ML Aggarwal Class 8 Maths Solutions Chapter 01 Rational Numbers

Exercise 1.1

Question 1. Add the following:
(i) \( \frac{4}{7} \) and \( \frac{5}{7} \)
(ii) \( \frac{7}{-13} \) and \( \frac{4}{-13} \)
Answer:
(i) When two fractions share the same denominator, combine the numerators and keep the denominator unchanged.
\( \frac{4}{7} + \frac{5}{7} = \frac{4 + 5}{7} = \frac{9}{7} \)
(ii) First, rewrite each fraction in standard form. For \( \frac{7}{-13} \), multiply both top and bottom by -1:
\( \frac{7}{-13} = \frac{-7}{13} \)
Similarly, \( \frac{4}{-13} = \frac{-4}{13} \)
Now add:
\( \frac{-7}{13} + \frac{-4}{13} = \frac{-7 - 4}{13} = \frac{-11}{13} \)
In simple words: When adding fractions with the same bottom number, just add the top numbers. If the denominator is negative, flip both the top and bottom to make it positive first.

Exam Tip: Always ensure the denominator is positive before adding. Group terms with matching denominators to simplify the process.

Question 2. Simplify:
(i) \( \frac{5}{11} + 4\frac{3}{9} \)
(ii) \( -\frac{4}{9} + 3\frac{5}{8} \)
Answer:
(i) Convert the mixed number to an improper fraction: \( 4\frac{3}{9} = \frac{39}{9} \)
Now rewrite the problem:
\( \frac{5}{11} + \frac{39}{9} \)
Find the LCM of 11 and 9, which is 99. Convert both fractions:
\( \frac{5}{11} = \frac{5 \times 9}{11 \times 9} = \frac{45}{99} \)
\( \frac{39}{9} = \frac{39 \times 11}{9 \times 11} = \frac{429}{99} \)
Add them:
\( \frac{45}{99} + \frac{429}{99} = \frac{474}{99} \)
Reduce by dividing both by 3:
\( \frac{474 \div 3}{99 \div 3} = \frac{158}{33} \)
(ii) Convert the mixed number: \( 3\frac{5}{8} = \frac{29}{8} \)
Rewrite: \( -\frac{4}{9} + \frac{29}{8} \)
The LCM of 9 and 8 is 72. Convert both:
\( -\frac{4}{9} = \frac{-4 \times 8}{9 \times 8} = \frac{-32}{72} \)
\( \frac{29}{8} = \frac{29 \times 9}{8 \times 9} = \frac{261}{72} \)
Add:
\( \frac{-32}{72} + \frac{261}{72} = \frac{229}{72} \)
In simple words: Change mixed numbers into improper fractions first. Then find a common bottom number for both fractions. Add the top numbers and reduce the answer if possible.

Exam Tip: Always convert mixed numbers to improper fractions before finding the LCM. Check if your final answer can be reduced further by finding the GCD.

Question 3. Verify commutative property of addition for the following pairs of rational numbers.
(i) \( -\frac{4}{3} \) and \( \frac{3}{7} \)
(ii) \( -\frac{2}{-5} \) and \( \frac{1}{3} \)
(iii) \( \frac{9}{11} \) and \( \frac{2}{13} \)
Answer:
(i) Calculate \( -\frac{4}{3} + \frac{3}{7} \):
LCM of 3 and 7 is 21.
\( -\frac{4}{3} + \frac{3}{7} = \frac{-28 + 9}{21} = \frac{-19}{21} \)
Now calculate \( \frac{3}{7} + (-\frac{4}{3}) \):
\( \frac{3}{7} + (-\frac{4}{3}) = \frac{9 - 28}{21} = \frac{-19}{21} \)
Both give the same result, confirming the commutative property: \( -\frac{4}{3} + \frac{3}{7} = \frac{3}{7} + (-\frac{4}{3}) \)
(ii) Rewrite \( -\frac{2}{-5} = \frac{2}{5} \)
Calculate \( \frac{2}{5} + \frac{1}{3} \):
LCM of 5 and 3 is 15.
\( \frac{2}{5} + \frac{1}{3} = \frac{6 + 5}{15} = \frac{11}{15} \)
Now calculate \( \frac{1}{3} + \frac{2}{5} \):
\( \frac{1}{3} + \frac{2}{5} = \frac{5 + 6}{15} = \frac{11}{15} \)
Verified: \( \frac{2}{5} + \frac{1}{3} = \frac{1}{3} + \frac{2}{5} \)
(iii) Calculate \( \frac{9}{11} + \frac{2}{13} \):
LCM of 11 and 13 is 143.
\( \frac{9}{11} + \frac{2}{13} = \frac{117 + 22}{143} = \frac{139}{143} \)
Now calculate \( \frac{2}{13} + \frac{9}{11} \):
\( \frac{2}{13} + \frac{9}{11} = \frac{22 + 117}{143} = \frac{139}{143} \)
Verified: \( \frac{9}{11} + \frac{2}{13} = \frac{2}{13} + \frac{9}{11} \)
In simple words: The commutative property says that changing the order of two numbers when adding them does not change the answer. No matter which fraction you add first, you always get the same result.

Exam Tip: Show both orders of addition clearly and ensure your LCM calculations are correct. This property holds for all rational numbers.

Question 4. Find the additive inverse of the following rational numbers:
(i) \( \frac{2}{-3} \)
(ii) \( \frac{-7}{-12} \)
Answer:
(i) The additive inverse of a number is the value that, when added to the original number, gives zero. For \( \frac{2}{-3} \):
First rewrite in standard form: \( \frac{2}{-3} = -\frac{2}{3} \)
The additive inverse is: \( -(-\frac{2}{3}) = \frac{2}{3} \)
(ii) Rewrite \( \frac{-7}{-12} \) in standard form: \( \frac{-7}{-12} = \frac{7}{12} \)
The additive inverse is: \( -\frac{7}{12} \)
In simple words: To find the additive inverse, just flip the sign. If the number is positive, make it negative. If it is negative, make it positive. When you add a number and its additive inverse, you always get zero.

Exam Tip: Always write fractions in standard form (positive denominator) before finding the additive inverse. The additive inverse is unique for each rational number.

Question 5. Verify that \( -(-x) = x \) for
(i) \( x = \frac{10}{13} \)
(ii) \( x = -\frac{15}{17} \)
Answer:
(i) Given \( x = \frac{10}{13} \):
\( -x = -\frac{10}{13} \)
\( -(-x) = -(-\frac{10}{13}) = \frac{10}{13} = x \)
Verified: \( -(-x) = x \)
(ii) Given \( x = -\frac{15}{17} \):
\( -x = \frac{15}{17} \)
\( -(-x) = -(\frac{15}{17}) = -\frac{15}{17} = x \)
Verified: \( -(-x) = x \)
In simple words: The opposite of the opposite of any number always brings you back to that same number. This works for all rational numbers, whether positive or negative.

Exam Tip: This property (called the double negative rule) is fundamental to algebra. It works universally for all rational, real, and even complex numbers.

Question 6. Using appropriate properties of addition, find the following:
(i) \( \frac{4}{5} + \frac{11}{7} + (-\frac{7}{5}) + (-\frac{2}{7}) \)
(ii) \( \frac{3}{7} + \frac{4}{9} + (-\frac{5}{21}) + \frac{2}{3} \)
Answer:
(i) Rearrange by grouping fractions with the same denominator (using the commutative and associative properties):
\( [\frac{4}{5} + (-\frac{7}{5})] + [\frac{11}{7} + (-\frac{2}{7})] \)
\( = \frac{4 - 7}{5} + \frac{11 - 2}{7} \)
\( = \frac{-3}{5} + \frac{9}{7} \)
Find the LCM of 5 and 7, which is 35:
\( = \frac{-21}{35} + \frac{45}{35} = \frac{24}{35} \)
(ii) Regroup:\br />\( [\frac{3}{7} + (-\frac{5}{21})] + [\frac{4}{9} + \frac{2}{3}] \)
For the first group, LCM of 7 and 21 is 21:
\( \frac{9}{21} + (-\frac{5}{21}) = \frac{4}{21} \)
For the second group, LCM of 9 and 3 is 9:
\( \frac{4}{9} + \frac{6}{9} = \frac{10}{9} \)
Now add \( \frac{4}{21} + \frac{10}{9} \). LCM of 21 and 9 is 63:
\( \frac{12}{63} + \frac{70}{63} = \frac{82}{63} \)
Reduce: \( = 1\frac{19}{63} \)
In simple words: Group fractions that have matching bottom numbers. Add those pairs first, then add the final results together. This makes the work much easier.

Exam Tip: Use the commutative property to rearrange terms so matching denominators sit together. This reduces calculation errors and saves time.

