Rational Numbers

Q1: Multiple Choice Type:

i. The number which on multiplying with \(5\frac{2}{3}\) gives \(-1\frac{2}{3}\) is:
Let the required number be \(x\)
We are given: \[ x \times 5\frac{2}{3} = -1\frac{2}{3} \] Convert mixed numbers to improper fractions: \[ x \times \frac{17}{3} = \frac{-5}{3} \] Now solve for \(x\): \[ x = \frac{-5}{3} \div \frac{17}{3} \] \[ x = \frac{-5}{3} \times \frac{3}{17} \] \[ x = \frac{-15}{51} = \frac{-5}{17} \] Correct Answer: (c) \(-\frac{5}{17}\)

ii. If a, b and c are three rational numbers, we have:
We apply the distributive property of multiplication over subtraction: \[ a \times (b – c) = a \times b – a \times c \] Correct Answer: (c)

iii. The product of a positive rational number and its reciprocal is:
Let the number be \(\frac{a}{b}\), where \(a > 0,\ b > 0\)
Then its reciprocal is \(\frac{b}{a}\)

Product: \[ \frac{a}{b} \times \frac{b}{a} = \frac{ab}{ba} = 1 \] Correct Answer: (c) 1

iv. The area of a rectangular paper is \(7\frac{1}{3}\) cm2. If its length \(4\frac{2}{5}\) cm, its breadth is:
Let breadth = \(x\)
Area = Length × Breadth \[ 7\frac{1}{3} = 4\frac{2}{5} \times x \] Convert mixed numbers to improper fractions: \[ \frac{22}{3} = \frac{22}{5} \times x \] Now solve for \(x\): \[ x = \frac{22}{3} \div \frac{22}{5} = \frac{22}{3} \times \frac{5}{22} = \frac{5}{3} = 1\frac{2}{3} \text{ cm} \] Correct Answer: (b) \(1\frac{2}{3}\) cm

v. The product of a rational number \(\frac{2}{7}\) and its additive inverse is:
Additive inverse of \(\frac{2}{7}\) is \(-\frac{2}{7}\)
\[ \frac{2}{7} \times \left(-\frac{2}{7}\right) = \frac{-4}{49} \] Correct Answer: (b) \(-\frac{4}{49}\)

Q2: Evaluate

i. \(\frac{-14}{5}\times\frac{-6}{7}\)
Step-by-step: \[ \frac{-14}{5} \times \frac{-6}{7} = \frac{(-14) \times (-6)}{5 \times 7} = \frac{84}{35} \] Simplify: \[ \frac{84}{35} = \frac{12}{5} \] Answer: \(\frac{12}{5}\)

ii. \(\frac{7}{6}\times\frac{-18}{91}\)
Step-by-step: \[ \frac{7}{6} \times \frac{-18}{91} = \frac{7 \times (-18)}{6 \times 91} = \frac{-126}{546} \] Simplify:
Divide numerator and denominator by 42: \[ \frac{-126 \div 42}{546 \div 42} = \frac{-3}{13} \] Answer: \(\frac{-3}{13}\)

iii. \(\frac{-125}{72}\times\frac{9}{-5}\)
Step-by-step:
Two negatives make a positive: \[ \frac{-125}{72} \times \frac{9}{-5} = \frac{125 \times 9}{72 \times 5} = \frac{1125}{360} \] Simplify:
Divide numerator and denominator by 45: \[ \frac{1125 \div 45}{360 \div 45} = \frac{25}{8} \] Answer: \(\frac{25}{8}\)

iv. \(\frac{-11}{9}\times\frac{-51}{44}\)
Step-by-step:
Two negatives make a positive: \[ \frac{-11}{9} \times \frac{-51}{-44} = \frac{-561}{396} \] Simplify:
Divide numerator and denominator by 33: \[ \frac{-561 \div 33}{396 \div 33} = \frac{-17}{12} \] Answer: \(\frac{-17}{12}\)

v. \(-\frac{16}{5}\times\frac{20}{8}\)
Step-by-step: \[ -\frac{16}{5} \times \frac{20}{8} = \frac{-320}{40} = -8 \] Answer: \(-8\)

