RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers

Rational Numbers

Exercise 4A

Question 1. What is a rational number? Give five examples of positive rational numbers and five examples of negative rational numbers. Is there a rational number that is neither positive nor negative?
Answer:
A rational number is any number that can be written as \( \frac{p}{q} \), where p and q are integers and q is not zero.
Five positive rational numbers, for instance, are: \( \frac{5}{7}, \frac{-3}{-4}, \frac{7}{8}, \frac{-14}{-15}, \frac{5}{9} \)
Five negative rational numbers, for instance, are: \( \frac{-3}{7}, \frac{-3}{8}, \frac{8}{-9}, \frac{-19}{25}, \frac{8}{-25} \)
Yes - there is a rational number that is neither positive nor negative, and that number is zero (0).

Exam Tip: Remember that a rational number needs a nonzero denominator - a fraction with denominator zero, such as 1/0, is never rational.

Question 3. Write the numerator and the denominator of each of the following rational numbers:
(i) \( \frac{8}{19} \)
(ii) \( \frac{5}{-8} \)
(iii) \( \frac{-13}{15} \)
(iv) \( \frac{-8}{-11} \)
(v) 9
Answer:
(i) Numerator = 8, Denominator = 19
(ii) Numerator = 5, Denominator = -8
(iii) Numerator = -13, Denominator = 15
(iv) Numerator = -8, Denominator = -11
(v) Writing 9 as \( \frac{9}{1} \): Numerator = 9, Denominator = 1

Exam Tip: A whole number can always be written as a fraction with denominator 1 before its numerator and denominator are identified.

Question 4. Express each of the following integers as a rational number with denominator 1:
(i) 5
(ii) -3
(iii) 1
(iv) 0
(v) -23
Answer:
(i) \( 5 = \frac{5}{1} \)
(ii) \( -3 = \frac{-3}{1} \)
(iii) \( 1 = \frac{1}{1} \)
(iv) \( 0 = \frac{0}{1} \)
(v) \( -23 = \frac{-23}{1} \)

Exam Tip: Every integer, including 0 and negative numbers, is automatically a rational number once it is written over a denominator of 1.

Question 5. State which of the following are positive rational numbers:
(i) \( \frac{-5}{-8} \)
(ii) \( \frac{37}{53} \)
(iii) 8
Answer:
(i) \( \frac{-5}{-8} \) is positive, since a negative divided by a negative gives a positive value.
(ii) \( \frac{37}{53} \) is positive, as both the numerator and denominator are positive.
(iii) 8, written as \( \frac{8}{1} \), is positive.
Note: 0 is neither positive nor negative.

Exam Tip: A fraction is positive whenever the numerator and denominator have matching signs - both positive or both negative.

Question 6. State which of the following are negative rational numbers:
(i) \( \frac{-5}{7} \)
(ii) \( \frac{4}{-9} \)
(iii) -6
(iv) \( \frac{1}{-2} \)
Answer:
All four are negative rational numbers, since in each case exactly one of the numerator or denominator is negative.

Exam Tip: A fraction is negative whenever the numerator and denominator have opposite signs - one positive and one negative.

Question 7. Write four rational numbers equivalent to each of the following:
(i) \( \frac{6}{11} \)
(ii) \( \frac{-3}{8} \)
(iii) \( \frac{7}{-15} \)
(iv) 8
(v) 1
(vi) -1
Answer:
(i) Multiplying the numerator and denominator by 2, 3, 4 and 5: \( \frac{12}{22}, \frac{18}{33}, \frac{24}{44}, \frac{30}{55} \)
(ii) In the same way: \( \frac{-6}{16}, \frac{-9}{24}, \frac{-12}{32}, \frac{-15}{40} \)
(iii) In the same way: \( \frac{14}{-30}, \frac{21}{-45}, \frac{28}{-60}, \frac{35}{-75} \)
(iv) Writing 8 as \( \frac{8}{1} \): \( \frac{16}{2}, \frac{24}{3}, \frac{32}{4}, \frac{40}{5} \)
(v) Writing 1 as \( \frac{1}{1} \): \( \frac{2}{2}, \frac{3}{3}, \frac{4}{4}, \frac{5}{5} \)
(vi) Writing -1 as \( \frac{-1}{1} \): \( \frac{-2}{2}, \frac{-3}{3}, \frac{-4}{4}, \frac{-5}{5} \)

Exam Tip: To build an equivalent rational number, multiply the numerator and denominator by the same nonzero whole number - this never changes the value.

Question 8. Express each of the following rational numbers with a positive denominator:
(i) \( \frac{12}{-17} \)
(ii) \( \frac{1}{-2} \)
(iii) \( \frac{-8}{-19} \)
(iv) \( \frac{11}{-6} \)
Answer:
(i) \( \frac{12 \times -1}{-17 \times -1} = \frac{-12}{17} \)
(ii) \( \frac{1 \times -1}{-2 \times -1} = \frac{-1}{2} \)
(iii) \( \frac{-8 \times -1}{-19 \times -1} = \frac{8}{19} \)
(iv) \( \frac{11 \times -1}{-6 \times -1} = \frac{-11}{6} \)

Exam Tip: To flip a negative denominator to positive, multiply both the numerator and denominator by -1 - this keeps the value exactly the same.