Question 7. Fill in the blanks:
(i) \( (-\frac{4}{9}) + (\frac{2}{7}) \) is a __________ number
(ii) \( (\frac{43}{89}) + (-\frac{51}{47}) = \) __________ \( + (\frac{43}{89}) \)
(iii) \( \frac{2}{7} + \) __________ \( = \frac{2}{7} = 0 + \) __________
(iv) \( \frac{4}{11} + \{(-\frac{7}{12}) + \frac{9}{10}\} = \{(\frac{4}{11}) + (-\frac{7}{12})\} + \) __________
(v) \( \frac{5}{9} + \) __________ \( = 0 = (-\frac{5}{9}) + \) __________
Answer:
(i) rational
(ii) \( (-\frac{51}{47}) \) (This demonstrates the commutative property)
(iii) \( 0 \); \( \frac{2}{7} \) (This demonstrates the identity property where 0 is the additive identity)
(iv) \( \frac{9}{10} \) (This demonstrates the associative property)
(v) \( (-\frac{5}{9}) \); \( \frac{5}{9} \) (This demonstrates the existence of additive inverse)
In simple words: These blanks test your knowledge of addition properties: the sum of two rational numbers is always rational; you can swap the order (commutative); adding zero keeps the number the same (identity); changing how you group terms does not change the result (associative); and every number has an opposite (inverse).

Exam Tip: Write the property name next to each answer. This shows the examiner that you understand not just the calculation, but the mathematical principle behind it.

Question 8. If \( a = -\frac{11}{27} \), \( b = \frac{4}{9} \) and \( c = -\frac{5}{18} \), then verify that \( a + (b + c) = (a + b) + c \)
Answer:
Given: \( a = -\frac{11}{27} \), \( b = \frac{4}{9} \), \( c = -\frac{5}{18} \)

Calculate L.H.S. (Left Hand Side): \( a + (b + c) \)
First find \( b + c = \frac{4}{9} + (-\frac{5}{18}) \)
LCM of 9 and 18 is 18:
\( = \frac{8}{18} - \frac{5}{18} = \frac{3}{18} \)
Now add \( a \):
\( a + (b + c) = -\frac{11}{27} + \frac{3}{18} \)
LCM of 27 and 18 is 54:
\( = \frac{-22}{54} + \frac{9}{54} = \frac{-13}{54} \)

Calculate R.H.S. (Right Hand Side): \( (a + b) + c \)
First find \( a + b = -\frac{11}{27} + \frac{4}{9} \)
LCM of 27 and 9 is 27:
\( = \frac{-11}{27} + \frac{12}{27} = \frac{1}{27} \)
Now add \( c \):
\( (a + b) + c = \frac{1}{27} + (-\frac{5}{18}) \)
LCM of 27 and 18 is 54:
\( = \frac{2}{54} - \frac{15}{54} = \frac{-13}{54} \)

Since L.H.S. = R.H.S. = \( \frac{-13}{54} \), the associative property of addition is verified.
In simple words: When you add three numbers, it does not matter which two you add first. You always get the same final answer. This is the associative property.

Exam Tip: Show all LCM calculations clearly and work through one side completely before moving to the other. Always verify both sides are identical before concluding.

Exercise 1.2

Question 1. Subtract:
(i) \( 2\frac{3}{5} \) from \( -\frac{3}{7} \)
(ii) \( -\frac{4}{9} \) from \( 3\frac{5}{8} \)
(iii) \( -3\frac{1}{5} \) from \( -4\frac{7}{9} \)
Answer:
(i) Convert \( 2\frac{3}{5} = \frac{13}{5} \)
Subtract: \( -\frac{3}{7} - \frac{13}{5} \)
LCM of 7 and 5 is 35:
\( = \frac{-15}{35} - \frac{91}{35} = \frac{-106}{35} = -3\frac{1}{35} \)
(ii) Convert \( 3\frac{5}{8} = \frac{29}{8} \)
Subtract: \( \frac{29}{8} - (-\frac{4}{9}) = \frac{29}{8} + \frac{4}{9} \)
LCM of 8 and 9 is 72:
\( = \frac{261}{72} + \frac{32}{72} = \frac{293}{72} = 4\frac{5}{72} \)
(iii) Convert \( -3\frac{1}{5} = -\frac{16}{5} \) and \( -4\frac{7}{9} = -\frac{43}{9} \)
Subtract: \( -\frac{43}{9} - (-\frac{16}{5}) = -\frac{43}{9} + \frac{16}{5} \)
LCM of 9 and 5 is 45:
\( = \frac{-215}{45} + \frac{144}{45} = \frac{-71}{45} = -1\frac{26}{45} \)
In simple words: Convert all mixed numbers to improper fractions first. When subtracting a negative, change it to adding a positive. Find a common denominator, then subtract the numerators.

Exam Tip: Always change "subtract a negative" to "add a positive" before doing any calculations. This prevents sign errors.

Question 2. Sum of two rational numbers is \( \frac{3}{5} \). If one of them is \( -\frac{2}{7} \), find the other.
Answer:
Let the unknown number be \( x \).
Given: \( x + (-\frac{2}{7}) = \frac{3}{5} \)
To find \( x \), subtract \( -\frac{2}{7} \) from both sides:
\( x = \frac{3}{5} - (-\frac{2}{7}) = \frac{3}{5} + \frac{2}{7} \)
LCM of 5 and 7 is 35:
\( = \frac{21}{35} + \frac{10}{35} = \frac{31}{35} \)
The other number is \( \frac{31}{35} \).
In simple words: If you know the sum and one number, subtract the known number from the sum to find the missing number.

Exam Tip: Set up the equation clearly and remember that subtracting a negative is the same as adding a positive.

Question 3. What rational number should be added to \( -\frac{5}{11} \) to get \( -\frac{7}{8} \)?
Answer:
Let the required number be \( y \).
According to the problem: \( -\frac{5}{11} + y = -\frac{7}{8} \)
Solve for \( y \) by subtracting \( -\frac{5}{11} \) from both sides:
\( y = -\frac{7}{8} - (-\frac{5}{11}) = -\frac{7}{8} + \frac{5}{11} \)
LCM of 8 and 11 is 88:
\( = \frac{-77}{88} + \frac{40}{88} = \frac{-37}{88} \)
Therefore, \( -\frac{37}{88} \) should be added to \( -\frac{5}{11} \) to get \( -\frac{7}{8} \).
In simple words: Set up an equation with the unknown number. Then rearrange to find what that number must be by subtracting the known value from the target value.

Exam Tip: Always verify your answer by adding the number you found back to the original to check if you get the target result.

Question 4. What rational number should be subtracted from \( -4\frac{3}{5} \) to get \( -3\frac{1}{2} \)?
Answer:
Convert to improper fractions: \( -4\frac{3}{5} = -\frac{23}{5} \) and \( -3\frac{1}{2} = -\frac{7}{2} \)
Let the required number be \( z \).
Set up: \( -\frac{23}{5} - z = -\frac{7}{2} \)
Rearrange to find \( z \):
\( z = -\frac{23}{5} - (-\frac{7}{2}) = -\frac{23}{5} + \frac{7}{2} \)
LCM of 5 and 2 is 10:
\( = \frac{-46}{10} + \frac{35}{10} = \frac{-11}{10} = -1\frac{1}{10} \)
Therefore, \( -\frac{11}{10} \) should be subtracted from \( -4\frac{3}{5} \) to get \( -3\frac{1}{2} \).
In simple words: When you subtract a negative number, it acts like adding a positive. Set up the equation, rearrange, and solve step by step.

Exam Tip: Be very careful with signs. Subtracting a negative becomes adding a positive, which students often overlook.

Question 5. Subtract the sum of \( -\frac{5}{7} \) and \( -\frac{8}{3} \) from the sum of \( \frac{5}{2} \) and \( -\frac{11}{12} \).
Answer:
Step 1: Find the sum of \( -\frac{5}{7} \) and \( -\frac{8}{3} \)
LCM of 7 and 3 is 21:
\( (-\frac{5}{7}) + (-\frac{8}{3}) = \frac{-15}{21} + \frac{-56}{21} = \frac{-71}{21} \)

Step 2: Find the sum of \( \frac{5}{2} \) and \( -\frac{11}{12} \)
LCM of 2 and 12 is 12:
\( \frac{5}{2} + (-\frac{11}{12}) = \frac{30}{12} - \frac{11}{12} = \frac{19}{12} \)

Step 3: Subtract the first sum from the second sum:
\( \frac{19}{12} - (-\frac{71}{21}) = \frac{19}{12} + \frac{71}{21} \)
LCM of 12 and 21 is 84:
\( = \frac{133}{84} + \frac{284}{84} = \frac{417}{84} = 4\frac{81}{84} \)
In simple words: First, add the two negative fractions together. Then, add the two other fractions together. Finally, subtract the first result from the second result. Watch out - subtracting a negative becomes addition.

Exam Tip: Break the problem into three steps. Calculate each sum separately, then perform the final subtraction. Double-check your LCM calculations.

