CBSE Class 8 Mathematics Rational Numbers Worksheet Set 08

Chapter 1 - Rational Numbers

Question 1. Fill in the blanks:
a) Rational numbers are numbers of the form \(\frac{p}{q}\) where p,q are integers and \(q \neq 0\).
b) Rational numbers are not closed under division (specifically division by zero).
c) 0 is called the additive identity of rational numbers.
d) Zero has no reciprocal.
e) The numbers 1 and -1 are their own reciprocals.
f) \(\frac{3}{10}\) is the multiplicative inverse of \(3\frac{1}{3}\).
g) The rational number that is equal to its negative is 0.
h) There are infinitely many rational numbers between any two given rational numbers.
i) Nine times the reciprocal of a number is 3. The number is 3.
j) \(\frac{7}{-7} = -1\).
Answer: Filled in all the blanks correctly based on the properties of rational numbers.
In simple words: These blanks test the fundamental definitions, identities, and properties of rational numbers.

Exam Tip: Memorize standard properties like additive identity (0) and multiplicative identity (1) to quickly solve objective questions.

Question 2. Write the additive inverse of
a) \(\frac{2}{-9}\)
b) \(\frac{-3}{5}\)
Answer:
a) The additive inverse of a number \(a\) is \(-a\).
Given number = \(\frac{2}{-9} = -\frac{2}{9}\)
Additive inverse = \(-\left(-\frac{2}{9}\right) = \frac{2}{9}\)
b) Given number = \(\frac{-3}{5}\)
Additive inverse = \(-\left(\frac{-3}{5}\right) = \frac{3}{5}\)
In simple words: To find the additive inverse, just change the sign of the given rational number.

Exam Tip: The sum of a number and its additive inverse is always equal to 0.

Question 3. Write the multiplicative inverse of
a) -1
b) \(\frac{-13}{19}\)
Answer:
a) Given number = -1
Multiplicative inverse (reciprocal) = \(\frac{1}{-1} = -1\)
b) Given number = \(\frac{-13}{19}\)
Multiplicative inverse (reciprocal) = \(\frac{19}{-13} = -\frac{19}{13}\)
In simple words: The multiplicative inverse is obtained by flipping the numerator and denominator while keeping the sign same.

Exam Tip: The product of a number and its multiplicative inverse is always equal to 1.

Question 4. Verify that -(-a) = a is true for \(a = \frac{-19}{21}\)
Answer:
Given \(a = \frac{-19}{21}\)
LHS = \(-(-a) = -\left(-\left(\frac{-19}{21}\right)\right)\)
First, evaluate the inner bracket: \(-\left(\frac{-19}{21}\right) = \frac{19}{21}\)
Now, apply the outer negative sign: \(-\left(\frac{19}{21}\right) = -\frac{19}{21}\)
RHS = \(a = -\frac{19}{21}\)
Since LHS = RHS, the property -(-a) = a is verified.
In simple words: Taking the negative of a negative number gives back the original positive or negative number itself.

Exam Tip: Double negative signs cancel each other out to form a positive sign, but count the total number of negative signs carefully.

Question 5. Verify the property: \(a \times (b + c) = (a \times b) + (a \times c)\) by taking \(a = \frac{-5}{2}, b = -2, c = \frac{11}{3}\)
Answer:
Given values: \(a = \frac{-5}{2}\), \(b = -2\), \(c = \frac{11}{3}\)
LHS = \(a \times (b + c)\)
\(= \frac{-5}{2} \times \left(-2 + \frac{11}{3}\right)\)
\(= \frac{-5}{2} \times \left(\frac{-6 + 11}{3}\right)\)
\(= \frac{-5}{2} \times \left(\frac{5}{3}\right) = \frac{-25}{6}\)
RHS = \((a \times b) + (a \times c)\)
\(= \left(\frac{-5}{2} \times -2\right) + \left(\frac{-5}{2} \times \frac{11}{3}\right)\)
\(= \frac{10}{2} + \left(\frac{-55}{6}\right) = 5 - \frac{55}{6}\)
\(= \frac{30 - 55}{6} = \frac{-25}{6}\)
Since LHS = RHS, the distributive property is verified.
In simple words: Multiplication distributes over addition, meaning you can multiply outside terms with each inside term separately and add them.

Exam Tip: Take the LCM correctly when adding fractions inside the bracket to avoid calculation errors.

Question 6. Find five rational numbers between \(\frac{-1}{2}\) and 2.
Answer:
Convert the given rational numbers to have a common denominator.
Denominator = 2
\(\frac{-1}{2}\) and \(2 = \frac{4}{2}\)
To find 5 rational numbers, multiply the numerator and denominator of both numbers by a larger factor, say 6:
\(\frac{-1 \times 6}{2 \times 6} = \frac{-6}{12}\)
\(\frac{4 \times 6}{2 \times 6} = \frac{24}{12}\)
Now, choose any five rational numbers between \(\frac{-6}{12}\) and \(\frac{24}{12}\):
\(\frac{-5}{12}, \frac{-4}{12}, \frac{0}{12}, \frac{1}{12}, \frac{2}{12}\) (or any other valid choices)
In simple words: Make denominators equal and choose numbers lying between the expanded numerators.

Exam Tip: There are infinitely many rational numbers between any two numbers; any correct set between the limits is accepted.

