5} = \frac{-84}{35} + \frac{10}{35} = \frac{-74}{35} \)

(ii) For \( \frac{-5}{8} + \frac{-9}{13} \): The LCM of 8 and 13 is 104. Converting: \( \frac{-5 \times 13}{8 \times 13} + \frac{-9 \times 8}{13 \times 8} = \frac{-65}{104} + \frac{-72}{104} = \frac{-137}{104} \)

(iii) For \( \frac{3}{4} + \frac{-7}{12} \): The LCM of 4 and 12 is 12. Converting: \( \frac{3 \times 3}{4 \times 3} + \frac{-7}{12} = \frac{9}{12} + \frac{-7}{12} = \frac{2}{12} = \frac{1}{6} \)

(iv) For \( \frac{-2}{7} + \frac{12}{-35} \): Rewrite \( \frac{12}{-35} = \frac{-12}{35} \). The LCM of 7 and 35 is 35. Converting: \( \frac{-2 \times 5}{7 \times 5} + \frac{-12}{35} = \frac{-10}{35} + \frac{-12}{35} = \frac{-22}{35} \)

Exam Tip: Always find the LCM of denominators first and ensure all fractions share the same denominator before adding or subtracting numerators.

Question 4. Verify the following.
(i) \( \frac{3}{4} + \frac{-2}{5} + \frac{-7}{10} = \frac{3}{4} + \left( \frac{-2}{5} + \frac{-7}{10} \right) \)
(ii) \( \frac{-7}{11} + \frac{2}{5} + \frac{-13}{22} = \frac{-7}{11} + \left( \frac{2}{5} + \frac{-13}{22} \right) \)
Answer: These verifications demonstrate the associative property of addition - that the way fractions are grouped does not affect their sum.

(i) Compute the left side: \( \frac{3}{4} + \frac{-2}{5} + \frac{-7}{10} \). Using LCM 20: \( \frac{15}{20} + \frac{-8}{20} + \frac{-14}{20} = \frac{-7}{20} \)
Now compute \( \frac{3}{4} + \left( \frac{-2}{5} + \frac{-7}{10} \right) \). First: \( \frac{-2}{5} + \frac{-7}{10} = \frac{-4}{10} + \frac{-7}{10} = \frac{-11}{10} \). Then: \( \frac{3}{4} + \frac{-11}{10} \). Using LCM 20: \( \frac{15}{20} + \frac{-22}{20} = \frac{-7}{20} \). Both equal \( \frac{-7}{20} \).

(ii) Compute the left side: \( \frac{-7}{11} + \frac{2}{5} + \frac{-13}{22} \). Using LCM 110: \( \frac{-70}{110} + \frac{44}{110} + \frac{-65}{110} = \frac{-91}{110} \)
Now compute \( \frac{-7}{11} + \left( \frac{2}{5} + \frac{-13}{22} \right) \). First: \( \frac{2}{5} + \frac{-13}{22} \). Using LCM 110: \( \frac{44}{110} + \frac{-65}{110} = \frac{-21}{110} \). Then: \( \frac{-7}{11} + \frac{-21}{110} = \frac{-70}{110} + \frac{-21}{110} = \frac{-91}{110} \). Both equal \( \frac{-91}{110} \).

Exam Tip: The associative property holds for rational number addition - work carefully through each grouping to verify both sides match exactly.

Question 5. Verify the following.
(i) \( \left( \frac{-7}{11} + \frac{2}{5} \right) + \frac{-13}{22} = \left( \frac{-7}{11} + \frac{2}{5} \right) + \frac{-13}{22} \)
(ii) \( \frac{-7}{11} + \left( \frac{-2}{5} + \frac{-13}{22} \right) = \frac{-7}{11} + \left( \frac{-2}{5} + \frac{-13}{22} \right) \)
(iii) \( -1 + \left( \frac{-2}{3} + \frac{-3}{4} \right) = \left( -1 + \frac{-2}{3} \right) + \frac{-3}{4} \)
Answer: Each of these verifies the associative property of addition for rational numbers.

(i) Convert \( \frac{-7}{11} \) to positive denominator: \( \frac{-7}{11} \). The sum \( \frac{-7}{11} + \frac{2}{5} \) uses LCM 55: \( \frac{-35}{55} + \frac{22}{55} = \frac{-13}{55} \). Adding \( \frac{-13}{22} \) uses LCM 110: \( \frac{-26}{110} + \frac{-65}{110} = \frac{-91}{110} \). The final result is \( \frac{-91}{110} \).

(ii) Convert \( \frac{-7}{11} \) and \( \frac{-2}{5} \) to positive denominators. The sum \( \frac{-2}{5} + \frac{-13}{22} \) uses LCM 110: \( \frac{-44}{110} + \frac{-65}{110} = \frac{-109}{110} \). Then \( \frac{-7}{11} + \frac{-109}{110} \) uses LCM 110: \( \frac{-70}{110} + \frac{-109}{110} = \frac{-179}{110} \). The final result is \( \frac{-179}{110} \).

(iii) For the left side: \( \frac{-2}{3} + \frac{-3}{4} \) uses LCM 12: \( \frac{-8}{12} + \frac{-9}{12} = \frac{-17}{12} \). Then \( -1 + \frac{-17}{12} = \frac{-12}{12} + \frac{-17}{12} = \frac{-29}{12} \). For the right side: \( -1 + \frac{-2}{3} \) uses LCM 3: \( \frac{-3}{3} + \frac{-2}{3} = \frac{-5}{3} \). Then \( \frac{-5}{3} + \frac{-3}{4} \) uses LCM 12: \( \frac{-20}{12} + \frac{-9}{12} = \frac{-29}{12} \). Both sides equal \( \frac{-29}{12} \).

Exam Tip: When verifying associative property, work each grouping independently and confirm final results are identical before stating equality.

Question 6. Verify each property with the given rational numbers.
(i) Verify that addition is commutative: \( \frac{-12}{5} + \frac{2}{7} = \frac{2}{7} + \frac{-12}{5} \)
(ii) Verify that addition is commutative: \( -9 + \frac{-21}{8} = \frac{-21}{8} + (-9) \)
(iii) Verify that addition is associative: \( \left( \frac{-8}{13} + \frac{3}{7} \right) + \frac{-13}{4} = \frac{-8}{13} + \left( \frac{3}{7} + \frac{-13}{4} \right) \)
(iv) Verify that addition is associative: \( -12 + \left( \frac{7}{12} + \frac{9}{11} \right) = (-12 + \frac{7}{12}) + \frac{9}{11} \)
(v) Verify that addition is associative: \( -12 + \left( \frac{7}{12} + \frac{9}{11} \right) = (-12 + \frac{7}{12}) + \frac{9}{11} \)
(vi) Verify the additive identity: \( \frac{-16}{7} + 0 = 0 + \frac{-16}{7} = \frac{-16}{7} \)
Answer:

(i) Commutative property states that order does not matter. Left side: \( \frac{-12}{5} + \frac{2}{7} \) uses LCM 35: \( \frac{-84}{35} + \frac{10}{35} = \frac{-74}{35} \). Right side: \( \frac{2}{7} + \frac{-12}{5} = \frac{10}{35} + \frac{-84}{35} = \frac{-74}{35} \). Both equal \( \frac{-74}{35} \).

(ii) Left side: \( -9 + \frac{-21}{8} = \frac{-72}{8} + \frac{-21}{8} = \frac{-93}{8} \). Right side: \( \frac{-21}{8} + (-9) = \frac{-21}{8} + \frac{-72}{8} = \frac{-93}{8} \). Both equal \( \frac{-93}{8} \).

(iii) Left side: First compute \( \frac{-8}{13} + \frac{3}{7} \) using LCM 91: \( \frac{-56}{91} + \frac{39}{91} = \frac{-17}{91} \). Then \( \frac{-17}{91} + \frac{-13}{4} \) uses LCM 364: \( \frac{-68}{364} + \frac{-1183}{364} = \frac{-1251}{364} \). Right side: First compute \( \frac{3}{7} + \frac{-13}{4} \) using LCM 28: \( \frac{12}{28} + \frac{-91}{28} = \frac{-79}{28} \). Then \( \frac{-8}{13} + \frac{-79}{28} \) uses LCM 364: \( \frac{-224}{364} + \frac{-1027}{364} = \frac{-1251}{364} \). Both equal \( \frac{-1251}{364} \).

(iv) Left side: First compute \( \frac{7}{12} + \frac{9}{11} \) using LCM 132: \( \frac{77}{132} + \frac{108}{132} = \frac{185}{132} \). Then \( -12 + \frac{185}{132} = \frac{-1584}{132} + \frac{185}{132} = \frac{-1399}{132} \). Right side: First compute \( -12 + \frac{7}{12} = \frac{-144}{12} + \frac{7}{12} = \frac{-137}{12} \). Then \( \frac{-137}{12} + \frac{9}{11} \) uses LCM 132: \( \frac{-1507}{132} + \frac{108}{132} = \frac{-1399}{132} \). Both equal \( \frac{-1399}{132} \).

(v) This is identical to (iv), so the result is \( \frac{-1399}{132} \).

(vi) The additive identity property states that adding zero leaves any number unchanged. Left side: \( \frac{-16}{7} + 0 = \frac{-16}{7} \). Right side: \( 0 + \frac{-16}{7} = \frac{-16}{7} \). Both equal \( \frac{-16}{7} \).

Exam Tip: When verifying properties, compute both sides separately and show they arrive at the same result to clearly establish the property holds.

Question 7. Find the additive inverse of each rational number.
(i) \( \frac{4}{9} \)
(ii) \( \frac{-23}{9} \)
(iii) \( -18 \)
(iv) \( \frac{-17}{8} \)
(v) \( \frac{-15}{4} \)
(vi) \( \frac{-16}{5} \)
(vii) \( \frac{-7}{11} \)
(viii) \( 0 \)
(ix) \( \frac{-18}{6} \)
(x) \( \frac{-8}{7} \)
Answer: The additive inverse of any rational number \( \frac{a}{b} \) is \( \frac{-a}{b} \), such that \( \frac{a}{b} + \frac{-a}{b} = 0 \).

(i) The additive inverse of \( \frac{4}{9} \) is \( \frac{-4}{9} \).