Question 6. If \( x = -\frac{4}{7} \) and \( y = \frac{2}{5} \), then verify that \( x - y \neq y - x \)
Answer:
Calculate \( x - y \):
\( -\frac{4}{7} - \frac{2}{5} \)
LCM of 7 and 5 is 35:
\( = \frac{-20}{35} - \frac{14}{35} = \frac{-34}{35} \)

Calculate \( y - x \):
\( \frac{2}{5} - (-\frac{4}{7}) = \frac{2}{5} + \frac{4}{7} \)
LCM of 5 and 7 is 35:
\( = \frac{14}{35} + \frac{20}{35} = \frac{34}{35} \)

Since \( -\frac{34}{35} \neq \frac{34}{35} \), we have verified that \( x - y \neq y - x \).
In simple words: Subtraction is not commutative. The order matters. Switching the order gives you a different answer - in fact, it gives you the opposite sign.

Exam Tip: When verifying inequalities, show both calculations fully and clearly state they are not equal. This property is called "non-commutativity of subtraction."

Question 7. If \( x = \frac{4}{9} \), \( y = -\frac{7}{12} \) and \( z = -\frac{2}{3} \), then verify that \( x - (y - z) \neq (x - y) - z \)
Answer:
Calculate L.H.S.: \( x - (y - z) \)
First find \( y - z = -\frac{7}{12} - (-\frac{2}{3}) = -\frac{7}{12} + \frac{2}{3} \)
LCM of 12 and 3 is 12:
\( = \frac{-7}{12} + \frac{8}{12} = \frac{1}{12} \)
Now subtract from \( x \):
\( x - (y - z) = \frac{4}{9} - \frac{1}{12} \)
LCM of 9 and 12 is 36:
\( = \frac{16}{36} - \frac{3}{36} = \frac{13}{36} \)

Calculate R.H.S.: \( (x - y) - z \)
First find \( x - y = \frac{4}{9} - (-\frac{7}{12}) = \frac{4}{9} + \frac{7}{12} \)
LCM of 9 and 12 is 36:
\( = \frac{16}{36} + \frac{21}{36} = \frac{37}{36} \)
Now subtract \( z \):
\( (x - y) - z = \frac{37}{36} - (-\frac{2}{3}) = \frac{37}{36} + \frac{2}{3} \)
LCM of 36 and 3 is 36:
\( = \frac{37}{36} + \frac{24}{36} = \frac{61}{36} \)

Since \( \frac{13}{36} \neq \frac{61}{36} \), we have verified that \( x - (y - z) \neq (x - y) - z \).
In simple words: Subtraction is not associative. Changing which operation you do first gives a completely different answer. Always work from left to right, or use parentheses to show what order you want.

Exam Tip: Be extremely careful with parentheses and the order of operations. Show all intermediate steps so the examiner can follow your logic.

Question 8. Which of the following statements is true / false?
(i) \( \frac{2}{3} - \frac{4}{5} \) is not a rational number.
(ii) \( -\frac{5}{7} \) is the additive inverse of \( \frac{5}{7} \).
(iii) 0 is the additive inverse of itself.
(iv) Commutative property holds for subtraction of rational numbers.
(v) Associative property does not hold for subtraction of rational numbers.
(vi) 0 is the identity element for subtraction of rational numbers.
Answer:
(i) False. Calculate \( \frac{2}{3} - \frac{4}{5} \). LCM of 3 and 5 is 15: \( \frac{10}{15} - \frac{12}{15} = -\frac{2}{15} \), which is a rational number.
(ii) True. The additive inverse of \( \frac{5}{7} \) is the number that, when added to it, gives zero. \( \frac{5}{7} + (-\frac{5}{7}) = 0 \), so \( -\frac{5}{7} \) is correct.
(iii) True. Adding 0 to itself: \( 0 + 0 = 0 \). The additive inverse of 0 is 0 itself.
(iv) False. Let \( a = \frac{5}{4} \) and \( b = \frac{3}{4} \). Then \( a - b = \frac{2}{4} \) but \( b - a = -\frac{2}{4} \). Since \( \frac{2}{4} \neq -\frac{2}{4} \), the property does not hold.
(v) True. From the calculation in Question 7 above, we showed that \( x - (y - z) \neq (x - y) - z \). The associative property does not work for subtraction.
(vi) False. An identity element is a value that, when used in an operation, leaves the number unchanged. For subtraction, \( a - 0 = a \) (true), but \( 0 - a = -a \) (not equal to \( a \) unless \( a = 0 \)). Since both directions must work, 0 is not an identity element for subtraction.
In simple words: Subtraction is very different from addition. Two numbers subtracted in different orders give opposite answers. Also, subtracting from zero is not the same as subtracting zero from a number.

Exam Tip: State whether each is true or false clearly, then justify with either a calculation or a counterexample. This shows complete understanding, not just memory.

Exercise 1.3

Question 1. Multiply and express the result in the lowest form:
(i) \( \frac{6}{-7} \times \frac{14}{30} \)
(ii) \( 6\frac{2}{3} \times 1\frac{2}{7} \)
(iii) \( \frac{25}{-9} \times \frac{-3}{10} \)
Answer:
(i) Multiply the numerators and denominators:
\( \frac{6}{-7} \times \frac{14}{30} = \frac{6 \times 14}{-7 \times 30} = \frac{84}{-210} \)
Find the GCD of 84 and 210, which is 42:
\( = \frac{84 \div 42}{-210 \div 42} = \frac{2}{-5} \)
Rewrite with a positive denominator:
\( = -\frac{2}{5} \)
(ii) Convert to improper fractions: \( 6\frac{2}{3} = \frac{20}{3} \) and \( 1\frac{2}{7} = \frac{9}{7} \)
Multiply:
\( \frac{20}{3} \times \frac{9}{7} = \frac{20 \times 9}{3 \times 7} = \frac{180}{21} \)
Find the GCD of 180 and 21, which is 3:
\( = \frac{180 \div 3}{21 \div 3} = \frac{60}{7} = 8\frac{4}{7} \)
(iii) Multiply:
\( \frac{25}{-9} \times \frac{-3}{10} = \frac{25 \times (-3)}{-9 \times 10} = \frac{-75}{-90} \)
Find the GCD of 75 and 90, which is 15:
\( = \frac{-75 \div (-15)}{-90 \div (-15)} = \frac{5}{6} \)
In simple words: Multiply the top numbers together and the bottom numbers together. Then reduce by dividing both by their greatest common divisor. Always keep the denominator positive.

Exam Tip: You can simplify before multiplying by canceling common factors from numerators and denominators. This makes the arithmetic easier and reduces mistakes.

Question 2. Verify commutative property of multiplication for the following pairs of rational numbers:
(i) \( \frac{4}{5} \) and \( -\frac{7}{8} \)
(ii) \( 13\frac{1}{3} \) and \( 1\frac{1}{8} \)
(iii) \( -\frac{7}{-20} \) and \( \frac{5}{-14} \)
Answer:
(i) Calculate \( \frac{4}{5} \times (-\frac{7}{8}) \):
\( = \frac{4 \times (-7)}{5 \times 8} = \frac{-28}{40} \)
Now calculate \( (-\frac{7}{8}) \times \frac{4}{5} \):
\( = \frac{(-7) \times 4}{8 \times 5} = \frac{-28}{40} \)
Both give the same result, so the property is verified: \( \frac{4}{5} \times (-\frac{7}{8}) = (-\frac{7}{8}) \times \frac{4}{5} \)
(ii) Convert: \( 13\frac{1}{3} = \frac{40}{3} \) and \( 1\frac{1}{8} = \frac{9}{8} \)
Calculate \( \frac{40}{3} \times \frac{9}{8} \):
\( = \frac{40 \times 9}{3 \times 8} = \frac{360}{24} = 15 \)
Calculate \( \frac{9}{8} \times \frac{40}{3} \):
\( = \frac{9 \times 40}{8 \times 3} = \frac{360}{24} = 15 \)
Both give 15, so the property is verified: \( \frac{40}{3} \times \frac{9}{8} = \frac{9}{8} \times \frac{40}{3} \)
(iii) Simplify \( -\frac{7}{-20} = \frac{7}{20} \)
Calculate \( \frac{7}{20} \times \frac{5}{-14} \):
\( = \frac{7 \times 5}{20 \times (-14)} = \frac{35}{-280} = -\frac{1}{8} \)
Calculate \( \frac{5}{-14} \times \frac{7}{20} \):
\( = \frac{5 \times 7}{-14 \times 20} = \frac{35}{-280} = -\frac{1}{8} \)
Both give \( -\frac{1}{8} \), so the property is verified: \( \frac{7}{20} \times \frac{5}{-14} = \frac{5}{-14} \times \frac{7}{20} \)
In simple words: The commutative property says that changing the order of two numbers in multiplication does not change the answer. You get the same product regardless of which fraction you multiply first.