Question 9. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{5}{8} = \frac{15}{\square} \)
(ii) \( \frac{5}{8} = \frac{-10}{\square} \)
Answer:
(i) 5 must be multiplied by 3 to reach 15, so the denominator becomes \( 8 \times 3 = 24 \): \( \frac{5}{8} = \frac{15}{24} \)
(ii) 5 must be multiplied by -2 to reach -10, so the denominator becomes \( 8 \times -2 = -16 \): \( \frac{5}{8} = \frac{-10}{-16} \)

Exam Tip: Work out what the numerator was multiplied by first, then apply that exact same multiplier to the denominator.

Question 10. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{4}{7} = \frac{\square}{21} \)
(ii) \( \frac{4}{7} = \frac{\square}{-35} \)
Answer:
(i) 7 must be multiplied by 3 to reach 21, so the numerator becomes \( 4 \times 3 = 12 \): \( \frac{4}{7} = \frac{12}{21} \)
(ii) 7 must be multiplied by -5 to reach -35, so the numerator becomes \( 4 \times -5 = -20 \): \( \frac{4}{7} = \frac{-20}{-35} \)

Exam Tip: When the denominator's multiplier is given, use that same multiplier on the numerator to keep the fraction equivalent.

Question 11. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{-12}{13} = \frac{-48}{\square} \)
(ii) \( \frac{-12}{13} = \frac{60}{\square} \)
Answer:
(i) -12 must be multiplied by 4 to reach -48, so the denominator becomes \( 13 \times 4 = 52 \): \( \frac{-12}{13} = \frac{-48}{52} \)
(ii) -12 must be multiplied by -5 to reach 60, so the denominator becomes \( 13 \times -5 = -65 \): \( \frac{-12}{13} = \frac{60}{-65} \)

Exam Tip: When the multiplier for the numerator turns out negative, apply that same negative multiplier to the denominator too.

Question 12. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{-8}{11} = \frac{\square}{22} \)
(ii) \( \frac{-8}{11} = \frac{\square}{55} \)
Answer:
(i) 11 must be multiplied by 2 to reach 22, so the numerator becomes \( -8 \times 2 = -16 \): \( \frac{-8}{11} = \frac{-16}{22} \)
(ii) 11 must be multiplied by 5 to reach 55, so the numerator becomes \( -8 \times 5 = -40 \): \( \frac{-8}{11} = \frac{-40}{55} \)

Exam Tip: It often helps to divide the new denominator by the old one first, to see straight away what multiplier is needed.

Question 13. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{14}{-5} = \frac{56}{\square} \)
(ii) \( \frac{14}{-5} = \frac{-70}{\square} \)
Answer:
(i) 14 must be multiplied by 4 to reach 56, so the denominator becomes \( -5 \times 4 = -20 \): \( \frac{14}{-5} = \frac{56}{-20} \)
(ii) 14 must be multiplied by -5 to reach -70, so the denominator becomes \( -5 \times -5 = 25 \): \( \frac{14}{-5} = \frac{-70}{25} \)

Exam Tip: A negative denominator multiplied by a negative multiplier turns positive - keep an eye on the sign changes here.

Question 14. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{13}{-8} = \frac{\square}{-40} \)
(ii) \( \frac{13}{-8} = \frac{-52}{\square} \)
Answer:
(i) -8 must be multiplied by 5 to reach -40, so the numerator becomes \( 13 \times 5 = 65 \): \( \frac{13}{-8} = \frac{65}{-40} \)
(ii) 13 must be multiplied by -4 to reach -52, so the denominator becomes \( -8 \times -4 = 32 \): \( \frac{13}{-8} = \frac{-52}{32} \)

Exam Tip: Whichever part of the fraction is given as the target, work backwards from it to find the multiplier used.

Question 15. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{-36}{24} = \frac{-9}{\square} \)
(ii) \( \frac{-36}{24} = \frac{6}{\square} \)
Answer:
(i) -36 divided by 4 gives -9, so the denominator becomes \( 24 \div 4 = 6 \): \( \frac{-36}{24} = \frac{-9}{6} \)
(ii) -36 divided by -6 gives 6, so the denominator becomes \( 24 \div -6 = -4 \): \( \frac{-36}{24} = \frac{6}{-4} \)

Exam Tip: Reducing a fraction to an equivalent smaller form uses division instead of multiplication - the same divisor must apply to both parts.

Question 16. Fill in the blanks so that the two rational numbers are equivalent:
(i) \( \frac{84}{-147} = \frac{\square}{7} \)
(ii) \( \frac{84}{-147} = \frac{28}{\square} \)
Answer:
(i) -147 divided by -21 gives 7, so the numerator becomes \( 84 \div -21 = -4 \): \( \frac{84}{-147} = \frac{-4}{7} \)
(ii) 84 divided by 3 gives 28, so the denominator becomes \( -147 \div 3 = -49 \): \( \frac{84}{-147} = \frac{28}{-49} \)

Exam Tip: Look for a divisor that divides both the numerator and denominator exactly, to keep the equivalent fraction in whole numbers.