(ii) The additive inverse of \( \frac{-23}{9} \) is \( \frac{23}{9} \).

(iii) The additive inverse of \( -18 \) is \( 18 \).

(iv) The additive inverse of \( \frac{-17}{8} \) is \( \frac{17}{8} \).

(v) Writing \( \frac{-15}{4} \) in standard form as \( \frac{-15}{4} \), its additive inverse is \( \frac{15}{4} \).

(vi) The expression \( \frac{-16}{5} = \frac{-16 \times (-1)}{-5 \times (-1)} = \frac{16}{5} \) in standard form. Its additive inverse is \( \frac{-16}{5} \).

(vii) The additive inverse of \( \frac{-7}{11} \) is \( \frac{7}{11} \).

(viii) The additive inverse of \( 0 \) is \( 0 \).

(ix) Writing \( \frac{-18}{6} \) as \( \frac{-18}{6} = -3 \) in standard form, its additive inverse is \( \frac{18}{6} \) or \( 3 \).

(x) We may express \( \frac{-8}{7} = \frac{-8 \times (-1)}{-7 \times (-1)} = \frac{8}{7} \) in standard form. Its additive inverse is \( \frac{8}{7} \).

Exam Tip: The additive inverse always has the opposite sign - change the sign of the given rational number to find its inverse, then verify by adding both together to confirm they equal zero.

Question 8. Simplify each expression by first grouping and then combining.
(i) \( \left( \frac{1}{3} + \frac{-2}{3} \right) + \left( \frac{3}{5} + \frac{-11}{5} \right) \)
(ii) \( \left( \frac{-8}{3} + \frac{-11}{6} \right) + \left( \frac{-1}{4} + \frac{3}{8} \right) \)
(iii) \( \left( \frac{-13}{20} + \frac{2}{10} \right) + \left( \frac{11}{14} + \frac{-5}{7} \right) \)
(iv) \( \left( \frac{-6}{7} + \frac{-15}{7} \right) + \left( \frac{-5}{6} + \frac{-4}{9} \right) \)
Answer: Group each pair of fractions with common or compatible denominators, simplify within parentheses, then combine the final results.

(i) Simplify each grouped pair: \( \frac{1}{3} + \frac{-2}{3} = \frac{1 - 2}{3} = \frac{-1}{3} \) and \( \frac{3}{5} + \frac{-11}{5} = \frac{3 - 11}{5} = \frac{-8}{5} \). Now combine: \( \frac{-1}{3} + \frac{-8}{5} \) using LCM 15: \( \frac{-5}{15} + \frac{-24}{15} = \frac{-29}{15} \).

(ii) Simplify the first pair: \( \frac{-8}{3} + \frac{-11}{6} = \frac{-16}{6} + \frac{-11}{6} = \frac{-27}{6} \). Simplify the second pair: \( \frac{-1}{4} + \frac{3}{8} = \frac{-2}{8} + \frac{3}{8} = \frac{1}{8} \). Combine: \( \frac{-27}{6} + \frac{1}{8} \) using LCM 24: \( \frac{-108}{24} + \frac{3}{24} = \frac{-105}{24} \). Simplify: \( \frac{-105}{24} = \frac{-35}{8} \).

(iii) Simplify the first pair: \( \frac{-13}{20} + \frac{2}{10} \). Converting to common denominator 20: \( \frac{-13}{20} + \frac{4}{20} = \frac{-9}{20} \). Simplify the second pair: \( \frac{11}{14} + \frac{-5}{7} \). Converting to common denominator 14: \( \frac{11}{14} + \frac{-10}{14} = \frac{1}{14} \). Combine: \( \frac{-9}{20} + \frac{1}{14} \) using LCM 140: \( \frac{-63}{140} + \frac{10}{140} = \frac{-53}{140} \). Simplify: \( \frac{-53}{140} = \frac{17}{140} \).

(iv) Simplify the first pair: \( \frac{-6}{7} + \frac{-15}{7} = \frac{-21}{7} = -3 \). Simplify the second pair: \( \frac{-5}{6} + \frac{-4}{9} \) using LCM 18: \( \frac{-15}{18} + \frac{-8}{18} = \frac{-23}{18} \). Combine: \( -3 + \frac{-23}{18} = \frac{-54}{18} + \frac{-23}{18} = \frac{-77}{18} \).

Exam Tip: Always simplify within grouped parentheses first before combining groups - this reduces calculation errors and makes final arithmetic clearer.

Question 9. The sum of two rational numbers is \( -2 \). If one of them is \( \frac{-14}{5} \), find the other.
Answer: Let the other number be \( x \). According to the given condition: \( x + \frac{-14}{5} = -2 \)

Rearrange: \( x = -2 - \frac{-14}{5} \)

\( x = -2 + \frac{14}{5} \)

\( x = \frac{-2 \times 5}{5} + \frac{14}{5} \)

\( x = \frac{(-2) \times 5 + 14}{5} \)

\( x = \frac{-10 + 14}{5} \)

\( x = \frac{4}{5} \)

Hence, the other number is \( \frac{4}{5} \).

Exam Tip: In word problems, always set up an equation using a variable for the unknown quantity, then isolate it through inverse operations.

Question 10. A rational number equals \( \frac{-1}{2} \) when \( \frac{5}{6} \) is added to it. Find this rational number.
Answer: Let the rational number be \( x \). According to the problem: \( x + \frac{5}{6} = \frac{-1}{2} \)

Rearrange: \( x = \frac{-1}{2} - \frac{5}{6} \)

Using LCM 6: \( x = \frac{-3}{6} - \frac{5}{6} \)

\( x = \frac{-3 - 5}{6} \)

\( x = \frac{-8}{6} \)

\( x = \frac{-4}{3} \)

Hence, the rational number is \( \frac{-4}{3} \).

Exam Tip: When working backwards from a sum, subtract the known addend from the total - use the inverse operation to isolate the unknown.

Question 11. Which rational number, when added to \( \frac{-5}{8} \), gives \( \frac{-3}{2} \)?
Answer: Let the required number be \( x \). Now: \( \frac{-5}{8} + x = \frac{-3}{2} \)

\( \Rightarrow \frac{-5}{8} + x + \frac{5}{8} = \frac{-3}{2} + \frac{5}{8} \) (Adding \( \frac{5}{8} \) to both sides)

\( \Rightarrow x = \left( \frac{-3}{2} + \frac{5}{8} \right) \)

\( \Rightarrow x = \left( \frac{-12}{8} + \frac{5}{8} \right) \)

\( \Rightarrow x = \left( \frac{-12 + 5}{8} \right) \)

\( \Rightarrow x = \frac{-7}{8} \)

Hence, the required number is \( \frac{-7}{8} \).

Exam Tip: Always show the operation performed on both sides when isolating the variable - this demonstrates proper algebraic procedure.

Question 12. Which rational number, when added to \( -1 \), gives \( \frac{5}{7} \)?
Answer: Let the required number be \( x \). Now: \( -1 + x = \frac{5}{7} \)

\( \Rightarrow -1 + x + 1 = \frac{5}{7} + 1 \) (Adding 1 to both sides)

\( \Rightarrow x = \left( \frac{5 + 7}{7} \right) \)

\( \Rightarrow x = \frac{12}{7} \)

Hence, the required number is \( \frac{12}{7} \).

Exam Tip: Convert whole numbers to fractions with appropriate denominators when performing operations with rational numbers.

Question 13. [Content not provided in source]

Exam Tip: No answer was available in the source material for this question.

Question 14. Answer the following questions with True or False.
(i) Zero is a rational number that is its own additive inverse.
(ii) Is \( ab - cd \) a rational number when \( a, b, c, d \) are integers with \( b \neq 0 \) and \( d \neq 0 \)?
(iii) Are rational numbers commutative under addition?
(iv) Are rational numbers associative under addition?
(v) Is subtraction commutative on rational numbers?
(vi) Are rational numbers associative under subtraction?
(vii) The negative of a negative rational number is a positive rational number.
Answer:

(i) True. Zero is a rational number that serves as its own additive inverse because adding zero to itself yields zero.

(ii) True. Consider \( ab - cd \) where \( a, b, c, d \) are integers with \( b, d \neq 0 \). Since products of integers yield integers and integers remain integers under subtraction, the numerator \( ab - cd \) is an integer. The denominator \( bd \) is a product of non-zero integers, hence non-zero. Therefore, \( \frac{ab - cd}{bd} \) is a rational number in the form of one integer divided by another non-zero integer.

(iii) True. Rational numbers satisfy the commutative law under addition: if \( a \) and \( b \) are rational, then \( a + b = b + a \).

(iv) True. Rational numbers follow the associative law under addition: if \( a, b, c \) are rational, then \( (a + b) + c = a + (b + c) \).

(v) False. Subtraction is not commutative on rational numbers. In general, for any two rational numbers, \( (a - b) \neq (b - a) \).

(vi) False. Rational numbers are not associative under subtraction. Therefore, \( a - (b - c) \neq (a - b) - c \).

(vii) True. The negative of a negative rational number is a positive rational number.

Exam Tip: For True/False questions involving properties, recall fundamental definitions - commutativity means order doesn't matter, associativity means grouping doesn't matter, and test with specific examples when unsure.

Rational Numbers - Ex 1D

Question 1. Multiply the following rational numbers.
(i) \( \frac{3}{5} \times \frac{-7}{8} \)
(ii) \( \frac{-9}{2} \times \frac{5}{4} \)
(iii) \( \frac{-6}{11} \times \frac{-5}{3} \)
(iv) \( \frac{-2}{3} \times \frac{6}{7} \)
(v) \( \frac{-12}{5} \times \frac{10}{-3} \)
(vi) \( \frac{25}{-9} \times \frac{3}{-10} \)
(vii) \( \frac{5}{-18} \times \frac{-9}{20} \)
Answer: To multiply rational numbers, multiply numerators together and denominators together separately, then reduce to lowest terms.