Exam Tip: Show both multiplication orders clearly and simplify to the same lowest form in both cases to demonstrate they are equal.

Question 3. Verify the following and name the property also:
(i) \( \frac{3}{5} \times (-\frac{4}{7} \times -\frac{8}{9}) = (\frac{3}{5} \times -\frac{4}{7}) \times -\frac{8}{9} \)
(ii) \( \frac{5}{9} \times (-\frac{3}{2} + \frac{7}{5}) = \frac{5}{9} \times -\frac{3}{2} + \frac{5}{9} \times \frac{7}{5} \)
Answer:
(i) Calculate L.H.S.: \( \frac{3}{5} \times (-\frac{4}{7} \times -\frac{8}{9}) \)
First: \( (-\frac{4}{7}) \times (-\frac{8}{9}) = \frac{32}{63} \)
Then: \( \frac{3}{5} \times \frac{32}{63} = \frac{96}{315} \)
Calculate R.H.S.: \( (\frac{3}{5} \times -\frac{4}{7}) \times -\frac{8}{9} \)
First: \( \frac{3}{5} \times (-\frac{4}{7}) = -\frac{12}{35} \)
Then: \( (-\frac{12}{35}) \times (-\frac{8}{9}) = \frac{96}{315} \)
Both equal \( \frac{96}{315} \), so the equation is verified.
Property name: Associative property of multiplication
(ii) Calculate L.H.S.: \( \frac{5}{9} \times (-\frac{3}{2} + \frac{7}{5}) \)
First: \( -\frac{3}{2} + \frac{7}{5} = \frac{-15 + 14}{10} = -\frac{1}{10} \)
Then: \( \frac{5}{9} \times (-\frac{1}{10}) = -\frac{5}{90} = -\frac{1}{18} \)
Calculate R.H.S.: \( \frac{5}{9} \times (-\frac{3}{2}) + \frac{5}{9} \times \frac{7}{5} \)
First part: \( \frac{5}{9} \times (-\frac{3}{2}) = -\frac{15}{18} \)
Second part: \( \frac{5}{9} \times \frac{7}{5} = \frac{7}{9} \)
Add: \( -\frac{15}{18} + \frac{7}{9} = -\frac{15}{18} + \frac{14}{18} = -\frac{1}{18} \)
Both equal \( -\frac{1}{18} \), so the equation is verified.
Property name: Distributive property of multiplication over addition
In simple words: The associative property says that how you group numbers when multiplying does not change the answer. The distributive property says that multiplying a sum is the same as multiplying each number separately and then adding the results.

Exam Tip: Name the property by its correct mathematical term. Use the names "Associative" and "Distributive" precisely - they are not interchangeable.

Question 4. Find the multiplication inverse of the following:
(i) 12
(ii) \( \frac{2}{3} \)
(iii) \( -\frac{4}{7} \)
(iv) \( -\frac{3}{8} \times \left(-\frac{7}{13}\right) \)
Answer:
(i) The reciprocal of 12 is \( \frac{1}{12} \).
(ii) The reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \).
(iii) The reciprocal of \( -\frac{4}{7} \) is \( -\frac{7}{4} \).
(iv) First, work out \( -\frac{3}{8} \times \left(-\frac{7}{13}\right) = \frac{21}{104} \). The reciprocal of \( \frac{21}{104} \) is \( \frac{104}{21} \).
In simple words: To find the reciprocal (or multiplicative inverse) of a number, flip it upside down - put the numerator as the denominator and the denominator as the numerator.

Exam Tip: Remember that the reciprocal of a whole number like 12 is a fraction with 1 in the numerator. The signs stay the same when you flip - negative stays negative.

Question 5. Using the appropriate properties of operations of rational numbers, evaluate the following:
(i) \( \frac{2}{5} \times -\frac{3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5} \)
(ii) \( \frac{8}{9} \times \frac{4}{5} + \frac{5}{6} - \frac{9}{5} \times \frac{8}{9} \)
(iii) \( -\frac{3}{7} \times \frac{14}{15} \times \frac{7}{12} \times \left(-\frac{30}{35}\right) \)
Answer:
(i) Rearranging: \( \frac{2}{5} \times -\frac{3}{7} - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14} \). Extract the common factor \( -\frac{3}{7} \): \( -\frac{3}{7}\left(\frac{2}{5} + \frac{3}{5}\right) - \frac{1}{14} = -\frac{3}{7} \times 1 - \frac{1}{14} = -\frac{3}{7} - \frac{1}{14} \). Finding the LCM: \( \frac{-6-1}{14} = -\frac{7}{14} = -\frac{1}{2} \).

(ii) Rearranging: \( \frac{8}{9} \times \frac{4}{5} - \frac{9}{5} \times \frac{8}{9} + \frac{5}{6} \). Extract the common factor \( \frac{8}{9} \): \( \frac{8}{9}\left(\frac{4}{5} - \frac{9}{5}\right) + \frac{5}{6} = \frac{8}{9} \times \left(-1\right) + \frac{5}{6} = -\frac{8}{9} + \frac{5}{6} \). Finding the LCM: \( \frac{-16+15}{18} = -\frac{1}{18} \).

(iii) Grouping the terms: \( \left(-\frac{3}{7} \times \frac{14}{15}\right) \times \left(\frac{7}{12} \times -\frac{30}{35}\right) = -\frac{2}{5} \times -\frac{1}{2} = \frac{1}{5} \).
In simple words: Look for common factors in different parts of the problem and pull them out. This makes the problem smaller and easier to work with. Use properties like grouping and regrouping to simplify step by step.

Exam Tip: Always look for ways to factor out common terms before doing multiplication or division - this saves time and reduces mistakes. Show each step clearly to get full marks.

Question 6. If p = -\frac{8}{27}, q = \frac{3}{4} and r = -\frac{12}{15}, then verify that (i) p × (q × r) = (p × q) × r (ii) p × (q - r) = p × q - p × r
Answer:
(i) Verification of Associative Property:
L.H.S. = p × (q × r) = \( -\frac{8}{27} \times \left(\frac{3}{4} \times -\frac{12}{15}\right) = -\frac{8}{27} \times -\frac{3}{5} = \frac{24}{135} = \frac{8}{45} \).

R.H.S. = (p × q) × r = \( \left(-\frac{8}{27} \times \frac{3}{4}\right) \times -\frac{12}{15} = -\frac{2}{9} \times -\frac{12}{15} = \frac{8}{45} \).

Therefore, L.H.S. = R.H.S., proving the associative property holds.

(ii) Verification of Distributive Property:
L.H.S. = p × (q - r) = \( -\frac{8}{27} \times \left\{\frac{3}{4} - \left(-\frac{12}{15}\right)\right\} \). Finding the LCM: \( -\frac{8}{27} \times \frac{45+48}{60} = -\frac{8}{27} \times \frac{93}{60} = -\frac{62}{135} \).

R.H.S. = p × q - p × r = \( -\frac{8}{27} \times \frac{3}{4} - \left(-\frac{8}{27} \times -\frac{12}{15}\right) = -\frac{2}{9} - \frac{32}{135} = \frac{-30-32}{135} = -\frac{62}{135} \).

Therefore, L.H.S. = R.H.S., proving the distributive property holds.
In simple words: These are important rules: grouping factors in any way gives the same answer (associative), and spreading multiplication over subtraction works like spreading it over addition (distributive). These rules always work with rational numbers.

Exam Tip: Always show both sides (L.H.S. and R.H.S.) separately and clearly state "Therefore, L.H.S. = R.H.S." at the end - this is what examiners are looking for to award full marks for verification questions.

Question 7. Fill in the following blanks:
(i) \( \frac{2}{3} \times -\frac{4}{5} \) is a ....... number.
(ii) \( \frac{54}{81} \times -\frac{63}{108} = ......... \times \frac{54}{81} \)
(iii) \( \frac{4}{5} \times 1 = ...... = 1 \times ...... \)
(iv) \( \frac{5}{-12} \times ...... = 1 = -\frac{12}{5} \times ...... \)
(v) \( \frac{3}{7} \times \left(-\frac{2}{8} \times .....\right) = \left(\frac{3}{7} \times -\frac{2}{8}\right) \times \frac{5}{9} \)
(vi) \( -\frac{8}{9} \times \left\{\frac{4}{13} + \frac{5}{17}\right\} = -\frac{8}{9} \times \frac{4}{13} + ......... \)
(vii) \( -\frac{6}{13} \times \left\{\frac{8}{9} - \frac{4}{7}\right\} = -\frac{6}{13} \times ....... - \left(-\frac{6}{13}\right) \times \frac{4}{7} \)
(viii) \( \frac{16}{23} \times ......... = 0 \)
(ix) The reciprocal of 0 is ..........
(x) The numbers ....... and ....... are their own reciprocals.
(xi) If y be the reciprocal of x, then the reciprocal of y² in terms of x will be ......
(xii) The product of a non-zero rational number and its reciprocal is ........
(xiii) The reciprocal of a negative rational number is ........
Answer:
(i) rational
(ii) \( -\frac{63}{108} \)
(iii) \( \frac{4}{5} \); \( \frac{4}{5} \)
(iv) \( -\frac{12}{5} \); \( \frac{5}{-12} \)
(v) \( \frac{5}{9} \)
(vi) \( -\frac{8}{9} \times \frac{5}{17} \)
(vii) \( \frac{8}{9} \)
(viii) 0
(ix) not defined
(x) 1; -1
(xi) x²
(xii) 1
(xiii) a negative rational number
In simple words: These blanks test important facts about rational numbers: which operations give rational answers, what special numbers are involved (like 0 and 1), and what properties hold (like commutativity and distributivity).