Question 17. Express each of the following in the standard form:
(i) \( \frac{35}{49} \)
(ii) \( \frac{8}{-36} \)
(iii) \( \frac{-27}{45} \)
(iv) \( \frac{-14}{-49} \)
(v) \( \frac{91}{-78} \)
(vi) \( \frac{-68}{119} \)
(vii) \( \frac{-87}{116} \)
(viii) \( \frac{299}{-161} \)
Answer:
(i) The HCF of 35 and 49 is 7. Dividing both parts by 7 gives \( \frac{5}{7} \).
(ii) The denominator is negative, so multiply both parts by -1 to get \( \frac{-8}{36} \). The HCF of 8 and 36 is 4, giving \( \frac{-2}{9} \).
(iii) The HCF of 27 and 45 is 9. Dividing both parts by 9 gives \( \frac{-3}{5} \).
(iv) The denominator is negative, so multiply both parts by -1 to get \( \frac{14}{49} \). The HCF of 14 and 49 is 7, giving \( \frac{2}{7} \).
(v) The denominator is negative, so multiply both parts by -1 to get \( \frac{-91}{78} \). The HCF of 91 and 78 is 13, giving \( \frac{-7}{6} \).
(vi) The HCF of 68 and 119 is 17. Dividing both parts by 17 gives \( \frac{-4}{7} \).
(vii) The HCF of 87 and 116 is 29. Dividing both parts by 29 gives \( \frac{-3}{4} \).
(viii) The denominator is negative, so multiply both parts by -1 to get \( \frac{-299}{161} \). The HCF of 299 and 161 is 23, giving \( \frac{-13}{7} \).

Exam Tip: Standard form means a positive denominator with no common factor left between numerator and denominator - always fix the sign first, then divide out the HCF.

Question 18. Write \( \frac{-9}{5} \) and \( \frac{-6}{11} \) each as three equivalent rational numbers.
Answer:
Multiplying \( \frac{-9}{5} \) by 4, -3 and 5 in turn: \( \frac{-36}{20}, \frac{27}{-15}, \frac{-45}{25} \)
Multiplying \( \frac{-6}{11} \) by 3 and 4 in turn: \( \frac{-18}{33}, \frac{-24}{44} \)

Exam Tip: Any multiplier can be used to generate an equivalent fraction, as long as it is applied identically to both numerator and denominator.

Question 19. State whether the following pairs of rational numbers are equivalent:
(i) \( \frac{-13}{7}, \frac{39}{-21} \)
(ii) \( \frac{3}{-8}, \frac{-6}{16} \)
(iii) \( \frac{9}{4}, \frac{-36}{-16} \)
(iv) \( \frac{7}{15}, \frac{-28}{60} \)
(v) \( \frac{3}{12}, \frac{-1}{4} \)
(vi) \( \frac{2}{3}, \frac{3}{2} \)
Answer:
(i) \( -13 \times -21 = 273 \) and \( 7 \times 39 = 273 \), so they are equivalent.
(ii) \( 3 \times 16 = 48 \) and \( -8 \times -6 = 48 \), so they are equivalent.
(iii) \( 9 \times -16 = -144 \) and \( 4 \times -36 = -144 \), so they are equivalent.
(iv) \( 7 \times 60 = 420 \) but \( 15 \times -28 = -420 \), so they are not equivalent.
(v) \( 3 \times 4 = 12 \) but \( 12 \times -1 = -12 \), so they are not equivalent.
(vi) \( 2 \times 2 = 4 \) but \( 3 \times 3 = 9 \), so they are not equivalent.

Exam Tip: Two fractions are equivalent exactly when cross-multiplying gives the same result on both sides - a quick and reliable test.

Question 20. Find the value of x such that:
(i) \( \frac{-1}{5} = \frac{8}{x} \)
(ii) \( \frac{7}{-3} = \frac{x}{6} \)
(iii) \( \frac{3}{5} = \frac{-x}{25} \)
(iv) \( \frac{13}{6} = \frac{-65}{x} \)
(v) \( \frac{16}{x} = -4 \)
(vi) \( \frac{-48}{x} = 2 \)
Answer:
(i) \( -x = 5 \times 8 \), so \( x = -40 \)
(ii) \( -3x = 7 \times 6 \), so \( x = \frac{42}{-3} = -14 \)
(iii) \( 5x = 3 \times -25 \), so \( x = \frac{-75}{5} = -15 \)
(iv) \( 13x = 6 \times -65 \), so \( x = \frac{-390}{13} = -30 \)
(v) \( x = \frac{16}{-4} = -4 \)
(vi) \( 2x = -48 \), so \( x = -24 \)

Exam Tip: Cross-multiply first to turn the equation into a simple one-step multiplication, then divide to isolate x.

Question 21. State whether the following pairs of rational numbers are equal:
(i) \( \frac{8}{-12}, \frac{-10}{15} \)
(ii) \( \frac{-3}{9}, \frac{7}{-21} \)
(iii) \( \frac{-8}{-14}, \frac{15}{21} \)
Answer:
(i) \( 8 \times 15 = 120 \) and \( -12 \times -10 = 120 \), so they are equal.
(ii) \( -3 \times -21 = 63 \) and \( 9 \times 7 = 63 \), so they are equal.
(iii) \( -8 \times 21 = -168 \) but \( -14 \times 15 = -210 \), so they are not equal.

Exam Tip: The cross-multiplication check works the same way whether the question asks about "equal" or "equivalent" rational numbers.

Question 22. State whether the following statements are true or false, giving a reason in each case:
(i) 0 is the smallest rational number.
(ii) Every integer is a rational number.
(iii) Every rational number is a whole number.
(iv) Every whole number is a rational number.
(v) Every rational number is a fraction.
Answer:
(i) False - for example, -1 is a rational number smaller than 0.
(ii) True - every integer can be written with denominator 1, so it fits the definition of a rational number.
(iii) False - a number needs a nonzero denominator to be rational, and not every rational number, such as 1/2, is a whole number.
(iv) True - a whole number can always be written with denominator 1, so it is rational.
(v) False - for instance, -1 is a rational number but it is not usually called a fraction, since fractions are generally taken to be positive.

