CBSE Class 8 Mathematics Rational Numbers Assignment Set 03

Multiply:

Question. \( \frac{7}{11} \) by \( \frac{5}{4} \)
Answer: \( \frac{35}{44} \)

Question. \( \frac{5}{7} \) by \( \frac{-3}{4} \)
Answer: \( -\frac{15}{28} \)

Question. \( \frac{-2}{9} \) by \( \frac{5}{11} \)
Answer: \( -\frac{10}{99} \)

Question. \( \frac{-3}{17} \) by \( \frac{-5}{-4} \)
Answer: \( -\frac{15}{68} \)

Question. \( \frac{9}{-7} \) by \( \frac{36}{-11} \)
Answer: \( \frac{324}{77} \)

Question. \( \frac{-11}{13} \) by \( \frac{-21}{7} \)
Answer: \( \frac{33}{13} \)

Question. \( \frac{3}{-5} \) by \( -\frac{4}{7} \)
Answer: \( \frac{12}{35} \)

Question. \( \frac{15}{-11} \) by 7
Answer: \( -\frac{105}{11} \)

Multiply:

Question. \( \frac{-5}{17} \) by \( \frac{51}{-60} \)
Answer: \( \frac{1}{4} \)

Question. \( \frac{-6}{11} \) by \( \frac{-55}{36} \)
Answer: \( \frac{5}{6} \)

Question. \( \frac{-8}{25} \) by \( \frac{-5}{16} \)
Answer: \( \frac{1}{10} \)

Question. \( \frac{6}{7} \) by \( \frac{-49}{36} \)
Answer: \( -\frac{7}{6} \)

Question. \( \frac{8}{-9} \) by \( \frac{-7}{-16} \)
Answer: \( -\frac{7}{18} \)

Question. \( \frac{-8}{9} \) by \( \frac{3}{64} \)
Answer: \( -\frac{1}{24} \)

Simplify each of the following and express the result as a rational number in standard form:

Question. \( \frac{-16}{21} \times \frac{14}{5} \)
Answer: \( -\frac{32}{15} \)

Question. \( \frac{7}{6} \times \frac{-3}{28} \)
Answer: \( -\frac{1}{8} \)

Question. \( \frac{-19}{36} \times 16 \)
Answer: \( -\frac{76}{9} \)

Question. \( \frac{-13}{9} \times \frac{27}{-26} \)
Answer: \( \frac{3}{2} \)

Question. \( \frac{-9}{16} \times \frac{-64}{-27} \)
Answer: \( -\frac{4}{3} \)

Question. \( \frac{-50}{7} \times \frac{14}{3} \)
Answer: \( -\frac{100}{3} \)

Question. \( \frac{-11}{9} \times \frac{-81}{-88} \)
Answer: \( -\frac{9}{8} \)

Question. \( \frac{-5}{9} \times \frac{72}{-25} \)
Answer: \( \frac{8}{5} \)

Simplify:

Question. \( \left( \frac{25}{8} \times \frac{2}{5} \right) - \left( \frac{3}{5} \times \frac{-10}{9} \right) \)
Answer: \( \frac{5}{4} - \left( -\frac{2}{3} \right) = \frac{5}{4} + \frac{2}{3} = \frac{15+8}{12} = \frac{23}{12} \)

Question. \( \left( \frac{1}{2} \times \frac{1}{4} \right) + \left( \frac{1}{2} \times 6 \right) \)
Answer: \( \frac{1}{8} + 3 = \frac{25}{8} \)

Question. \( \left( -5 \times \frac{2}{15} \right) - \left( -6 \times \frac{2}{9} \right) \)
Answer: \( -\frac{2}{3} - \left( -\frac{4}{3} \right) = \frac{2}{3} \)

Question. \( \left( \frac{-9}{4} \times \frac{5}{3} \right) + \left( \frac{13}{2} \times \frac{5}{6} \right) \)
Answer: \( -\frac{15}{4} + \frac{65}{12} = \frac{-45+65}{12} = \frac{20}{12} = \frac{5}{3} \)

Question. \( \left( \frac{-4}{3} \times \frac{12}{-5} \right) + \left( \frac{3}{7} \times \frac{21}{15} \right) \)
Answer: \( \frac{16}{5} + \frac{3}{5} = \frac{19}{5} \)

Question. \( \left( \frac{13}{5} \times \frac{8}{3} \right) - \left( \frac{-5}{2} \times \frac{11}{3} \right) \)
Answer: \( \frac{104}{15} - \left( -\frac{55}{6} \right) = \frac{104}{15} + \frac{55}{6} = \frac{208+275}{30} = \frac{483}{30} = \frac{161}{10} \)

Question. \( \left( \frac{13}{7} \times \frac{11}{26} \right) - \left( \frac{-4}{3} \times \frac{5}{6} \right) \)
Answer: \( \frac{11}{14} - \left( -\frac{10}{9} \right) = \frac{11}{14} + \frac{10}{9} = \frac{99+140}{126} = \frac{239}{126} \)

Question. \( \left( \frac{8}{5} \times \frac{-3}{2} \right) + \left( \frac{-3}{10} \times \frac{11}{16} \right) \)
Answer: \( -\frac{12}{5} + \left( -\frac{33}{160} \right) = \frac{-384-33}{160} = -\frac{417}{160} \)

EXERCISE # 2

Give examples of:

Question. The rational number that does not have a reciprocal.
Answer: 0

Question. The rational numbers that are equal to their reciprocals.
Answer: 1 and – 1

Question. The rational number that is equal to its negative.
Answer: 0

Fill in the blanks.

Question. Zero has ______ reciprocal.
Answer: No

Question. The numbers______ and _____ are their own reciprocals.
Answer: 1, – 1

Question. The reciprocal of – 5 is _______.
Answer: \( -\frac{1}{5} \)

Question. Reciprocal of \( \frac{1}{x} \), where x ≠ 0 is______.
Answer: x

Question. The product of two rational numbers is always a______.
Answer: Rational Number

Question. The reciprocal of a positive rational number is ________.
Answer: positive

Question. Represent \( \frac{7}{4} \) and \( \frac{-5}{6} \) on the number line.
Answer: Refer to diagram in key.

Question. Represent \( \frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11} \) on the number line.
Answer: Refer to diagram in key.