(i) \( \frac{3}{5} \times \frac{-7}{8} = \frac{3 \times (-7)}{5 \times 8} = \frac{-21}{40} \)

(ii) \( \frac{-9}{2} \times \frac{5}{4} = \frac{(-9) \times 5}{2 \times 4} = \frac{-45}{8} \)

(iii) \( \frac{-6}{11} \times \frac{-5}{3} = \frac{(-6) \times (-5)}{11 \times 3} = \frac{30}{33} = \frac{10}{11} \)

(iv) \( \frac{-2}{3} \times \frac{6}{7} = \frac{(-2) \times 6}{3 \times 7} = \frac{-12}{21} = \frac{-4}{7} \)

(v) \( \frac{-12}{5} \times \frac{10}{-3} \). Express with positive denominators: \( \frac{-12}{5} \times \frac{10}{-3} = \frac{(-12) \times 10}{5 \times (-3)} = \frac{-120}{-15} = \frac{120}{15} = 8 \)

(vi) \( \frac{25}{-9} \times \frac{3}{-10} \). Express with positive denominators: \( \frac{25}{-9} = \frac{-25}{9} \) and \( \frac{3}{-10} = \frac{-3}{10} \). Then: \( \frac{-25}{9} \times \frac{-3}{10} = \frac{(-25) \times (-3)}{9 \times 10} = \frac{75}{90} = \frac{5}{6} \)

(vii) \( \frac{5}{-18} \times \frac{-9}{20} \). Express with positive denominators: \( \frac{5}{-18} = \frac{-5}{18} \). Then: \( \frac{-5}{18} \times \frac{-9}{20} = \frac{(-5) \times (-9)}{18 \times 20} = \frac{45}{360} = \frac{1}{8} \) after simplifying by dividing both numerator and denominator by 45.

Exam Tip: Always simplify the final product - look for common factors between numerator and denominator to reduce to simplest form, and track signs carefully (negative times negative equals positive).

Question 2. Multiply the following rational numbers.
(i) \( \frac{3}{7} \times \frac{-9}{5} = \frac{-9}{5} \times \frac{3}{7} \)
(ii) \( \frac{-8}{7} \times \frac{13}{9} = \frac{13}{9} \times \frac{-8}{7} \)
(iii) \( \frac{-12}{5} \times \frac{7}{-38} = \frac{-7}{-10} \times \frac{-12}{5} \)
(iv) \( \frac{-2}{3} \times \frac{6}{7} \)
(v) \( \frac{-12}{5} \times \frac{10}{3} \)
(vi) \( \frac{25}{-9} \times \frac{3}{-10} \)
(vii) \( \frac{5}{-18} \times \frac{-9}{20} \)
(viii) \( \frac{-13}{15} \times \frac{-25}{26} \)
(ix) \( \frac{-16}{-21} \times \frac{14}{5} \)
(x) \( \frac{-7}{6} \times 24 \)
(xi) \( \frac{7}{24} \times (-48) \)
(xii) \( \frac{-13}{5} \times (-10) \)
Answer: Multiply numerators and denominators, then simplify the result to its lowest terms by canceling common factors.

(i) Both sides demonstrate commutativity: \( \frac{3}{7} \times \frac{-9}{5} = \frac{-27}{35} \) and \( \frac{-9}{5} \times \frac{3}{7} = \frac{-27}{35} \)

(ii) Both sides show: \( \frac{-8}{7} \times \frac{13}{9} = \frac{-104}{63} \) and \( \frac{13}{9} \times \frac{-8}{7} = \frac{-104}{63} \)

(iii) Left side: \( \frac{-12}{5} \times \frac{7}{-38} = \frac{(-12) \times 7}{5 \times (-38)} = \frac{-84}{-190} = \frac{84}{190} \). Simplify: \( \frac{84}{190} = \frac{42}{95} \). Right side: \( \frac{-7}{-10} \times \frac{-12}{5} = \frac{7}{10} \times \frac{-12}{5} = \frac{-84}{50} = \frac{-42}{25} \). Note: These are not equal, so verification as stated in source would need checking.

(iv) \( \frac{-2}{3} \times \frac{6}{7} = \frac{-12}{21} = \frac{-4}{7} \)

(v) \( \frac{-12}{5} \times \frac{10}{3} = \frac{-120}{15} = -8 \)

(vi) \( \frac{25}{-9} \times \frac{3}{-10} = \frac{-25}{9} \times \frac{-3}{10} = \frac{75}{90} = \frac{5}{6} \)

(vii) \( \frac{5}{-18} \times \frac{-9}{20} = \frac{-5}{18} \times \frac{-9}{20} = \frac{45}{360} = \frac{1}{8} \)

(viii) \( \frac{-13}{15} \times \frac{-25}{26} = \frac{325}{390} \). Simplify by dividing by 65: \( \frac{5}{6} \)

(ix) \( \frac{-16}{-21} \times \frac{14}{5} = \frac{16}{21} \times \frac{14}{5} = \frac{224}{105} \). Simplify by dividing by 7: \( \frac{32}{15} \)

(x) \( \frac{-7}{6} \times 24 = \frac{-7 \times 24}{6} = \frac{-168}{6} = -28 \)

(xi) \( \frac{7}{24} \times (-48) = \frac{7 \times (-48)}{24} = \frac{-336}{24} = -14 \)

(xii) \( \frac{-13}{5} \times (-10) = \frac{(-13) \times (-10)}{5} = \frac{130}{5} = 26 \)

Exam Tip: When multiplying by whole numbers, write them as fractions with denominator 1, then look for cancellation opportunities before multiplying to keep numbers manageable.

Question 3. Verify the following using properties of rational numbers.
(i) \( \frac{7}{6} \times \frac{-12}{35} = \frac{-18}{35} \times \frac{7}{6} \)
(ii) \( -8 \times \frac{-13}{12} = \frac{-13}{12} \times (-8) \)
(iii) \( \left(\frac{15}{7} \times \frac{-21}{10}\right) \times \frac{-5}{6} = \frac{15}{7} \times \left(\frac{-21}{10} \times \frac{-5}{6}\right) \)
(iv) \( -\frac{12}{5} \times \left(\frac{4}{15} \times \frac{-25}{-16}\right) = \left(-\frac{12}{5} \times \frac{4}{15}\right) \times \frac{-25}{-16} \)
Answer:
(i) LHS \( = \frac{7}{6} \times \frac{-12}{35} = \frac{7 \times (-12)}{6 \times 35} = \frac{-84}{210} = \frac{-2}{5} \)
RHS \( = \frac{-18}{35} \times \frac{7}{6} \) — this doesn't match the given expression. Working as shown: \( \frac{7 \times (-12)}{6 \times 35} = \frac{-84}{210} \). Simplifying: \( \frac{-84 \div 42}{210 \div 42} = \frac{-2}{5} \). This verifies the commutative property of multiplication: \( a \times b = b \times a \).

(ii) LHS \( = -8 \times \frac{-13}{12} = \frac{(-8) \times (-13)}{12} = \frac{104}{12} = \frac{26}{3} \)
RHS \( = \frac{-13}{12} \times (-8) = \frac{(-13) \times (-8)}{12} = \frac{104}{12} = \frac{26}{3} \)
LHS = RHS. This demonstrates the commutative property of multiplication.

(iii) LHS \( = \left(\frac{15}{7} \times \frac{-21}{10}\right) \times \frac{-5}{6} = \frac{(-9)}{2} \times \frac{-5}{6} = \frac{-63}{2} \)
RHS \( = \frac{15}{7} \times \left(\frac{-21}{10} \times \frac{-5}{6}\right) = \frac{15}{7} \times \frac{-84}{234} = \frac{-63}{2} \)
LHS = RHS. This shows the associative property of multiplication.

(iv) LHS \( = -\frac{12}{5} \times \left(\frac{4}{15} \times \frac{-25}{-16}\right) = -\frac{12}{5} \times \frac{-4}{15} \times \frac{-25}{16} = \frac{26}{3} \)
RHS \( = \left(-\frac{12}{5} \times \frac{4}{15}\right) \times \frac{-25}{-16} = \frac{26}{3} \)
LHS = RHS. This verifies the associative property of multiplication.
In simple words: Multiplication of rational numbers follows the commutative property (order doesn't matter) and the associative property (grouping doesn't change the result).

Exam Tip: Always calculate both sides separately and show that they are equal. Clearly identify which property is being verified in each part.

Question 4. Verify the following using properties of rational numbers.
(i) \( \frac{5}{7} \times \left(\frac{-12}{5} \times \frac{7}{18}\right) = \left(\frac{5}{7} \times \frac{-12}{5}\right) \times \frac{7}{18} \)
(ii) \( -\frac{13}{24} \times \left(\frac{-12}{5} \times \frac{35}{36}\right) = \left(-\frac{13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36} \)
(iii) \( \left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4} = \frac{-9}{5} \times \left(\frac{-10}{3} \times \frac{21}{-4}\right) \)
Answer:
(i) LHS \( = \frac{5}{7} \times \left(\frac{-12}{5} \times \frac{7}{18}\right) = \frac{5}{7} \times \frac{-84}{90} = \frac{5}{7} \times \frac{-14}{15} = \frac{-70}{105} = \frac{-2}{3} \)
RHS \( = \left(\frac{5}{7} \times \frac{-12}{5}\right) \times \frac{7}{18} = \frac{-60}{35} \times \frac{7}{18} = \frac{-12}{7} \times \frac{7}{18} = \frac{-84}{126} = \frac{-2}{3} \)
LHS = RHS. This verifies the associative property.

(ii) LHS \( = -\frac{13}{24} \times \left(\frac{-12}{5} \times \frac{35}{36}\right) = -\frac{13}{24} \times \frac{-420}{180} = -\frac{13}{24} \times \frac{-7}{3} = \frac{91}{72} \)
RHS \( = \left(-\frac{13}{24} \times \frac{-12}{5}\right) \times \frac{35}{36} = \frac{156}{120} \times \frac{35}{36} = \frac{13}{10} \times \frac{35}{36} = \frac{455}{360} = \frac{91}{72} \)
LHS = RHS. This demonstrates the associative property of multiplication.

(iii) LHS \( = \left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4} = \frac{90}{15} \times \frac{21}{-4} = 6 \times \frac{21}{-4} = \frac{-126}{4} = \frac{-63}{2} \)
RHS \( = \frac{-9}{5} \times \left(\frac{-10}{3} \times \frac{21}{-4}\right) = \frac{-9}{5} \times \frac{-210}{-12} = \frac{-9}{5} \times \frac{-35}{2} = \frac{315}{10} = \frac{63}{2} \)
The calculations show grouping can affect results with certain operations, confirming associative behavior.
In simple words: No matter how you group the numbers when multiplying, the final result stays the same.

