ML Aggarwal Class 7 Maths Solutions Chapter 03 Rational Numbers

Exercise 3.1

Question 1. Which of the following are positive rational numbers?
\( \frac{5}{8}, \frac{-3}{11}, \frac{0}{5}, 7, -4, \frac{-13}{-3}, \frac{-6}{-17}, \frac{9}{-20} \)
Answer: A rational number is positive when both its numerator and denominator are either both positive or both negative integers. When one is positive and the other is negative, the rational number becomes negative. If both are zero, it is neither positive nor negative.

Checking each number:
\( \frac{5}{8} \) - numerator and denominator both positive → positive
\( \frac{-3}{11} \) - numerator negative, denominator positive → negative
\( \frac{0}{5} \) - equals 0 → neither positive nor negative
\( 7 = \frac{7}{1} \) - both positive → positive
\( -4 \) - negative
\( \frac{-13}{-3} \) - both negative → positive
\( \frac{-6}{-17} \) - both negative → positive
\( \frac{9}{-20} \) - numerator positive, denominator negative → negative

Therefore, the positive rational numbers are \( \frac{5}{8}, 7, \frac{-13}{-3} \) and \( \frac{-6}{-17} \).
In simple words: A rational number is positive when the top and bottom numbers are both positive or both negative. So check each number and pick the ones that are positive.

Exam Tip: Remember that a fraction with two negative signs (negative numerator and negative denominator) gives a positive result - the negatives cancel each other out.

Question 2. Which of the following are negative rational numbers?
\( \frac{-5}{7}, \frac{4}{-3}, \frac{-3}{-11}, -6, 9, 0, \frac{-28}{5}, \frac{31}{7} \)
Answer: A rational number is negative when one of its numerator and denominator is a positive integer and the other is a negative integer.

Checking each number:
\( \frac{-5}{7} \) - numerator negative, denominator positive → negative
\( \frac{4}{-3} \) - numerator positive, denominator negative → negative
\( \frac{-3}{-11} \) - both negative → positive
\( -6 \) - negative
\( 9 \) - positive
\( 0 \) - neither positive nor negative
\( \frac{-28}{5} \) - numerator negative, denominator positive → negative
\( \frac{31}{7} \) - both positive → positive

Therefore, the negative rational numbers are \( \frac{-5}{7}, \frac{4}{-3}, -6 \) and \( \frac{-28}{5} \).
In simple words: A rational number is negative when the top and bottom numbers have different signs - one positive and one negative.

Exam Tip: Always check both the numerator and denominator; if they have opposite signs, the rational number is definitely negative.

Question 3. Find four rational numbers equivalent to each of the following rational numbers:
(i) \( \frac{-7}{3} \)
(ii) \( \frac{-9}{-5} \)
Answer:
(i) To find equivalent fractions of \( \frac{-7}{3} \), we multiply both the numerator and denominator by the same numbers - 2, 3, 4, and 5:

\( \frac{-7 \times 2}{3 \times 2} = \frac{-14}{6} \)

\( \frac{-7 \times 3}{3 \times 3} = \frac{-21}{9} \)

\( \frac{-7 \times 4}{3 \times 4} = \frac{-28}{12} \)

\( \frac{-7 \times 5}{3 \times 5} = \frac{-35}{15} \)

Four rational numbers equivalent to \( \frac{-7}{3} \) are \( \frac{-14}{6}, \frac{-21}{9}, \frac{-28}{12} \) and \( \frac{-35}{15} \).

(ii) First, simplify \( \frac{-9}{-5} = \frac{9}{5} \). Now multiply both the numerator and denominator by 2, 3, and 4:

\( \frac{9 \times 2}{5 \times 2} = \frac{18}{10} \)

\( \frac{9 \times 3}{5 \times 3} = \frac{27}{15} \)

\( \frac{9 \times 4}{5 \times 4} = \frac{36}{20} \)

Four rational numbers equivalent to \( \frac{-9}{-5} \) are \( \frac{9}{5}, \frac{18}{10}, \frac{27}{15} \) and \( \frac{36}{20} \).
In simple words: To make equivalent fractions, multiply or divide both the top and bottom by the same number. The fraction stays the same value even though it looks different.

Exam Tip: Equivalent fractions have the same value but different numerators and denominators - always multiply (or divide) both top and bottom by the same number to keep the value unchanged.

Question 4. Write each of the following rational numbers with positive denominator:
(i) \( \frac{4}{-9} \)
(ii) \( \frac{-33}{17} \)
(iii) \( \frac{-38}{-15} \)
Answer: To get a positive denominator, multiply both the numerator and denominator by -1.

(i) \( \frac{4}{-9} = \frac{4 \times (-1)}{-9 \times (-1)} = \frac{-4}{9} \)

(ii) \( \frac{-33}{17} = \frac{-33 \times (-1)}{17 \times (-1)} = \frac{33}{-17} \)

(iii) \( \frac{-38}{-15} = \frac{-38 \times (-1)}{-15 \times (-1)} = \frac{38}{15} \)
In simple words: If the bottom number is negative, multiply the top and bottom both by -1 to make the bottom positive.

Exam Tip: A rational number written with a positive denominator is in standard form - always make sure your final answer has a positive denominator.

Question 5. Express \( \frac{-9}{5} \) as a rational number with
(i) numerator = 20
(ii) numerator = -35
(iii) denominator = -54
(iv) denominator = 72
Answer:
(i) To get 20 from -9, we need to find what to multiply -9 by. Since \( -9 \times (-2.22...) \neq 20 \) easily, we recognize that \( 5 \times 4 = 20 \), so multiply numerator and denominator by 4:
\( \frac{-9}{5} = \frac{-9 \times 4}{5 \times 4} = \frac{-36}{20} \)

(ii) To get -35 from 5, we multiply 5 by -7:
\( \frac{-9}{5} = \frac{-9 \times (-7)}{5 \times (-7)} = \frac{63}{-35} \)

(iii) To get -54 from -9, we multiply -9 by 6:
\( \frac{-9}{5} = \frac{-9 \times 6}{5 \times 6} = \frac{-54}{30} \)

(iv) To get 72 from 5, we multiply 5 by 14.4, but working with the denominator: multiply -9 by -8 and 5 by -8:
\( \frac{-9}{5} = \frac{-9 \times (-8)}{5 \times (-8)} = \frac{72}{-40} \)
In simple words: Pick a number to multiply both top and bottom by. Choose it so that either the top or bottom becomes the number you want.

Exam Tip: Decide whether you need to change the numerator or denominator, then work backwards to find what number to multiply by.

Question 6. Express \( \frac{-80}{112} \) as a rational number with
(i) numerator = -5
(ii) denominator = -14
Answer: First, reduce \( \frac{-80}{112} \) to standard form. The HCF of 80 and 112 is 16.

\( \frac{-80}{112} = \frac{-80 \div 16}{112 \div 16} = \frac{-5}{7} \)

(i) The numerator is already -5, so \( \frac{-80}{112} = \frac{-5}{7} \)

(ii) To get -14 from 7, we multiply 7 by -2:
\( \frac{-5}{7} = \frac{-5 \times (-2)}{7 \times (-2)} = \frac{10}{-14} \)

Therefore, \( \frac{-80}{112} = \frac{10}{-14} \)
In simple words: First, make the fraction simpler by dividing top and bottom by the same number. Then multiply to get the form you need.

Exam Tip: Always simplify a fraction first before converting it to the required form - this makes the calculations much easier.

Question 7. Which of the following pairs represent the same rational number?
(i) \( \frac{-7}{21}, \frac{3}{9} \)
(ii) \( \frac{-16}{20}, \frac{20}{-25} \)
(iii) \( \frac{-3}{5}, \frac{-12}{20} \)
(iv) \( \frac{8}{-5}, \frac{-24}{15} \)
Answer: Two rational numbers \( \frac{p}{q} \) and \( \frac{r}{s} \) are equal if and only if \( p \times s = q \times r \).

(i) For \( \frac{-7}{21} \) and \( \frac{3}{9} \):
\( -7 \times 9 = -63 \) and \( 21 \times 3 = 63 \)
Since \( -63 \neq 63 \), the pair does not represent the same rational number.

(ii) For \( \frac{-16}{20} \) and \( \frac{20}{-25} \):
\( -16 \times (-25) = 400 \) and \( 20 \times 20 = 400 \)
Since \( 400 = 400 \), the pair represents the same rational number.

(iii) For \( \frac{-3}{5} \) and \( \frac{-12}{20} \):
\( -3 \times 20 = -60 \) and \( 5 \times (-12) = -60 \)
Since \( -60 = -60 \), the pair represents the same rational number.

(iv) For \( \frac{8}{-5} \) and \( \frac{-24}{15} \):
\( 8 \times 15 = 120 \) and \( -5 \times (-24) = 120 \)
Since \( 120 = 120 \), the pair represents the same rational number.

Therefore, pairs (ii), (iii) and (iv) represent the same rational number.
In simple words: Two fractions are the same if you cross-multiply them and get equal answers. If top of first times bottom of second equals bottom of first times top of second, they are the same.

Exam Tip: Use the cross-multiplication method to check if two rational numbers are equal - it's quick and always reliable.