Exam Tip: Memorise the special cases: any number times 0 equals 0, the reciprocal of 0 does not exist, and 1 and -1 are their own reciprocals. These appear frequently on exams.

Question 8. Is 4/5 the multiplicative inverse of -5/4? Why or why not?
Answer: No, \( \frac{4}{5} \) is not the multiplicative inverse of \( -\frac{5}{4} \). The multiplicative inverse of \( \frac{4}{5} \) is \( \frac{5}{4} \) (not \( -\frac{5}{4} \)). Two numbers are multiplicative inverses only if their product equals 1. Since \( \frac{4}{5} \times -\frac{5}{4} = -1 \) (not 1), these two numbers are not inverses of each other. The correct inverse of \( \frac{4}{5} \) must have the same sign and have the numerator and denominator flipped.
In simple words: Multiplicative inverses must multiply to give 1, not -1. Flipping a fraction gives you its reciprocal only if the signs are the same.

Exam Tip: Always verify by multiplying the two numbers - if you get 1, they are reciprocals; if you get anything else (like -1), they are not.

Question 9. Using distributivity, find:
(i) \( \left\{\frac{7}{5} \times \left(-\frac{3}{12}\right)\right\} + \left\{\frac{7}{5} \times \frac{5}{12}\right\} \)
(ii) \( \left\{\frac{9}{16} \times \frac{4}{12}\right\} + \left\{\frac{9}{16} \times \left(-\frac{3}{9}\right)\right\} \)
Answer:
(i) Factoring out the common term \( \frac{7}{5} \): \( \frac{7}{5} \times \left(-\frac{3}{12} + \frac{5}{12}\right) = \frac{7}{5} \times \frac{2}{12} = \frac{7}{30} \).

(ii) Factoring out the common term \( \frac{9}{16} \): \( \frac{9}{16} \times \left(\frac{4}{12} - \frac{3}{9}\right) = \frac{9}{16} \times \left(\frac{1}{3} - \frac{1}{3}\right) = \frac{9}{16} \times 0 = 0 \).
In simple words: Distributive property lets you pull out a common factor and add (or subtract) what is left. This makes the calculation faster and easier than doing each part separately.

Exam Tip: Always look for a common factor that appears in multiple terms - pulling it out reduces the work. Show the factoring step clearly to demonstrate understanding of the distributive property.

Question 10. Find the sum of additive inverse and multiplication inverse of 9.
Answer: The additive inverse (opposite) of 9 is -9. The multiplicative inverse (reciprocal) of 9 is \( \frac{1}{9} \). Adding these together: \( -9 + \frac{1}{9} = \frac{-81+1}{9} = -\frac{80}{9} \).
In simple words: The additive inverse is what you add to get zero. The multiplicative inverse is what you multiply by to get one. Here we add these two different numbers together.

Exam Tip: Do not confuse additive and multiplicative inverses - they are completely different operations. Find each one separately, then add them as instructed.

Question 11. Find the product of additive inverse and multiplicative inverse of -3/7
Answer: The additive inverse of \( -\frac{3}{7} \) is \( \frac{3}{7} \). The multiplicative inverse of \( -\frac{3}{7} \) is \( -\frac{7}{3} \). Multiplying these together: \( \frac{3}{7} \times \left(-\frac{7}{3}\right) = -1 \).
In simple words: When you flip a fraction upside down and keep the sign, you get the multiplicative inverse. When you change the sign, you get the additive inverse. Multiplying the additive inverse by the multiplicative inverse of the same number always gives -1.

Exam Tip: There is a pattern here: for any rational number a, the product of its additive inverse (-a) and multiplicative inverse (1/a) equals -1. This is a useful shortcut.

Exercise 1.4

Question 1. Find the value of the following:
(i) \( -\frac{3}{7} \div 4 \)
(ii) \( 4\frac{5}{8} \div \left(-\frac{4}{9}\right) \)
(iii) \( -\frac{8}{9} \div -\frac{3}{5} \)
Answer:
(i) \( -\frac{3}{7} \div 4 = -\frac{3}{7} \times \frac{1}{4} = -\frac{3}{28} \).

(ii) Convert \( 4\frac{5}{8} \) to an improper fraction: \( \frac{37}{8} \). Then \( \frac{37}{8} \div \left(-\frac{4}{9}\right) = \frac{37}{8} \times \frac{9}{-4} = \frac{333}{-32} = -\frac{333}{32} = -10\frac{13}{32} \).

(iii) \( -\frac{8}{9} \div -\frac{3}{5} = -\frac{8}{9} \times \frac{5}{-3} = \frac{40}{27} = 1\frac{13}{27} \).
In simple words: To divide by a fraction, flip that fraction upside down and multiply instead. This is called multiplying by the reciprocal.

Exam Tip: Always change division into multiplication by the reciprocal before computing. Do not forget to convert mixed numbers to improper fractions first.

Question 2. State whether the following statements are true or false:
(i) \( -\frac{9}{13} \div \frac{2}{7} \) is a rational number.
(ii) \( \frac{4}{13} \div \frac{11}{12} = \frac{11}{12} \div \frac{4}{13} \)
(iii) \( -\frac{3}{4} \div \left(\frac{5}{9} \div -\frac{4}{11}\right) = \left(-\frac{3}{4} \div \frac{5}{9}\right) \div -\frac{4}{11} \)
(iv) \( \frac{13}{14} \div -\frac{5}{7} \neq -\frac{5}{7} \div \frac{13}{14} \)
(v) \( \left(-7 \div \frac{4}{5}\right) \div -\frac{9}{10} \neq -7 \div \left(\frac{4}{5} \div -\frac{9}{10}\right) \)
(vi) \( -\frac{7}{24} \div \frac{6}{11} \) is not a rational number.
Answer:
(i) True - Division of two rational numbers always gives a rational number.

(ii) False - Commutative property does not apply to division. The order matters: \( \frac{4}{13} \div \frac{11}{12} = \frac{48}{143} \) while \( \frac{11}{12} \div \frac{4}{13} = \frac{143}{48} \), which are not equal.

(iii) False - Associative property does not apply to division. Changing the grouping changes the result.

(iv) True - Since division is not commutative, these two expressions are indeed not equal.

(v) True - Since division is not associative, changing the grouping with parentheses changes the final answer.

(vi) False - This is a rational number. The division of any two non-zero rational numbers produces a rational number.
In simple words: Division is different from addition and multiplication - it does not obey the commutative property (order does not give the same result) and does not obey the associative property (grouping changes the answer).

Exam Tip: Always test both sides before declaring an answer - do the calculation for "left side" and separately for "right side" to confirm they are equal or not equal.

Question 3. The product of two rational numbers is -11/12. If one of them is 2 4/9, find the other.
Answer: Convert the mixed number: \( 2\frac{4}{9} = \frac{22}{9} \). Since the product of two numbers is \( -\frac{11}{12} \) and one number is \( \frac{22}{9} \), the other number is calculated as: \( -\frac{11}{12} \div \frac{22}{9} = -\frac{11}{12} \times \frac{9}{22} = -\frac{3}{8} \). Check: \( \frac{22}{9} \times -\frac{3}{8} = -\frac{11}{12} \) ✓
In simple words: If you know the product of two numbers and you know one of them, divide the product by the known number to find the other.

Exam Tip: Always verify your answer by multiplying the two numbers together - the product should equal what the question stated.

Question 4. By what rational number should -7/12 be multiplied to get the product as 5/14?
Answer: To find the number that \( -\frac{7}{12} \) should be multiplied by, divide the desired product by \( -\frac{7}{12} \): \( \frac{5}{14} \div -\frac{7}{12} = \frac{5}{14} \times \frac{12}{-7} = \frac{60}{-98} = -\frac{30}{49} \). Check: \( -\frac{7}{12} \times -\frac{30}{49} = \frac{5}{14} \) ✓
In simple words: If a number times something equals a result, divide the result by the number to find what it was multiplied by.