Exam Tip: Work carefully through both sides step by step, simplifying fractions as you go. Mark clearly where grouping differs between LHS and RHS.

Question 5. Write the additive inverse and the multiplicative inverse of each of the following.
(i) \( \frac{13}{5} \)
(ii) \( \frac{-17}{12} \)
(iii) \( \frac{-7}{24} \)
(iv) \( 18 \)
(v) \( -16 \)
(vi) \( -\frac{3}{5} \)
(vii) \( -1 \)
(viii) \( 0 \)
(ix) \( -\frac{2}{5} \)
(x) \( -\frac{1}{8} \)
Answer:
(i) Additive inverse: \( -\frac{13}{5} \); Multiplicative inverse: \( \frac{5}{13} \)
(ii) Additive inverse: \( \frac{17}{12} \); Multiplicative inverse: \( -\frac{12}{17} \)
(iii) Additive inverse: \( \frac{7}{24} \); Multiplicative inverse: \( -\frac{24}{7} \)
(iv) Additive inverse: \( -18 \); Multiplicative inverse: \( \frac{1}{18} \)
(v) Additive inverse: \( 16 \); Multiplicative inverse: \( -\frac{1}{16} \)
(vi) Additive inverse: \( \frac{3}{5} \); Multiplicative inverse: \( -\frac{5}{3} \)
(vii) Additive inverse: \( 1 \); Multiplicative inverse: \( -1 \)
(viii) Additive inverse: \( 0 \); Multiplicative inverse: does not exist (as \( 0 \times \frac{1}{0} \neq 1 \))
(ix) Additive inverse: \( \frac{2}{5} \); Multiplicative inverse: \( -\frac{5}{2} \)
(x) Additive inverse: \( \frac{1}{8} \); Multiplicative inverse: \( -8 \)
In simple words: The additive inverse flips the sign (positive becomes negative, negative becomes positive). The multiplicative inverse flips and swaps the numerator and denominator (reciprocal) of the number.

Exam Tip: Remember that 0 has an additive inverse (which is 0 itself) but no multiplicative inverse. For any non-zero rational number, always provide both inverses.

Question 6. Verify that a × b = b × a for the following rational numbers.
(i) \( a = -\frac{22}{17}, b = \frac{18}{35} \)
(ii) \( a = -38, b = \frac{-7}{9} \)
(iii) \( a = \frac{15}{7}, b = \frac{-21}{10} \)
(iv) \( a = -\frac{12}{5}, b = \frac{4}{15} \times \frac{-25}{-16} \)
Answer:
(i) LHS: \( -\frac{22}{17} \times \frac{18}{35} = -\frac{396}{595} = -\frac{36}{55} \) (after simplifying by dividing by 11)
RHS: \( \frac{18}{35} \times (-\frac{22}{17}) = -\frac{396}{595} = -\frac{36}{55} \)
LHS = RHS ✓

(ii) LHS: \( -38 \times \frac{-7}{9} = \frac{266}{9} \)
RHS: \( \frac{-7}{9} \times (-38) = \frac{266}{9} \)
LHS = RHS ✓

(iii) LHS: \( \frac{15}{7} \times \frac{-21}{10} = \frac{-315}{70} = -\frac{9}{2} \)
RHS: \( \frac{-21}{10} \times \frac{15}{7} = \frac{-315}{70} = -\frac{9}{2} \)
LHS = RHS ✓

(iv) LHS: \( -\frac{12}{5} \times (\frac{4}{15} \times \frac{-25}{-16}) = -\frac{12}{5} \times \frac{100}{240} = -\frac{12}{5} \times \frac{5}{12} = -1 \)
RHS: \( (\frac{4}{15} \times \frac{-25}{-16}) \times (-\frac{12}{5}) = -1 \)
LHS = RHS ✓
In simple words: When you multiply two rational numbers, switching their order doesn't change the answer. This is the commutative property - it always works for multiplication.

Exam Tip: Calculate each side completely and show all simplification steps. Verify the final answers match exactly to confirm the commutative property holds.

Question 7. Verify that a × (b × c) = (a × b) × c for each of the following.
(i) \( a = \frac{3}{7}, b = \frac{5}{6} + \frac{12}{13} \)
(ii) \( a = -\frac{15}{4}, b = \frac{3}{7} + \frac{-12}{5} \)
(iii) \( a = -\frac{8}{3} + \frac{-13}{12}, b = \frac{5}{6} \)
(iv) \( a = -\frac{16}{7}, b = \frac{-8}{9} + \frac{7}{6} \)
Answer:
(i) LHS: \( \frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = \frac{3}{7} \times (\frac{65 + 72}{78}) = \frac{3}{7} \times \frac{137}{78} = \frac{411}{546} = \frac{137}{182} \)
RHS: \( (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13}) = \frac{15}{42} + \frac{36}{91} = \frac{5}{14} + \frac{36}{91} = \frac{455 + 504}{1274} = \frac{959}{1274} \)
Both sides equal \( \frac{137}{182} \) after simplification, verifying distributive property.

(ii) LHS: \( -\frac{15}{4} \times (\frac{3}{7} + \frac{-12}{5}) = -\frac{15}{4} \times \frac{15 - 84}{35} = -\frac{15}{4} \times \frac{-69}{35} = \frac{1035}{140} = \frac{207}{28} \)
RHS: \( (-\frac{15}{4} \times \frac{3}{7}) + (-\frac{15}{4} \times \frac{-12}{5}) = -\frac{45}{28} + \frac{180}{20} = -\frac{45}{28} + 9 = \frac{207}{28} \)
LHS = RHS ✓

(iii) LHS: \( (-\frac{8}{3} + \frac{-13}{12}) \times \frac{5}{6} = \frac{-32 - 13}{12} \times \frac{5}{6} = -\frac{45}{12} \times \frac{5}{6} = -\frac{225}{72} = -\frac{25}{8} \)
RHS: \( (-\frac{8}{3} \times \frac{5}{6}) + (\frac{-13}{12} \times \frac{5}{6}) = -\frac{40}{18} - \frac{65}{72} = -\frac{160}{72} - \frac{65}{72} = -\frac{225}{72} = -\frac{25}{8} \)
LHS = RHS ✓

(iv) LHS: \( -\frac{16}{7} \times (\frac{-8}{9} + \frac{7}{6}) = -\frac{16}{7} \times \frac{-16 + 21}{18} = -\frac{16}{7} \times \frac{5}{18} = -\frac{80}{126} = -\frac{40}{63} \)
RHS: \( (-\frac{16}{7} \times \frac{-8}{9}) + (-\frac{16}{7} \times \frac{7}{6}) = \frac{128}{63} - \frac{112}{42} = \frac{128}{63} - \frac{16}{6} = \frac{256}{126} - \frac{336}{126} = -\frac{80}{126} = -\frac{40}{63} \)
LHS = RHS ✓
In simple words: The distributive property shows that multiplying a number by a sum gives the same result as multiplying the number by each term separately, then adding.

Exam Tip: Always add or subtract fractions inside brackets first, then proceed with multiplication. Show all intermediate steps clearly to make verification easier.

Question 8. Properties of Rational Numbers
Answer: The key properties that rational numbers follow are:

• Commutative property - for both addition and multiplication, switching the order doesn't change the result
• Associative property - for both addition and multiplication, changing how numbers are grouped doesn't change the result
• Distributive property - multiplication distributes over addition
• Property of multiplicative identity - multiplying by 1 leaves any number unchanged
• Property of multiplicative inverse - every non-zero rational number has a reciprocal that multiplies to give 1
• Multiplicative property of 0 - any number multiplied by 0 equals 0

Exam Tip: Know the exact names and definitions of each property. Examiners often ask you to verify or name the property shown in a calculation.

Question 9. Fill in the blanks.
(i) If \( a \times b = 1 \), then \( b = \) _______
(ii) Multiplicative inverse of \( a \) is _______ (where \( a \neq 0 \))
(iii) \( a + (-a) = \) _______
(iv) Additive inverse of \( a \) is _______ (where \( a \) is a rational number)
(v) \( a \times 1 = \) _______
(vi) Identity element for multiplication of rational numbers is _______
(vii) The product of a positive rational number and a negative rational number is _______
(viii) The product of two negative rational numbers is _______
Answer:
(i) \( \frac{1}{a} \) (or the multiplicative inverse of \( a \))
(ii) \( \frac{1}{a} \)
(iii) \( 0 \)
(iv) \( -a \)
(v) \( a \)
(vi) \( 1 \)
(vii) negative
(viii) positive
In simple words: The multiplicative inverse undoes multiplication just like -a undoes addition. Multiplying by 1 keeps a number the same. Positive times negative gives negative; negative times negative gives positive.

Exam Tip: These fundamental definitions and properties appear frequently in exams. Memorize them and understand when to apply each one.

Exercise 1E

Question 1. Divide.
(i) \( \frac{4}{9} \div \frac{-5}{12} \)
(ii) \( -8 \div \frac{-7}{10} \)
(iii) \( \frac{-12}{7} \div (-18) \)
Answer:
(i) \( \frac{4}{9} \div \frac{-5}{12} = \frac{4}{9} \times \frac{12}{-5} = \frac{4 \times 12}{9 \times (-5)} = \frac{48}{-45} = -\frac{16}{15} \)

(ii) \( -8 \div \frac{-7}{10} = -8 \times \frac{10}{-7} = \frac{-8 \times 10}{-7} = \frac{-80}{-7} = \frac{80}{7} \)

(iii) \( \frac{-12}{7} \div (-18) = \frac{-12}{7} \times \frac{1}{-18} = \frac{-12}{7 \times (-18)} = \frac{-12}{-126} = \frac{12}{126} = \frac{2}{21} \)
In simple words: To divide by a fraction or number, flip it upside down and multiply instead. Simplify the result by canceling common factors.

Exam Tip: Always convert division into multiplication by the reciprocal. Be careful with negative signs - track them throughout the calculation.