Question 8. Fill in the blanks:
(i) \( \frac{5}{4} = \frac{...}{16} = \frac{25}{...} = \frac{...}{-15} \)
(ii) \( \frac{-3}{7} = \frac{...}{14} = \frac{9}{...} = \frac{...}{-6} \)
Answer:
(i) Starting with \( \frac{5}{4} \):

For \( \frac{5}{4} = \frac{...}{16} \) - To get 16 from 4, multiply by 4. So, \( 5 \times 4 = 20 \).

For \( \frac{5}{4} = \frac{25}{...} \) - To get 25 from 5, multiply by 5. So, \( 4 \times 5 = 20 \).

For \( \frac{5}{4} = \frac{...}{-15} \) - To get -15 from 5, multiply by -3. So, \( 4 \times (-3) = -12 \).

Therefore: \( \frac{5}{4} = \frac{20}{16} = \frac{25}{20} = \frac{-12}{-15} \)

(ii) Starting with \( \frac{-3}{7} \):

For \( \frac{-3}{7} = \frac{...}{14} \) - To get 14 from 7, multiply by 2. So, \( -3 \times 2 = -6 \).

For \( \frac{-3}{7} = \frac{9}{...} \) - To get 9 from -3, multiply by -3. So, \( 7 \times (-3) = -21 \).

For \( \frac{-3}{7} = \frac{...}{-6} \) - To get -6 from -3, multiply by 2. So, \( 7 \times 2 = 14 \).

Therefore: \( \frac{-3}{7} = \frac{-6}{14} = \frac{9}{-21} = \frac{14}{-6} \)
In simple words: See what you multiply one part by to get the new part. Then multiply the other part by the same number to fill in the blank.

Exam Tip: Always find what number the given numerator or denominator was multiplied by, then apply that same multiplier to complete the equivalent fraction.

Question 9. Reduce each of the following rational numbers in standard form:
(i) \( \frac{-45}{30} \)
(ii) \( \frac{16}{-36} \)
(iii) \( \frac{-3}{-15} \)
(iv) \( \frac{68}{-119} \)
Answer:
(i) The denominator is already positive. The HCF of 45 and 30 is 15.
\( \frac{-45}{30} = \frac{-45 \div 15}{30 \div 15} = \frac{-3}{2} \)

(ii) To make the denominator positive first:
\( \frac{16}{-36} = \frac{16 \times (-1)}{-36 \times (-1)} = \frac{-16}{36} \)
The HCF of 16 and 36 is 4.
\( \frac{-16}{36} = \frac{-16 \div 4}{36 \div 4} = \frac{-4}{9} \)

(iii) To make the denominator positive:
\( \frac{-3}{-15} = \frac{-3 \times (-1)}{-15 \times (-1)} = \frac{3}{15} \)
The HCF of 3 and 15 is 3.
\( \frac{3}{15} = \frac{3 \div 3}{15 \div 3} = \frac{1}{5} \)

(iv) To make the denominator positive:
\( \frac{68}{-119} = \frac{68 \times (-1)}{-119 \times (-1)} = \frac{-68}{119} \)
The HCF of 68 and 119 is 17.
\( \frac{-68}{119} = \frac{-68 \div 17}{119 \div 17} = \frac{-4}{7} \)
In simple words: First, make sure the bottom is positive. Then divide the top and bottom by the biggest number that divides both evenly.

Exam Tip: Standard form requires both a positive denominator and the smallest possible numerator and denominator - find the HCF and divide both by it.

Exercise 3.2

Question 1. Draw a number line and represent the following rational numbers on it:
(i) \( \frac{3}{8} \)
(ii) \( \frac{-3}{4} \)
(iii) \( \frac{-7}{8} \)
(iv) \( \frac{-17}{8} \)
Answer:
(i) To mark \( \frac{3}{8} \) on the number line, divide the space between 0 and 1 into 8 equal sections. Then mark the 3rd section from 0 to the right.

Number line:
(ii) To mark \( \frac{-3}{4} \) on the number line, divide the space between 0 and -1 into 4 equal sections. Then mark the 3rd section from 0 to the left. Number line:
(iii) To mark \( \frac{-7}{8} \) on the number line, divide the space between 0 and -1 into 8 equal sections. Then mark the 7th section from 0 to the left. Number line:
(iv) Since \( \frac{-17}{8} = -2\frac{1}{8} \), it lies between -2 and -3. Divide the space between -2 and -3 into 8 equal sections and mark the 1st section from -2 to the left. Number line:

In simple words: On a number line, positive numbers go to the right of zero and negative numbers go to the left. Divide the space into equal parts to show fractions.

Exam Tip: Always divide the interval into as many equal parts as the denominator shows, then count from zero to place the point accurately.

Question 2. The points P, Q, R, S, T, U, A and B on the number line are such that TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S respectively.
Answer: Looking at the number line, A and B are located at 2 and 3. The gap AB is divided into 3 equal sections by points P and Q, so each part equals \( \frac{1}{3} \).

\( P = 2 + \frac{1}{3} = \frac{7}{3} \)

\( Q = 2 + \frac{2}{3} = \frac{8}{3} \)

Also, T and U are at -1 and -2. The gap TU is divided into 3 equal sections by points R and S, so each part equals \( \frac{1}{3} \).

\( R = -1 - \frac{1}{3} = \frac{-4}{3} \)

\( S = -1 - \frac{2}{3} = \frac{-5}{3} \)

Therefore, P, Q, R and S represent \( \frac{7}{3}, \frac{8}{3}, \frac{-4}{3} \) and \( \frac{-5}{3} \) respectively.
In simple words: When points divide a space into equal parts on a number line, add or subtract each part's distance from the starting point to find the position of each new point.

Exam Tip: When finding the position of a point between two integers, always divide the space into equal parts as indicated and count carefully from the starting point.

Question 3. State whether the following statements are true or false:
(i) The rational number \( \frac{-13}{-5} \) lies to the right of zero on the number line.
(ii) The rational numbers \( \frac{-5}{-7} \) and \( \frac{7}{-9} \) lie on opposite sides of zero on the number line.
(iii) The rational numbers \( \frac{-17}{6} \) and \( \frac{8}{-15} \) lie on opposite sides of zero on the number line.
Answer:
(i) \( \frac{-13}{-5} = \frac{13}{5} \), which is a positive rational number, so it lies to the right of zero on the number line.
Therefore, the statement is True.

(ii) \( \frac{-5}{-7} = \frac{5}{7} \) is positive (lies to the right of zero) and \( \frac{7}{-9} = \frac{-7}{9} \) is negative (lies to the left of zero). Since one is positive and one is negative, they lie on opposite sides of zero.
Therefore, the statement is True.

(iii) \( \frac{-17}{6} \) has a negative numerator and positive denominator, making it negative (lies to the left of zero). Also, \( \frac{8}{-15} = \frac{-8}{15} \) has a positive numerator and negative denominator, making it negative (lies to the left of zero). Since both are negative, they lie on the same side of zero, not opposite sides.
Therefore, the statement is False.
In simple words: A rational number is positive if both top and bottom have the same sign (both positive or both negative), and negative if they have different signs. Positive numbers are right of zero, negative numbers are left of zero.

Exam Tip: Always reduce a rational number to standard form first to clearly see if it is positive or negative, then determine which side of zero it lies on.

Question 4. Find:
(i) \( \frac{5}{63} - \left(\frac{-6}{21}\right) \)
(ii) \( \frac{-6}{13} - \left(\frac{-7}{15}\right) \)
(iii) \( 3\frac{1}{8} - \left(-1\frac{5}{6}\right) \)
Answer:
(i) \( \frac{5}{63} - \left(\frac{-6}{21}\right) = \frac{5}{63} + \frac{6}{21} \)

The LCM of 63 and 21 is 63.

\( \Rightarrow \frac{5 \times 1}{63 \times 1} + \frac{6 \times 3}{21 \times 3} = \frac{5}{63} + \frac{18}{63} = \frac{5 + 18}{63} = \frac{23}{63} \)

\( \therefore \frac{5}{63} - \left(\frac{-6}{21}\right) = \frac{23}{63} \)

(ii) \( \frac{-6}{13} - \left(\frac{-7}{15}\right) = \frac{-6}{13} + \frac{7}{15} \)

The LCM of 13 and 15 is 195.

\( \Rightarrow \frac{-6 \times 15}{13 \times 15} + \frac{7 \times 13}{15 \times 13} = \frac{-90}{195} + \frac{91}{195} = \frac{-90 + 91}{195} = \frac{1}{195} \)

\( \therefore \frac{-6}{13} - \left(\frac{-7}{15}\right) = \frac{1}{195} \)

(iii) \( 3\frac{1}{8} - \left(-1\frac{5}{6}\right) = \frac{25}{8} - \left(\frac{-11}{6}\right) = \frac{25}{8} + \frac{11}{6} \)

The LCM of 8 and 6 is 24.

\( \Rightarrow \frac{25 \times 3}{8 \times 3} + \frac{11 \times 4}{6 \times 4} = \frac{75}{24} + \frac{44}{24} = \frac{75 + 44}{24} = \frac{119}{24} \)

\( \therefore 3\frac{1}{8} - \left(-1\frac{5}{6}\right) = \frac{119}{24} \)
In simple words: When you subtract a negative number, flip it to addition instead. Then find the lowest number both denominators divide into evenly, rewrite each fraction with that bottom number, and add the tops together.

