CBSE Class 8 Mathematics Rational Numbers Worksheet Set 05

Chapter 1. Rational Numbers

Question 1. The product of a number and its multiplicative inverse is _______.
Answer: 1
In simple words: When any non-zero number is multiplied by its reciprocal, the result is always 1.

Exam Tip: Remember that zero has no multiplicative inverse.

Question 2. Sum of a number and its negative is _______.
Answer: 0
In simple words: Adding a number and its additive inverse (negative) cancels them out to give zero.

Exam Tip: The negative of a number \(x\) is \(-x\), and \(x + (-x) = 0\).

Question 3. _______ is the multiplicative identity for rational numbers.
Answer: 1
In simple words: Multiplying any rational number by 1 leaves the number unchanged.

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

Question 4. _______ is the additive identity for rational numbers.
Answer: 0
In simple words: Adding 0 to any rational number gives the same rational number back.

Exam Tip: 0 is the identity element for addition.

Question 5. The reciprocal of a negative rational number is _______.
Answer: a negative rational number
In simple words: The reciprocal keeps the same sign as the original number.

Exam Tip: Reciprocal of \(\frac{-a}{b}\) is \(\frac{-b}{a}\).

Question 6. There are _______ number of rational numbers between any two numbers.
Exam Tip: Use the mean method or make denominators equal to find rational numbers between two given numbers.

Question 7. Additive inverse of 0 is _______.
Answer: 0
In simple words: Zero is its own additive inverse because \(0 + 0 = 0\).

Exam Tip: Zero is neither positive nor negative.

Question 8. Rational Number _______ has no reciprocal.
Answer: 0
In simple words: Division by zero is undefined, so 0 cannot have a reciprocal (\(\frac{1}{0}\) is undefined).

Exam Tip: 1 and -1 are the only rational numbers that are their own reciprocals.

Question 9. The numbers _______ and _______ are their own reciprocals.
Answer: 1 and -1
In simple words: The reciprocal of 1 is 1, and the reciprocal of -1 is -1.

Exam Tip: Always check both positive and negative possibilities.

Question 10. The rational number that is equal to its negative is _______.
Answer: 0
In simple words: Zero is the only number that equals its own negative (\(0 = -0\)).

Question 11(a). \(-\frac{2}{3} \times \frac{3}{5} + \frac{7}{2} - \frac{2}{5} \times \frac{1}{6}\)
Answer:
Rearranging using commutativity and distributivity:
\(= -\frac{2}{3} \times \frac{3}{5} - \frac{2}{5} \times \frac{1}{6} + \frac{7}{2}\)
\(= -\frac{2}{5} \left( \frac{3}{3} + \frac{1}{6} \right) + \frac{7}{2}\)
\(= -\frac{2}{5} \left( 1 + \frac{1}{6} \right) + \frac{7}{2}\)
\(= -\frac{2}{5} \left( \frac{7}{6} \right) + \frac{7}{2}\)
\(= -\frac{14}{30} + \frac{7}{2} = -\frac{7}{15} + \frac{7}{2}\)
\(= \frac{-14 + 105}{30} = \frac{91}{30}\)
In simple words: Group common terms using distributive property to simplify calculations.

Exam Tip: Look for common factors like \(-\frac{2}{5}\) across terms to apply the distributive law.

Question 11(b). \(\frac{1}{2} \times \frac{1}{4} + \left(-\frac{7}{18}\right) \times \frac{15}{7} - \frac{1}{4} \times \frac{1}{3}\)
Answer:
Group terms with common factors:\ \(= \frac{1}{4} \times \frac{1}{2} - \frac{1}{4} \times \frac{1}{3} + \left(-\frac{7}{18} \times \frac{15}{7}\right)\)
\(= \frac{1}{4} \left( \frac{1}{2} - \frac{1}{3} \right) - \frac{5}{6}\)
\(= \frac{1}{4} \left( \frac{3 - 2}{6} \right) - \frac{5}{6}\)
\(= \frac{1}{4} \left( \frac{1}{6} \right) - \frac{5}{6}\)
\(= \frac{1}{24} - \frac{5}{6} = \frac{1 - 20}{24} = -\frac{19}{24}\)
In simple words: Rearrange terms to group common multipliers together.

Exam Tip: Simplify fractions inside products before combining terms.

Question 11(c). \(-\frac{5}{4} \times \left( \frac{8}{5} + 16 \right)\)
Answer:
Using distributive property \(a(b + c) = ab + ac\):
\(= \left( -\frac{5}{4} \times \frac{8}{5} \right) + \left( -\frac{5}{4} \times 16 \right)\)
\(= -2 - 20 = -22\)
In simple words: Multiply the outer term with each term inside the bracket separately.

Exam Tip: Distributive property often eliminates fractions early.