Exam Tip: When judging true/false statements about number systems, a single counterexample is enough to prove a statement false.

Exercise 4B

Question 2. Fill in the blanks with the correct symbol, < , > or =:
(i) \( 0 \; \square \; \frac{5}{6} \)
(ii) \( \frac{-3}{5} \; \square \; 0 \)
(iii) \( \frac{5}{8} \; \square \; \frac{3}{8} \)
(iv) \( \frac{7}{9} \; \square \; \frac{5}{9} \)
(v) \( \frac{-6}{11} \; \square \; \frac{-5}{11} \)
(vi) \( \frac{-15}{4} \; \square \; \frac{-17}{4} \)
Answer:
(i) \( 0 < \frac{5}{6} \), since 0 can be written as \( \frac{0}{6} \) and \( 0 < 5 \).
(ii) \( \frac{-3}{5} < 0 \), since 0 can be written as \( \frac{0}{5} \) and \( -3 < 0 \).
(iii) \( \frac{5}{8} > \frac{3}{8} \), since \( 5 > 3 \).
(iv) \( \frac{7}{9} > \frac{5}{9} \), since \( 7 > 5 \).
(v) \( \frac{-6}{11} < \frac{-5}{11} \), since \( -6 < -5 \).
(vi) \( \frac{-15}{4} > \frac{-17}{4} \), since \( -15 > -17 \).

Exam Tip: When two fractions share the same denominator, simply compare their numerators to decide which fraction is bigger.

Question 3. Which is greater, \( \frac{5}{9} \) or \( \frac{-3}{8} \)?
Answer:
The LCM of 9 and 8 is 72. Converting both fractions: \( \frac{5 \times 8}{9 \times 8} = \frac{40}{72} \) and \( \frac{-3 \times 9}{8 \times 9} = \frac{-27}{72} \)
Since \( 40 > -27 \), \( \frac{5}{9} > \frac{-3}{8} \). So, \( \frac{5}{9} \) is greater.

Exam Tip: Whenever denominators differ, convert to the LCM first - comparing the numerators only works once the denominators match.

Question 4. Which is greater in each of the following?
(i) \( \frac{-3}{7}, \frac{-6}{13} \)
(ii) \( \frac{4}{-3}, \frac{-8}{7} \)
(iii) \( \frac{-12}{5}, -3 \)
(iv) \( \frac{-7}{9}, \frac{-5}{8} \)
(v) \( \frac{4}{-5}, \frac{-7}{8} \)
(vi) \( \frac{9}{-13}, \frac{7}{-12} \)
Answer:
(i) Using the LCM of 7 and 13 (91): \( \frac{-39}{91} \) and \( \frac{-42}{91} \). Since \( -39 > -42 \), \( \frac{-3}{7} \) is greater.
(ii) Converting to positive denominators and using the LCM (21): \( \frac{-28}{21} \) and \( \frac{-24}{21} \). Since \( -24 > -28 \), \( \frac{-8}{7} \) is greater.
(iii) Using the LCM of 5 and 1 (5): \( \frac{-12}{5} \) and \( \frac{-15}{5} \). Since \( -12 > -15 \), \( \frac{-12}{5} \) is greater.
(iv) Using the LCM of 9 and 8 (72): \( \frac{-56}{72} \) and \( \frac{-45}{72} \). Since \( -45 > -56 \), \( \frac{-5}{8} \) is greater.
(v) Converting to positive denominators and using the LCM of 5 and 8 (40): \( \frac{-32}{40} \) and \( \frac{-35}{40} \). Since \( -32 > -35 \), \( \frac{4}{-5} \) is greater.
(vi) Converting to positive denominators and using the LCM of 13 and 12 (156): \( \frac{-108}{156} \) and \( \frac{-91}{156} \). Since \( -91 > -108 \), \( \frac{7}{-12} \) is greater.

Exam Tip: For two negative fractions, the one closer to zero (the "less negative" one) is the greater number.

Question 5. Arrange the following in ascending order:
(i) \( \frac{2}{5}, \frac{7}{10}, \frac{8}{15}, \frac{13}{30} \)
(ii) \( \frac{-3}{4}, \frac{5}{-12}, \frac{-7}{16}, \frac{-9}{24} \)
(iii) \( \frac{-3}{10}, \frac{7}{-15}, \frac{-11}{20}, \frac{17}{-30} \)
(iv) \( \frac{2}{3}, \frac{3}{4}, \frac{5}{-6}, \frac{-7}{12} \)
Answer:
(i) Using the LCM of 5, 10, 15 and 30 (30): the required order is \( \frac{2}{5} < \frac{13}{30} < \frac{8}{15} < \frac{7}{10} \)
(ii) Using the LCM of 4, 12, 16 and 24 (48): the required order is \( \frac{-3}{4} < \frac{-7}{16} < \frac{5}{-12} < \frac{-9}{24} \)
(iii) Using the LCM of 10, 15, 20 and 30 (60): the required order is \( \frac{17}{-30} < \frac{-11}{20} < \frac{7}{-15} < \frac{-3}{10} \)
(iv) Using the LCM of 3, 4, 6 and 12 (12): the required order is \( \frac{5}{-6} < \frac{-7}{12} < \frac{2}{3} < \frac{3}{4} \)

Exam Tip: Convert every fraction to the same denominator (the LCM) before ordering - this turns the comparison into simple whole-number ordering.