Exam Tip: Always convert negative denominators to positive form first. Remember that subtracting a negative is the same as adding its opposite.

Question 5. Multiply:
(i) \( \frac{5}{7} \times \frac{3}{4} \)
(ii) \( \frac{-2}{3} \times \frac{9}{-4} \)
(iii) \( 2\frac{1}{5} \times 3\frac{2}{3} \)
Answer:
(i) \( \frac{5}{7} \times \frac{3}{4} = \frac{5 \times 3}{7 \times 4} = \frac{15}{28} \)

\( \therefore \frac{5}{7} \times \frac{3}{4} = \frac{15}{28} \)

(ii) \( \frac{-2}{3} \times \frac{9}{-4} = \frac{-2 \times 9}{3 \times (-4)} = \frac{-18}{-12} = \frac{18}{12} = \frac{3}{2} \)

\( \therefore \frac{-2}{3} \times \frac{9}{-4} = \frac{3}{2} \)

(iii) \( 2\frac{1}{5} \times 3\frac{2}{3} = \frac{11}{5} \times \frac{11}{3} = \frac{11 \times 11}{5 \times 3} = \frac{121}{15} \)

\( \therefore 2\frac{1}{5} \times 3\frac{2}{3} = \frac{121}{15} \)
In simple words: Multiply the top numbers together and the bottom numbers together. If either fraction has a negative, count how many negatives there are - if it's an even count, your answer is positive; if it's odd, it stays negative.

Exam Tip: Before multiplying, cancel any common factors between numerators and denominators across fractions - this makes the numbers smaller and reduces your chance of arithmetic errors.

Question 6. Divide:
(i) \( \frac{5}{9} \div \frac{3}{2} \)
(ii) \( \frac{-4}{15} \div \frac{2}{-3} \)
(iii) \( 2\frac{1}{4} \div 1\frac{1}{2} \)
Answer:
(i) \( \frac{5}{9} \div \frac{3}{2} = \frac{5}{9} \times \frac{2}{3} = \frac{5 \times 2}{9 \times 3} = \frac{10}{27} \)

\( \therefore \frac{5}{9} \div \frac{3}{2} = \frac{10}{27} \)

(ii) \( \frac{-4}{15} \div \frac{2}{-3} = \frac{-4}{15} \times \frac{-3}{2} = \frac{(-4) \times (-3)}{15 \times 2} = \frac{12}{30} = \frac{2}{5} \)

\( \therefore \frac{-4}{15} \div \frac{2}{-3} = \frac{2}{5} \)

(iii) \( 2\frac{1}{4} \div 1\frac{1}{2} = \frac{9}{4} \div \frac{3}{2} = \frac{9}{4} \times \frac{2}{3} = \frac{9 \times 2}{4 \times 3} = \frac{18}{12} = \frac{3}{2} \)

\( \therefore 2\frac{1}{4} \div 1\frac{1}{2} = \frac{3}{2} \)
In simple words: To divide by a fraction, flip the second fraction upside down and multiply instead. This is much easier than trying to divide directly.

Exam Tip: Always write mixed numbers as improper fractions before you start the division. Double-check your sign - with two negatives, the answer is positive.

Question 1. Add the following pairs of rational numbers:
(i) \( \frac{3}{11}, \frac{-5}{11} \)
(ii) \( \frac{4}{9}, \frac{-9}{5} \)
(iii) \( \frac{5}{-7}, \frac{-2}{-7} \)
(iv) \( \frac{-2}{5}, \frac{3}{4} \)
Answer:
(i) Solving:
\( \frac{3}{11} + \frac{-5}{11} = \frac{3 + (-5)}{11} = \frac{-2}{11} \)

\( \therefore \frac{3}{11} + \frac{-5}{11} = \frac{-2}{11} \)

(ii) \( \frac{5}{-9} = \frac{-5}{9} \)

\( \frac{4}{9} + \frac{-5}{9} = \frac{4 + (-5)}{9} = \frac{-1}{9} \)

\( \therefore \frac{4}{9} + \frac{-9}{5} = \frac{-1}{9} \)

(iii) \( \frac{5}{-7} = \frac{-5}{7} \) and \( \frac{-2}{-7} = \frac{2}{7} \)

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

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

(iv) The LCM of 5 and 4 is 20.

\( \frac{-2 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5} = \frac{-8}{20} + \frac{15}{20} = \frac{-8 + 15}{20} = \frac{7}{20} \)

\( \therefore \frac{-2}{5} + \frac{3}{4} = \frac{7}{20} \)
In simple words: If both fractions have the same bottom number, just add or subtract the tops. If the bottoms are different, find the smallest number that both divide into evenly, rewrite the fractions, then add the tops.

Exam Tip: Always rewrite negative denominators as positive before adding. For fractions with different denominators, finding the LCM correctly is the key - take time on that step.

Question 2. Find the sum:
(i) \( \frac{27}{-4} + \frac{-15}{8} \)
(ii) \( \frac{-1}{18} + \frac{-3}{8} \)
(iii) \( -3\frac{1}{6} + 2\frac{3}{8} \)
(iv) \( -2\frac{4}{5} + 4\frac{3}{10} \)
Answer:
(i) \( \frac{27}{-4} = \frac{-27}{4} \). The LCM of 4 and 8 is 8.

\( \Rightarrow \frac{-27 \times 2}{4 \times 2} + \frac{-15 \times 1}{8 \times 1} = \frac{-54}{8} + \frac{-15}{8} = \frac{-54 + (-15)}{8} = \frac{-69}{8} \)

\( \therefore \frac{27}{-4} + \frac{-15}{8} = \frac{-69}{8} \)

(ii) The LCM of 18 and 8 is 72.

\( \Rightarrow \frac{-1 \times 4}{18 \times 4} + \frac{-3 \times 9}{8 \times 9} = \frac{-4}{72} + \frac{-27}{72} = \frac{-4 + (-27)}{72} = \frac{-31}{72} \)

\( \therefore \frac{-1}{18} + \frac{-3}{8} = \frac{-31}{72} \)

(iii) \( -3\frac{1}{6} = \frac{-19}{6} \) and \( 2\frac{3}{8} = \frac{19}{8} \). The LCM of 6 and 8 is 24.

\( \Rightarrow \frac{-19 \times 4}{6 \times 4} + \frac{19 \times 3}{8 \times 3} = \frac{-76}{24} + \frac{57}{24} = \frac{-76 + 57}{24} = \frac{-19}{24} \)

\( \therefore -3\frac{1}{6} + 2\frac{3}{8} = \frac{-19}{24} \)

(iv) \( -2\frac{4}{5} = \frac{-14}{5} \) and \( 4\frac{3}{10} = \frac{43}{10} \). The LCM of 5 and 10 is 10.

\( \Rightarrow \frac{-14 \times 2}{5 \times 2} + \frac{43 \times 1}{10 \times 1} = \frac{-28}{10} + \frac{43}{10} = \frac{-28 + 43}{10} = \frac{15}{10} = \frac{3}{2} = 1\frac{1}{2} \)

\( \therefore -2\frac{4}{5} + 4\frac{3}{10} = 1\frac{1}{2} \)
In simple words: Convert mixed numbers to improper fractions first. Make sure both denominators are positive. Find the LCM, rewrite with that common denominator, then add the numerators.

Exam Tip: When the final fraction can be simplified or turned back into a mixed number, do so - the answer should be in simplest form.

Question 3. Subtract:
(i) \( \frac{-6}{13} \) from \( \frac{4}{13} \)
(ii) \( \frac{-1}{2} \) from \( \frac{-2}{3} \)
(iii) \( \frac{5}{9} \) from \( \frac{-2}{3} \)
Answer:
(i) Solving:
\( \frac{4}{13} - \frac{-6}{13} = \frac{4}{13} + \frac{6}{13} = \frac{4 + 6}{13} = \frac{10}{13} \)

\( \therefore \) The result is \( \frac{10}{13} \).

(ii) The LCM of 3 and 2 is 6.

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

\( \therefore \) The result is \( \frac{-1}{6} \).

(iii) The LCM of 3 and 9 is 9.

\( \Rightarrow \frac{-2}{3} - \frac{5}{9} = \frac{-2 \times 3}{3 \times 3} - \frac{5 \times 1}{9 \times 1} = \frac{-6}{9} - \frac{5}{9} = \frac{-6 - 5}{9} = \frac{-11}{9} \)

\( \therefore \) The result is \( \frac{-11}{9} \).
In simple words: Subtracting a negative number flips to adding. If denominators differ, find the LCM, rewrite both fractions, then subtract the numerators.

Exam Tip: Pay close attention to the order - "subtract A from B" means B minus A, not the other way around. A single sign mistake changes the entire answer.

Question 5. Add the following rational numbers:
(i) \( \frac{-6}{13} - \left(\frac{-7}{15}\right) \)
(ii) \( 3\frac{1}{8} - \left(-1\frac{5}{6}\right) \)
Answer:
(i) To find \( \frac{-6}{13} - \left(\frac{-7}{15}\right) = \frac{-6}{13} + \frac{7}{15} \)

Rewrite with a common denominator. The LCM of 13 and 15 is 195.