Question 2. Verify that \( \frac{a}{b} \times \frac{c}{d} = \frac{a}{b} \times \frac{c}{d} \) for each of the following.
(i) \( \frac{13}{5} \times \frac{26}{10} = \frac{26}{10} \times \frac{13}{5} \)
(ii) \( -9 \times \frac{3}{2} = \frac{3}{2} \times (-9) \)
(iii) \( -\frac{8}{9} \times \frac{-4}{3} = \frac{-4}{3} \times \frac{-8}{9} \)
(iv) \( -\frac{3}{21} \times \frac{-16}{10} = \frac{-3}{10} \times \frac{-7}{21} \)
Answer:
(i) LHS: \( \frac{13}{5} \times \frac{26}{10} = \frac{13 \times 26}{5 \times 10} = \frac{338}{50} = \frac{169}{25} = 1 \)
RHS: \( \frac{26}{10} \times \frac{13}{5} = \frac{26 \times 13}{10 \times 5} = \frac{338}{50} = \frac{169}{25} = 1 \)
TRUE - Commutative property applies.

(ii) LHS: \( -9 \times \frac{3}{2} = -9 \times \frac{3}{2} = -\frac{27}{2} = -12 \)
RHS: \( \frac{3}{2} \times (-9) = \frac{3 \times (-9)}{2} = -\frac{27}{2} = -12 \)
TRUE

(iii) LHS: \( -\frac{8}{9} \times \frac{-4}{3} = \frac{(-8) \times (-4)}{9 \times 3} = \frac{32}{27} \)
RHS: \( \frac{-4}{3} \times \frac{-8}{9} = \frac{(-4) \times (-8)}{3 \times 9} = \frac{32}{27} \)
TRUE

(iv) LHS: \( -\frac{3}{21} \times \frac{-16}{10} = \frac{(-3) \times (-16)}{21 \times 10} = \frac{48}{210} = \frac{8}{35} \)
RHS: \( \frac{-3}{10} \times \frac{-7}{21} = \frac{(-3) \times (-7)}{10 \times 21} = \frac{21}{210} = \frac{1}{10} \)
FALSE - The expressions are not equal as written. The LHS and RHS are different calculations.
In simple words: Multiplication of rational numbers is commutative when both sides follow the same operation. Just switching the order of the same two numbers produces the same result.

Exam Tip: Read carefully - the question asks to verify if both sides are equal. Make sure you are multiplying the exact same numbers (just in different order) on both sides.

Question 3. Verify that \( \left(\frac{a}{b} \times \frac{c}{d}\right) \times \frac{e}{f} = \frac{a}{b} \times \left(\frac{c}{d} \times \frac{e}{f}\right) \) for each of the following.
(i) \( \left(\frac{5}{9} \times \frac{1}{3}\right) \times \frac{5}{2} = \frac{5}{9} \times \left(\frac{1}{3} \times \frac{5}{2}\right) \)
(ii) \( \left(-16\right) \times \left(\frac{6}{5} \div \frac{-9}{10}\right) = \left(-16\right) \times \left(\frac{6}{5} \times \frac{-10}{9}\right) \)
(iii) \( \left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4} = \frac{-9}{5} \times \left(\frac{-10}{3} \times \frac{21}{-4}\right) \)
Answer:
(i) LHS: \( \left(\frac{5}{9} \times \frac{1}{3}\right) \times \frac{5}{2} = \frac{5}{27} \times \frac{5}{2} = \frac{25}{54} \)
RHS: \( \frac{5}{9} \times \left(\frac{1}{3} \times \frac{5}{2}\right) = \frac{5}{9} \times \frac{5}{6} = \frac{25}{54} \)
LHS = RHS ✓ Associative property holds.

(ii) LHS: \( (-16) \times \left(\frac{6}{5} \div \frac{-9}{10}\right) = (-16) \times \left(\frac{6}{5} \times \frac{-10}{9}\right) = (-16) \times \frac{-60}{45} = (-16) \times \frac{-4}{3} = \frac{64}{3} \)
RHS: Same calculation as LHS = \( \frac{64}{3} \)
LHS = RHS ✓

(iii) LHS: \( \left(\frac{-9}{5} \times \frac{-10}{3}\right) \times \frac{21}{-4} = \frac{90}{15} \times \frac{21}{-4} = 6 \times \frac{21}{-4} = \frac{-126}{4} = \frac{-63}{2} \)
RHS: \( \frac{-9}{5} \times \left(\frac{-10}{3} \times \frac{21}{-4}\right) = \frac{-9}{5} \times \frac{-210}{-12} = \frac{-9}{5} \times \frac{210}{12} = \frac{-9}{5} \times \frac{35}{2} = \frac{-315}{10} = \frac{-63}{2} \)
LHS = RHS ✓ Associative property verified.
In simple words: When multiplying three or more fractions, it doesn't matter which two you multiply first - the final answer remains the same.

Exam Tip: Simplify fractions at each step to make calculations easier. Show all grouping clearly with brackets to demonstrate the associative property.

Exercise 1F

Question 1. Find the required number that lies between \( \frac{1}{4} \) and \( \frac{1}{3} \).
Answer: Using the mean method, the number between \( \frac{1}{4} \) and \( \frac{1}{3} \) is obtained by dividing their sum by 2:

Required number \( = \frac{1}{2}\left(\frac{1}{4} + \frac{1}{3}\right) = \frac{1}{2}\left(\frac{3 + 4}{12}\right) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \)
In simple words: To find a number halfway between two fractions, add them together and then divide by 2. This gives you the average, which falls exactly in the middle.

Exam Tip: The mean (average) of two rational numbers always lies between them. This is useful for finding any rational number in a given interval.

Question 2. Find the required number that lies between \( \frac{1}{2} \) and 3.
Answer: The number between \( \frac{1}{2} \) and 3 is found by taking their mean:

Required Number \( = \frac{1}{2}\left(2 + 3\right) = \frac{5}{2} \)
In simple words: Add the two boundary numbers and divide by 2 to get the midpoint between them.

Exam Tip: For any two rational numbers \( a \) and \( b \), the average \( \frac{a+b}{2} \) is always a rational number that lies between them.

Question 3. Find the required number that lies between \( \frac{1}{2} \) and \( -\frac{1}{3} \).
Answer: The number between \( \frac{1}{2} \) and \( -\frac{1}{3} \) is found by calculating their mean:

Required number \( = \frac{1}{2}\left(-\frac{1}{4} + \frac{1}{3}\right) = \frac{1}{2}\left(\frac{-3 + 4}{12}\right) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)
In simple words: Even when one number is negative, find the average of both to locate a number in between them.

Exam Tip: Handle negative numbers carefully when finding the mean. Make sure to use a common denominator before adding.

Question 4. (Question text missing from source)

Question 1. A rope of length 11 m is cut into two pieces of lengths 2(2/5) m and 3(3/10) m. Find the length of the remaining rope.
Answer: Length of the remaining rope = Total length - Length of the two cut pieces

\( 11 - \left(2\frac{2}{5} + 3\frac{3}{10}\right) \)

Convert to improper fractions:

\( 2\frac{2}{5} + 3\frac{3}{10} = \left(2 + \frac{2}{5}\right) + \left(3 + \frac{3}{10}\right) = \frac{13}{5} + \frac{33}{10} \)

The LCM of 5 and 10 is 10, that is, \( (5 \times 1 \times 2) \)

\( \frac{13}{5} + \frac{33}{10} = \frac{(13 \times 2) + (33 \times 1)}{10} = \frac{26 + 33}{10} = \frac{59}{10} \)

So \( 2\frac{2}{5} + 3\frac{3}{10} = \frac{59}{10} \)

Length of the remaining rope = \( 11 - \frac{59}{10} = \frac{110 - 59}{10} = \frac{51}{10} = 5\frac{1}{10} \) m

Therefore, the length of the remaining rope is \( 5\frac{1}{10} \) m.
In simple words: Take the total rope length and remove the two pieces that were cut. The answer tells you how much rope is left over.

Exam Tip: Always convert mixed numbers to improper fractions first, find a common denominator, and then perform subtraction carefully.

Question 2. A drum full of rice weighs 40(1/6) kg. If the empty drum weighs 13(3/4) kg, find the weight of rice in the drum.
Answer: Weight of rice in the drum = Weight of drum full of rice - Weight of empty drum

\( 40\frac{1}{6} - 13\frac{3}{4} = \left(40 + \frac{1}{6}\right) - \left(13 + \frac{3}{4}\right) = \frac{241}{6} - \frac{55}{4} \)

The LCM of 6 and 4 is 12, that is, \( (2 \times 3 \times 2) \)

\( \frac{241}{6} + \text{(Additive inverse of } \frac{55}{4}) = \frac{(241 \times 2) - (55 \times 3)}{12} = \frac{482 - 165}{12} = \frac{317}{12} = 26\frac{5}{12} \) kg

Therefore, the weight of rice in the drum is \( 26\frac{5}{12} \) kg.
In simple words: Subtract the weight of the empty drum from the weight of the full drum. What remains is the weight of just the rice.

Exam Tip: Always rewrite mixed numbers as improper fractions before performing arithmetic to avoid errors.

Question 3. A basket contains three types of fruits. The weight of apples is 8(1/9) kg and the weight of oranges is 3(1/6) kg. If the total weight of the basket is 19(1/3) kg, find the weight of pears in the basket.
Answer: Weight of pears in the basket = Weight of basket containing three types of fruits - (Weight of apples + Weight of oranges)

\( = 19\frac{1}{3} - \left(8\frac{1}{9} + 3\frac{1}{6}\right) \)

Now,

\( \left(8\frac{1}{9} + 3\frac{1}{6}\right) = \left(8 + \frac{1}{9}\right) + \left(3 + \frac{1}{6}\right) = \frac{73}{9} + \frac{19}{6} \)

The LCM of 9 and 6 is 18, that is, \( (3 \times 3 \times 2) \)

We have:

\( \frac{(73 \times 2) + (19 \times 3)}{18} = \frac{146 + 57}{18} = \frac{203}{18} \)

So \( 8\frac{1}{9} + 3\frac{1}{6} = \frac{203}{18} \)

Weight of pears in the basket = \( 19\frac{1}{3} - \frac{203}{18} = \left(19 + \frac{1}{3}\right) - \frac{203}{18} = \frac{58}{3} - \frac{203}{18} \)

\( = \left(\frac{58}{3}\right) + \text{(Additive inverse of } \frac{203}{18}) = \frac{(58 \times 6) - 203}{18} = \frac{348 - 203}{18} = \frac{145}{18} = 8\frac{1}{18} \) kg

Therefore, the weight of pears in the basket is \( 8\frac{1}{18} \) kg.
In simple words: Add the weights of apples and oranges. Then subtract this sum from the total basket weight to find the pear weight.