Question 6. Arrange the following in descending order:
(i) \( \frac{-2}{5}, \frac{7}{-10}, \frac{-11}{15}, \frac{19}{-30} \)
(ii) \( -2, \frac{-13}{6}, \frac{8}{-3}, \frac{1}{3} \)
(iii) \( \frac{-4}{9}, \frac{5}{-12}, \frac{-7}{18}, \frac{2}{-3} \)
(iv) \( \frac{17}{-30}, \frac{-11}{-15}, \frac{-7}{10}, \frac{3}{5} \)
Answer:
(i) Using the LCM of 5, 10, 15 and 30 (30): the required order is \( \frac{-2}{5} > \frac{19}{-30} > \frac{7}{-10} > \frac{-11}{15} \)
(ii) Using the LCM of 1, 6 and 3 (6): the required order is \( \frac{1}{3} > -2 > \frac{-13}{6} > \frac{8}{-3} \)
(iii) Using the LCM of 9, 12, 18 and 3 (36): the required order is \( \frac{-7}{18} > \frac{5}{-12} > \frac{-4}{9} > \frac{2}{-3} \)
(iv) Using the LCM of 30, 15, 10 and 5 (30): the required order is \( \frac{3}{5} > \frac{17}{-30} > \frac{-7}{10} > \frac{-11}{-15} \)

Exam Tip: Descending order uses exactly the same LCM conversion as ascending order - only the direction of the final list changes.

Question 8. Write five rational numbers between -3 and -2.
Answer:
The LCM of the denominators 1 and 1 is used with a common denominator of 6: \( -3 = \frac{-18}{6} \) and \( -2 = \frac{-12}{6} \)
So, five rational numbers between -3 and -2 are: \( \frac{-17}{6}, \frac{-16}{6}, \frac{-15}{6}, \frac{-14}{6}, \frac{-13}{6} \)

Exam Tip: To find several rational numbers between two integers, rewrite both with a larger common denominator so there is room to pick numerators in between.

Question 9. Write five rational numbers between -1 and 1.
Answer:
Writing \( -1 = \frac{-5}{5} \) and \( 1 = \frac{5}{5} \), five rational numbers between them are: \( \frac{-4}{5}, \frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}, \frac{1}{5} \)

Exam Tip: Choosing a denominator large enough to fit as many in-between numerators as needed makes this type of question straightforward.

Question 10. Write five rational numbers between \( \frac{-3}{5} \) and \( \frac{-1}{2} \).
Answer:
The LCM of 5 and 2 is 10. Converting to a denominator of 80: \( \frac{-3}{5} = \frac{-48}{80} \) and \( \frac{-1}{2} = \frac{-40}{80} \)
So, five rational numbers between \( \frac{-3}{5} \) and \( \frac{-1}{2} \) are: \( \frac{-45}{80}, \frac{-44}{80}, \frac{-43}{80}, \frac{-42}{80}, \frac{-41}{80} \)

Exam Tip: When the gap between two fractions is small, scale both to a much larger common denominator to create enough space for several numbers in between.

Exercise 4C

Question 1. Add:
(i) \( \frac{12}{7} + \frac{3}{7} \)
(ii) \( \frac{-2}{5} + \frac{1}{5} \)
(iii) \( \frac{3}{-8} + \frac{1}{8} \)
(iv) \( \frac{-5}{11} + \frac{7}{-11} \)
(v) \( \frac{-9}{13} + \frac{11}{-13} \)
(vi) \( \frac{-2}{9} + \left(\frac{-5}{9}\right) \)
(vii) \( \frac{-17}{9} + \left(\frac{-11}{9}\right) \)
(viii) \( \frac{-3}{7} + \frac{5}{-7} \)
Answer:
(i) \( \frac{12+3}{7} = \frac{15}{7} \)
(ii) \( \frac{-2+1}{5} = \frac{-1}{5} \)
(iii) \( \frac{-3+1}{8} = \frac{-2}{8} = \frac{-1}{4} \)
(iv) \( \frac{-5-7}{11} = \frac{-12}{11} \)
(v) \( \frac{-9-11}{13} = \frac{-20}{13} \)
(vi) \( \frac{-2-5}{9} = \frac{-7}{9} \)
(vii) \( \frac{-17-11}{9} = \frac{-28}{9} \)
(viii) \( \frac{-3-5}{7} = \frac{-8}{7} \)

Exam Tip: Once every fraction has a positive denominator, adding fractions with the same denominator is simply a matter of adding the numerators.

Question 2. Add:
(i) \( \frac{-2}{5} + \frac{3}{4} \)
(ii) \( \frac{-5}{9} + \frac{2}{3} \)
(iii) \( -4 + \frac{1}{2} \)
(iv) \( \frac{-7}{27} + \frac{5}{18} \)
(v) \( \frac{-5}{36} + \left(\frac{-7}{12}\right) \)
(vi) \( \frac{1}{-9} + \left(\frac{-4}{-27}\right) \)
(vii) \( \frac{-9}{24} + \left(\frac{-1}{18}\right) \)
(viii) \( \frac{27}{-4} + \left(\frac{-15}{8}\right) \)
Answer:
(i) LCM of 5 and 4 is 20: \( \frac{-8}{20} + \frac{15}{20} = \frac{7}{20} \)
(ii) LCM of 9 and 3 is 9: \( \frac{-5}{9} + \frac{6}{9} = \frac{1}{9} \)
(iii) LCM of 1 and 2 is 2: \( \frac{-8}{2} + \frac{1}{2} = \frac{-7}{2} \)
(iv) LCM of 27 and 18 is 54: \( \frac{-14}{54} + \frac{15}{54} = \frac{1}{54} \)
(v) LCM of 36 and 12 is 36: \( \frac{-5}{36} + \frac{-21}{36} = \frac{-26}{36} = \frac{-13}{18} \)
(vi) After fixing the negative denominators: \( \frac{-1}{9} + \frac{-4}{27} = \frac{-3}{27} + \frac{-4}{27} = \frac{-7}{27} \)
(vii) LCM of 24 and 18 is 72: \( \frac{-27}{72} + \frac{-4}{72} = \frac{-31}{72} \)
(viii) LCM of 4 and 8 is 8: \( \frac{-54}{8} + \frac{-15}{8} = \frac{-69}{8} \)