\( \frac{-6 \times 15}{13 \times 15} + \frac{7 \times 13}{15 \times 13} = \frac{-90}{195} + \frac{91}{195} = \frac{-90 + 91}{195} = \frac{1}{195} \)

(ii) Convert the mixed numbers to improper fractions: \( 3\frac{1}{8} = \frac{25}{8} \) and \( -1\frac{5}{6} = \frac{-11}{6} \)

So, \( \frac{25}{8} - \left(\frac{-11}{6}\right) = \frac{25}{8} + \frac{11}{6} \)

The LCM of 8 and 6 is 24.

\( \frac{25 \times 3}{8 \times 3} + \frac{11 \times 4}{6 \times 4} = \frac{75}{24} + \frac{44}{24} = \frac{75 + 44}{24} = \frac{119}{24} = 4\frac{23}{24} \)

In simple words: When subtracting a negative number, change it to addition of the positive number. Find a common denominator, then add or subtract the numerators.

Exam Tip: Always remember that subtracting a negative is the same as adding a positive. Use LCM correctly and simplify your final answer if possible.

Question 6. The sum of two rational numbers is \( \frac{2}{5} \). If one of them is \( \frac{-4}{7} \), find the other number.
Answer: Let the other number be \( x \).

\( x + \frac{-4}{7} = \frac{2}{5} \)

\( x = \frac{2}{5} - \frac{-4}{7} = \frac{2}{5} + \frac{4}{7} \)

The LCM of 5 and 7 is 35.

\( x = \frac{2 \times 7}{5 \times 7} + \frac{4 \times 5}{7 \times 5} = \frac{14}{35} + \frac{20}{35} = \frac{14 + 20}{35} = \frac{34}{35} \)

In simple words: Move the known number to the other side by reversing its operation. Convert to a common denominator, then add the fractions.

Exam Tip: Check your answer by adding it back to the given number - you should get the sum stated in the problem.

Question 7. What rational number should be added to \( \frac{-5}{12} \) to get \( \frac{-7}{8} \)?
Answer: Let the required number be \( x \).

\( \frac{-5}{12} + x = \frac{-7}{8} \)

\( x = \frac{-7}{8} - \frac{-5}{12} = \frac{-7}{8} + \frac{5}{12} \)

The LCM of 8 and 12 is 24.

\( x = \frac{-7 \times 3}{8 \times 3} + \frac{5 \times 2}{12 \times 2} = \frac{-21}{24} + \frac{10}{24} = \frac{-21 + 10}{24} = \frac{-11}{24} \)

In simple words: The number we need to add is found by subtracting the original number from the target result.

Exam Tip: Verify by adding your answer to the starting number to confirm you get the final result.

Question 8. What rational number should be subtracted from \( \frac{-2}{3} \) to get \( \frac{-5}{6} \)?
Answer: Let the required number be \( x \).

\( \frac{-2}{3} - x = \frac{-5}{6} \)

\( x = \frac{-2}{3} - \frac{-5}{6} = \frac{-2}{3} + \frac{5}{6} \)

The LCM of 3 and 6 is 6.

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

In simple words: To find what number to subtract, rearrange the equation so that number stands alone on one side, then solve.

Exam Tip: Remember that when a subtraction moves to the other side of the equals sign, it becomes an addition operation.

Question 9. Find the product:
(i) \( \frac{2}{3} \times \frac{-7}{8} \)
(ii) \( \frac{-6}{7} \times \frac{5}{7} \)
(iii) \( \frac{-2}{9} \times (-5) \)
(iv) \( \frac{-5}{11} \times \left(\frac{-11}{5}\right) \)
(v) \( \frac{8}{35} \times \frac{-32}{21} \)
(vi) \( \frac{-105}{128} \times \left(-1\frac{29}{35}\right) \)
Answer:
(i) \( \frac{2}{3} \times \frac{-7}{8} = \frac{2 \times (-7)}{3 \times 8} = \frac{-14}{24} = \frac{-7}{12} \)

(ii) \( \frac{-6}{7} \times \frac{5}{7} = \frac{-6 \times 5}{7 \times 7} = \frac{-30}{49} \)

(iii) \( \frac{-2}{9} \times (-5) = \frac{-2}{9} \times \frac{-5}{1} = \frac{-2 \times (-5)}{9 \times 1} = \frac{10}{9} \)

(iv) \( \frac{-5}{11} \times \frac{-11}{5} = \frac{-5 \times (-11)}{11 \times 5} = \frac{55}{55} = 1 \)

(v) \( \frac{8}{35} \times \frac{-32}{21} = \frac{8 \times (-32)}{35 \times 21} = \frac{-256}{735} = \frac{-3}{20} \)

(vi) First convert \( -1\frac{29}{35} = \frac{-64}{35} \). Then \( \frac{-105}{128} \times \frac{-64}{35} = \frac{-105 \times (-64)}{128 \times 35} = \frac{6720}{4480} = \frac{3}{2} = 1\frac{1}{2} \)

In simple words: Multiply the numerators together and the denominators together. A negative times a negative gives a positive result.

Exam Tip: Cancel common factors before multiplying to make calculations simpler. Watch the signs - two negatives make a positive.

Question 10. Find the value of:
(i) \( (-6) \div \frac{2}{5} \)
(ii) \( \frac{-1}{10} \div \frac{-8}{5} \)
(iii) \( \frac{-65}{14} \div \frac{13}{-7} \)
(iv) \( (-6) \div 3\frac{3}{5} \)
(v) \( \frac{-48}{49} \div \frac{72}{-35} \)
(vi) \( 3\frac{1}{7} \div \left(\frac{-33}{34}\right) \)
Answer:
(i) \( (-6) \div \frac{2}{5} = -6 \times \frac{5}{2} = \frac{-6 \times 5}{2} = \frac{-30}{2} = -15 \)

(ii) \( \frac{-1}{10} \div \frac{-8}{5} = \frac{-1}{10} \times \frac{5}{-8} = \frac{-1 \times 5}{10 \times (-8)} = \frac{-5}{-80} = \frac{1}{16} \)

(iii) \( \frac{-65}{14} \div \frac{13}{-7} = \frac{-65}{14} \times \frac{-7}{13} = \frac{-65 \times (-7)}{14 \times 13} = \frac{455}{182} = \frac{5}{2} = 2\frac{1}{2} \)

(iv) Convert \( 3\frac{3}{5} = \frac{18}{5} \). Then \( (-6) \div \frac{18}{5} = -6 \times \frac{5}{18} = \frac{-6 \times 5}{18} = \frac{-30}{18} = -1\frac{2}{3} \)

(v) \( \frac{-48}{49} \div \frac{72}{-35} = \frac{-48}{49} \times \frac{-35}{72} = \frac{-48 \times (-35)}{49 \times 72} = \frac{1680}{3528} = \frac{10}{21} \)

(vi) Convert \( 3\frac{1}{7} = \frac{22}{7} \). Then \( \frac{22}{7} \div \frac{-33}{34} = \frac{22}{7} \times \frac{34}{-33} = \frac{22 \times 34}{7 \times (-33)} = \frac{748}{-231} = -3\frac{5}{21} \)

In simple words: Division of fractions is done by flipping the divisor and changing division to multiplication.

Exam Tip: Always convert mixed numbers to improper fractions before dividing. Remember to flip the fraction that comes after the division sign.

Question 11. The product of two rational numbers is \( \frac{18}{35} \). If one of them is \( \frac{-2}{5} \), find the other number.
Answer: Let the other number be \( x \).

\( x \times \frac{-2}{5} = \frac{18}{35} \)

\( x = \frac{18}{35} \div \frac{-2}{5} = \frac{18}{35} \times \frac{5}{-2} = \frac{18 \times 5}{35 \times (-2)} = \frac{90}{-70} = \frac{-9}{7} = -1\frac{2}{7} \)

In simple words: To find the missing number in a multiplication, divide the product by the known number.

Exam Tip: Check your answer by multiplying it with the given number - the result should equal the stated product.

Question 12. Find the value of:
(i) \( \left(\frac{13}{21} \div \frac{39}{42}\right) \times \left(\frac{-3}{5}\right) \)
(ii) \( \left(-5\frac{5}{21}\right) \div \left(\frac{7}{11} \times \frac{5}{12}\right) \)
Answer:
(i) First solve the bracket: \( \frac{13}{21} \div \frac{39}{42} = \frac{13}{21} \times \frac{42}{39} = \frac{13 \times 42}{21 \times 39} = \frac{546}{819} = \frac{2}{3} \)

Now, \( \frac{2}{3} \times \frac{-3}{5} = \frac{2 \times (-3)}{3 \times 5} = \frac{-6}{15} = \frac{-2}{5} \)

(ii) First solve the bracket: \( \frac{7}{11} \times \frac{5}{12} = \frac{7 \times 5}{11 \times 12} = \frac{35}{132} \)

Also, \( -5\frac{5}{21} = \frac{-110}{21} \). Now, \( \frac{-110}{21} \div \frac{35}{132} = \frac{-110}{21} \times \frac{132}{35} = \frac{-110 \times 132}{21 \times 35} = \frac{-14520}{735} = \frac{-968}{49} = -19\frac{37}{49} \)

In simple words: Work through the brackets first, then carry out the remaining operations in order from left to right.

Exam Tip: Always complete operations inside brackets before moving to operations outside them. This avoids errors and keeps your working organized.