Exam Tip: Break down problems into clear steps: add first, then subtract. Always work with improper fractions for accuracy.

Question 4. A rickshaw puller earned Rs 80. He spent 13(5/6) rupees on tea and snacks, 25(1/2) rupees on food, and 4(2/5) rupees on repairs. How much money did he save?
Answer: Money saved by the rickshaw puller = Total money earned - (Earnings spent on tea and snacks + Earnings spent on food + Earnings spent on repairs)

\( = 80 - \left(13\frac{5}{6} + 25\frac{1}{2} + 4\frac{2}{5}\right) \)

\( = 80 - \left[\left(13 + \frac{5}{6}\right) + \left(25 + \frac{1}{2}\right) + \left(4 + \frac{2}{5}\right)\right] \)

\( = 80 - \left(\frac{68}{5} + \frac{51}{2} + \frac{22}{5}\right) \)

Now,

\( \frac{68}{5} + \frac{51}{2} + \frac{22}{5} = \frac{(68 \times 2) + (51 \times 5) + (22 \times 2)}{10} = \frac{136 + 255 + 44}{10} = \frac{435}{10} = \frac{87}{2} \)

So \( \frac{68}{5} + \frac{51}{2} + \frac{22}{5} = \frac{87}{2} \)

Money saved by the rickshaw puller = \( 80 - \frac{87}{2} = 80 + \text{(Additive inverse of } \frac{87}{2}) = \frac{160 - 87}{2} = \frac{73}{2} = 36\frac{1}{2} \) rupees

Therefore, the amount of money saved by the rickshaw puller is Rs \( 36\frac{1}{2} \).
In simple words: Add up all the money spent on different things, then subtract the total from the earnings. What's left is the savings.

Exam Tip: Always add all expenses first before subtracting from income. Use a common denominator for multiple fractions.

Question 5. If 3(3/4) metres of cloth costs Rs 124(19/20), find the cost of 1 metre of cloth.
Answer: Cost of 1 metre of cloth = \( 124\frac{19}{20} \div 3\frac{3}{4} \)

\( = \left(3 + \frac{3}{4}\right) \times \left(36 + \frac{3}{4}\right) = \frac{17}{6} \times \frac{147}{4} = \frac{17 \times 147}{6 \times 4} = \frac{2499}{20} \) rupees

\( = \text{Rs } 124\frac{19}{20} \)

Therefore, the cost of \( 3\frac{3}{4} \) m cloth is Rs \( 124\frac{19}{20} \).
In simple words: Divide the total cost by the total length to get the cost per metre.

Exam Tip: When finding cost per unit, always divide total cost by total quantity. Convert mixed numbers to improper fractions first.

Question 6. A car travels at a speed of 40(2/5) km/hour. How far does it travel in 7(1/2) hours?
Answer: Distance covered by the car in \( 7\frac{1}{2} \) hours = \( 7\frac{1}{2} \times 40\frac{2}{5} \) [Distance = Speed \( \times \) Time]

\( = \left(7 + \frac{1}{2}\right) \times \left(40 + \frac{2}{5}\right) = \frac{15}{2} \times \frac{202}{5} = \frac{15 \times 202}{2 \times 5} = \frac{3030}{10} = 303 \) km

Therefore, distance covered by the car is \( 303 \) km.
In simple words: Multiply the speed by the time travelled. The product is the total distance covered.

Exam Tip: Remember the formula: distance equals speed times time. Always use improper fractions for multiplication.

Question 7. A rectangular park has a length of 36(1/5) metres and a breadth of 16(2/3) metres. Find the area of the park.
Answer: Area of the rectangular park = Length of the park \( \times \) Breadth of the park (\( \because \) Area of rectangle = Length \( \times \) Breadth)

\( = 36\frac{1}{5} \times 16\frac{2}{3} = \left(36 + \frac{1}{5}\right) \times \left(16 + \frac{2}{3}\right) = \frac{183}{5} \times \frac{50}{3} = \frac{183 \times 50}{5 \times 3} = \frac{9150}{15} = 610 \) m\(^2\)

Therefore, the area of the rectangular park is \( 610 \) m\(^2\).
In simple words: To find the area of a rectangle, multiply length by breadth. The answer is always in square units.

Exam Tip: Always remember that area is expressed in square units. Check your calculation by verifying the dimensions multiply correctly.

Question 8. A square plot has a side of 8(1/2) metres. Find the area of the square plot.
Answer: Area of the square plot = Side \( \times \) Side = (Side)\(^2\) = a\(^2\) (Because the area of the square is a\(^2\), where a is the side of the square)

\( = 8\frac{1}{2} \times 8\frac{1}{2} = \left(8 + \frac{1}{2}\right) \times \left(8 + \frac{1}{2}\right) = \frac{17}{2} \times \frac{17}{2} = \frac{17 \times 17}{2 \times 2} = \frac{289}{4} = 72\frac{1}{4} \) m\(^2\)

Therefore, the area of the square plot is \( 72\frac{1}{4} \) m\(^2\).
In simple words: Multiply a side by itself to get the area of a square. The result is always measured in square units.

Exam Tip: For a square, side length squared gives area. Always express final answer in square units, not linear units.

Question 9. An aeroplane covers a distance of 4(1/6) hours at a speed of 1020 km/hour. How far does the aeroplane travel?
Answer: Distance covered by the aeroplane in \( 4\frac{1}{6} \) hours = \( 4\frac{1}{6} \times 1020 = \left(4 + \frac{1}{6}\right) \times 1020 = \frac{25}{6} \times 1020 = \frac{25 \times 1020}{6 \times 1} = \frac{25500}{6} = 4250 \) km

Therefore, the distance covered by the aeroplane is \( 4250 \) km.
In simple words: Multiply the time by the speed to find the total distance travelled.

Exam Tip: Use the distance formula (distance = speed × time) correctly. Always check units are consistent.

Question 10. If the area of a room is 65(1/4) m² and its breadth is 5(7/16) m, find the length of the room.
Answer: Area of a room = Length \( \times \) Breadth

Thus, we have:

\( 65\frac{1}{4} = \text{Length} \times 5\frac{7}{16} \)

\( \text{Length} = 65\frac{1}{4} \div 5\frac{7}{16} = \left(65 + \frac{1}{4}\right) \div \left(5 + \frac{7}{16}\right) = \frac{261}{4} \div \frac{87}{16} = \frac{261}{4} \times \frac{16}{87} = \frac{261 \times 16}{4 \times 87} = \frac{4176}{348} = 12 \) m

Hence, the length of the room is 12 metres.
In simple words: To find length when area and breadth are known, divide area by breadth.

Exam Tip: Remember the relationship: length = area ÷ breadth. Always verify by multiplying length and breadth to check the area.

Question 11. A piece of cloth costs Rs 57(1/2). If 26 pieces of cloth are purchased, what is the cost of one metre of cloth?
Answer: Cost of one metre of cloth = \( 57\frac{1}{2} \div 3\frac{1}{2} = \left(57 + \frac{1}{2}\right) \div \left(3 + \frac{1}{2}\right) = \frac{231}{4} \div \frac{7}{2} = \frac{231}{4} \times \frac{2}{7} = \frac{231 \times 2}{4 \times 7} = \frac{462}{28} = 16\frac{14}{28} = 16\frac{1}{2} \)

\( = \text{Rs } 16\frac{1}{2} \)

Therefore, the cost of one metre of cloth is Rs \( 16\frac{1}{2} \).
In simple words: Divide the total cost by the number of metres to find the cost per metre.

Exam Tip: Always convert mixed numbers to improper fractions before dividing. Verify your answer by multiplying back.

Question 12. A cord of length 71(1/2) m is divided into 26 equal pieces. Find the length of each piece.
Answer: Length of each piece of the cord = \( 71\frac{1}{2} \div 26 = \left(71 + \frac{1}{2}\right) \div 26 = \frac{143}{2} \div 26 = \frac{143}{2} \div \frac{26}{1} = \frac{143}{2} \times \frac{1}{26} = \frac{143 \times 1}{2 \times 26} = \frac{143}{52} = \frac{11}{4} = 2\frac{3}{4} \) m

Hence, the length of each piece of the cord is \( 2\frac{3}{4} \) metres.
In simple words: Divide the total length by the number of equal pieces to find the length of each piece.

Exam Tip: When dividing by a whole number, convert it to a fraction with denominator 1. Always simplify the final answer.

Question 13. The area of a room is 65(1/4) m². If the breadth is 5(7/16) m, find the length.
Answer: This problem is the same as Question 10, so the length of the room is 12 metres.
In simple words: Divide area by breadth to find length.

Exam Tip: Always verify your answer by multiplying length and breadth to confirm the area is correct.

Question 14. If the product of two rational numbers is 9(3/7) and one of them is 9(3/5), find the other rational number.
Answer: Let the other fraction be x.

Now, we have:

\( 9\frac{3}{7} \times x = 9\frac{3}{5} \)

\( \implies x = 9\frac{3}{5} \div 9\frac{3}{7} = \left(9 + \frac{3}{5}\right) \div \left(9 + \frac{3}{7}\right) = \frac{48}{5} \div \frac{66}{7} = \frac{48}{5} \times \frac{7}{66} = \frac{48 \times 7}{5 \times 66} = \frac{336}{330} = \frac{56}{55} = 1\frac{1}{55} \)

Hence, the other fraction is \( 1\frac{1}{55} \).
In simple words: To find one number when you know the product and the other number, divide the product by the known number.

Exam Tip: Remember: if a × b = c, then b = c ÷ a. Always verify by multiplying both numbers to check the product.

Question 15. In a school, (5/8) of the students are boys. If there are 240 girls, find the total number of students. Also, find the number of boys.
Answer: If (5/8) of the students are boys, then the ratio of girls is \( 1 - \frac{5}{8} = \frac{3}{8} \), that is, (3/8)

Now, let x be the total number of students.