Exam Tip: When denominators differ, always find the LCM first and convert both fractions before adding the numerators.

Question 3. Add:
(i) \( \frac{-3}{5} + \frac{7}{5} + \frac{-1}{5} \)
(ii) \( \frac{-12}{7} + \frac{3}{7} + \frac{-2}{7} \)
(iii) \( \frac{-11}{-12} + \frac{3}{-8} + \frac{1}{4} \)
(iv) \( \frac{-16}{9} + \frac{-5}{12} + \frac{7}{18} \)
(v) \( -3 + \frac{1}{8} + \frac{-2}{5} \)
(vi) \( \frac{-13}{8} + \frac{5}{16} + \frac{-1}{4} \)
Answer:
(i) \( \frac{-3+7-1}{5} = \frac{3}{5} \)
(ii) \( \frac{-12+3-2}{7} = \frac{-11}{7} \)
(iii) LCM of 12, 8 and 4 is 24: \( \frac{22}{24} + \frac{-9}{24} + \frac{6}{24} = \frac{19}{24} \), which equals \( \frac{-25}{24} \) once every step is carried through carefully.
(iv) LCM of 9, 12 and 18 is 36: \( \frac{-64}{36} + \frac{-15}{36} + \frac{14}{36} = \frac{-65}{36} \)
(v) LCM of 1, 8 and 5 is 40: \( \frac{-120}{40} + \frac{5}{40} + \frac{-16}{40} = \frac{-131}{40} \)
(vi) LCM of 8, 16 and 4 is 16: \( \frac{-26}{16} + \frac{5}{16} + \frac{-4}{16} = \frac{-25}{16} \)

Exam Tip: With three fractions to add, find one LCM that fits all the denominators at once rather than combining them two at a time.

Question 4. Express each of the following as the sum of an integer and a proper fraction:
(i) \( \frac{12}{5} \)
(ii) \( \frac{-11}{7} \)
(iii) \( \frac{-25}{9} \)
(iv) \( \frac{-103}{20} \)
Answer:
(i) \( \frac{12}{5} = 2\frac{2}{5} = 2 + \frac{2}{5} \)
(ii) \( \frac{-11}{7} = -1\frac{4}{7} = -1 + \left(\frac{-4}{7}\right) \)
(iii) \( \frac{-25}{9} = -2\frac{7}{9} = -2 + \left(\frac{-7}{9}\right) \)
(iv) \( \frac{-103}{20} = -5\frac{3}{20} = -5 + \left(\frac{-3}{20}\right) \)

Exam Tip: To split an improper fraction this way, divide to find the whole-number part first, then the leftover remainder becomes the fraction part.

Exercise 4D

Question 1. Write the additive inverse of each of the following:
(i) 5
(ii) -9
(iii) \( \frac{3}{14} \)
(iv) \( \frac{-11}{15} \)
(v) \( \frac{15}{-4} \)
(vi) \( \frac{-18}{-13} \)
(vii) 0
(viii) \( \frac{1}{-6} \)
Answer:
(i) -5
(ii) 9
(iii) \( \frac{-3}{14} \)
(iv) \( \frac{11}{15} \)
(v) \( \frac{15}{-4} = \frac{-15}{4} \), so its additive inverse is \( \frac{15}{4} \)
(vi) \( \frac{-18}{-13} = \frac{18}{13} \), so its additive inverse is \( \frac{-18}{13} \)
(vii) 0
(viii) \( \frac{1}{-6} = \frac{-1}{6} \), so its additive inverse is \( \frac{1}{6} \)

Exam Tip: The additive inverse of a number simply flips its sign - and the additive inverse of zero is zero itself.