Question 13. Find the reciprocal of the following:
(i) \( \frac{3}{13} \div \frac{-4}{65} \)
(ii) \( \left(-5 \times \frac{12}{15}\right) - \left(-3 \times \frac{2}{9}\right) \)
Answer:
(i) First find the value: \( \frac{3}{13} \div \frac{-4}{65} = \frac{3}{13} \times \frac{65}{-4} = \frac{3 \times 65}{13 \times (-4)} = \frac{195}{-52} = \frac{-15}{4} \)

The reciprocal of \( \frac{-15}{4} \) is \( \frac{-4}{15} \).

(ii) First find the value: \( -5 \times \frac{12}{15} = \frac{-60}{15} = -4 \)

\( -3 \times \frac{2}{9} = \frac{-6}{9} = \frac{-2}{3} \)

So, \( -4 - \left(\frac{-2}{3}\right) = -4 + \frac{2}{3} = \frac{-12}{3} + \frac{2}{3} = \frac{-12 + 2}{3} = \frac{-10}{3} \)

The reciprocal of \( \frac{-10}{3} \) is \( \frac{-3}{10} \).

In simple words: Find the value of the expression first. The reciprocal is obtained by flipping the fraction - numerator becomes denominator and denominator becomes numerator.

Exam Tip: Make sure to simplify the expression completely before finding its reciprocal. A reciprocal of a/b is always b/a.

Mental Maths

Question 1. Fill in the blanks:
(i) Two rational numbers are called equivalent if they have ..... value.
(ii) The rational number \( \frac{-5}{-11} \) lies to the ..... of zero on the number line.
(iii) The standard form of the rational number \( \frac{-12}{14} \) is .....
(iv) The multiplicative inverse of \( -3\frac{1}{5} \) is .....
(v) If p and q are positive integers, then \( \frac{p}{q} \) is a ..... rational number and \( \frac{p}{-q} \) is a ..... rational number.
Answer:
(i) Two rational numbers are called equivalent if they have the same value.

(ii) \( \frac{-5}{-11} = \frac{5}{11} \), which is positive, so it lies to the right of zero on the number line.

(iii) Making the denominator positive, \( \frac{-12}{14} = \frac{12}{-14} \). The HCF of 14 and 12 is 2, so \( \frac{-12}{14} = \frac{-7}{6} \). The standard form is \( \frac{-7}{6} \).

(iv) \( -3\frac{1}{5} = \frac{-16}{5} \). The multiplicative inverse is \( \frac{-5}{16} \).

(v) If p and q are positive integers, then \( \frac{p}{q} \) is a positive rational number and \( \frac{p}{-q} \) is a negative rational number.

In simple words: Equivalent fractions show the same value in different forms. A positive divided by a negative always gives a negative result.

Exam Tip: Always ensure the denominator is positive in standard form. A rational number's sign depends on whether numerator and denominator have the same or different signs.

Question 2. State whether the following statements are true (T) or false (F):
(i) Zero is the smallest rational number.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every fraction is a rational number.
(v) Every rational number is a fraction.
(vi) The reciprocal of -1 is -1.
(vii) The difference of two rational numbers is always a rational number.
(viii) The quotient of two integers is always a rational number.
(ix) The value of a rational number remains the same if both its numerator and denominator are multiplied (or divided) by the same (non-zero) integer.
Answer:
(i) False. Negative rational numbers are smaller than zero, so zero is not the smallest rational number.

(ii) True. Every integer can be written in the form \( \frac{p}{q} \) with \( q = 1 \).

(iii) False. For example, \( \frac{2}{3} \) is a rational number but not an integer.

(iv) True. Every fraction is a rational number.

(v) False. For example, \( \frac{-2}{3} \) is a rational number but not a fraction (the numerator is negative).

(vi) True. The reciprocal of -1 is -1.

(vii) True. The difference of two rational numbers is always a rational number.

(viii) False. Division by zero is not defined, so the quotient of two integers is not always a rational number.

(ix) True. The value of a rational number remains the same if both numerator and denominator are multiplied (or divided) by the same non-zero integer.

In simple words: Rational numbers include integers. A fraction has a positive numerator and denominator. Division by zero is never allowed.

Exam Tip: Understand the distinction between "integer," "fraction," and "rational number" - they are related but not identical concepts.

Multiple Choice Questions

Question 3. The rational number \( \frac{-110}{132} \) when reduced to standard form is
(1) \( \frac{-10}{12} \)
(2) \( \frac{-6}{5} \)
(3) \( \frac{-5}{6} \)
(4) \( \frac{110}{-132} \)
Answer: (3) \( \frac{-5}{6} \)

Making the denominator positive, \( \frac{-110}{132} = \frac{-110}{132} \). The HCF of 110 and 132 is 22.

\( \frac{-110}{132} = \frac{-110 \div 22}{132 \div 22} = \frac{-5}{6} \)

In simple words: Find the greatest common factor of the numerator and denominator, then divide both by it to get the simplest form.

Exam Tip: Always reduce a fraction to its simplest form by dividing by the HCF. The standard form has a positive denominator.

Question 4. Which of the following is not equal to the others?
(1) \( \frac{-21}{56} \)
(2) \( \frac{-56}{21} \)
(3) \( \frac{-6}{5} \)
(4) \( \frac{6}{-5} \)
Answer: (2) \( \frac{-56}{21} \)

Simplify each option:
- Option (1): \( \frac{-21}{56} = \frac{-3}{8} \)
- Option (2): \( \frac{-56}{21} \) does not simplify to match the others
- Option (3): \( \frac{-6}{5} \) (already in simplest form)
- Option (4): \( \frac{6}{-5} = \frac{-6}{5} \)

Options (3) and (4) are equal. Option (1) is different from both. Option (2) is also different. However, options (1) and (2) are reciprocals of each other. The question asks which is NOT equal to the others - option (2) stands alone as distinctly different.

In simple words: Reduce each fraction to its simplest form, then compare them to find which one does not match the rest.

Exam Tip: When comparing rational numbers, always simplify them first. Two fractions are equal if they reduce to the same simplified form.

Question 4. Which of the following is not equal to \( \frac{8}{-3} \)?
(1) \( \frac{40}{-15} \)
(2) \( \frac{-56}{21} \)
(3) \( \frac{56}{-21} \)
(4) \( \frac{48}{18} \)
Answer: (4) \( \frac{48}{18} \)
In simple words: When you reduce all four choices to their simplest form, three of them become \( \frac{8}{-3} \), but choice (4) becomes \( \frac{8}{3} \) - a positive fraction instead of a negative one, so it is different.

Exam Tip: Always reduce fractions to standard form before comparing them - two fractions that look different can be equal once simplified.

Question 5. The multiplicative inverse of \( \frac{-4}{9} \) is
(1) \( \frac{4}{9} \)
(2) \( \frac{-9}{4} \)
(3) \( \frac{9}{4} \)
(4) \( \frac{4}{-9} \)
Answer: (2) \( \frac{-9}{4} \)
In simple words: To get the multiplicative inverse, flip the fraction upside down. \( \frac{-4}{9} \) flipped becomes \( \frac{9}{-4} \), which equals \( \frac{-9}{4} \).

Exam Tip: The multiplicative inverse (reciprocal) simply swaps the numerator and denominator - watch the negative sign carefully to make sure it goes to the right place.

Question 6. The reciprocal of the rational number \( -2\frac{3}{7} \) is
(1) \( -\frac{17}{7} \)
(2) \( \frac{7}{17} \)
(3) \( -\frac{7}{17} \)
(4) none of these
Answer: (3) \( -\frac{7}{17} \)
In simple words: First change the mixed number \( -2\frac{3}{7} \) to an improper fraction, which gives \( \frac{-17}{7} \). Then flip it to get the reciprocal: \( \frac{7}{-17} \) or \( -\frac{7}{17} \).

Exam Tip: Always convert mixed numbers to improper fractions first, then find the reciprocal by flipping - this reduces mistakes with signs.

Question 7. The product of rational number \( \frac{-2}{5} \) and its multiplicative inverse is
(1) 1
(2) 0
(3) \( \frac{4}{25} \)
(4) \( \frac{2}{5} \)
Answer: (1) 1
In simple words: When you multiply any non-zero rational number by its multiplicative inverse (its reciprocal), you always get 1. This happens because flipping and multiplying cancels everything out.

Exam Tip: This is a key property to remember - the product of a rational number and its multiplicative inverse is always 1, never any other number.

Question 8. The product of rational number \( \frac{-2}{3} \) and its additive inverse is
(1) 1
(2) \( \frac{2}{3} \)
(3) \( \frac{4}{9} \)
(4) \( \frac{-4}{9} \)
Answer: (4) \( \frac{-4}{9} \)
In simple words: The additive inverse of \( \frac{-2}{3} \) is \( \frac{2}{3} \) (flip the sign). Multiply them: \( \frac{-2}{3} \times \frac{2}{3} = \frac{-4}{9} \).

Exam Tip: Do not confuse additive inverse (change the sign) with multiplicative inverse (flip the fraction) - they are different operations with different results.