Q3: Multiply

i. \(\frac{5}{6}\ and\ \frac{8}{9}\)
Step-by-step: \[ \frac{5}{6} \times \frac{8}{9} = \frac{5 \times 8}{6 \times 9} = \frac{40}{54} \] Simplify: \[ \frac{40}{54} = \frac{20}{27} \] (Divide both by 2)
Answer: \(\frac{20}{27}\)

ii. \(\frac{2}{7}\ and\ \frac{-14}{9}\)
Step-by-step: \[ \frac{2}{7} \times \frac{-14}{9} = \frac{2 \times -14}{7 \times 9} = \frac{-28}{63} \] Simplify: \[ \frac{-28}{63} = \frac{-4}{9} \] (Divide both by 7)
Answer: \(\frac{-4}{9}\)

iii. \(\frac{-7}{8}\ and\ 4\)
Step-by-step: \[ \frac{-7}{8} \times 4 = \frac{-7 \times 4}{8} = \frac{-28}{8} = \frac{-7}{2} \] Answer: \(\frac{-7}{2}\)

iv. \(\frac{36}{-7}\ and\ \frac{-9}{28}\)
Step-by-step:
Two negatives make a positive: \[ \frac{36}{-7} \times \frac{-9}{28} = \frac{36 \times 9}{7 \times 28} = \frac{324}{196} \] Simplify:
Divide numerator and denominator by 4: \[ \frac{324 \div 4}{196 \div 4} = \frac{81}{49} \] Answer: \(\frac{81}{49}\)

v. \(\frac{-7}{10}\ and\ \frac{-8}{15}\)
Step-by-step: \[ \frac{-7}{10} \times \frac{-8}{15} = \frac{56}{150} \] Simplify: \[ \frac{56}{150} = \frac{28}{75} \] (Divide by 2)
Answer: \(\frac{28}{75}\)

Q4: Evaluate

i. \(\left(\frac{2}{-3}\times\frac{5}{4}\right)+\left(\frac{5}{9}\times\frac{3}{-10}\right)\)
Step-by-step: \[ \frac{2}{-3} \times \frac{5}{4} = \frac{-10}{12} = \frac{-5}{6} \] \[ \frac{5}{9} \times \frac{3}{-10} = \frac{-15}{90} = \frac{-1}{6} \] Now add both: \[ \frac{-5}{6} + \frac{-1}{6} = \frac{-6}{6} = -1 \] Answer: \(-1\)

ii. \(\left(2\times\frac{1}{4}\right)-\left(\frac{-18}{7}\times\frac{-7}{15}\right)\)
Step-by-step: \[ 2 \times \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \] \[ \frac{-18}{7} \times \frac{-7}{15} = \frac{126}{105} = \frac{6}{5} \] (Simplified)
Now subtract: \[ \frac{1}{2} – \frac{6}{5} = \frac{5 – 12}{10} = \frac{-7}{10} \] Answer: \(\frac{-7}{10}\)

iii. \(\left(-5\times\frac{2}{15}\right)-\left(-6\times\frac{2}{9}\right)\)
Step-by-step: \[ -5 \times \frac{2}{15} = \frac{-10}{15} = \frac{-2}{3} \] \[ -6 \times \frac{2}{9} = \frac{-12}{9} = \frac{-4}{3} \] Now subtract: \[ \frac{-2}{3} – (\frac{-4}{3}) = \frac{-2}{3} + \frac{4}{3} = \frac{2}{3} \] Answer: \(\frac{2}{3}\)

iv. \(\left(\frac{8}{5}\times\frac{-3}{2}\right)+\left(\frac{-3}{10}\times\frac{9}{16}\right)\)
Step-by-step: \[ \frac{8}{5} \times \frac{-3}{2} = \frac{-24}{10} = \frac{-12}{5} \] \[ \frac{-3}{10} \times \frac{9}{16} = \frac{-27}{160} \] Now add:
Convert both to common denominator (LCM = 160): \[ \frac{-12}{5} = \frac{-384}{160} \] \[ \frac{-384}{160} + \frac{-27}{160} = \frac{-411}{160} \] Answer: \(\frac{-411}{160}\)