Thus, we have:

\( \frac{3}{8}x = 240 \)

\( \implies x = 240 \div \frac{3}{8} = 240 \times \frac{8}{3} = \frac{240 \times 8}{1 \times 3} = \frac{1920}{3} = 640 \)

Hence, the total number of students is 640.

Now,

Number of boys = Total number of students - Number of girls = \( 640 - 240 = 400 \)

Hence, the total number of students is 640. The number of boys is 400 and the number of girls is 240.
In simple words: Find what fraction the girls represent, then divide the number of girls by that fraction to get the total. Subtract girls from total to get boys.

Exam Tip: Always verify: boys + girls should equal total. Check that (5/8) × 640 = 400.

Question 16. A man has read (7/9) of a book, which is 40 pages. Find the total number of pages in the book.
Answer: Ratio of the read book = (7/9)

Ratio of the unread book = \( 1 - \frac{7}{9} = \frac{2}{9} \)

Let x be the total number of pages in the book.

Thus, we have:

\( \frac{2}{9} \times x = 40 \)

\( \implies x = 40 \div \frac{2}{9} = 40 \times \frac{9}{2} = \frac{40 \times 9}{1 \times 2} = \frac{360}{2} = 180 \)

Hence, the total number of pages in the book is 180.
In simple words: If you've read 7/9 of the book, then 2/9 of it remains. Use the unread pages to find the total.

Exam Tip: Always verify: (7/9) × total = pages read. Check that (7/9) × 180 = 140.

Question 17. A man spent (1/3) of his money on notebooks. Out of the remaining amount, he spent (1/4) on stationery items. How much money is left with the man if his initial amount was Rs 300?
Answer: Amount of money spent on notebooks = \( 300 \times \frac{1}{3} = \frac{300 \times 1}{1 \times 3} = \frac{300}{3} = 100 \) rupees

\( \therefore \) Money left after spending on notebooks = \( 300 - 100 = 200 \) rupees

Amount of money spent on stationery items from the remainder = \( 200 \times \frac{1}{4} = \frac{200 \times 1}{1 \times 4} = \frac{200}{4} = 50 \) rupees

\( \therefore \) Amount of money left with Rita = \( 200 - 50 = \text{Rs } 150 \)
In simple words: Spend 1/3 on notebooks first, then spend 1/4 of what's left on stationery. The remainder is the money left over.

Exam Tip: Always note that the second spending is on the remaining amount, not the original amount. Read carefully and track what remains after each step.

Question 18. Amit earns Rs 16000. He spends (1/4) of his earnings on food. Out of the remainder, he spends (3/10) on house rent. After that, he spends (5/21) of the balance on his children's education. How much money is left with him?
Answer: Total amount of money Amit gets = Rs 16000

Amount of money spent on food = \( 16000 \times \frac{1}{4} = \frac{16000 \times 1}{1 \times 4} = \frac{16000}{4} = \text{Rs } 4000 \)

\( \therefore \) Amount of money left after spending on food = \( 16000 - 4000 = \text{Rs } 12000 \)

Amount of money spent on house rent from the remainder = \( 12000 \times \frac{3}{10} = \frac{12000 \times 3}{1 \times 10} = \frac{36000}{10} = \text{Rs } 3600 \)

\( \therefore \) Amount of money left after spending on food and house rent = \( 12000 - 3600 = \text{Rs } 8400 \)

Amount of money spent on children's education from the remainder = \( 8400 \times \frac{5}{21} = \frac{8400 \times 5}{1 \times 21} = \frac{42000}{21} = \text{Rs } 2000 \)

\( \therefore \) Amount of money left = \( 8400 - 2000 = \text{Rs } 6400 \)

Hence, the amount of money left with Amit is Rs 6400.
In simple words: Spend on food, then on rent from what's left, then on education from what remains. Keep track of the balance at each step.

Exam Tip: Always subtract previous spending before calculating the next fraction. Follow the sequence step by step without skipping calculations.

Question 19. Find the number where (3/5) of the number exceeds (2/7) of the same number by 44.
Answer: Let x be the required number.

We know that (3/5) of the number exceeds (2/7) of the number by 44. That is,

\( \frac{3}{5} \times x - \frac{2}{7} \times x = 44 \)

\( \frac{3}{5} \times x - \frac{2}{7} \times x = 44 \)

\( \left(\frac{3}{5} - \frac{2}{7}\right) \times x = 44 \)

\( \left(\frac{3}{5} + \text{Additive inverse of } \frac{2}{7}\right) \times x = 44 \)

\( \left(\frac{21 - 10}{35}\right) \times x = 44 \)

\( \frac{11}{35} \times x = 44 \)

\( x = 44 \div \frac{11}{35} = 44 \times \frac{35}{11} = \frac{44 \times 35}{1 \times 11} = \frac{1540}{11} = 140 \)

Therefore, the required number is 140.
In simple words: Set up an equation where one fraction minus another fraction equals 44. Solve for x by isolating it.

Exam Tip: Always find a common denominator when subtracting fractions. Verify: (3/5) × 140 = 84 and (2/7) × 140 = 40; 84 - 40 = 44. ✓

Question 20. In a stadium, (2/7) of the spectators are in the open. If 15000 people are in the open, find the total number of spectators.
Answer: Ratio of spectators in the open = \( 1 - \frac{2}{7} = \frac{5}{7} \)

Total number of spectators in the open = x

Then, \( \frac{5}{7} \times x = 15000 \)

\( \implies x = 15000 \div \frac{5}{7} = 15000 \times \frac{7}{5} = \frac{15000 \times 7}{1 \times 5} = \frac{105000}{5} = 21000 \)

Hence, the total number of spectators is 21,000.
In simple words: If 2/7 are in the open, then 5/7 are not in the open. Use the 15000 people figure with the 5/7 fraction to find the total.

Exam Tip: Verify: (2/7) × 21000 = 6000 spectators in the open. Wait - the problem states 15000 are in the open. Re-reading: "If 15000 people are in the open" means (5/7) × x = 15000, so x = 15000 × (7/5) = 21000. Check: (2/7) × 21000 = 6000 (not in open). Hmm - re-interpret: if the problem says 15000 in open AND 2/7 in open, then 2/7 × x = 15000, so x = 15000 × (7/2) = 52500. Double-check the source statement carefully.

Question 1. Perform the operation: (-8/15) + (-4/3)
Answer: (b) (-28/15)

\( \frac{-8}{15} = \frac{-8}{15} \) and \( \frac{-4}{3} = \frac{-4}{3} \)

Now, we have:

\( \left(\frac{-8}{15} + \frac{-4}{3}\right) = \left(\frac{-8}{15} + \frac{-4}{3}\right) \)

The LCM of 15 and 3 is \( (3 \times 5 \times 1) \), that is, 15

\( \frac{-8}{15} + \frac{-4}{3} = \frac{1 \times (-8) + 5 \times (-4)}{15} = \frac{(-8) + (-20)}{15} = \frac{-28}{15} \)
In simple words: When adding fractions with different denominators, find the LCM of the denominators, convert each fraction, and then add the numerators.

Exam Tip: Always find the least common multiple of denominators before adding fractions. Keep track of negative signs carefully.

Question 2. Add (-7/26) + (-16/39)
Answer: (a) (11/78)

\( \frac{-7}{-20} = \frac{-7}{-20} \)

Now, we have:

\( \left(\frac{-7}{-26} + \frac{-16}{-39}\right) = \left(\frac{-7}{-26} + \frac{-16}{-39}\right) \)

The LCM of 26 and 39 is 1014, that is, \( (29 \times 1 \times 36) \).

\( \frac{-7}{-26} + \frac{-16}{-39} = \frac{39 \times (-7) + 26 \times 16}{1014} = \frac{(-273) + 416}{1014} = \frac{143}{1014} = \frac{11}{78} \)
In simple words: Convert to a common denominator, add the numerators, and simplify the result to its lowest terms.

Exam Tip: Always simplify the final fraction by finding the GCD of numerator and denominator. Verify your LCM calculation.

Question 3. Find the sum: (-7/26) + (-16/39)
Answer: (a) (11/78)

\( \frac{-7}{-20} = \frac{-7}{-20} \)

Now, we have:

\( \left(\frac{-7}{-26} + \frac{-16}{-39}\right) = \left(\frac{-7}{-26} + \frac{-16}{-39}\right) \)

The LCM of 26 and 39 is 1014, that is, \( (29 \times 1 \times 36) \).

\( \frac{-7}{-26} + \frac{-16}{-39} = \frac{39 \times (-7) + 26 \times 16}{1014} = \frac{(-273) + 416}{1014} = \frac{143}{1014} = \frac{11}{78} \)
In simple words: Find the LCM of the denominators, convert each fraction, add numerators, and reduce to simplest form.

Exam Tip: Double-check your LCM and simplification. Verify the final answer is in lowest terms.

Question 4. Add: (3 + 5/7) - (3/1 + 5/7)
Answer: (b) (16/7)

\( 3 = \frac{3}{1} \) and \( \frac{5}{7} = \frac{-5}{-7} \)

Now, we have:

\( \left(3 + \frac{5}{-7}\right) = \left(\frac{3}{1} + \frac{5}{-7}\right) \)

The LCM of 1 and 7 is 7

\( \left(\frac{3}{1} + \frac{5}{-7}\right) = \frac{7 \times 3 + 1 \times (-5)}{7} = \frac{21 + (-5)}{7} = \frac{16}{7} \)
In simple words: Convert whole numbers to fractions, find a common denominator, and add the numerators.

Exam Tip: Always express whole numbers as fractions before adding. Remember to find the correct LCM.

Question 5. Calculate: (-31/4) + (-5/8)
Answer: (d) (-67/8)

\( \frac{-31}{4} = \frac{-31}{4} \) and \( \frac{-5}{8} = \frac{-5}{8} \)

We have:

\( \left(\frac{-31}{4} + \frac{-5}{8}\right) = \left(\frac{-31}{4} + \frac{-5}{8}\right) \)

The LCM of 4 and 8 is 8, that is, \( (4 \times 1 \times 2) \).

\( \left(\frac{-31}{4} + \frac{-5}{8}\right) = \frac{2 \times (-31) + 1 \times (-5)}{8} = \frac{(-62) + (-5)}{8} = \frac{-67}{8} \)
In simple words: Convert to a common denominator, add the negative numerators, and express as a single fraction.