Question 9. The sum of rational number \( \frac{-1}{3} \) and its reciprocal is
(1) 0
(2) 1
(3) \( \frac{-10}{3} \)
(4) \( \frac{-3}{10} \)
Answer: (3) \( \frac{-10}{3} \)
In simple words: The reciprocal of \( \frac{-1}{3} \) is -3. Add them together: \( \frac{-1}{3} + (-3) = \frac{-1}{3} + \frac{-9}{3} = \frac{-10}{3} \).

Exam Tip: Convert whole numbers to fractions with the same denominator before adding - this makes combining them much simpler.

Question 10. \( \frac{-3}{5} - \left(\frac{-2}{15}\right) \) is equal to
(1) \( \frac{-11}{5} \)
(2) \( \frac{-1}{15} \)
(3) \( \frac{-7}{15} \)
(4) \( \frac{7}{15} \)
Answer: (3) \( \frac{-7}{15} \)
In simple words: First, subtract becomes add: \( \frac{-3}{5} + \frac{2}{15} \). Find a common denominator (15), then combine: \( \frac{-9}{15} + \frac{2}{15} = \frac{-7}{15} \).

Exam Tip: Subtracting a negative is the same as adding a positive - convert the operation first, then find the common denominator.

Question 11. \( \left(-5\frac{1}{3}\right) \times \left(-1\frac{7}{8}\right) \) is equal to
(1) 10
(2) -10
(3) \( 5\frac{7}{24} \)
(4) \( -5\frac{7}{24} \)
Answer: (1) 10
In simple words: Change both mixed numbers to improper fractions: \( -5\frac{1}{3} = \frac{-16}{3} \) and \( -1\frac{7}{8} = \frac{-15}{8} \). Multiply them: \( \frac{-16}{3} \times \frac{-15}{8} = \frac{240}{24} = 10 \). Two negatives make a positive.

Exam Tip: Remember that multiplying two negative numbers gives a positive result - negative times negative equals positive.

Question 12. \( \left(-2\frac{1}{3}\right) \div 2\frac{11}{12} \) is equal to
(1) \( -\frac{4}{5} \)
(2) \( \frac{4}{5} \)
(3) \( \frac{4}{11} \)
(4) \( -\frac{4}{11} \)
Answer: (1) \( -\frac{4}{5} \)
In simple words: Convert to improper fractions: \( -2\frac{1}{3} = \frac{-7}{3} \) and \( 2\frac{11}{12} = \frac{35}{12} \). Dividing means multiply by the reciprocal: \( \frac{-7}{3} \times \frac{12}{35} = \frac{-84}{105} = -\frac{4}{5} \).

Exam Tip: Division of fractions always means "multiply by the flip" - flip the second fraction and multiply, then simplify by cancelling common factors.

Question 13. In the standard form of a rational number, the denominator is always
(1) 0
(2) a negative integer
(3) 1
(4) a positive integer
Answer: (4) a positive integer
In simple words: Standard form means the fraction is simplified and written so that the bottom number is always positive. Any negative sign goes in the numerator, not the denominator.

Exam Tip: To write in standard form, always move the negative sign to the top (numerator) and make the bottom (denominator) positive.

Question 14. The sum of two rational numbers is -1. If one of them is \( \frac{-5}{7} \), then the other is
(1) \( \frac{5}{7} \)
(2) \( \frac{-2}{7} \)
(3) \( \frac{12}{7} \)
(4) \( \frac{-12}{7} \)
Answer: (2) \( \frac{-2}{7} \)
In simple words: Let the unknown number be x. Set up: \( x + \frac{-5}{7} = -1 \)
\( \implies \) \( x = -1 - \frac{-5}{7} = -1 + \frac{5}{7} = \frac{-7}{7} + \frac{5}{7} = \frac{-2}{7} \)

Exam Tip: Set up an equation when a problem says "sum" or "product" with one number unknown - rearrange to solve for the missing number.

Question 15. Statement I: \( \frac{-2}{-3} = \frac{2}{3} = \frac{4}{6} \)
Statement II: The rational number \( \frac{-2}{-3} \) is in the standard form.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (1) Statement I is true but statement II is false.
In simple words: Statement I shows correct simplifications - all three forms equal \( \frac{2}{3} \). Statement II is wrong because \( \frac{-2}{-3} \) has a negative denominator, so it is not in standard form. Standard form requires a positive denominator.

Exam Tip: For assertion-reason questions, check each statement separately before choosing your answer - one can be true while the other is false.

Question 16. Statement I: \( \frac{304}{12589} > \frac{4782}{-844921} \)
Statement II: The two numbers \( \frac{304}{12589} \) and \( \frac{4782}{-844921} \) are rational numbers.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (3) Both Statement I and statement II are true.
In simple words: Statement I is correct because \( \frac{304}{12589} \) is positive (both parts positive) while \( \frac{4782}{-844921} \) is negative (one part negative), and positive numbers are always greater than negative numbers. Statement II is correct because both are written as a fraction with integer numerator and non-zero integer denominator, which is the definition of a rational number.

Exam Tip: Any positive rational number is greater than any negative rational number - this is a quick way to compare without finding exact decimal values.

Question 17. Statement I: The multiplicative inverse of 0 is 0.
Statement II: The additive inverse of 0 is 0.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (2) Statement I is false but statement II is true.
In simple words: Statement I is false because no number times 0 equals 1, so 0 has no multiplicative inverse. Statement II is true because 0 added to 0 gives 0, so 0 is its own additive inverse.

Exam Tip: Zero is special - it has an additive inverse (itself) but NO multiplicative inverse (you cannot divide by zero).

Question 18. Statement I: All positive rational numbers are greater than 0.
Statement II: 0 is the smallest rational number.
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both Statement I and statement II are true.
(4) Both Statement I and statement II are false.
Answer: (1) Statement I is true but statement II is false.
In simple words: Statement I is true - by definition, all positive rational numbers are bigger than zero. Statement II is false because negative rational numbers (like \( -1, -\frac{1}{2} \)) are smaller than 0, so zero is not the smallest.

Exam Tip: There is no "smallest" or "largest" rational number - there are always smaller and larger ones you can find. Negative numbers go on forever in the downward direction.

Check Your Progress

Question 1. Write five rational numbers equivalent to \( \frac{-11}{5} \).
Answer: First rewrite \( \frac{-11}{5} \) with a positive denominator: \( \frac{-11}{5} = \frac{-11}{5} \).

Now multiply the top and bottom by 2, 3, 4, and 5 to get five equivalent rational numbers:

\( \frac{-11 \times 2}{5 \times 2} = \frac{-22}{10} \)

\( \frac{-11 \times 3}{5 \times 3} = \frac{-33}{15} \)

\( \frac{-11 \times 4}{5 \times 4} = \frac{-44}{20} \)

\( \frac{-11 \times 5}{5 \times 5} = \frac{-55}{25} \)

Therefore, five rational numbers equivalent to \( \frac{-11}{5} \) are \( \frac{-5}{11} \), \( \frac{-22}{10} \), \( \frac{-33}{15} \), \( \frac{-44}{20} \), and \( \frac{-55}{25} \).
In simple words: To write equal rational numbers, multiply or divide both the top and bottom by the same number. This keeps the value the same even though the fraction looks different.

Exam Tip: You can generate endless equivalent fractions by scaling both numerator and denominator by any non-zero integer - choose multipliers that are easy to work with.

Question 2. Express \( \frac{-9}{-15} \) as a rational number with:
(i) denominator 5
(ii) numerator -12
(iii) denominator 30
Answer: First reduce \( \frac{-9}{-15} \) to standard form.

\( \frac{-9}{-15} = \frac{9}{15} = \frac{-3}{5} \) (dividing by HCF 3).

(i) The denominator is already 5, so \( \frac{-9}{-15} = \frac{-3}{5} \).

(ii) To get -12 in the numerator from -3, multiply by 4: \( \frac{-3}{5} = \frac{-3 \times 4}{5 \times 4} = \frac{-12}{20} \), so \( \frac{-9}{-15} = \frac{-12}{20} \).

(iii) To get 30 in the denominator from 5, multiply by 6: \( \frac{-3}{5} = \frac{-3 \times 6}{5 \times 6} = \frac{-18}{30} \), so \( \frac{-9}{-15} = \frac{-18}{30} \).
In simple words: Simplify first, then scale up to get the required numerator or denominator by multiplying both top and bottom by the same number.

Exam Tip: Always reduce to standard form first before scaling - this saves you from working with large numbers and reduces errors.

Question 3. Write each of the following numbers in standard form:
(i) \( \frac{78}{-91} \)
(ii) \( \frac{-216}{162} \)
(iii) \( \frac{-520}{-195} \)
Answer:
(i) First make the denominator positive: \( \frac{78}{-91} = \frac{-78}{91} \). The HCF of 78 and 91 is 13.

\( \frac{-78}{91} = \frac{-78 \div 13}{91 \div 13} = \frac{-6}{7} \).

Therefore, the standard form of \( \frac{78}{-91} \) is \( \frac{-6}{7} \).

(ii) The denominator is already positive. The HCF of 216 and 162 is 54.

\( \frac{-216}{162} = \frac{-216 \div 54}{162 \div 54} = \frac{-4}{3} \).

Therefore, the standard form of \( \frac{-216}{162} \) is \( \frac{-4}{3} \).

(iii) Make the denominator positive: \( \frac{-520}{-195} = \frac{520}{195} \). The HCF of 520 and 195 is 65.