Q5: Multiply each rational number, given below, by one (1):

i. \(\frac{7}{-5}\)
Step-by-step:
Any number multiplied by 1 gives the number itself: \[ \frac{7}{-5} \times 1 = \frac{7}{-5} \] Answer: \(\frac{7}{-5}\)

ii. \(\frac{-3}{-4}\)
Step-by-step: \[ \frac{-3}{-4} \times 1 = \frac{-3}{-4} = \frac{3}{4} \] (Negative signs cancel)
Answer: \(\frac{3}{4}\)

iii. 0
Step-by-step:
Zero multiplied by any number is always zero: \[ 0 \times 1 = 0 \] Answer: \(0\)

iv. \(\frac{-8}{13}\)
Step-by-step: \[ \frac{-8}{13} \times 1 = \frac{-8}{13} \] Answer: \(\frac{-8}{13}\)

v. \(\frac{-6}{-7}\)
Step-by-step: \[ \frac{-6}{-7} \times 1 = \frac{-6}{-7} = \frac{6}{7} \] (Negative signs cancel)
Answer: \(\frac{6}{7}\)

Q6: For each pair of rational numbers. given below, verify that the multiplication is commutative:

i. \(\frac{-1}{5}\ and\ \frac{2}{9}\)
Check: \(\frac{-1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{-1}{5}\)

LHS: \[ \frac{-1}{5} \times \frac{2}{9} = \frac{-2}{45} \] RHS: \[ \frac{2}{9} \times \frac{-1}{5} = \frac{-2}{45} \] Yes, multiplication is commutative.

ii. \(\frac{5}{-3}\ and\frac{13}{-11}\)
Check: \(\frac{5}{-3} \times \frac{13}{-11} = \frac{13}{-11} \times \frac{5}{-3}\)

LHS: \[ \frac{5}{-3} \times \frac{13}{-11} = \frac{65}{33} \] RHS: \[ \frac{13}{-11} \times \frac{5}{-3} = \frac{65}{33} \] Yes, multiplication is commutative.

iii. \(3\ and\frac{-8}{9}\)
Check: \(3 \times \frac{-8}{9} = \frac{-8}{9} \times 3\)

LHS: \[ 3 \times \frac{-8}{9} = \frac{-24}{9} \] RHS: \[ \frac{-8}{9} \times 3 = \frac{-24}{9} \] Yes, multiplication is commutative.

iv. \(0\ and\frac{-12}{17}\)
Check: \(0 \times \frac{-12}{17} = \frac{-12}{17} \times 0\)

LHS: \[ 0 \times \frac{-12}{17} = 0 \] RHS: \[ \frac{-12}{17} \times 0 = 0 \] Yes, multiplication is commutative.

Q7: Write the reciprocal (multiplicative inverse) of each rational number given below:

i. 5
Reciprocal of a number \(x\) is \(\frac{1}{x}\) \[ \text{Reciprocal of } 5 = \frac{1}{5} \] Answer: \(\frac{1}{5}\)

ii. -3 \[ \text{Reciprocal of } -3 = \frac{1}{-3} = -\frac{1}{3} \] Answer: \(-\frac{1}{3}\)

iii. \(\frac{5}{11}\) \[ \text{Reciprocal of } \frac{5}{11} = \frac{11}{5} \] Answer: \(\frac{11}{5}\)

iv. \(\frac{-7}{-8}\)
Note: \(\frac{-7}{-8} = \frac{7}{8}\) (negative signs cancel out) \[ \text{Reciprocal of } \frac{7}{8} = \frac{8}{7} \] Answer: \(\frac{8}{7}\)

v. \(\frac{-8}{-7}\)
Note: \(\frac{-8}{-7} = \frac{8}{7}\) \[ \text{Reciprocal of } \frac{8}{7} = \frac{7}{8} \] Answer: \(\frac{7}{8}\)