Question. Write five rational numbers which are smaller than 2.
Answer: Some of these are 1, \( \frac{1}{2} \), 0, –1, \( -\frac{1}{2} \)

Question. Find ten rational numbers between \( \frac{-2}{5} \) and \( \frac{1}{2} \).
Answer: \( -\frac{7}{20}, -\frac{6}{20}, -\frac{5}{20}, -\frac{4}{20}, -\frac{3}{20}, -\frac{2}{20}, -\frac{1}{20}, 0, ..., \frac{1}{20}, \frac{2}{20} \)

Click on link below to download CBSE Class 8 Mathematics Rational Numbers Assignment Set C

1. Rational Numbers

Question 1. Find the reciprocal of -2.
Answer: The reciprocal (multiplicative inverse) of a non-zero rational number \(a\) is \(\frac{1}{a}\).
Given number = \(-2\)
Reciprocal = \(\frac{1}{-2} = -\frac{1}{2}\)
In simple words: To find the reciprocal of a number, flip the numerator and denominator.

Exam Tip: The product of a number and its reciprocal is always 1.

Question 2. The sum of two numbers is \(\frac{5}{9}\). If one of the numbers is \(\frac{1}{3}\), find the other.
Answer: Let the other number be \(x\).
According to the question:
\(x + \frac{1}{3} = \frac{5}{9}\)
\(x = \frac{5}{9} - \frac{1}{3}\)
Taking the LCM of 9 and 3, which is 9:
\(x = \frac{5}{9} - \frac{3}{9}\)
\(x = \frac{5 - 3}{9} = \frac{2}{9}\)
In simple words: Subtract the known number from the total sum to find the missing number.

Exam Tip: Always make the denominators equal using LCM before adding or subtracting fractions.

Question 3. Write the additive inverse of \(\frac{2}{3}\) and show that their sum is zero.
Answer: The additive inverse of a number \(a\) is \(-a\) such that \(a + (-a) = 0\).
Additive inverse of \(\frac{2}{3}\) is \(-\frac{2}{3}\).
Sum = \(\frac{2}{3} + \left(-\frac{2}{3}\right) = \frac{2 - 2}{3} = \frac{0}{3} = 0\)
In simple words: The additive inverse is the same number with the opposite sign.

Exam Tip: A number plus its additive inverse always equals zero.

Question 4. Simplify \(\frac{7}{6} \times \frac{-3}{28}\) and find its reciprocal.
Answer: First, simplify the given expression:
\(\frac{7}{6} \times \frac{-3}{28} = \frac{7 \times (-3)}{6 \times 28}\)
Canceling common factors: \(7\) and \(28\) cancel to give \(4\); \(3\) and \(6\) cancel to give \(2\).
\(= \frac{1 \times (-1)}{2 \times 4} = -\frac{1}{8}\)
Now, find the reciprocal of \(-\frac{1}{8}\):
Reciprocal = \(\frac{1}{-\frac{1}{8}} = -8\)
In simple words: Multiply the numerators together and denominators together, simplify, then flip the result.

Exam Tip: Cancel common factors diagonally before multiplying to make calculations easier.

Question 5. The product of two rational numbers is 15. If one of the numbers is -10, find the other.
Answer: Let the other number be \(x\).
According to the question:
\(x \times (-10) = 15\)
\(x = \frac{15}{-10}\)
Simplifying by dividing numerator and denominator by 5:
\(x = -\frac{3}{2}\)
In simple words: Divide the total product by the known number to find the other number.

Exam Tip: When shifting a multiplier to the other side, it goes to the denominator with its sign.

Question 6. Write any three rational numbers between -5 and 0.
Answer: We can write -5 and 0 with a common denominator, or simply choose numbers between them.
Let us write -5 as \(-\frac{15}{3}\) and 0 as \(\frac{0}{3}\).
Three rational numbers between -5 and 0 are: \(-1, -2, -3\) (or fractional equivalents like \(-\frac{1}{2}, -1, -\frac{3}{2}\)).
In simple words: Any numbers lying between -5 and 0 on the number line are valid.

Exam Tip: There are infinitely many rational numbers between any two rational numbers.

Question 7. True or False: 1 is the only rational number that is equal to its reciprocal.
Answer: False.
Reason: Both 1 and -1 are equal to their own reciprocals, because \(\frac{1}{1} = 1\) and \(\frac{1}{-1} = -1\).
In simple words: Both 1 and -1 have reciprocals equal to themselves.

Exam Tip: Read "only" carefully in mathematical statements; exceptions often make them false.

Question 8. Simplify: \(\frac{-4}{13} - \frac{-3}{26}\)
Answer: Given expression: \(\frac{-4}{13} - \left(-\frac{3}{26}\right)\)
\(= \frac{-4}{13} + \frac{3}{26}\)
The LCM of 13 and 26 is 26.
\(= \frac{-4 \times 2}{13 \times 2} + \frac{3}{26}\)
\(= \frac{-8}{26} + \frac{3}{26}\)
\(= \frac{-8 + 3}{26} = -\frac{5}{26}\)
In simple words: Convert subtraction of a negative into addition, then make denominators equal.

Exam Tip: Watch out for double negative signs: subtracting a negative is the same as adding a positive.

Question 9. Verify \(\left[\frac{-2}{5} + \frac{3}{5}\right] + \frac{1}{2} = \frac{-2}{5} + \left[\frac{3}{5} + \frac{1}{2}\right]\).
Answer: This demonstrates the associative property of addition.
Left Hand Side (LHS):
\(\left[\frac{-2}{5} + \frac{3}{5}\right] + \frac{1}{2}\)
\(= \frac{-2 + 3}{5} + \frac{1}{2}\)
\(= \frac{1}{5} + \frac{1}{2}\)
LCM of 5 and 2 is 10.
\(= \frac{2}{10} + \frac{5}{10} = \frac{7}{10}\)
Right Hand Side (RHS):
\(\frac{-2}{5} + \left[\frac{3}{5} + \frac{1}{2}\right]\)
LCM of 5 and 2 inside bracket is 10.
\(= \frac{-2}{5} + \left[\frac{6}{10} + \frac{5}{10}\right]\)
\(= \frac{-2}{5} + \frac{11}{10}\)
\(= \frac{-4}{10} + \frac{11}{10} = \frac{7}{10}\)
Since LHS = RHS = \(\frac{7}{10}\), the statement is verified.
In simple words: Changing the grouping of numbers in addition does not change the result.