\( \frac{520}{195} = \frac{520 \div 65}{195 \div 65} = \frac{8}{3} \).

Therefore, the standard form of \( \frac{-520}{-195} \) is \( \frac{8}{3} \).
In simple words: To get standard form: (1) make sure the bottom is positive, (2) divide both top and bottom by their biggest common factor, and (3) check that you cannot simplify any further.

Exam Tip: Finding the HCF (Highest Common Factor) is the key step - use prime factorization or the division method if you are unsure, then divide both parts by it to fully simplify.

Question 4. Which of the following are pairs of equivalent rational numbers?
(i) \( \frac{-4}{13}, \frac{60}{-195} \)
(ii) \( \frac{7}{-15}, \frac{-35}{-75} \)
(iii) \( \frac{16}{-20}, \frac{-56}{70} \)
Answer: To check if two rational numbers are equivalent, reduce each to its simplest form and see if they match. For (i): The first fraction simplifies to \( \frac{-4}{13} \). The second can be rewritten as \( \frac{-60}{195} \), which reduces to \( \frac{-4}{13} \) when divided by the HCF of 15. Since both equal \( \frac{-4}{13} \), they are equivalent. For (ii): The first becomes \( \frac{-7}{15} \) when written with a positive denominator. The second simplifies to \( \frac{-75}{-35} = \frac{75}{35} = \frac{15}{7} \) when reduced. Since \( \frac{-7}{15} \neq \frac{15}{7} \), these are not equivalent. For (iii): Writing with positive denominators: \( \frac{16}{-20} = \frac{-16}{20} = \frac{-4}{5} \) and \( \frac{-56}{70} = \frac{-4}{5} \) when divided by the HCF of 14. Both equal \( \frac{-4}{5} \), so they are equivalent. Therefore, pairs (i) and (iii) are equivalent rational numbers.
In simple words: Two fractions are the same if they reduce to the same number when made simpler. Check (i) and (iii) - they both work this way.

Exam Tip: Always express both fractions with positive denominators first, then divide by their HCF to find the simplest form - this makes comparison clear and avoids confusion.

Question 5. Arrange the following rational numbers in ascending order:
(i) \( \frac{-5}{6}, \frac{-17}{18}, \frac{23}{-24}, \frac{-11}{-13} \)
(ii) \( \frac{-25}{6}, \frac{15}{-4}, \frac{-17}{8}, \frac{-53}{12} \)
Answer:
(i) Start by writing each with a positive denominator: \( \frac{-5}{6}, \frac{-17}{18}, \frac{-23}{24}, \frac{11}{13} \). The first three are negative and the last is positive. Find the LCM of the denominators 6, 18, and 24, which is 72. Convert each to an equivalent fraction with denominator 72: \( \frac{-5 \times 12}{72} = \frac{-60}{72} \), \( \frac{-17 \times 4}{72} = \frac{-68}{72} \), \( \frac{-23 \times 3}{72} = \frac{-69}{72} \). Since -69 < -68 < -60, the order of the negative fractions is \( \frac{-23}{24}, \frac{-17}{18}, \frac{-5}{6} \), and the positive fraction \( \frac{11}{13} \) is the largest. Therefore, the ascending order is \( \frac{-23}{24}, \frac{-17}{18}, \frac{-5}{6}, \frac{-11}{-13} \).

(ii) Rewrite with positive denominators: \( \frac{-25}{6}, \frac{-15}{4}, \frac{-17}{8}, \frac{-53}{12} \). All are negative. The LCM of 6, 4, 8, and 12 is 24. Convert to equivalent fractions: \( \frac{-25 \times 4}{24} = \frac{-100}{24} \), \( \frac{-15 \times 6}{24} = \frac{-90}{24} \), \( \frac{-17 \times 3}{24} = \frac{-51}{24} \), \( \frac{-53 \times 2}{24} = \frac{-106}{24} \). Since -106 < -100 < -90 < -51, the ascending order is \( \frac{-53}{12}, \frac{-25}{6}, \frac{15}{-4}, \frac{-17}{8} \).
In simple words: To arrange fractions, make all denominators the same using the LCM. Then compare numerators - a smaller (more negative) numerator means a smaller fraction.

Exam Tip: Always convert negative denominators to positive form first - this prevents sign errors. Use the LCM method consistently to avoid mixing different denominators.

Question 6. Arrange the rational numbers \( \frac{-7}{10}, \frac{5}{-8}, \frac{-3}{2}, \frac{4}{-1}, \frac{-3}{5} \) in descending order.
Answer: First, express each with a positive denominator: \( \frac{-7}{10}, \frac{-5}{8}, \frac{-3}{2}, \frac{-4}{1}, \frac{-3}{5} \). All are negative. The LCM of 10, 8, 2, 1, and 5 is 120. Convert to equivalent fractions with denominator 120: \( \frac{-7 \times 12}{120} = \frac{-84}{120} \), \( \frac{-5 \times 15}{120} = \frac{-75}{120} \), \( \frac{-3 \times 40}{120} = \frac{-120}{120} \) (wait, this is -3/2, so it should be \( \frac{-3 \times 60}{120} = \frac{-180}{120} \), let me recalculate: 2 goes into 120 sixty times, so \( \frac{-3 \times 60}{120} = \frac{-180}{120} \)), \( \frac{-4 \times 120}{120} = \frac{-480}{120} \), \( \frac{-3 \times 24}{120} = \frac{-72}{120} \). Since -30 > -72 > -75 > -80 > -84, in descending order (largest to smallest): \( \frac{-1}{4}, \frac{-3}{5}, \frac{5}{-8}, \frac{-3}{2}, \frac{-7}{10} \).
In simple words: To arrange in descending order, find the largest (least negative) fraction first. A numerator that is less negative is larger.

Exam Tip: In descending order, remember that among negative numbers, the one closest to zero is the largest - so -1 is greater than -100.

Question 7. Find the sum:
(i) \( \frac{-2}{3} + \left( \frac{-5}{-7} \right) \)
(ii) \( -1\frac{1}{12} + \frac{-5}{9} \)
(iii) \( 2\frac{2}{5} + \left( -4\frac{3}{10} \right) \)
Answer:
(i) Simplify the second fraction: \( \frac{-5}{-7} = \frac{5}{7} \). Now add \( \frac{-2}{3} + \frac{5}{7} \). The LCM of 3 and 7 is 21. Convert: \( \frac{-2 \times 7}{21} + \frac{5 \times 3}{21} = \frac{-14}{21} + \frac{15}{21} = \frac{-14 + 15}{21} = \frac{1}{21} \). Therefore, \( \frac{-2}{3} + \left( \frac{-5}{-7} \right) = \frac{1}{21} \).

(ii) Convert the mixed number: \( -1\frac{1}{12} = \frac{-13}{12} \). Add \( \frac{-13}{12} + \frac{-5}{9} \). The LCM of 12 and 9 is 36. Convert: \( \frac{-13 \times 3}{36} + \frac{-5 \times 4}{36} = \frac{-39}{36} + \frac{-20}{36} = \frac{-39 - 20}{36} = \frac{-59}{36} = -1\frac{23}{36} \). Therefore, \( -1\frac{1}{12} + \frac{-5}{9} = -1\frac{23}{36} \).

(iii) Convert the mixed numbers: \( 2\frac{2}{5} = \frac{12}{5} \) and \( -4\frac{3}{10} = \frac{-43}{10} \). Add \( \frac{12}{5} + \frac{-43}{10} \). The LCM of 5 and 10 is 10. Convert: \( \frac{12 \times 2}{10} + \frac{-43 \times 1}{10} = \frac{24}{10} + \frac{-43}{10} = \frac{24 - 43}{10} = \frac{-19}{10} = -1\frac{9}{10} \). Therefore, \( 2\frac{2}{5} + \left( -4\frac{3}{10} \right) = -1\frac{9}{10} \).
In simple words: To add fractions, make the denominators the same by finding the LCM. Then add the numerators and keep the denominator.

Exam Tip: Always convert mixed numbers to improper fractions before adding. Watch for negative signs - adding a negative number is the same as subtracting.

Question 8. Subtract:
(i) \( \frac{-11}{24} \) from \( \frac{-5}{36} \)
(ii) \( \frac{-8}{15} \) from \( -1\frac{2}{5} \)
(iii) \( -2\frac{2}{9} \) from \( -3\frac{5}{12} \)
Answer:
(i) Rewrite as: \( \frac{-5}{36} - \frac{-11}{24} = \frac{-5}{36} + \frac{11}{24} \). The LCM of 36 and 24 is 72. Convert: \( \frac{-5 \times 2}{72} + \frac{11 \times 3}{72} = \frac{-10}{72} + \frac{33}{72} = \frac{-10 + 33}{72} = \frac{23}{72} \). Therefore, the result is \( \frac{23}{72} \).

(ii) Convert: \( -1\frac{2}{5} = \frac{-7}{5} \). Now find \( \frac{-7}{5} - \frac{-8}{15} = \frac{-7}{5} + \frac{8}{15} \). The LCM of 5 and 15 is 15. Convert: \( \frac{-7 \times 3}{15} + \frac{8 \times 1}{15} = \frac{-21}{15} + \frac{8}{15} = \frac{-21 + 8}{15} = \frac{-13}{15} \). Therefore, the result is \( \frac{-13}{15} \).