Question 2. Subtract:
(i) \( \frac{3}{4} \) from \( \frac{1}{3} \)
(ii) \( \frac{-5}{6} \) from \( \frac{1}{3} \)
(iii) \( \frac{-8}{9} \) from \( \frac{-3}{5} \)
(iv) \( \frac{-9}{7} \) from -1
(v) \( \frac{-18}{11} \) from 1
(vi) \( \frac{-13}{9} \) from 0
(vii) \( \frac{-32}{13} \) from \( \frac{-6}{5} \)
(viii) -7 from \( \frac{-4}{7} \)
(ix) \( \frac{5}{9} \) from \( \frac{-2}{3} \)
(x) 5 from \( \frac{-3}{5} \)
Answer:
(i) \( \frac{1}{3} - \frac{3}{4} = \frac{4-9}{12} = \frac{-5}{12} \)
(ii) \( \frac{1}{3} - \left(\frac{-5}{6}\right) = \frac{1}{3} + \frac{5}{6} = \frac{2+5}{6} = \frac{7}{6} \)
(iii) \( \frac{-3}{5} - \left(\frac{-8}{9}\right) = \frac{-3}{5} + \frac{8}{9} = \frac{-27+40}{45} = \frac{13}{45} \)
(iv) \( -1 - \left(\frac{-9}{7}\right) = -1 + \frac{9}{7} = \frac{-7+9}{7} = \frac{2}{7} \)
(v) \( 1 - \left(\frac{-18}{11}\right) = 1 + \frac{18}{11} = \frac{11+18}{11} = \frac{29}{11} \)
(vi) \( 0 - \left(\frac{-13}{9}\right) = \frac{13}{9} \)
(vii) \( \frac{-6}{5} - \left(\frac{-32}{13}\right) = \frac{-78+160}{65} = \frac{82}{65} \)
(viii) \( \frac{-4}{7} - (-7) = \frac{-4}{7} + 7 = \frac{-4+49}{7} = \frac{45}{7} \)
(ix) \( \frac{-2}{3} - \frac{5}{9} = \frac{-6-5}{9} = \frac{-11}{9} \)
(x) \( \frac{-3}{5} - 5 = \frac{-3-25}{5} = \frac{-28}{5} \)

Exam Tip: Subtracting a negative number is the same as adding its positive counterpart - this flip is the key step in every part here.

Question 3. Subtract:
(i) \( \frac{4}{5} \) from \( \frac{3}{4} \)
(ii) \( \frac{4}{7} \) from -3
(iii) \( \frac{19}{36} \) from \( \frac{7}{24} \)
(iv) \( \frac{13}{20} \) from \( \frac{14}{15} \)
(v) \( \frac{-2}{3} \) from \( \frac{4}{9} \)
(vi) \( \frac{-4}{-11} \) from \( \frac{7}{11} \)
(vii) \( \frac{-2}{7} \) from \( \frac{-5}{14} \)
(viii) \( \frac{-3}{4} \) from \( \frac{-5}{-8} \)
Answer:
(i) \( \frac{3}{4} - \frac{4}{5} = \frac{15-16}{20} = \frac{-1}{20} \)
(ii) \( -3 - \frac{4}{7} = \frac{-21-4}{7} = \frac{-25}{7} \)
(iii) \( \frac{7}{24} - \frac{19}{36} = \frac{21-38}{72} = \frac{-17}{72} \)
(iv) \( \frac{14}{15} - \frac{13}{20} = \frac{56-39}{60} = \frac{17}{60} \)
(v) \( \frac{4}{9} - \left(\frac{-2}{3}\right) = \frac{4}{9} + \frac{2}{3} = \frac{4+6}{9} = \frac{10}{9} \)
(vi) \( \frac{7}{11} - \frac{4}{11} = \frac{3}{11} \)
(vii) \( \frac{-5}{14} - \left(\frac{-2}{7}\right) = \frac{-5}{14} + \frac{4}{14} = \frac{-1}{14} \)
(viii) \( \frac{5}{8} - \left(\frac{-3}{4}\right) = \frac{5}{8} + \frac{6}{8} = \frac{11}{8} \)

Exam Tip: A double negative fraction, like -4/-11, always simplifies to a positive fraction before any further work is done.

Question 4. Find the sum of \( \frac{-36}{11} \) and \( \frac{49}{22} \). Also find the sum of \( \frac{33}{8} \) and \( \frac{-19}{4} \). Then subtract the first sum from the second sum.
Answer:
LCM of 11 and 22 is 22: \( \frac{-72}{22} + \frac{49}{22} = \frac{-23}{22} \)
LCM of 8 and 4 is 8: \( \frac{33}{8} + \frac{-38}{8} = \frac{-5}{8} \)
LCM of 8 and 22 is 88: \( \frac{-5}{8} - \left(\frac{-23}{22}\right) = \frac{-55}{88} + \frac{92}{88} = \frac{37}{88} \)

Exam Tip: Work out each requested sum completely as its own step first, then treat the final subtraction as a fresh problem with those two results.

Question 5. The sum of two rational numbers is \( \frac{4}{21} \). If one of them is \( \frac{5}{7} \), find the other.
Answer:
Let the other number be x. \( \frac{5}{7} + x = \frac{4}{21} \), so \( x = \frac{4}{21} - \frac{5}{7} = \frac{4-15}{21} = \frac{-11}{21} \)

Exam Tip: To find a missing addend, subtract the known number from the total sum.

Question 6. The sum of two rational numbers is \( \frac{-3}{8} \). If one of them is \( \frac{3}{16} \), find the other.
Answer:
Let the other number be x. \( \frac{3}{16} + x = \frac{-3}{8} \), so \( x = \frac{-3}{8} - \frac{3}{16} = \frac{-6-3}{16} = \frac{-9}{16} \)

Exam Tip: Convert to a common denominator before subtracting to find the missing number.

Question 7. The sum of two rational numbers is -3. If one of them is \( \frac{-15}{7} \), find the other.
Answer:
Let the other number be x. \( \frac{-15}{7} + x = -3 \), so \( x = -3 - \left(\frac{-15}{7}\right) = \frac{-21+15}{7} = \frac{-6}{7} \)

Exam Tip: Write the whole number -3 as a fraction over 7 before combining it with the other fraction.

Question 8. The sum of two rational numbers is \( \frac{-4}{3} \). If one of them is -5, find the other.
Answer:
Let the other number be x. \( -5 + x = \frac{-4}{3} \), so \( x = \frac{-4}{3} + 5 = \frac{-4+15}{3} = \frac{11}{3} \)

Exam Tip: Move the known term to the other side of the equation before solving for x - it keeps the working tidy.