Exam Tip: Solve brackets first following the BODMAS rule.

Question 10. Verify: \(\left[\frac{1}{2} \div \frac{-1}{3}\right] \div \frac{2}{5} \neq \frac{1}{2} \div \left[\frac{-1}{3} \div \frac{2}{5}\right]\)
Answer: Division is not associative for rational numbers.
Left Hand Side (LHS):
\(\left[\frac{1}{2} \div \frac{-1}{3}\right] \div \frac{2}{5}\)
\(= \left[\frac{1}{2} \times \frac{3}{-1}\right] \div \frac{2}{5}\)
\(= -\frac{3}{2} \div \frac{2}{5}\)
\(= -\frac{3}{2} \times \frac{5}{2} = -\frac{15}{4}\)
Right Hand Side (RHS):
\(\frac{1}{2} \div \left[\frac{-1}{3} \div \frac{2}{5}\right]\)
\(= \frac{1}{2} \div \left[\frac{-1}{3} \times \frac{5}{2}\right]\)
\(= \frac{1}{2} \div \left(-\frac{5}{6}\right)\)
\(= \frac{1}{2} \times \left(-\frac{6}{5}\right) = -\frac{6}{10} = -\frac{3}{5}\)
Since LHS (\(-\frac{15}{4}\)) \(\neq\) RHS (\(-\frac{3}{5}\)), the inequality is verified.
In simple words: Division does not follow the associative property.

Exam Tip: Convert division to multiplication by taking the reciprocal of the divisor.

Question 11. Verify the associative property of multiplication in \(\frac{-7}{3} \times \left(\frac{5}{4} \times \frac{2}{9}\right) = \left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}\).
Answer:
Left Hand Side (LHS):
\(\frac{-7}{3} \times \left(\frac{5}{4} \times \frac{2}{9}\right)\)
Inside bracket: \(\frac{5 \times 2}{4 \times 9} = \frac{10}{36} = \frac{5}{18}\)
\(= \frac{-7}{3} \times \frac{5}{18} = \frac{-35}{54}\)
Right Hand Side (RHS):
\(\left(\frac{-7}{3} \times \frac{5}{4}\right) \times \frac{2}{9}\)
Inside bracket: \(\frac{-35}{12}\)
\(= \frac{-35}{12} \times \frac{2}{9} = \frac{-70}{108} = -\frac{35}{54}\)
Since LHS = RHS = \(-\frac{35}{54}\), the associative property of multiplication is verified.
In simple words: Changing the grouping of numbers in multiplication gives the same product.

Exam Tip: Simplify fractions at each step to keep numbers small and manageable.

Question 12. Find \(\frac{3}{7} + \frac{6}{11} + \left(\frac{-8}{21}\right)\).
Answer: Given expression: \(\frac{3}{7} + \frac{6}{11} - \frac{8}{21}\)
Denominators are 7, 11, and 21. The LCM of 7, 11, and 21 is \(21 \times 11 = 231\).
Convert each fraction to have denominator 231:
\(\frac{3}{7} = \frac{3 \times 33}{7 \times 33} = \frac{99}{231}\)
\(\frac{6}{11} = \frac{6 \times 21}{11 \times 21} = \frac{126}{231}\)
\(-\frac{8}{21} = -\frac{8 \times 11}{21 \times 11} = -\frac{88}{231}\)
Now, add them up:
\(= \frac{99 + 126 - 88}{231}\)
\(= \frac{225 - 88}{231} = \frac{137}{231}\)
In simple words: Find the LCM of all denominators, convert fractions, and combine numerators.

Exam Tip: Always look for common factors among denominators to simplify LCM calculation.

Question 13. Simplify: \(\left[\frac{1}{2} \div \left(\frac{-1}{2}\right)\right] \div \frac{2}{5}\)
Answer:
First, solve the inner bracket:
\(\frac{1}{2} \div \left(-\frac{1}{2}\right) = \frac{1}{2} \times \left(-\frac{2}{1}\right) = -1\)
Now, substitute back into the expression:
\(= -1 \div \frac{2}{5}\)
\(= -1 \times \frac{5}{2} = -\frac{5}{2}\)
In simple words: Divide the first two numbers inside the brackets, then divide that result by the last number.

Exam Tip: Any non-zero number divided by itself gives 1.

Question 14. Find a rational number between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Answer: Method 1: Using the mean formula \(\frac{a + b}{2}\).
\(\text{Rational number} = \frac{\frac{1}{4} + \frac{1}{2}}{2}\)
\(= \frac{\frac{1}{4} + \frac{2}{4}}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{8}\)
In simple words: Find the average of the two numbers to get a number right in the middle.

Exam Tip: You can also make denominators equal (e.g., \(\frac{2}{8}\) and \(\frac{4}{8}\)) and pick a number in between (\(\frac{3}{8}\)).

Question 15. Find two rational numbers between -2 and 0.
Answer: Write -2 and 0 as rational numbers with denominator 3:
\(-2 = -\frac{6}{3}\) and \(0 = \frac{0}{3}\)
Two rational numbers between them are \(-\frac{5}{3}, -\frac{4}{3}\) (or simply \(-1\) and \(-1.5\)).
In simple words: Convert the integers into fractions with the same denominator to easily spot numbers in between.

Exam Tip: Multiplying numerator and denominator by a larger number gives more intermediate slots.

Question 16. Find \(\frac{-6}{5} \times \frac{3}{8} \times \frac{15}{24} \times \frac{-16}{9}\)
Answer:
Given expression: \(\frac{-6}{5} \times \frac{3}{8} \times \frac{15}{24} \times \frac{-16}{9}\)
Simplify fractions where possible:
\(\frac{15}{24} = \frac{5}{8}\)
Substitute and cancel terms:
\(= \frac{-6}{5} \times \frac{3}{8} \times \frac{5}{8} \times \frac{-16}{9}\)
Cancel 5 in numerator and denominator.
Cancel \(-6\) and \(9\) by 3: \(-2\) and \(3\).
Cancel \(3\) in numerator with \(3\) in denominator.
Cancel \(8 \times 8 = 64\) or cross-cancel \(-16\) with \(8\).
Let us do step-by-step multiplication of numerators and denominators:
Numerator = \((-6) \times 3 \times 15 \times (-16) = 4320\)
Denominator = \(5 \times 8 \times 24 \times 9 = 8640\)
\(= \frac{4320}{8640} = \frac{1}{2}\)
In simple words: Multiply all numerators together and all denominators together, then simplify the fraction.