Exam Tip: Set up the equation clearly: if a × x = b, then x = b ÷ a. This pattern works for all such questions.

Question 5. By what rational number should -3 is divided to get -9/13?
Answer: To find the divisor, use: \( -3 \div x = -\frac{9}{13} \). Rearranging: \( x = -3 \div -\frac{9}{13} = -3 \times \frac{13}{-9} = \frac{-3 \times 13}{-9} = \frac{13}{3} = 4\frac{1}{3} \). Check: \( -3 \div \frac{13}{3} = -3 \times \frac{3}{13} = -\frac{9}{13} \) ✓
In simple words: If a number divided by something gives a result, multiply the result by the original number's reciprocal to find the divisor. Or rearrange: if a ÷ x = b, then x = a ÷ b.

Exam Tip: When the divisor is unknown, rearrange the equation by dividing the original number by the result.

Question 6. Divide the sum of -13/8 and 5/12 by their difference.
Answer: Sum: \( -\frac{13}{8} + \frac{5}{12} \). Find LCM of 8 and 12, which is 24: \( -\frac{39}{24} + \frac{10}{24} = -\frac{29}{24} \).

Difference: \( -\frac{13}{8} - \frac{5}{12} = -\frac{39}{24} - \frac{10}{24} = -\frac{49}{24} \).

Dividing sum by difference: \( -\frac{29}{24} \div -\frac{49}{24} = -\frac{29}{24} \times \frac{24}{-49} = \frac{29}{49} \).
In simple words: First, find both the sum and the difference using the same common denominator. Then divide the sum by the difference using the reciprocal method.

Exam Tip: When two numbers are added and subtracted, always use the same common denominator for both operations to avoid mistakes.

Question 7. Divide the sum of 8/3 and 4/7 by the product of -3/7 and 14/9.
Answer: Sum: \( \frac{8}{3} + \frac{4}{7} = \frac{56+12}{21} = \frac{68}{21} \).

Product: \( -\frac{3}{7} \times \frac{14}{9} = -\frac{42}{63} = -\frac{2}{3} \).

Dividing: \( \frac{68}{21} \div -\frac{2}{3} = \frac{68}{21} \times -\frac{3}{2} = -\frac{204}{42} = -\frac{34}{7} = -4\frac{6}{7} \).
In simple words: Compute the sum and the product separately, then divide one by the other. Simplify fractions before multiplying to make the numbers smaller.

Exam Tip: Cancel common factors between numerator and denominator before doing final multiplication - this keeps numbers manageable and reduces arithmetic errors.

Question 8. If p = -3/2, q = 4/5 and r = -7/12, then verify that (p ÷ q) ÷ r ≠ p ÷ (q ÷ r)
Answer:
L.H.S. = (p ÷ q) ÷ r:
\( \left(-\frac{3}{2} \div \frac{4}{5}\right) \div -\frac{7}{12} = \left(-\frac{3}{2} \times \frac{5}{4}\right) \div -\frac{7}{12} = -\frac{15}{8} \div -\frac{7}{12} = -\frac{15}{8} \times -\frac{12}{7} = \frac{180}{56} = \frac{45}{14} \).

R.H.S. = p ÷ (q ÷ r):
\( -\frac{3}{2} \div \left(\frac{4}{5} \div -\frac{7}{12}\right) = -\frac{3}{2} \div \left(\frac{4}{5} \times -\frac{12}{7}\right) = -\frac{3}{2} \div -\frac{48}{35} = -\frac{3}{2} \times -\frac{35}{48} = \frac{105}{96} = \frac{35}{32} \).

Since \( \frac{45}{14} \neq \frac{35}{32} \), we have verified that L.H.S. ≠ R.H.S. Division is not associative.
In simple words: The associative property works for addition and multiplication, but not for division or subtraction. The order in which you group terms matters when dividing.

Exam Tip: For "not equal" proofs, always compute both sides fully and clearly show that the final answers differ - do not leave them in different forms that might look equal.

Exercise 1.5

Question 1. Represent the following rational numbers on the number line.
(i) \( \frac{11}{4} \)
(ii) \( 4\frac{3}{5} \)
(iii) \( -\frac{9}{7} \)
(iv) \( -\frac{2}{-5} \)
Answer:
(i) \( \frac{11}{4} = 2\frac{3}{4} = 2.75 \). Mark the point between 2 and 3, specifically 3 divisions of the way from 2 to 3.

(ii) \( 4\frac{3}{5} = 4.6 \). Mark the point between 4 and 5, specifically at 3 divisions of the way from 4 to 5.

(iii) \( -\frac{9}{7} = -1\frac{2}{7} \approx -1.29 \). Mark the point between -2 and -1, specifically 2 divisions of the way from -2 to -1.

(iv) \( -\frac{2}{-5} = \frac{2}{5} = 0.4 \). Mark the point between 0 and 1, specifically 2 divisions of the way from 0 to 1.
In simple words: Convert each fraction to a mixed number or decimal. Then find where it falls on the number line - whole number part tells you which integers it sits between, and the fraction part tells you exactly where between them.

Exam Tip: Always simplify negative signs first (like changing \( -\frac{2}{-5} \) to \( \frac{2}{5} \)). Then convert to a mixed number to see the whole part and the fractional part clearly.

Question 2. Write the rational numbers for each point labeled with a letter:
Answer:
(i) A = \( \frac{3}{7} \); B = \( \frac{7}{7} = 1 \); C = \( \frac{8}{7} = 1\frac{1}{7} \); D = \( \frac{12}{7} = 1\frac{5}{7} \); E = \( \frac{13}{7} = 1\frac{6}{7} \)

(ii) P = \( -\frac{3}{8} \); Q = \( -\frac{4}{8} = -\frac{1}{2} \); R = \( -\frac{7}{8} \); S = \( -\frac{11}{8} = -1\frac{3}{8} \); T = \( -\frac{12}{8} = -\frac{3}{2} = -1\frac{1}{2} \)
In simple words: Count the tick marks from 0 to find the position of each letter. Each tick represents one equal part of the denominator. Express as a single fraction or convert to a mixed number if it is bigger than one.

Exam Tip: Reduce fractions to simplest form and express improper fractions as mixed numbers for the final answer - this shows you fully understand the number line.

Question 3. Find twenty rational numbers between -3/7 and 2/3
Answer: Find the LCM of 7 and 3, which is 21. Convert both fractions:
\( -\frac{3}{7} = -\frac{9}{21} \) and \( \frac{2}{3} = \frac{14}{21} \).

Any twenty rational numbers between \( -\frac{9}{21} \) and \( \frac{14}{21} \) are: \( -\frac{8}{21}, -\frac{7}{21}, -\frac{6}{21}, -\frac{5}{21}, -\frac{4}{21}, -\frac{3}{21}, -\frac{2}{21}, -\frac{1}{21}, 0, \frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}, \frac{6}{21}, \frac{7}{21}, \frac{8}{21}, \frac{9}{21}, \frac{10}{21}, \frac{11}{21} \)
In simple words: Convert both fractions to the same denominator. Then you can easily list all the fractions between them by changing only the numerator while keeping the denominator fixed.

Exam Tip: Using a common denominator makes it clear that you can fit as many numbers as you want between any two rational numbers - there is always room for more.

Question 4. Find six rational numbers between -1/2 and 5/4
Answer: Find the LCM of 2 and 4, which is 4. Convert: \( -\frac{1}{2} = -\frac{2}{4} \).

Six rational numbers between \( -\frac{2}{4} \) and \( \frac{5}{4} \) are: \( -\frac{1}{4}, 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4} \)

(Note: \( \frac{2}{4} = \frac{1}{2} \) and \( \frac{4}{4} = 1 \) can also be written in simplest form.)
In simple words: Using a common denominator, list six numerators that fall strictly between -2 and 5. Any such choice works as long as there are six of them.

Exam Tip: You can list more or fewer numbers - the question asks for "six", so provide exactly six. Writing them in simplest form is a bonus but not always required.

Question 5. Find three rational numbers between -2 and -1
Answer: Use the midpoint method to find rational numbers between two given numbers.

First rational number = \( \frac{1}{2}[(-2) + (-1)] = \frac{1}{2} \times (-3) = -\frac{3}{2} \).

Second rational number = \( \frac{1}{2}\left[(-2) + \left(-\frac{3}{2}\right)\right] = \frac{1}{2} \times -\frac{7}{2} = -\frac{7}{4} \).

Third rational number = \( \frac{1}{2}\left[\left(-\frac{3}{2}\right) + (-1)\right] = \frac{1}{2} \times -\frac{5}{2} = -\frac{5}{4} \).

Therefore, three rational numbers are: \( -\frac{7}{4}, -\frac{3}{2}, -\frac{5}{4} \) (in order from left to right: \( -1.75, -1.5, -1.25 \)).
In simple words: Find the average (midpoint) of two numbers to get a number between them. Then find the midpoint between one of the original numbers and the midpoint you just found, and repeat. This method guarantees numbers in the correct range.