Question 7. Arrange in ascending order \(\frac{-3}{4}, \frac{2}{-12}, \frac{-9}{16}, \frac{7}{-24}\)
Answer:
First, rewrite all rational numbers with positive denominators:
\(\frac{-3}{4}, \frac{-2}{12}, \frac{-9}{16}, \frac{-7}{24}\)
Simplify fractions where possible:
\(\frac{-3}{4}, \frac{-1}{6}, \frac{-9}{16}, \frac{-7}{24}\)
Find the LCM of denominators (4, 6, 16, 24) = 48
Convert each fraction to have 48 as the denominator:
\(\frac{-3 \times 12}{4 \times 12} = \frac{-36}{48}\)
\(\frac{-1 \times 8}{6 \times 8} = \frac{-8}{48}\)
\(\frac{-9 \times 3}{16 \times 3} = \frac{-27}{48}\)
\(\frac{-7 \times 2}{24 \times 2} = \frac{-14}{48}\)
Compare the numerators (-36, -8, -27, -14):
\(-36 < -27 < -14 < -8\)
Therefore, the ascending order is:
\(\frac{-36}{48} < \frac{-27}{48} < \frac{-14}{48} < \frac{-8}{48}\)
Original form: \(\frac{-3}{4}, \frac{-9}{16}, \frac{7}{-24}, \frac{2}{-12}\)
In simple words: Make all denominators equal using LCM, then compare numerators from smallest to largest.

Exam Tip: For negative numbers, the number with the larger absolute value is actually smaller.

Question 8. Represent \(\frac{-3}{4}, \frac{7}{6}, \frac{9}{-11}\) on the number line.
Answer:
1. For \(\frac{-3}{4}\): It lies between 0 and -1 on the negative side. Divide the unit space between 0 and -1 into 4 equal parts and take the 3rd mark to the left of 0.
2. For \(\frac{7}{6} = 1\frac{1}{6}\): It lies between 1 and 2 on the positive side. Divide the unit space between 1 and 2 into 6 equal parts and take the 1st mark after 1.
3. For \(\frac{9}{-11} = -\frac{9}{11}\): It lies between 0 and -1 on the negative side. Divide the unit space between 0 and -1 into 11 equal parts and take the 9th mark to the left of 0.
In simple words: Convert improper fractions to mixed fractions to locate the whole number integers they fall between, then split the segment according to the denominator.

Exam Tip: Always make the denominator positive before plotting on the number line.

Question 9. Find \(\frac{3}{7} + \left(\frac{-6}{11}\right) + \left(\frac{-8}{21}\right) + \left(\frac{5}{22}\right)\)
Answer:
Group terms with common or related denominators:
\(= \left[\frac{3}{7} + \left(\frac{-8}{21}\right)\right] + \left[\left(\frac{-6}{11}\right) + \frac{5}{22}\right]\)
For the first group, LCM of 7 and 21 is 21:
\(\frac{3 \times 3}{7 \times 3} - \frac{8}{21} = \frac{9 - 8}{21} = \frac{1}{21}\)
For the second group, LCM of 11 and 22 is 22:
\(\frac{-6 \times 2}{11 \times 2} + \frac{5}{22} = \frac{-12 + 5}{22} = \frac{-7}{22}\)
Now add both results: \(\frac{1}{21} - \frac{7}{22}\)
LCM of 21 and 22 = \(21 \times 22 = 462\)
\(= \frac{1 \times 22}{462} - \frac{7 \times 21}{462}\)
\(= \frac{22 - 147}{462} = \frac{-125}{462}\)
In simple words: Rearrange fractions with related denominators together to simplify addition and subtraction.

Exam Tip: Use commutativity and associativity to group terms intelligently rather than finding a massive LCM for all four denominators at once.

Question 10. Using appropriate properties, find
i) \(\frac{2}{5} \times \frac{-3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5}\)
ii) \(\frac{1}{2} \times \frac{1}{4} - \frac{1}{2} \times \frac{3}{5} + \frac{7}{2} \times \frac{1}{4}\)
Answer:
i) Rearrange using commutative property:
\(= \frac{2}{5} \times \frac{-3}{7} - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14}\)
Take out common term \(\frac{3}{7}\) using distributive property:
\(= \frac{-3}{7} \times \left(\frac{2}{5} + \frac{3}{5}\right) - \frac{1}{14}\)
\(= \frac{-3}{7} \times \left(\frac{5}{5}\right) - \frac{1}{14}\)
\(= \frac{-3}{7} \times 1 - \frac{1}{14} = \frac{-3}{7} - \frac{1}{14}\)
Make denominators equal (LCM = 14):
\(= \frac{-3 \times 2}{14} - \frac{1}{14} = \frac{-6 - 1}{14} = \frac{-7}{14} = \frac{-1}{2}\)

ii) Rearrange terms using commutative and distributive properties:
\(= \left(\frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{7}{2} \times \frac{1}{4}\right) - \left(\frac{1}{2} \times \frac{3}{5}\right)\)
Factor out \(\frac{1}{4}\) from the first two terms:
\(= \frac{1}{4} \times \left(\frac{1}{2} + \frac{7}{2}\right) - \left(\frac{1}{2} \times \frac{3}{5}\right)\)
\(= \frac{1}{4} \times \left(\frac{8}{2}\right) - \frac{3}{10}\)
\(= \frac{1}{4} \times 4 - \frac{3}{10}\)
\(= 1 - \frac{3}{10} = \frac{10 - 3}{10} = \frac{7}{10}\)
In simple words: Use rearrangement and distributive properties to factor out common multipliers, making complex arithmetic much easier.

Exam Tip: Always look for common factors across different terms to apply the distributive property before performing direct multiplication.