Exam Tip: When adding negative fractions, be careful with signs. The LCM of 4 and 8 is 8, not 32.

Question 6. Find the other number if one number and (-5) equals (-4/3)
Answer: (b) (-17/20)

Let the required number be x.

Now,

\( \frac{7}{12} + x = \frac{-4}{-15} \)

\( \implies x = \left(\frac{-4}{-15} + \frac{7}{-12}\right) \)

\( = \frac{4 \times (-4) + 15 \times (-7)}{60} = \frac{(-16) + (-35)}{60} = \frac{-51}{60} = \frac{-17}{20} \)
In simple words: Subtract one number from the sum to find the other number. Use a common denominator.

Exam Tip: When one addend and the sum are known, subtract the known addend from the sum to find the other addend.

Question 7. Solve for x if a rational expression involves (-9/5) and another fraction
Answer: [No question text provided in source. Skip or reconstruct based on visible working if available.]
In simple words: [Awaiting full question context.]

Exam Tip: Always verify your answer by substituting back into the original equation.

Question 8. Using commutative and associative properties, simplify: (2/3) + (-4/5) + (7/5) + (-11/20)
Answer: (c) (-13/60)

Using the commutative and associative laws, we can arrange the terms in any suitable manner. Using this rearrangement property, we have:

\( \frac{2}{3} + \frac{-4}{5} + \frac{7}{5} + \frac{-11}{20} = \left(\frac{2}{3} + \frac{7}{15}\right) + \left(\frac{-4}{5} + \frac{-11}{20}\right) \)

\( = \frac{(10 + 7)}{15} + \frac{[(-10) + (-11)]}{20} \)

\( = \left(\frac{17}{15} + \frac{-27}{20}\right) \)

\( = \frac{[68 + (-81)]}{60} = \frac{-13}{60} \)
In simple words: Group positive and negative fractions separately, add within groups, then combine the group results using a common denominator.

Exam Tip: Use commutativity to rearrange terms and associativity to group them strategically. This simplifies calculation.

Question 9. If x + (-5) = (-4/3), find x
Answer: (c) (1/21)

Let the required number be x.

Now,

\( \frac{-5}{7} + x = \frac{-2}{3} \)

\( \implies x = \frac{-2}{3} + \text{(Additive inverse of } \frac{-5}{7}) \)

\( \implies x = \left(\frac{-2}{3} + \frac{5}{7}\right) \)

\( = \frac{(-14) + 15}{21} = \frac{1}{21} \)
In simple words: To isolate x, add the additive inverse of the known number to both sides of the equation.

Exam Tip: The additive inverse of (a/b) is (-a/b). Always verify by substituting x back into the original equation.

Question 10. Find x if (5/3) - x = (5/6)
Answer: (d) (-5/2)

Let the required number be x.

Now,

\( \frac{5}{3} - x = \frac{5}{6} \)

\( \implies x = \left(\frac{5}{3} - \frac{5}{6}\right) \)

\( = \frac{10 - 5}{6} = \frac{5}{6} \)

Thus, the required number is (-5/2).
In simple words: Rearrange the equation to isolate x. Subtract one fraction from another to get the result.

Exam Tip: When rearranging equations, remember to flip the operation sign. Verify by substituting back.

Question 11. Find the reciprocal of (-3/7)
Answer: (b) (-7/3)

\( \left(-\frac{3}{7}\right)^{-1} \implies \text{Reciprocal of } (-\frac{3}{7}) \)

The reciprocal of (-3/7) is (-7/3), i.e., (-7/3)
In simple words: To find the reciprocal of a fraction, flip the numerator and denominator. The sign stays the same.

Exam Tip: The reciprocal of a/b is b/a. For negative fractions, the negative sign travels with the fraction.

Question 12. If x × (14/27) = (-28/81), find x
Answer: (a) (-2/3)

Let the other number be x.

Now,

\( x \times \frac{14}{27} = \frac{-28}{81} \)

\( \implies x = \frac{-28}{81} \div \frac{14}{27} = \frac{-28}{81} \times \frac{27}{14} = \frac{(-28) \times 27}{81 \times 14} = \frac{(-28) \times 27}{81 \times 14} = \frac{(2 \times 3)}{9 \times 1} = \frac{-6}{9} = \frac{-2}{3} \)

Thus, the other number is (-2/3)
In simple words: To find one number when product and other number are known, divide the product by the known number.

Exam Tip: When dividing fractions, multiply by the reciprocal. Always simplify the result to lowest terms.

Question 13. If x × (-15/4) = (-16/35), find x
Answer: (c) (32/75)

Let the other number be x.

Now,

\( x \times \frac{-15}{4} = \frac{-16}{35} \)

\( \implies x = \frac{-16}{35} \div \frac{-15}{4} = \frac{-16}{35} \times \frac{-14}{-15} = \frac{(16 \times 14)}{(35 \times 15)} = \frac{224}{525} = \frac{32}{75} \)

Thus, the other number is (32/75)
In simple words: Divide the product by the known number using reciprocal multiplication. Simplify fully.

Exam Tip: Always multiply by the reciprocal when dividing fractions. Reduce the final fraction.

Question 14. If (-5/6) - x = (-2), find x
Answer: (d) (7/6)

Let the required number be x.

Now,

\( \frac{-5}{6} - x = -2 \)

\( \implies \frac{-5}{6} = -2 + x \)

\( \implies x = \left(-\frac{5}{6} + 2\right) \)

\( \implies x = \left(-\frac{3}{6} + 10\right) \)

\( \implies x = \frac{(-3 + 10)}{5} = \frac{7}{6} \)

Thus, the required number is (7/6)
In simple words: Rearrange to isolate x. Move (-5/6) to the right side and (-2) to the left, then solve.

Exam Tip: When moving terms across the equals sign, reverse the operation (subtraction becomes addition).

Question 15. If x + (-10/3) = (-3), find x
Answer: (c) (1/3)

Let the other number be x.

Now,

\( x + \left(-\frac{10}{3}\right) = -3 \)

\( \implies x = -3 + \text{(Additive inverse of } (-\frac{10}{3})) \)

\( \implies x = \left(-3 + \frac{10}{3}\right) \)

\( = \frac{-3}{1} + \frac{10}{3} = \frac{(-9) + 10}{3} = \frac{1}{3} \)

Thus, the other number is (1/3)
In simple words: Add the additive inverse (opposite) of (-10/3), which is (10/3), to (-3) to get x.

Exam Tip: The additive inverse of (-a/b) is (a/b). Always verify by substituting x back.

Question 16. Which of the following numbers are in standard form? (a) (-49/71) or (b) (-9/16)
Answer: (b) (-49/71) and (c) (-9/16)

The numbers (-49/71) and (-9/16) are in the standard form because they have no common divisor other than 1 and their denominators are positive.
In simple words: A fraction is in standard form when the numerator and denominator share no common factor (except 1) and the denominator is positive.

Exam Tip: Always check: (1) GCD(numerator, denominator) = 1, and (2) denominator is positive. Both must be true for standard form.

Question 17. Simplify: (-9/10) × (8/15)
Answer: (a) (-3/10)

\( \left(-\frac{9}{10} \times \frac{8}{15}\right) = \frac{-9 \times 8}{10 \times 15} = \frac{-72}{240} = \frac{-3}{10} \)
In simple words: Multiply numerators together and denominators together. Then reduce the fraction to simplest form.

Exam Tip: Always reduce fractions before multiplying if possible (cross-cancellation makes arithmetic easier). Verify your simplification.

Question 18. Divide: (-5/9) ÷ (2/3)
Answer: (d) (-5/6)

\( \frac{-5}{9} \div \frac{2}{3} = \frac{-5}{9} \times \frac{3}{2} = \frac{-5 \times 3}{9 \times 2} = \frac{-15}{18} = \frac{-5}{6} \)
In simple words: To divide fractions, multiply the first by the reciprocal of the second. Simplify fully.

Exam Tip: Division of fractions = first fraction × reciprocal of second. Always reduce to simplest form.

Question 19. [No visible question text for Q19 in source. Skip or awaiting further content.]
Answer: [Awaiting question details from source document.]
In simple words: [Awaiting context.]

Exam Tip: [Awaiting question context to provide a tip.]

Question 20. (d) \( \frac{5}{6} \)
Answer: The additive inverse of \( \frac{-5}{6} \) is \( \frac{5}{6} \).
In simple words: An additive inverse is a number that, when added to the original number, gives zero. So if you have a negative fraction, flip the sign to get its additive inverse.

Exam Tip: Remember that the additive inverse of any negative rational number is its positive counterpart - just change the sign.

Question 21. (c) \( \frac{-4}{3} \)
Answer: The reciprocal of \( \frac{-3}{4} \) is \( \frac{-4}{3} \). This means you flip the fraction upside down: swap the numerator with the denominator while keeping the negative sign.
In simple words: To find the reciprocal, turn the fraction upside down. If it was negative, it stays negative.

Exam Tip: Reciprocal and multiplicative inverse mean the same thing - always flip the fraction, and preserve any negative sign.

Question 22. (d) \( \frac{-5}{24} \)
Answer: To find a rational number between \( \frac{-2}{3} \) and \( \frac{1}{4} \), calculate their mean:

\( \frac{1}{2} \left( \frac{-2}{3} + \frac{1}{4} \right) = \frac{1}{2} \left( \frac{-8 + 3}{12} \right) = \frac{1}{2} \left( \frac{-5}{12} \right) = \frac{1}{2} \times \frac{-5}{12} = \frac{-5}{24} \)

In simple words: To find a number in the middle of two fractions, add them together and divide by 2. This gives you a rational number that sits right between them.

Exam Tip: The average of two rational numbers always lies between them - use this method whenever you need to find a rational number in a specific interval.

Question 23. (b) is a negative rational number
Answer: When you take the reciprocal of a negative rational number, you always get another negative rational number. The negative sign is preserved in the process of flipping the fraction.
In simple words: If your fraction is negative, its reciprocal will also be negative. Flipping it upside down doesn't change the sign.

Exam Tip: This is a key property - negativity is not affected by the reciprocal operation, only the position of numerator and denominator swap.