(iii) Convert: \( -3\frac{5}{12} = \frac{-41}{12} \) and \( -2\frac{2}{9} = \frac{-20}{9} \). Now find \( \frac{-41}{12} - \frac{-20}{9} = \frac{-41}{12} + \frac{20}{9} \). The LCM of 12 and 9 is 36. Convert: \( \frac{-41 \times 3}{36} + \frac{20 \times 4}{36} = \frac{-123}{36} + \frac{80}{36} = \frac{-123 + 80}{36} = \frac{-43}{36} = -1\frac{7}{36} \). Therefore, the result is \( -1\frac{7}{36} \).
In simple words: Subtracting a negative number is the same as adding its opposite. Find a common denominator, then subtract the numerators.

Exam Tip: Convert mixed numbers to improper fractions first. Remember that subtracting a negative becomes addition - flip the sign of the second number.

Question 9. What should be subtracted from \( \frac{-3}{4} \) to get \( \frac{-5}{8} \)?
Answer: Let the required number be \( x \). Then \( \frac{-3}{4} - x = \frac{-5}{8} \). Rearranging: \( x = \frac{-3}{4} - \frac{-5}{8} = \frac{-3}{4} + \frac{5}{8} \). The LCM of 4 and 8 is 8. Convert: \( x = \frac{-3 \times 2}{8} + \frac{5 \times 1}{8} = \frac{-6}{8} + \frac{5}{8} = \frac{-6 + 5}{8} = \frac{-1}{8} \). Therefore, \( \frac{-1}{8} \) should be subtracted.
In simple words: To find the missing number, subtract the answer from the starting number. Use the same steps as regular fraction subtraction.

Exam Tip: Set up the equation carefully - the original number minus the unknown equals the result. Then rearrange to find the unknown.

Question 10. Find the following products:
(i) \( \frac{-3}{11} \times (-7) \)
(ii) \( \frac{-5}{12} \times \frac{16}{25} \)
(iii) \( \left( -2\frac{5}{8} \right) \times \left( 1\frac{3}{7} \right) \)
Answer:
(i) Write -7 as \( \frac{-7}{1} \). Then \( \frac{-3}{11} \times \frac{-7}{1} = \frac{(-3) \times (-7)}{11 \times 1} = \frac{21}{11} = 1\frac{10}{11} \). Therefore, \( \frac{-3}{11} \times (-7) = \frac{21}{11} \).

(ii) Multiply: \( \frac{-5}{12} \times \frac{16}{25} = \frac{(-5) \times 16}{12 \times 25} = \frac{-80}{300} \). Simplify by dividing by the HCF of 20: \( \frac{-80}{300} = \frac{-4}{15} \). Therefore, \( \frac{-5}{12} \times \frac{16}{25} = \frac{-4}{15} \).

(iii) Convert the mixed numbers: \( -2\frac{5}{8} = \frac{-21}{8} \) and \( 1\frac{3}{7} = \frac{10}{7} \). Multiply: \( \frac{-21}{8} \times \frac{10}{7} = \frac{(-21) \times 10}{8 \times 7} = \frac{-210}{56} \). Simplify by dividing by the HCF of 14: \( \frac{-210}{56} = \frac{-15}{4} = -3\frac{3}{4} \). Therefore, \( \left( -2\frac{5}{8} \right) \times \left( 1\frac{3}{7} \right) = -3\frac{3}{4} \).
In simple words: To multiply fractions, multiply numerators together and multiply denominators together. A negative times a positive is negative; a negative times a negative is positive.

Exam Tip: Cancel common factors before multiplying - this makes the numbers smaller and easier to work with. Always simplify the final answer.

Question 11. Find the value of:
(i) \( \frac{-8}{13} \div \frac{3}{-26} \)
(ii) \( 3\frac{1}{7} \div \frac{11}{-12} \)
(iii) \( \left( \frac{-3}{7} \times \frac{-2}{3} \right) \div \frac{16}{-21} \)
Answer:
(i) Rewrite the division as multiplication by the reciprocal: \( \frac{-8}{13} \div \frac{3}{-26} = \frac{-8}{13} \times \frac{-26}{3} = \frac{(-8) \times (-26)}{13 \times 3} = \frac{208}{39} \). Simplify by dividing by the HCF of 13: \( \frac{208}{39} = \frac{16}{3} = 5\frac{1}{3} \). Therefore, \( \frac{-8}{13} \div \frac{3}{-26} = \frac{16}{3} \).

(ii) Convert the mixed number: \( 3\frac{1}{7} = \frac{22}{7} \). Divide: \( \frac{22}{7} \div \frac{11}{-12} = \frac{22}{7} \times \frac{-12}{11} = \frac{22 \times (-12)}{7 \times 11} = \frac{-264}{77} \). Simplify by dividing by the HCF of 11: \( \frac{-264}{77} = \frac{-24}{7} = -3\frac{3}{7} \). Therefore, \( 3\frac{1}{7} \div \frac{11}{-12} = -3\frac{3}{7} \).

(iii) First, solve the bracket: \( \frac{-3}{7} \times \frac{-2}{3} = \frac{(-3) \times (-2)}{7 \times 3} = \frac{6}{21} = \frac{2}{7} \). Now divide: \( \frac{2}{7} \div \frac{16}{-21} = \frac{2}{7} \times \frac{-21}{16} = \frac{2 \times (-21)}{7 \times 16} = \frac{-42}{112} \). Simplify by dividing by the HCF of 14: \( \frac{-42}{112} = \frac{-3}{8} \). Therefore, \( \left( \frac{-3}{7} \times \frac{-2}{3} \right) \div \frac{16}{-21} = \frac{-3}{8} \).
In simple words: To divide fractions, flip the second fraction upside down and multiply. Follow the same rules for signs as multiplication.

Exam Tip: Always flip the divisor (the second fraction) and change division to multiplication. Simplify before multiplying if possible to reduce errors.

Question 12. From a rope 15 m long, \( 4\frac{1}{3} \) m is cut off and \( \frac{3}{5} \) of the remaining is cut off again. Find the length of the remaining part of the rope.
Answer: Convert the mixed number: \( 4\frac{1}{3} = \frac{13}{3} \) m. This is the first length cut off. Calculate the remaining length after the first cut: \( 15 - \frac{13}{3} = \frac{45}{3} - \frac{13}{3} = \frac{45 - 13}{3} = \frac{32}{3} \) m. Next, \( \frac{3}{5} \) of this remaining length is cut off. The part that stays is \( 1 - \frac{3}{5} = \frac{2}{5} \) of the remaining length. Calculate: \( \frac{2}{5} \times \frac{32}{3} = \frac{2 \times 32}{5 \times 3} = \frac{64}{15} = 4\frac{4}{15} \) m. Therefore, the length of the remaining part of the rope is \( 4\frac{4}{15} \) m.
In simple words: Subtract the first cut from the total length. Then find what fraction remains after the second cut, and multiply it by what was left.

Exam Tip: Read carefully - "of the remaining is cut off" means you must first find what is left, then take a fraction of that. It is not a fraction of the original rope.

Question 13. Perimeter of a rectangle is 2 m less than \( \frac{2}{5} \) of the perimeter of a square. If the perimeter of the square is 40 m, find the length and breadth of the rectangle given that the breadth is \( \frac{1}{3} \) of the length.
Answer: The perimeter of the square is 40 m. Calculate \( \frac{2}{5} \) of this: \( \frac{2}{5} \times 40 = 16 \) m. The perimeter of the rectangle is 2 m less than 16 m, so it is \( 16 - 2 = 14 \) m. Let the length be \( l \). Since the breadth is \( \frac{1}{3} \) of the length, the breadth is \( b = \frac{l}{3} \). Use the perimeter formula: \( 2(l + b) = 14 \). Substitute: \( 2 \left( l + \frac{l}{3} \right) = 14 \). Simplify: \( 2 \left( \frac{3l + l}{3} \right) = 14 \). This gives \( 2 \left( \frac{4l}{3} \right) = 14 \), so \( \frac{8l}{3} = 14 \). Solve: \( l = \frac{14 \times 3}{8} = \frac{42}{8} = \frac{21}{4} = 5\frac{1}{4} \) m. Actually, let me recalculate: \( 2(l + \frac{l}{3}) = 14 \), so \( l + \frac{l}{3} = 7 \), which gives \( \frac{3l + l}{3} = 7 \), so \( \frac{4l}{3} = 7 \), so \( l = \frac{21}{4} = 5\frac{1}{4} \) m. Wait, the source shows \( l = \frac{21}{4} \) m. Let me verify: if \( l = \frac{21}{4} \), then \( b = \frac{1}{3} \times \frac{21}{4} = \frac{21}{12} = \frac{7}{4} \) m. Check: \( 2(\frac{21}{4} + \frac{7}{4}) = 2 \times \frac{28}{4} = 2 \times 7 = 14 \) m. ✓ Therefore, the length of the rectangle is \( \frac{21}{4} \) m and the breadth is \( \frac{7}{4} \) m.
In simple words: First find the rectangle's perimeter using the square's measurements. Then set up an equation using the breadth - length relationship and solve for both dimensions.

Exam Tip: Always express the breadth in terms of the length when given a relationship between them. Substitute this expression into the perimeter formula to create a single equation with one unknown.