Exam Tip: Cross-canceling common factors before multiplying saves time and prevents arithmetic errors.

Question 17. By what number should \(\frac{3}{-14}\) be multiplied so as to get \(\frac{5}{12}\)?
Answer: Let the required number be \(x\).
\(x \times \left(\frac{3}{-14}\right) = \frac{5}{12}\)
\(x = \frac{5}{12} \div \left(\frac{3}{-14}\right)\)
\(x = \frac{5}{12} \times \left(\frac{-14}{3}\right)\)
\(x = \frac{5 \times (-14)}{12 \times 3} = \frac{-70}{36}\)
Simplifying by dividing by 2:
\(x = -\frac{35}{18}\)
In simple words: Divide the target product by the given number.

Exam Tip: Remember to change division to multiplication and flip the second fraction.

Question 18. Represent \(-\frac{7}{4}\) on the number line.
Answer:
1. Draw a straight line with arrows at both ends and mark an origin point 0.
2. Since \(-\frac{7}{4} = -1\frac{3}{4}\), the number lies between -1 and -2.
3. Divide the segment between 0 and -1 into 4 equal parts, and extend further left to divide the segment between -1 and -2 into 4 equal parts.
4. Count 7 parts to the left of 0 and mark the point \(-\frac{7}{4}\).

-7/4 In simple words: Convert the improper fraction to a mixed fraction to locate its position between integers.

Exam Tip: The denominator tells you how many equal parts each unit interval is divided into.

Question 19. Divide \(\frac{1}{2}\) by \(\left[\frac{-1}{3} + \frac{2}{5}\right]\).
Answer:
First, solve the expression inside the brackets:
\(\frac{-1}{3} + \frac{2}{5}\)
LCM of 3 and 5 is 15.
\(= \frac{-5}{15} + \frac{6}{15} = \frac{1}{15}\)
Now, perform the division:
\(\frac{1}{2} \div \frac{1}{15}\)
\(= \frac{1}{2} \times \frac{15}{1} = \frac{15}{2}\)
In simple words: Simplify the bracket first, then multiply the first fraction by the reciprocal of the bracket result.

Exam Tip: Always evaluate grouping symbols before performing operations outside.

Question 20. Multiply \(\frac{-7}{3} \times \left(\frac{-5}{4} \times \frac{2}{9}\right)\).
Answer:
First, multiply the terms inside the brackets:
\(\frac{-5}{4} \times \frac{2}{9} = \frac{-10}{36} = -\frac{5}{18}\)
Now, multiply by the outer term:
\(\frac{-7}{3} \times \left(-\frac{5}{18}\right)\)
\(= \frac{(-7) \times (-5)}{3 \times 18} = \frac{35}{54}\)
In simple words: Solve the multiplication inside the bracket first, then multiply by the remaining number.

Exam Tip: Negative multiplied by negative gives a positive result.

Question 21. Find \(\frac{-2}{3} - \left[\frac{-4}{5} - \frac{1}{2}\right]\).
Answer:
First, solve inside the square brackets:
\(\frac{-4}{5} - \frac{1}{2}\)
LCM of 5 and 2 is 10.
\(= \frac{-8}{10} - \frac{5}{10} = \frac{-13}{10}\)
Now, substitute back into the expression:
\(= \frac{-2}{3} - \left(\frac{-13}{10}\right)\)
\(= \frac{-2}{3} + \frac{13}{10}\)
LCM of 3 and 10 is 30.
\(= \frac{-2 \times 10}{3 \times 10} + \frac{13 \times 3}{10 \times 3}\)
\(= \frac{-20}{30} + \frac{39}{30} = \frac{19}{30}\)
In simple words: Evaluate the inner subtraction first, change the double negative to addition, and find a common denominator.

Exam Tip: Be careful with signs when subtracting a negative number.

Question 22. Using appropriate properties, find \(\frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}\).
Answer:
Rearrange using the commutative and distributive properties:
\(= \left(\frac{-2}{3} \times \frac{3}{5}\right) - \left(\frac{3}{5} \times \frac{1}{6}\right) + \frac{5}{2}\)
Factor out \(\frac{3}{5}\) using the distributive property:
\(= \frac{3}{5} \left(\frac{-2}{3} - \frac{1}{6}\right) + \frac{5}{2}\)
Solve inside the bracket (LCM of 3 and 6 is 6):
\(\frac{-2}{3} = \frac{-4}{6}\)
\(\frac{-4}{6} - \frac{1}{6} = \frac{-5}{6}\)
Substitute back:
\(= \frac{3}{5} \times \left(\frac{-5}{6}\right) + \frac{5}{2}\)
\(= -\frac{15}{30} + \frac{5}{2} = -\frac{1}{2} + \frac{5}{2}\)
\(= \frac{-1 + 5}{2} = \frac{4}{2} = 2\)
In simple words: Group terms with common factors and use the distributive property to simplify calculation.

Exam Tip: Spotting common multipliers like \(\frac{3}{5}\) avoids lengthy individual multiplications.

Question 23. Find three rational numbers between \(-\frac{5}{6}\) and \(-\frac{5}{8}\).
Answer:
First, make denominators equal using the LCM of 6 and 8, which is 24.
\(-\frac{5}{6} = -\frac{5 \times 4}{6 \times 4} = -\frac{20}{24}\)
\(-\frac{5}{8} = -\frac{5 \times 3}{8 \times 3} = -\frac{15}{24}\)
We need 3 rational numbers between \(-\frac{20}{24}\) and \(-\frac{15}{24}\).
These can be: \(-\frac{16}{24}, -\frac{17}{24}, -\frac{18}{24}\) (which simplify to \(-\frac{2}{3}, -\frac{17}{24}, -\frac{3}{4}\)).
In simple words: Make denominators equal and select three numerators lying between the two given numerators.

Exam Tip: If needed, multiply numerator and denominator by a larger factor to create enough gap between numerators.

Question 24. By what number should \(\frac{-33}{8}\) be divided to get \(-\frac{11}{2}\)?
Answer: Let the required number be \(x\).
\(\frac{-33}{8} \div x = -\frac{11}{2}\)
\(x = \frac{-33}{8} \div \left(-\frac{11}{2}\right)\)
\(x = \frac{-33}{8} \times \left(-\frac{2}{11}\right)\)
\(x = \frac{(-33) \times (-2)}{8 \times 11} = \frac{66}{88}\)
Simplifying by dividing by 22:
\(x = \frac{3}{4}\)
In simple words: Divide the first number by the result to find the divisor.