Exam Tip: The midpoint method always works and is more reliable than guessing. If you list numbers, make sure they truly fall between the given bounds by checking their decimal values.

Question 6. Write ten rational numbers which are greater than 0.
Answer: There are infinitely many positive rational numbers. Any ten examples are acceptable. A simple list: \( \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}, 4, \frac{9}{2}, 5 \).

Other valid examples: \( \frac{1}{10}, \frac{1}{5}, \frac{2}{5}, \frac{3}{10}, \frac{1}{100}, \frac{7}{6}, 0.5, 1.5, 10, 100 \).
In simple words: Any fraction or decimal that is bigger than zero is a positive rational number. You can pick any ten you like - mixing whole numbers, simple fractions, and improper fractions makes a good variety.

Exam Tip: This question has many correct answers - any ten positive rational numbers work. Choose ten that are easy to write and understand.

Question 7. Write five rational numbers which are smaller than -4
Answer: There are infinitely many rational numbers smaller than -4. Any five examples are acceptable. A simple list: \( -\frac{9}{2}, -5, -\frac{11}{2}, -6, -\frac{13}{2} \).

Other valid examples: \( -4.1, -4.5, -10, -100, -\frac{25}{6} \).
In simple words: Any fraction or decimal that is smaller (more negative) than -4 works. You can make numbers even more negative by using larger denominators or whole numbers further left on the number line.

Exam Tip: A reliable method: use \( -4 - \frac{1}{2}, -4 - 1, -4 - \frac{3}{2} \), etc. This guarantees all numbers are smaller than -4.

Question 8. Identify the rational number which is different from the other three. Explain your reasoning
\( \left(-\frac{5}{11}\right), \left(-\frac{1}{2}\right), \left(-\frac{4}{9}\right), \left(-\frac{7}{3}\right) \)
Answer: The different number is \( -\frac{7}{3} \). The reason: In \( -\frac{7}{3} \), the absolute value of the numerator (7) is greater than the absolute value of the denominator (3), making it an improper fraction with absolute value greater than 1. In contrast, the other three numbers - \( -\frac{5}{11}, -\frac{1}{2}, -\frac{4}{9} \) - are all proper fractions (numerator smaller than denominator in absolute value), so their absolute values are less than 1. Equivalently, \( -\frac{7}{3} \approx -2.33 \) (less than -1), while the other three fall between -1 and 0.
In simple words: One fraction is improper (bigger than 1 in size), while the others are proper (smaller than 1 in size). That makes one of them different from the rest.

Exam Tip: Always find the decimal or mixed number form to compare the numbers clearly - this makes it easier to spot which one does not fit the pattern.

Exercise 1.6

Question 1. In a bag, there are 20 kg of fruits. If 7 1/6 kg of these fruits be oranges and 8 2/3 kg of these are apples and rest are grapes. Find the mass of the grapes in the bag.
Answer: Total fruits in the bag = 20 kg. Oranges weigh 43/6 kg and apples weigh 26/3 kg. The remaining mass is found by subtracting both quantities from the total. First, combine the oranges and apples: (43/6) + (26/3) = (43 + 52)/6 = 95/6 kg. Then subtract from the total: 20 - (95/6) = (120 - 95)/6 = 25/6 kg = 4 1/6 kg. Therefore, the grapes weigh 4 1/6 kg.
In simple words: Add the mass of oranges and apples together. Then take away that total from 20 kg. The answer is the mass of grapes.

Exam Tip: Always convert mixed numbers to improper fractions first, find a common denominator before adding or subtracting, and simplify your final answer back to a mixed number if needed.

Question 2. The population of a city is 6,63,432. If 1/2 of the population are adult males and 1/3 of the population are adult females, then find the number of children in the city.
Answer: The total population is 6,63,432. Adult males make up 1/2 of this, which equals 3,31,716. Adult females are 1/3 of the total, which equals 2,21,144. To find the children, subtract both groups from the total: 6,63,432 - (3,31,716 + 2,21,144) = 6,63,432 - 5,52,860 = 1,10,572. Therefore, there are 1,10,572 children in the city.
In simple words: Find how many adult males and adult females there are. Add those numbers together. Subtract from the total population to get the number of children.

Exam Tip: Always double-check that your three groups add back to the original total population as a verification step.

Question 3. In an election of housing society, there are 30 voters. Each of them gives the vote. Three persons X, Y and Z are standing for the post of Secretary. If Mr X got 2/5 of the total votes and Mr Z got 1/3 of the total votes, then find the number of votes which Mr Y got.
Answer: There are 30 total votes cast. Mr X received 2/5 of these votes: (2/5) × 30 = 12 votes. Mr Z received 1/3 of the votes: (1/3) × 30 = 10 votes. The remaining votes go to Mr Y: 30 - (12 + 10) = 30 - 22 = 8 votes. Therefore, Mr Y got 8 votes.
In simple words: Multiply the total votes by each person's fraction. Add the votes X and Z got. Subtract that sum from the total to find Y's votes.

Exam Tip: Verify your answer by adding all three candidates' votes together - they must equal the total number of voters.

Question 4. A person earns Rs 100 in a day. If he spent Rs 14 2/7 on food and Rs 30 2/3 on petrol. How much did he save on that day?
Answer: The person's daily earnings are Rs 100. Food costs Rs 100/7 and petrol costs Rs 92/3. Total expenditure is: (100/7) + (92/3) = (300 + 644)/21 = 944/21. Savings are calculated by subtracting expenses from earnings: 100 - (944/21) = (2100 - 944)/21 = 1156/21 = 55 1/21 rupees. Therefore, the person saved Rs 55 1/21 on that day.
In simple words: Add up all the money spent on food and petrol. Subtract that total from Rs 100 earnings. The result is how much was saved.

Exam Tip: When working with mixed numbers and different denominators, always convert to improper fractions and find the least common denominator before performing operations.

Question 5. In an examination, 400 students appeared. If 2/3 of the boys and all 130 girls passed in the examination, then find how many boys failed in an examination?
Answer: The total students examined are 400. Since 130 are girls, the number of boys is 400 - 130 = 270. Among the boys, 2/3 passed: (2/3) × 270 = 180 boys passed. The boys who failed are: 270 - 180 = 90. Therefore, 90 boys failed in the examination.
In simple words: Find how many boys there are by subtracting girls from the total. Calculate how many boys passed using the fraction. Subtract the boys who passed from total boys to find those who failed.

Exam Tip: Ensure you identify which fraction applies to which group - here, the fraction applies only to boys, not to girls.

Question 6. A car is moving at the speed of 40 2/3 km/h. Find how much distance will it cover in 9/10 hrs?
Answer: The car's speed is 122/3 km/h. Distance equals speed multiplied by time. Distance = (122/3) × (9/10) = (122 × 9)/(3 × 10) = 1098/30 = 366/10 = 36.6 km = 36 3/5 km. Therefore, the car covers 36 3/5 km in 9/10 hours.
In simple words: Multiply the speed by the time to find the distance. Convert mixed numbers to improper fractions before multiplying, then simplify the result.

Exam Tip: Use the formula Distance = Speed × Time. Always simplify fractions completely and convert back to mixed numbers where appropriate.

Question 7. Find the area of a square lawn whose one side is 5 7/9 m long.
Answer: One side of the square lawn is 52/9 m. The area of a square is calculated by squaring the side length. Area = (52/9)² = (52 × 52)/(9 × 9) = 2704/81 = 33 31/81 sq. m. Therefore, the area of the square lawn is 33 31/81 sq. m.
In simple words: Multiply the side length by itself. Convert the mixed number to an improper fraction, square both numerator and denominator, then simplify.

Exam Tip: Remember that area is always in square units. Double-check your squaring by verifying the numerator and denominator calculations separately.

Question 8. Perimeter of a rectangle is 15 3/7 m. If the length is 4 2/7 m, find its breadth.
Answer: The perimeter is 108/7 m. Since perimeter = 2(length + breadth), we have length + breadth = (108/7) ÷ 2 = (108/7) × (1/2) = 54/7 m. The given length is 30/7 m. Therefore, breadth = (54/7) - (30/7) = 24/7 = 3 3/7 m. The breadth of the rectangle is 3 3/7 m.
In simple words: Divide the perimeter by 2 to get length plus breadth. Subtract the given length from this sum to find the breadth.

Exam Tip: Verify your answer by checking that 2(length + breadth) equals the original perimeter.