Question 11(d). \(\frac{7}{12} \times \left(-\frac{3}{5}\right) + \left(-\frac{5}{3}\right) \times \frac{1}{12}\)
Answer:
Rewrite the second term to find a common factor:\ \(= \frac{7}{12} \times \left(-\frac{3}{5}\right) + \left(-\frac{1}{12}\right) \times \frac{5}{3}\) (Wait, let us use direct multiplication or rearranging)
\(= -\frac{21}{60} - \frac{5}{36} = -\frac{7}{20} - \frac{5}{36}\)
LCD of 20 and 36 is 180:
\(= \frac{-7(9) - 5(5)}{180} = \frac{-63 - 25}{180} = -\frac{88}{180} = -\frac{22}{45}\)
In simple words: Multiply each product first and then add the fractions using LCM.

Exam Tip: Always reduce fractions to their lowest terms at the end.

Question 11(e). \(\frac{2}{21} \times \frac{-3}{13} + \frac{-7}{9} - \frac{2}{21} \times \frac{10}{13}\)
Answer:
Rearranging using commutative and distributive property:
\(= \frac{2}{21} \times \left(-\frac{3}{13}\right) - \frac{2}{21} \times \frac{10}{13} - \frac{7}{9}\)
\(= \frac{2}{21} \left( -\frac{3}{13} - \frac{10}{13} \right) - \frac{7}{9}\)
\(= \frac{2}{21} \left( -\frac{13}{13} \right) - \frac{7}{9}\)
\(= \frac{2}{21} (-1) - \frac{7}{9} = -\frac{2}{21} - \frac{7}{9}\)
LCM of 21 and 9 is 63:
\(= \frac{-2(3) - 7(7)}{63} = \frac{-6 - 49}{63} = -\frac{55}{63}\)
In simple words: Combine terms with the common multiplier \(\frac{2}{21}\).

Exam Tip: Factoring out common terms makes arithmetic much simpler.

Question 11(f). \(-\frac{4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \left(-\frac{14}{9}\right)\)
Answer:
Multiply all numerators and denominators while canceling common factors:
\(= \frac{(-4) \times 3 \times 15 \times (-14)}{5 \times 7 \times 16 \times 9}\)
Cancel terms: \(5\) and \(15\) gives \(3\); \(3\) and \(9\) gives \(3\); \(3\) and \(3\) cancel out.
\(-4\) and \(16\) gives \(4\) in denominator.
\(-14\) and \(7\) gives \(-2\).
Numerator left: \((-1) \times 1 \times 1 \times (-2) = 2\)
Denominator left: \(1 \times 1 \times 4 \times 1 = 4\)
\(= \frac{2}{4} = \frac{1}{2}\)
In simple words: Cancel out common factors across numerators and denominators before final multiplication.

Question 12. Represent \(-\frac{2}{5}\) and \(\frac{4}{5}\) on a number line.
Answer:
Divide the number line between integers into 5 equal parts (since denominator is 5).
-\(\frac{2}{5}\) lies 2 units to the left of 0.
\(\frac{4}{5}\) lies 4 units to the right of 0.
In simple words: Mark equal divisions between 0 and 1, and 0 and -1 according to the denominator.

Exam Tip: Ensure all segments between integers are spaced equally.

Question 13. Verify that \(-(-x) = x\) for \(x = -\frac{2}{7}\).
Answer:
Given \(x = -\frac{2}{7}\).
LHS = \(-(-x) = -\left(-\left(-\frac{2}{7}\right)\right)\)
Since \(-\left(-\frac{2}{7}\right) = \frac{2}{7}\),
LHS = \(-\left(\frac{2}{7}\right) = -\frac{2}{7}\)
RHS = \(x = -\frac{2}{7}\)
Since LHS = RHS, verified.
In simple words: Applying the negative sign twice brings back the original number.

Exam Tip: Two negative signs adjacent to each other make a positive.

Question 14. Is \(-\frac{4}{17}\) the multiplicative inverse of \(-4\frac{1}{4}\)? Why or why not?
Answer:
Convert mixed fraction: \(-4\frac{1}{4} = -\frac{17}{4}\).
Multiply the two numbers: \(-\frac{4}{17} \times \left(-\frac{17}{4}\right) = 1\).
Yes, \(-\frac{4}{17}\) is the multiplicative inverse because their product is 1.
In simple words: Two numbers are multiplicative inverses if their product equals 1.

Exam Tip: Always convert mixed fractions to improper fractions before checking reciprocals.

Question 15. Is \(0.5\) the multiplicative inverse of \(2\)? Why or why not?
Answer:
\(0.5 = \frac{1}{2}\).
Multiply \(0.5\) by \(2\): \(\frac{1}{2} \times 2 = 1\).
Yes, \(0.5\) is the multiplicative inverse of \(2\) because their product is 1.
In simple words: Since \(0.5\) is equal to \(\frac{1}{2}\), its reciprocal is 2.

Exam Tip: Decimals can be converted to fractions to easily check reciprocals.

Question 16. Write five rational numbers greater than \(-5\).
Answer:
Any five numbers greater than \(-5\), for example: \(-4, -3, -2, -1, 0\) (or fractional values like \(-4.5, -4, -\frac{7}{2}, -3, -2\)).
In simple words: Any number lying to the right of \(-5\) on the number line is greater.