Exam Tip: Be careful with the order in division; \(a \div x = b\) means \(x = a \div b\).

Question 25. Using the distributive property, find \(\frac{-3}{4} \times \frac{2}{3} + \left(\frac{-3}{4} \times \frac{-5}{6}\right)\).
Answer:
The common factor is \(\frac{-3}{4}\). Apply the distributive property \(a(b + c) = ab + ac\):
\(= \frac{-3}{4} \times \left[\frac{2}{3} + \left(\frac{-5}{6}\right)\right]\)
Simplify inside the bracket (LCM of 3 and 6 is 6):
\(\frac{2}{3} = \frac{4}{6}\)
\(\frac{4}{6} - \frac{5}{6} = -\frac{1}{6}\)
Substitute back:
\(= \frac{-3}{4} \times \left(-\frac{1}{6}\right)\)
\(= \frac{(-3) \times (-1)}{4 \times 6} = \frac{3}{24} = \frac{1}{8}\)
In simple words: Factor out the common term, add the remaining fractions inside the bracket, then multiply.

Exam Tip: Distributive property greatly simplifies calculations when the same factor appears in multiple terms.

Question 26. What should be subtracted from \(\frac{-2}{3}\) to get \(\frac{5}{6}\)?
Answer: Let the number to be subtracted be \(x\).
\(\frac{-2}{3} - x = \frac{5}{6}\)
\(-x = \frac{5}{6} - \left(\frac{-2}{3}\right)\)
\(-x = \frac{5}{6} + \frac{2}{3}\)
LCM of 6 and 3 is 6.
\(-x = \frac{5}{6} + \frac{4}{6} = \frac{9}{6} = \frac{3}{2}\)
Therefore, \(x = -\frac{3}{2}\)
In simple words: Set up the equation \(\text{Given} - x = \text{Result}\) and solve for \(x\).

Exam Tip: Double-check sign changes when moving terms across the equals sign.

Question 27. Find the sum of \(\frac{3}{7} + \left(\frac{-6}{11}\right) + \left(\frac{-8}{21}\right) + \frac{5}{22}\).
Answer:
Denominators are 7, 11, 21, and 22. Notice that 21 is \(3 \times 7\) and 22 is \(2 \times 11\).
The LCM of 7, 11, 21, and 22 is \(21 \times 22\) or LCM(7, 21) = 21 and LCM(11, 22) = 22, overall LCM = 462.
Convert each fraction:
\(\frac{3}{7} = \frac{3 \times 66}{462} = \frac{198}{462}\)
\(-\frac{6}{11} = -\frac{6 \times 42}{462} = -\frac{252}{462}\)
\(-\frac{8}{21} = -\frac{8 \times 22}{462} = -\frac{176}{462}\)
\(\frac{5}{22} = \frac{5 \times 21}{462} = \frac{105}{462}\)
Combine numerators:
\(= \frac{198 - 252 - 176 + 105}{462}\)
\(= \frac{303 - 428}{462} = -\frac{125}{462}\)
In simple words: Find the common LCM for all four denominators and combine all numerators.

Exam Tip: Group positive and negative numerators separately before final addition to avoid calculation errors.

Question 28. Find two rational numbers between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Answer:
Make denominators equal with a larger denominator to find two numbers easily. Multiply numerator and denominator by 3:
\(\frac{1}{4} = \frac{3}{12}\)
\(\frac{1}{2} = \frac{6}{12}\)
Two rational numbers between \(\frac{3}{12}\) and \(\frac{6}{12}\) are \(\frac{4}{12}\) and \(\frac{5}{12}\).
Simplifying: \(\frac{1}{3}\) and \(\frac{5}{12}\).
In simple words: Expand the fractions using a common denominator to easily fit intermediate numbers.

Exam Tip: Expanding denominators by multiplying by \(n+1\) (where \(n\) is the number of required rational numbers) always guarantees enough space.

Question 29. Find three rational numbers between \(\frac{2}{3}\) and \(\frac{4}{5}\).
Answer:
Make denominators equal using the LCM of 3 and 5, which is 15.
\(\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\)
\(\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}\)
Since there is only one integer between 10 and 12, multiply numerator and denominator by 4 to get more room:
\(\frac{10 \times 4}{15 \times 4} = \frac{40}{60}\)
\(\frac{12 \times 4}{5 \times 4} = \frac{48}{60}\)
Three rational numbers between \(\frac{40}{60}\) and \(\frac{48}{60}\) are \(\frac{41}{60}, \frac{42}{60}, \frac{43}{60}\).
In simple words: Expand the fractions further if the initial numerators do not have enough integers between them.

Exam Tip: Multiplying both fractions by 10/10 or 4/4 is a reliable way to generate multiple rational numbers.

Question 30. Find two rational numbers between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Answer:
Using the same method as Question 28:
\(\frac{1}{4} = \frac{3}{12}\), \(\frac{1}{2} = \frac{6}{12}\)
Two rational numbers are \(\frac{4}{12}\) and \(\frac{5}{12}\), which simplify to \(\frac{1}{3}\) and \(\frac{5}{12}\).
In simple words: Convert to equivalent fractions with denominator 12 and pick the numbers in between.

Exam Tip: Simplifying final answers to lowest terms is good practice.

Question 31. Find three rational numbers between \(\frac{3}{5}\) and \(\frac{3}{4}\).
Answer:
LCM of 5 and 4 is 20.
\(\frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20}\)
\(\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}\)
Multiply numerator and denominator by 2 to get enough numbers between 12 and 15:
\(\frac{12 \times 2}{20 \times 2} = \frac{24}{40}\)
\(\frac{15 \times 2}{20 \times 2} = \frac{30}{40}\)
Three rational numbers between \(\frac{24}{40}\) and \(\frac{30}{40}\) are \(\frac{25}{40}, \frac{26}{40}, \frac{27}{40}\) (or \(\frac{5}{8}, \frac{13}{20}, \frac{27}{40}\)).
In simple words: Equalize denominators, then scale up the fractions to find intermediate values.

Exam Tip: Scaling up ensures you can always find the exact number of requested rational numbers.