Question 9. Rahul had a rope of 325 1/5 m long. He cut off a 150 3/5 m long piece, then he divided the rest of the rope into 3 parts of equal length. Find the length of each part.
Answer: The original rope is 325 1/5 m = 1626/5 m long. Rahul cuts off a piece measuring 150 3/5 m = 753/5 m. The remaining rope is: (1626/5) - (753/5) = 873/5 = 174 3/5 m. He then divides this remaining length equally into 3 parts. Length of each part = (873/5) ÷ 3 = (873/5) × (1/3) = 873/15 = 58 1/5 m. Therefore, each part is 58 1/5 m long.
In simple words: Subtract the cut-off piece from the original rope length. Divide what remains by 3 to get each equal part.

Exam Tip: Convert all mixed numbers to improper fractions at the start to avoid arithmetic errors, and verify by multiplying each part's length by 3 to check it equals the remaining rope.

Question 10. If 3 1/2 litre of petrol costs Rs 270 3/8, then find the cost of 4 litre of petrol.
Answer: The cost of 7/2 litre is Rs 2163/8. To find the cost per litre, divide: cost per litre = (2163/8) ÷ (7/2) = (2163/8) × (2/7) = (2163 × 2)/(8 × 7). For 4 litres, multiply the cost per litre by 4: cost of 4 litre = (2163 × 2 × 4)/(8 × 7) = 17304/56 = Rs 309. Therefore, 4 litres of petrol cost Rs 309.
In simple words: First find the cost of one litre by dividing total cost by the number of litres. Then multiply that unit cost by 4 to get the cost of 4 litres.

Exam Tip: Breaking down to unit cost first makes the problem simpler and reduces chance of error in your calculations.

Question 11. Ramesh earns Rs 40,000 per month. He spends 3/8 of the income on food, 1/5 of the remaining on LIC premium and then 1/2 of the remaining on other expenses. Find how much money is left with him?
Answer: Ramesh's monthly income is Rs 40,000. Spending on food is (3/8) × 40,000 = Rs 15,000. Amount remaining = 40,000 - 15,000 = Rs 25,000. LIC premium is (1/5) × 25,000 = Rs 5,000. Amount remaining = 25,000 - 5,000 = Rs 20,000. Other expenses are (1/2) × 20,000 = Rs 10,000. Final remaining amount = 20,000 - 10,000 = Rs 10,000. Therefore, Ramesh has Rs 10,000 left.
In simple words: Subtract food costs from income. From what's left, subtract LIC premium. From what remains, subtract other expenses. The final answer is what's still left.

Exam Tip: Work step by step, calculating the "remaining" amount after each expense - the key phrase "of the remaining" means each fraction applies only to what's left from the previous step, not the original amount.

Question 12. A, B, C, D and E went to a restaurant for dinner. A paid 1/2 of the bill, B paid 1/5 of the bill and rest of the bill was shared equally by C, D and E. What fractions of the bill was paid by each?
Answer: Let the total bill be 1. A paid 1/2 and B paid 1/5. The remaining portion is: 1 - (1/2 + 1/5) = 1 - (5 + 2)/10 = 1 - 7/10 = 3/10. This remaining 3/10 is shared equally among C, D, and E. Each of them pays: (3/10) ÷ 3 = (3/10) × (1/3) = 1/10 of the bill. Therefore, A paid 1/2, B paid 1/5, and each of C, D, and E paid 1/10 of the bill.
In simple words: Find what portion A and B paid together. Subtract from the total bill to find the remaining portion. Divide that remaining portion equally by three to find each person's share.

Exam Tip: Verify by adding all five fractions: 1/2 + 1/5 + 1/10 + 1/10 + 1/10 = 5/10 + 2/10 + 1/10 + 1/10 + 1/10 = 10/10 = 1, confirming your answer is correct.

Question 13. 2/5 of total number of students of a school come by car while 1/4 of students come by bus to school. All the other students walk to school of which 1/3 walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in the school?
Answer: Let the total number of students be represented as 1. Students by car = 2/5 and by bus = 1/4. Students who walk = 1 - (2/5 + 1/4) = 1 - (8 + 5)/20 = 1 - 13/20 = 7/20. Of these walkers, 1/3 walk independently: (1/3) × (7/20) = 7/60 of all students. Since 224 students represent 7/60 of the total, the total students = 224 × (60/7) = 32 × 60 = 1,920. Therefore, 1,920 students study in the school.
In simple words: Find what fraction of students walk by subtracting car and bus students from the total. Find what fraction walk on their own by taking 1/3 of walkers. Use the given number to find the total.

Exam Tip: When a fraction of students equals a specific number, set up an equation and solve by multiplying by the reciprocal of the fraction.

Question 14. A mother and her two sons got a room constructed for Rs 60,000. The elder son contributes 3/8 of his mother's contribution while the younger son contributes 1/2 of his mother's share. How much do the three contribute individually?
Answer: Let the mother's contribution be 1. The elder son contributes 3/8 of this amount, and the younger son contributes 1/2 of this amount. The ratio of contributions is 1 - (3/8) - (1/2) = 8 - 3 - 4 = 1, which gives us 1: (3/8): (1/2) = 8: 3: 4. Sum of ratio parts = 8 + 3 + 4 = 15. Mother's share = (60,000 × 8)/15 = Rs 32,000. Elder son's share = (60,000 × 3)/15 = Rs 12,000. Younger son's share = (60,000 × 4)/15 = Rs 16,000. Therefore, the mother contributes Rs 32,000, the elder son Rs 12,000, and the younger son Rs 16,000.
In simple words: Express each person's share as a ratio based on the mother's amount. Add all ratio parts. Divide the total cost by the sum and multiply by each person's ratio part.

Exam Tip: Verify by adding all three contributions: 32,000 + 12,000 + 16,000 = 60,000, confirming the total cost is reached.

Question 15. In a class of 56 students, the number of boys is 2/5 th of the number of girls. Find the number of boys and girls.
Answer: Let the number of girls be 1. Then the number of boys = 2/5 of 1 = 2/5. The ratio of girls to boys is 1: (2/5) = 5: 2. Sum of ratio parts = 5 + 2 = 7. Number of girls = [56/(5 + 2)] × 5 = (56/7) × 5 = 8 × 5 = 40. Number of boys = (56/7) × 2 = 8 × 2 = 16. Therefore, there are 40 girls and 16 boys in the class.
In simple words: Express boys and girls as a ratio. Divide the total students by the sum of ratio parts. Multiply by each ratio part to find how many girls and boys.

Exam Tip: Always verify by adding the number of boys and girls to ensure they equal the total class strength of 56.

Question 16. A man donated 1/10 of his money to a school, 1/6 th of the remaining to a church and the remaining money he distributed equally among his three children. If each child gets Rs 50000, how much money did the man originally have?
Answer: Let the man's original money be 1. He donates 1/10 to school, leaving 9/10. From this, 1/6 goes to the church: (1/6) × (9/10) = 9/60 = 3/20. The money left for distribution is (9/10) - (3/20) = (18 - 3)/20 = 15/20. This is divided equally among three children, so each child receives (15/20) ÷ 3 = (15/20) × (1/3) = 15/60 = 1/4 of the original amount. Since each child gets Rs 50,000, the original amount = 50,000 × (4/1) = Rs 200,000. Therefore, the man originally had Rs 200,000.
In simple words: Subtract school donation from the total. Subtract church donation from what's left. Divide the final remainder among three children. Use the amount each child gets to find the original total.

Exam Tip: Work backward from the amount each child receives by multiplying by the reciprocal of the fraction they represent to find the original total.

Question 17. If 1/4 of a number is added to 1/3 of that number, the result is 15 greater than half of that number. Find the number.
Answer: Let the number be x. According to the given condition: (1/4)x + (1/3)x - (1/2)x = 15. Finding a common denominator of 12: (3x + 4x - 6x)/12 = 15, which simplifies to (1/12)x = 15. Solving: x = 15 × 12 = 180. Therefore, the required number is 180.
In simple words: Write an equation based on what the problem states. Combine like terms using a common denominator. Solve for the unknown number.

Exam Tip: Always verify your answer by substituting back: (1/4)(180) + (1/3)(180) = 45 + 60 = 105, and (1/2)(180) + 15 = 90 + 15 = 105. Both sides are equal, confirming the answer.

Question 18. A student was asked to multiply a given number by 4/5. By mistake, he divided the given number by 4/5. His answer was 36 more than the correct answer. What was the given number?
Answer: Let the given number be x. The correct operation is x × (4/5) = (4/5)x. The incorrect operation (division) gives x ÷ (4/5) = x × (5/4) = (5/4)x. The difference is: (5/4)x - (4/5)x = 36. Finding a common denominator of 20: (25x - 16x)/20 = 36, so 9x/20 = 36. Solving: 9x = 720, therefore x = 80. The given number is 80.
In simple words: Set up the correct and incorrect operations. Find their difference and set it equal to 36. Solve the resulting equation for the unknown number.

Exam Tip: Verify: Correct answer = (4/5) × 80 = 64. Incorrect answer = (5/4) × 80 = 100. Difference = 100 - 64 = 36, confirming the solution.