Q8: Find the reciprocal (multiplicative inverse) of:

i. \(\frac{3}{5}\times\frac{2}{3}\)
Step 1: Multiply the two rational numbers \[ \frac{3}{5} \times \frac{2}{3} = \frac{3 \times 2}{5 \times 3} = \frac{6}{15} = \frac{2}{5} \] Step 2: Find the reciprocal of the product \[ \text{Reciprocal of } \frac{2}{5} = \frac{5}{2} \] Answer: \(\frac{5}{2}\)

ii. \(\frac{-8}{3}\times\frac{13}{-7}\)
Step 1: Multiply the two rational numbers \[ \frac{-8}{3} \times \frac{13}{-7} = \frac{-8 \times 13}{3 \times -7} = \frac{-104}{-21} = \frac{104}{21} \] Step 2: Find the reciprocal of the product \[ \text{Reciprocal of } \frac{104}{21} = \frac{21}{104} \] Answer: \(\frac{21}{104}\)

iii. \(\frac{-3}{5}\times\frac{-1}{13}\)
Step 1: Multiply the two rational numbers \[ \frac{-3}{5} \times \frac{-1}{13} = \frac{3 \times 1}{5 \times 13} = \frac{3}{65} \] Step 2: Find the reciprocal of the product \[ \text{Reciprocal of } \frac{3}{65} = \frac{65}{3} \] Answer: \(\frac{65}{3}\)

Q9: Verify that \(\left(x+y\right)\times\ z=x\times\ z+y\times\ z\), if

i. \(x=\frac{4}{5},\ y=\frac{-2}{3}\ and\ z=-4\)
Step 1: Calculate LHS → \((x + y) \times z\) \[ x + y = \frac{4}{5} + \left(\frac{-2}{3}\right) = \frac{4}{5} – \frac{2}{3} \] LCM of 5 and 3 = 15 \[ \frac{4}{5} = \frac{12}{15}, \quad \frac{2}{3} = \frac{10}{15} \] \[ x + y = \frac{12}{15} – \frac{10}{15} = \frac{2}{15} \] \[ (x + y) \times z = \frac{2}{15} \times (-4) = \frac{-8}{15} \]Step 2: Calculate RHS → \(x \times z + y \times z\) \[ x \times z = \frac{4}{5} \times (-4) = \frac{-16}{5} \] \[ y \times z = \frac{-2}{3} \times (-4) = \frac{8}{3} \] Now add the results: \[ \frac{-16}{5} + \frac{8}{3} \] LCM = 15: \[ \frac{-16}{5} = \frac{-48}{15}, \quad \frac{8}{3} = \frac{40}{15} \] \[ \frac{-48}{15} + \frac{40}{15} = \frac{-8}{15} \]Answer: Verified, LHS = RHS = \(\frac{-8}{15}\)

ii. \(x=2\ ,\ y=\frac{4}{5}\ and\ z=\frac{3}{-10}\)
Step 1: Calculate LHS → \((x + y) \times z\) \[ x + y = 2 + \frac{4}{5} = \frac{10}{5} + \frac{4}{5} = \frac{14}{5} \] \[ (x + y) \times z = \frac{14}{5} \times \left(\frac{-3}{10}\right) = \frac{-42}{50} = \frac{-21}{25} \]Step 2: Calculate RHS → \(x \times z + y \times z\) \[ x \times z = 2 \times \left(\frac{-3}{10}\right) = \frac{-6}{10} = \frac{-3}{5} \] \[ y \times z = \frac{4}{5} \times \left(\frac{-3}{10}\right) = \frac{-12}{50} = \frac{-6}{25} \] Now add the results: \[ \frac{-3}{5} = \frac{-15}{25} \] \[ \frac{-15}{25} + \frac{-6}{25} = \frac{-21}{25} \]Answer: Verified, LHS = RHS = \(\frac{-21}{25}\)

Q10: Verify that \(x\times\left(y-z\right)=x\times\ y-x\times\ z\), if

i. \(x=\frac{4}{5},\ y=\frac{7}{4}\ and\ z=3\)
Step 1: Calculate LHS → \(x \times (y – z)\) \[ y – z = \frac{7}{4} – 3 = \frac{7}{4} – \frac{12}{4} = \frac{-5}{4} \] \[ x \times (y – z) = \frac{4}{5} \times \left(\frac{-5}{4}\right) = \frac{-20}{20} = -1 \]Step 2: Calculate RHS → \(x \times y – x \times z\) \[ x \times y = \frac{4}{5} \times \frac{7}{4} = \frac{28}{20} = \frac{7}{5} \] \[ x \times z = \frac{4}{5} \times 3 = \frac{12}{5} \] \[ x \times y – x \times z = \frac{7}{5} – \frac{12}{5} = \frac{-5}{5} = -1 \]Answer: Verified, LHS = RHS = -1