Question 32. Find three rational numbers between \(\frac{3}{5}\) and \(\frac{3}{4}\).
Answer:
Following the same steps as Question 31:
\(\frac{3}{5} = \frac{24}{40}\) and \(\frac{3}{4} = \frac{30}{40}\)
Three rational numbers between them are \(\frac{25}{40}, \frac{26}{40}, \frac{27}{40}\), which simplify to \(\frac{5}{8}, \frac{13}{20}, \frac{27}{40}\).
In simple words: Convert to equivalent fractions with a larger denominator and select numbers in between.

Exam Tip: Always reduce fractions to their simplest form if required.

Most Important Questions

Question 1. Using appropriate properties of rational numbers, find the sum \(\frac{3}{4} + \frac{4}{21} + \frac{7}{8} + \frac{5}{7}\).
Answer:
Group terms with common or related denominators:
\(= \left(\frac{3}{4} + \frac{7}{8}\right) + \left(\frac{4}{21} + \frac{5}{7}\right)\)
For the first group (LCM = 8):
\(\frac{6}{8} + \frac{7}{8} = \frac{13}{8}\)
For the second group (LCM = 21):
\(\frac{4}{21} + \frac{15}{21} = \frac{19}{21}\)
Now add both groups (LCM of 8 and 21 is 168):
\(= \frac{13 \times 21}{168} + \frac{19 \times 8}{168}\)
\(= \frac{273 + 152}{168} = \frac{425}{168}\)
In simple words: Group fractions with related denominators together to simplify addition.

Exam Tip: Rearranging terms using commutativity makes complex additions much easier.

Question 2. Using appropriate properties of rational numbers, find \(\frac{-5}{12} \times \frac{3}{16} - \frac{3}{8} \times \frac{7}{5} + \frac{5}{8} \times \frac{7}{7}\).
Answer:
Let us simplify each term:
Term 1: \(\frac{-5 \times 3}{12 \times 16} = \frac{-15}{192} = -\frac{5}{64}\)
Term 2: \(-\frac{3 \times 7}{8 \times 5} = -\frac{21}{40}\)
Term 3: \(\frac{5}{8} \times 1 = \frac{5}{8}\)
Combine using LCM of 64, 40, and 8, which is 320:
\(-\frac{25}{320} - \frac{168}{320} + \frac{200}{320}\)
\(= \frac{-25 - 168 + 200}{320} = \frac{-193 + 200}{320} = \frac{7}{320}\)
In simple words: Simplify each multiplication product first, then combine the fractions using a common denominator.

Exam Tip: Simplify individual fractions before finding the LCM.

Question 3. What is additive identity of rational number?
Answer: The additive identity for rational numbers is 0.
Reason: Adding 0 to any rational number \(a\) leaves the number unchanged (\(a + 0 = a\)).
In simple words: Zero is the additive identity because it does not change the value of a number when added.

Exam Tip: Do not confuse additive identity (0) with multiplicative identity (1).

Question 4. Find the additive inverse of (i) \(\frac{4}{21}\) (ii) \(-\frac{7}{11}\) (iii) \(\frac{9}{-2}\) (iv) \(2\frac{3}{7}\).
Answer:
(i) Additive inverse of \(\frac{4}{21}\) is \(-\frac{4}{21}\).
(ii) Additive inverse of \(-\frac{7}{11}\) is \(\frac{7}{11}\).
(iii) Additive inverse of \(\frac{9}{-2}\) (which is \(-\frac{9}{2}\)) is \(\frac{9}{2}\).
(iv) Additive inverse of \(2\frac{3}{7} = \frac{17}{7}\) is \(-\frac{17}{7}\) (or \(-2\frac{3}{7}\)).
In simple words: Simply change the sign of the number to find its additive inverse.

Exam Tip: Always convert mixed fractions into improper fractions before finding inverses if needed.

Question 5. What is multiplicative identity of rational number?
Answer: The multiplicative identity for rational numbers is 1.
Reason: Multiplying any rational number \(a\) by 1 leaves the number unchanged (\(a \times 1 = a\)).
In simple words: One is the multiplicative identity because multiplying by 1 keeps the number the same.

Exam Tip: Remember that 0 is additive identity and 1 is multiplicative identity.

Question 6. Find the multiplicative inverse of (i) \(\frac{3}{11}\) (ii) \(-2\frac{2}{7}\) (iii) \(\frac{-9}{-2}\) (iv) \(1\frac{8}{9}\).
Answer:
(i) Multiplicative inverse of \(\frac{3}{11}\) is \(\frac{11}{3}\).
(ii) \(-2\frac{2}{7} = -\frac{16}{7}\), so its multiplicative inverse is \(-\frac{7}{16}\).
(iii) \(\frac{-9}{-2} = \frac{9}{2}\), so its multiplicative inverse is \(\frac{2}{9}\).
(iv) \(1\frac{8}{9} = \frac{17}{9}\), so its multiplicative inverse is \(\frac{9}{17}\).
In simple words: Flip the fraction upside down while keeping the same sign to find the multiplicative inverse.

Exam Tip: The sign of a number never changes when finding its multiplicative inverse (reciprocal).

Question 7. Add \(\frac{4}{13}\) with the multiplicative inverse of \(5\frac{1}{5}\).
Answer:
First, convert \(5\frac{1}{5}\) to an improper fraction: \(\frac{26}{5}\).
Its multiplicative inverse is \(\frac{5}{26}\).
Now, add \(\frac{4}{13}\) to \(\frac{5}{26}\):
Make denominators equal (LCM of 13 and 26 is 26):
\(\frac{4}{13} = \frac{8}{26}\)
Sum = \(\frac{8}{26} + \frac{5}{26} = \frac{13}{26} = \frac{1}{2}\)
In simple words: Find the reciprocal of the mixed fraction, then add it to the given fraction using a common denominator.

Exam Tip: Always reduce the final fraction to its simplest form.

Question 8. Add \(-\frac{2}{5} \times \frac{15}{4}\) by the reciprocal of \(\frac{4}{13} \times \frac{-39}{5}\).
Answer:
First part: \(-\frac{2}{5} \times \frac{15}{4} = \frac{-30}{20} = -\frac{3}{2}\)
Second part: \(\frac{4}{13} \times \frac{-39}{5} = \frac{4 \times (-39)}{13 \times 5} = \frac{-156}{65} = -\frac{12}{5}\)
Reciprocal of the second part = \(-\frac{5}{12}\)
Now, add the first part and the reciprocal: 
\(-\frac{3}{2} + \left(-\frac{5}{12}\right)\)
LCM of 2 and 12 is 12.
\(= -\frac{18}{12} - \frac{5}{12} = -\frac{23}{12}\)
In simple words: Calculate both expressions, find the reciprocal of the second result, and add them together.