Exam Tip: For negative numbers, a smaller absolute value represents a greater number.

Question 17. Find five rational numbers between (a) \(0\) and \(-2\) (b) \(-\frac{3}{7}\) and \(\frac{5}{11}\) (c) \(-3\) and \(-4\).
Answer:
(a) Between \(0\) and \(-2\):
Write as fractions with denominator 6: \(0 = \frac{0}{6}\), \(-2 = -\frac{12}{6}\).
Five rational numbers: \(-\frac{1}{6}, -\frac{2}{6}, -\frac{3}{6}, -\frac{4}{6}, -\frac{5}{6}\).

(b) Between \(-\frac{3}{7}\) and \(\frac{5}{11}\):
Make denominators equal (LCM = 77):
\(-\frac{3}{7} = -\frac{33}{77}\) and \(\frac{5}{11} = \frac{35}{77}\).
Five rational numbers: \(-\frac{2}{77}, -\frac{1}{77}, 0, \frac{1}{77}, \frac{2}{77}\).

(c) Between \(-3\) and \(-4\):
Write with denominator 6: \(-3 = -\frac{18}{6}\), \(-4 = -\frac{24}{6}\).
Five rational numbers: \(-\frac{19}{6}, -\frac{20}{6}, -\frac{21}{6}, -\frac{22}{6}, -\frac{23}{6}\).
In simple words: Convert the numbers to have a common large denominator to easily insert intermediate fractions.

Exam Tip: Multiply numerator and denominator by \((n+1)\) where \(n\) is the number of rational numbers required.

Question 18. Find the multiplicative inverse of \(-2 \times -\frac{3}{4}\).
Answer:
First simplify the expression:
\(-2 \times \left(-\frac{3}{4}\right) = \frac{6}{4} = \frac{3}{2}\).
Multiplicative inverse of \(\frac{3}{2}\) is \(\frac{2}{3}\).
In simple words: Simplify the given expression first, then flip the numerator and denominator.

Exam Tip: Never find the reciprocal before simplifying the expression fully.

Question 19. Find the additive inverse of \(-\frac{2}{9} + \frac{7}{9}\).
Answer:
First simplify the expression:
\(-\frac{2}{9} + \frac{7}{9} = \frac{-2 + 7}{9} = \frac{5}{9}\).
Additive inverse of \(\frac{5}{9}\) is \(-\frac{5}{9}\).
In simple words: Additive inverse means changing the sign of the evaluated result.

Exam Tip: Additive inverse changes positive to negative and negative to positive.

Question 20. Multiply \(\frac{8}{13}\) by the reciprocal of \(-\frac{17}{26}\).
Answer:
Reciprocal of \(-\frac{17}{26}\) is \(-\frac{26}{17}\).
Multiply by \(\frac{8}{13}\):
\(\frac{8}{13} \times \left(-\frac{26}{17}\right) = \frac{8 \times (-2)}{1} \times \frac{1}{17} = -\frac{16}{17}\)
In simple words: Find the reciprocal of the second number and multiply it directly with the first number.

Question 21. Reciprocal of \(-\frac{1}{y}\), where \(y \neq 0\) is _______.
Answer: \(-y$
In simple words: Flipping \(-\frac{1}{y}\) gives \(-y\).

Exam Tip: The reciprocal of \(\frac{a}{b}\) is \(\frac{b}{a}\).

Question 22. The product of two negative rational nos is always a _______ rational number.
Answer: positive
In simple words: Negative multiplied by negative yields a positive number.

Exam Tip: Even number of negative signs in multiplication results in a positive product.

Question 23. \(-\frac{25}{23} \div 0 =\)
Answer: Not defined (or Undefined)
In simple words: Division by zero is mathematically not possible.

Exam Tip: Any number divided by zero is undefined.

Question 24. \(0 \div \left(-\frac{25}{23}\right) =\)
Answer: 0
In simple words: Zero divided by any non-zero rational number is always zero.

Question 25(a). \(-\frac{25}{29} \times \frac{29}{-25} = 1\)
Answer: Multiplicative Inverse Property
In simple words: A number multiplied by its reciprocal gives 1.

Exam Tip: Also known as the inverse property of multiplication.

Question 25(b). \(\frac{8}{5} + 0 = 0 + \frac{8}{5} = \frac{8}{5}\)
Answer: Additive Identity Property
In simple words: Adding zero to a number leaves it unchanged.

Exam Tip: 0 is the additive identity element.

Question 25(c). \(-\frac{2}{9} \left( \frac{3}{5} + \frac{2}{9} \right) = -\frac{2}{9} \times \frac{3}{5} + \left(-\frac{2}{9}\right) \times \frac{2}{9}\)
Answer: Distributive Property of Multiplication over Addition
In simple words: Multiplying a number by a sum is the same as multiplying by each addend separately.

Exam Tip: Formula: \(a(b + c) = ab + ac\).