ii. \(x=\frac{3}{4},\ y=\frac{8}{9}\ and\ z=-5\)
Step 1: Calculate LHS → \(x \times (y – z)\) \[ y – z = \frac{8}{9} – (-5) = \frac{8}{9} + 5 = \frac{8}{9} + \frac{45}{9} = \frac{53}{9} \] \[ x \times (y – z) = \frac{3}{4} \times \frac{53}{9} = \frac{159}{36} \]Step 2: Calculate RHS → \(x \times y – x \times z\) \[ x \times y = \frac{3}{4} \times \frac{8}{9} = \frac{24}{36} \] \[ x \times z = \frac{3}{4} \times (-5) = \frac{-15}{4} \]Now subtract: \[ \text{Convert } \frac{-15}{4} = \frac{-135}{36} \] \[ x \times y – x \times z = \frac{24}{36} – \left(\frac{-135}{36}\right) = \frac{24 + 135}{36} = \frac{159}{36} \]Answer: Verified, LHS = RHS = \(\frac{159}{36}\)

Q11: Name multiplication property of rational numbers shown below:

i. \(\frac{3}{5}\times\frac{-8}{9}=\frac{-8}{9}\times\frac{3}{5}\)
This shows that the order of multiplication does not affect the result.
Answer: Commutative Property of Multiplication

ii. \(\frac{-3}{4}\times\left(\frac{5}{7}\times\frac{-8}{15}\right)=\left(\frac{-3}{4}\times\frac{5}{7}\right)\times\frac{-8}{15}\)
This shows that how numbers are grouped (associativity) does not affect the product.
Answer: Associative Property of Multiplication

iii. \(\frac{4}{5}\times\left(\frac{3}{-8}+\frac{-4}{7}\right)=\frac{4}{5}\times\frac{3}{-8}+\frac{4}{5}\times\frac{-4}{7}\)
This shows that multiplication distributes over addition.
Answer: Distributive Property of Multiplication over Addition

iv. \(\frac{-7}{5}\times\frac{5}{-7}=1\)
The product of a number and its reciprocal is 1.
Answer: Multiplicative Inverse Property

v. \(\frac{8}{-9}\times1=1\times\frac{8}{-9}=\frac{8}{-9}\)
Multiplying any rational number by 1 leaves it unchanged.
Answer: Multiplicative Identity Property

Q12: Fill in the blanks:

i. The product of two positive rational numbers Is always ____________.
Answer: positive

ii. The product of two negative rational numbers Is always ____________.
Answer: positive

iii. If two rational numbers have opposite signs then their product is always ___________.
Answer: negative

iv. The reciprocal of a positive rational number is ____________ and the reciprocal of a negative rational number is __________.
Answer: positive, negative

v. Rational number 0 has ___________ reciprocal.
Answer: no

vi. The product of a non-zero rational number and its reciprocal is ___________.
Answer: 1

vii. The numbers ___________ and __________ are their own reciprocals.
Answer: 1 and -1

viii. If m is reciprocal of n, then the reciprocal of n is __________.
Answer: n

Q13: The length and breadth of a rectangular piece of paper are 9 cm and \(10\frac{2}{3} cm respectively. Find:

i. its area

Let length = \( 9 \) cm and breadth = \( 10\frac{2}{3} = \frac{32}{3} \) cm

Area of rectangle = Length × Breadth

\[ \text{Area} = 9 \times \frac{32}{3} = \frac{9 \times 32}{3} = \frac{288}{3} = 96 \text{ cm}^2 \]Answer: Area = 96 cm²

ii. its perimeter

Perimeter of rectangle = \(2 \times (\text{Length} + \text{Breadth})\)

Length = 9, Breadth = \(\frac{32}{3}\)

\[ \text{Perimeter} = 2 \times \left(9 + \frac{32}{3}\right) = 2 \times \left(\frac{27}{3} + \frac{32}{3}\right) = 2 \times \frac{59}{3} = \frac{118}{3} = 39\frac{1}{3} \text{ cm} \]Answer: Perimeter = \(39\frac{1}{3}\) cm

Q14: Find the area and the perimeter of a rectangular piece of land with length \(7\frac{2}{5}\) m and breadth \(4\frac{1}{6}\) m.