Exam Tip: Simplify multiplications before taking reciprocals.

Question 9. Is 0.15 the multiplicative inverse of \(-6\frac{2}{3}\)?
Answer:
Convert 0.15 to a fraction: \(0.15 = \frac{15}{100} = \frac{3}{20}\)
Convert \(-6\frac{2}{3}\) to an improper fraction: \(-\frac{20}{3}\)
Multiply them to check if the product is 1:
\(\frac{3}{20} \times \left(-\frac{20}{3}\right) = -1\)
Since the product is \(-1\) (not \(+1\)), 0.15 is not the multiplicative inverse of \(-6\frac{2}{3}\) (it is the additive inverse or negative reciprocal).
In simple words: Two numbers are multiplicative inverses only if their product is exactly 1.

Exam Tip: Always test the product of two numbers to verify if they are multiplicative inverses.

Question 10. Add the multiplicative inverses of \(\frac{3}{8}\) and \(\frac{7}{12}\).
Answer:
Multiplicative inverse of \(\frac{3}{8}\) is \(\frac{8}{3}\).
Multiplicative inverse of \(\frac{7}{12}\) is \(\frac{12}{7}\).
Sum = \(\frac{8}{3} + \frac{12}{7}\)
LCM of 3 and 7 is 21.
\(= \frac{8 \times 7}{21} + \frac{12 \times 3}{21}\)
\(= \frac{56 + 36}{21} = \frac{92}{21}\)
In simple words: Flip both fractions to find their reciprocals, then add them using a common denominator.

Exam Tip: Multiplicative inverse means reciprocal.

Question 11. Multiply the additive inverses of \(\frac{3}{8}\) and \(\frac{7}{12}\).
Answer:
Additive inverse of \(\frac{3}{8}\) is \(-\frac{3}{8}\).
Additive inverse of \(\frac{7}{12}\) is \(-\frac{7}{12}\).
Product = \(\left(-\frac{3}{8}\right) \times \left(-\frac{7}{12}\right)\)
\(= \frac{(-3) \times (-7)}{8 \times 12} = \frac{21}{96}\)
Simplifying by dividing numerator and denominator by 3:
\(= \frac{7}{32}\)
In simple words: Change the signs of both numbers to get their additive inverses, then multiply them.

Exam Tip: Additive inverse changes the sign; product of two negatives is positive.

Question 12. Verify \(-(-x) = x\) for \(x = \frac{7}{12}\).
Answer:
Given \(x = \frac{7}{12}\).
First negative: \(-x = -\frac{7}{12}\)
Second negative: \(-(-x) = -\left(-\frac{7}{12}\right) = \frac{7}{12}\)
Since \(-(-x) = x = \frac{7}{12}\), it is verified.
In simple words: Applying a negative sign twice brings you back to the original positive number.

Exam Tip: A negative of a negative is always a positive.

Question 13. For \(a = \frac{2}{3}, b = -\frac{5}{6},\) and \(c = \frac{1}{2}\), prove that \(a \times (b + c) = a \times b + a \times c\).
Answer: This verifies the distributive property of multiplication over addition.
Left Hand Side (LHS): \(a \times (b + c)\)
\(= \frac{2}{3} \times \left(-\frac{5}{6} + \frac{1}{2}\right)\)
LCM of 6 and 2 is 6:
\(-\frac{5}{6} + \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3}\)
\(= \frac{2}{3} \times \left(-\frac{1}{3}\right) = -\frac{2}{9}\)
Right Hand Side (RHS): \(a \times b + a \times c\)
\(= \left(\frac{2}{3} \times -\frac{5}{6}\right) + \left(\frac{2}{3} \times \frac{1}{2}\right)\)
\(= \left(-\frac{10}{18}\right) + \left(\frac{2}{6}\right)\)
\(= -\frac{5}{9} + \frac{1}{3}\)
LCM of 9 and 3 is 9:
\(= -\frac{5}{9} + \frac{3}{9} = -\frac{2}{9}\)
Since LHS = RHS = \(-\frac{2}{9}\), the distributive property is proven.
In simple words: Multiplying a number by a sum gives the same result as multiplying by each addend and adding the products.

Exam Tip: Always show both sides separately when asked to prove an identity.

Question 14. Using appropriate properties of rational numbers, find \(\frac{7}{11} + \left(\frac{5}{7} + \frac{3}{22}\right)\).
Answer:
Rearrange and group terms with common or related denominators using commutativity and associativity:
\(= \left(\frac{7}{11} + \frac{3}{22}\right) + \frac{5}{7}\)
For the first group (LCM of 11 and 22 is 22):
\(\frac{7 \times 2}{22} + \frac{3}{22} = \frac{14 + 3}{22} = \frac{17}{22}\)
Now add \(\frac{5}{7}\):
\(= \frac{17}{22} + \frac{5}{7}\)
LCM of 22 and 7 is 154:
\(= \frac{17 \times 7}{154} + \frac{5 \times 22}{154}\)
\(= \frac{119 + 110}{154} = \frac{229}{154}\)
In simple words: Group fractions with related denominators together to make addition easier.

Exam Tip: Look for denominators that are multiples of each other (like 11 and 22).

Question 15. Find \(-\frac{2}{5} \times \left(\frac{5}{3} \times \frac{15}{22}\right)\).
Answer:
First, simplify inside the brackets:
\(\frac{5}{3} \times \frac{15}{22} = \frac{5 \times 15}{3 \times 22} = \frac{75}{66} = \frac{25}{22}\)
Now, multiply by the outer term:
\(-\frac{2}{5} \times \frac{25}{22}\)
Cancel common factors: 2 and 22 cancel to give 1 and 11; 5 and 25 cancel to give 1 and 5.
\(= -\frac{1 \times 5}{1 \times 11} = -\frac{5}{11}\)
In simple words: Multiply the terms inside the bracket first, then cross-cancel and multiply by the remaining number.

Exam Tip: Cross-cancel numbers before multiplying to keep calculations simple.

Question 16. Represent the rational number \(\frac{12}{5}\) on the number line.
Answer:
1. Convert \(\frac{12}{5}\) to a mixed fraction: \(2\frac{2}{5}\).
2. Draw a number line with points marked at equal intervals representing integers 0, 1, 2, 3.
3. Divide each unit segment into 5 equal parts.
4. Count 12 parts to the right of 0 (or 2 full units plus 2 small parts) and mark \(\frac{12}{5}\).

12/5 3 In simple words: Convert the improper fraction to a mixed number to see which two integers it lies between.

Exam Tip: The denominator (5) indicates that each unit is divided into 5 equal subdivisions.

Question 17. Represent the rational number \(\frac{13}{6}\) on the number line.
Answer:
1. Convert to mixed fraction: \(\frac{13}{6} = 2\frac{1}{6}\).
2. The number lies between 2 and 3.
3. Divide the unit interval between 2 and 3 into 6 equal parts.
4. Mark 1 part to the right of 2 to represent \(\frac{13}{6}\).

13/6 3 In simple words: Locate the integer 2, then move 1 subdivision forward out of 6 equal parts towards 3.

Exam Tip: Clearly label integer points on the number line for accurate plotting.

Question 18. Represent the rational numbers \(\frac{2}{9}, \frac{5}{9}, \frac{8}{9}\) on a number line.
Answer:
1. All numbers lie between 0 and 1 since their numerators are smaller than the denominator 9.
2. Divide the segment between 0 and 1 into 9 equal parts.
3. Mark the 2nd, 5th, and 8th subdivisions respectively to represent \(\frac{2}{9}, \frac{5}{9},\) and \(\frac{8}{9}\).

In simple words: Since all fractions are between 0 and 1, divide the unit interval into 9 parts and plot accordingly.

Exam Tip: Multiple fractions with the same denominator can be plotted on the same single unit line segment.

Question 19. Find two rational numbers between 0 and \(\frac{7}{9}\).
Answer:
Write 0 as \(\frac{0}{9}\).
We need two rational numbers between \(\frac{0}{9}\) and \(\frac{7}{9}\).
Two such numbers are \(\frac{1}{9}\) and \(\frac{2}{9}\) (or any numbers up to \(\frac{6}{9}\)).
In simple words: Pick any two numerators between 0 and 7 while keeping the denominator as 9.

Exam Tip: Choosing numbers directly from numerators is the quickest method when one boundary is zero.

Question 20. Find five rational numbers between \(\frac{3}{5}\) and \(\frac{2}{3}\).
Answer:
Make denominators equal using the LCM of 5 and 3, which is 15.
\(\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}\)
\(\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\)
To find 5 rational numbers, multiply numerator and denominator by \(5 + 1 = 6\) (or 10):
\(\frac{9 \times 10}{15 \times 10} = \frac{90}{150}\)
\(\frac{10 \times 10}{15 \times 10} = \frac{100}{150}\)
Five rational numbers between \(\frac{90}{150}\) and \(\frac{100}{150}\) are: 
\(\frac{91}{150}, \frac{92}{150}, \frac{93}{150}, \frac{94}{150}, \frac{95}{150}\).
In simple words: Expand the equivalent fractions by a large factor to create enough room for 5 numbers in between.

Exam Tip: Multiplying by 10 or 100 makes generating multiple rational numbers straightforward.

Question 21. Find four rational numbers between \(\frac{7}{9}\) and \(\frac{1}{3}\).
Answer:
First, make denominators equal using the LCM of 9 and 3, which is 9.
\(\frac{1}{3} = \frac{3}{9}\) and \(\frac{7}{9} = \frac{7}{9}\)
Four rational numbers between \(\frac{3}{9}\) and \(\frac{7}{9}\) are: 
\(\frac{4}{9}, \frac{5}{9}, \frac{6}{9}\) (and we can expand to find more if needed, e.g., using denominator 18: \(\frac{7}{18}, \frac{9}{18}, \frac{11}{18}, \frac{13}{18}\)).
In simple words: Convert to a common denominator and pick four fractions lying between them.

Exam Tip: Check denominator compatibility before finding intermediate values.

Question 22. Find ten rational numbers between \(-\frac{1}{10}\) and \(-\frac{2}{5}\).
Answer:
Make denominators equal using the LCM of 10 and 5, which is 10.
\(-\frac{1}{10} = -\frac{1}{10}\)
\(-\frac{2}{5} = -\frac{2 \times 2}{5 \times 2} = -\frac{4}{10}\)
To find 10 rational numbers, multiply numerator and denominator by a larger factor, say 4:
\(-\frac{1}{10} = -\frac{4}{40}\)
\(-\frac{4}{10} = -\frac{16}{40}\)
Ten rational numbers between \(-\frac{4}{40}\) and \(-\frac{16}{40}\) are:
\(-\frac{5}{40}, -\frac{6}{40}, -\frac{7}{40}, -\frac{8}{40}, -\frac{9}{40}, -\frac{10}{40}, -\frac{11}{40}, -\frac{12}{40}, -\frac{13}{40}, -\frac{14}{40}\).
In simple words: Scale up the denominators sufficiently to fit 10 numbers between the numerators.

Exam Tip: When asked for 10 numbers, scaling by 10 or more guarantees enough space.

Question 23. Find five rational numbers less than -3.
Answer:
Write -3 as a fraction with denominator 1: \(-\frac{3}{1}\).
We can write -3 as \(-\frac{15}{5}\).
Five rational numbers less than -3 are: 
\(-4, -5, -6, -7, -8\) (or fractional equivalents like \(-\frac{16}{5}, -\frac{17}{5}, -\frac{18}{5}, -\frac{19}{5}, -\frac{20}{5}\)).
In simple words: Any numbers further to the left on the number line than -3 are less than -3.

Exam Tip: For negative numbers, larger absolute values represent smaller numbers.

Question 24. Find ten rational numbers greater than 4.
Answer:
Write 4 as a fraction with denominator 1: \(\frac{4}{1}\) or \(\frac{40}{10}\).
Ten rational numbers greater than 4 are:
\(4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5.0\) (or in fractional form: \(\frac{41}{10}, \frac{42}{10}, \frac{43}{10}, \dots, \frac{50}{10}\)).
In simple words: Any numbers to the right of 4 on the number line are greater than 4.

Exam Tip: Decimals provide an easy way to list numbers greater than an integer.