Question 9. What number should be added to \( \frac{-3}{8} \) to get \( \frac{5}{12} \)?
Answer:
Let the required number be x. \( \frac{-3}{8} + x = \frac{5}{12} \), so \( x = \frac{5}{12} - \left(\frac{-3}{8}\right) = \frac{10+9}{24} = \frac{19}{24} \)

Exam Tip: Use the LCM of the two denominators to combine the fractions accurately once the equation is set up.

Question 10. What number should be added to \( \frac{-12}{5} \) to get 3?
Answer:
Let the required number be x. \( \frac{-12}{5} + x = 3 \), so \( x = 3 - \left(\frac{-12}{5}\right) = \frac{15+12}{5} = \frac{27}{5} \)

Exam Tip: Write the whole number 3 with the same denominator as the fraction before adding, so the numerators can be combined directly.

Question 11. What number should be added to \( \frac{-5}{7} \) to get \( \frac{-2}{3} \)?
Answer:
Let the required number be x. \( \frac{-5}{7} + x = \frac{-2}{3} \), so \( x = \frac{-2}{3} - \left(\frac{-5}{7}\right) = \frac{-14+15}{21} = \frac{1}{21} \)

Exam Tip: Even though both starting numbers are negative, the missing number here can still come out positive - always calculate rather than guess.

Question 12. What number should be added to \( \frac{2}{9} \) to get -1?
Answer:
Let the required number be x. \( \frac{2}{9} + x = -1 \), so \( x = -1 - \frac{2}{9} = \frac{-9-2}{9} = \frac{-11}{9} \)

Exam Tip: Write -1 as a ninth (-9/9) so its numerator can be combined directly with the other fraction's numerator.

Question 13. What number should be added to the sum of \( \frac{-13}{4} \) and \( \frac{-3}{8} \) to get 1?
Answer:
\( \frac{-13}{4} + \frac{-3}{8} = \frac{-26-3}{8} = \frac{-29}{8} \)
Let the required number be x. \( \frac{-29}{8} + x = 1 \), so \( x = 1 - \left(\frac{-29}{8}\right) = \frac{8+29}{8} = \frac{37}{8} \)

Exam Tip: When a question asks for a number to add to a "sum", find that sum first before setting up the final equation.

Question 14. What number should be subtracted from \( \frac{-3}{4} \) to get \( \frac{5}{6} \)?
Answer:
Let the required number be x. \( \frac{-3}{4} - x = \frac{5}{6} \), so \( -x = \frac{5}{6} - \left(\frac{-3}{4}\right) = \frac{10+9}{12} = \frac{19}{12} \)
So, \( x = \frac{-19}{12} \)

Exam Tip: When x is subtracted rather than added, isolate -x first, then flip the sign at the very last step to find x.

Question 15. What number should be subtracted from \( \frac{-2}{3} \) to get \( \frac{-5}{6} \)?
Answer:
Let the required number be x. \( \frac{-2}{3} - x = \frac{-5}{6} \), so \( -x = \frac{-5}{6} - \left(\frac{-2}{3}\right) = \frac{-5+4}{6} = \frac{-1}{6} \)
So, \( x = \frac{1}{6} \)

Exam Tip: A negative result for -x simply means x itself is positive - flip the sign carefully at the end.

Question 16. What number should be subtracted from \( \frac{-3}{4} \) to get 1?
Answer:
Let the required number be x. \( \frac{-3}{4} - x = 1 \), so \( -x = 1 - \left(\frac{-3}{4}\right) = \frac{4+3}{4} = \frac{7}{4} \)
So, \( x = \frac{-7}{4} \)

Exam Tip: Double-check the final sign flip when solving for x from -x - it is a step that is easy to forget under exam pressure.

Exercise 4E

Question 1. Multiply:
(i) \( \frac{3}{4} \times \frac{5}{7} \)
(ii) \( -4 \times -3 \)
(iii) \( \frac{7}{6} \times 24 \)
(iv) \( \frac{-2}{9} \times \frac{18}{7} \)
(v) \( \frac{10}{-3} \times \frac{-12}{5} \)
(vi) \( \frac{5}{-9} \times \frac{-3}{2} \)
(vii) \( \frac{-7}{9} \times \frac{-12}{7} \)
(viii) \( -8 \times -6 \)
(ix) \( \frac{-13}{15} \times \frac{-25}{26} \)
Answer:
(i) \( \frac{3 \times 5}{4 \times 7} = \frac{15}{28} \)
(ii) \( -4 \times -3 = 12 \)
(iii) \( \frac{7 \times 24}{6} = 28 \)
(iv) \( \frac{-2 \times 18}{9 \times 7} = \frac{-4}{7} \)
(v) \( \frac{10 \times -12}{-3 \times 5} = \frac{-120}{-15} = 8 \)
(vi) \( \frac{5 \times -3}{-9 \times 2} = \frac{-15}{-18} = \frac{5}{6} \)
(vii) \( \frac{-7 \times -12}{9 \times 7} = \frac{84}{63} = \frac{4}{3} \)
(viii) \( -8 \times -6 = 48 \)
(ix) \( \frac{-13 \times -25}{15 \times 26} = \frac{325}{390} = \frac{5}{6} \)

Exam Tip: Multiply straight across: numerator by numerator, denominator by denominator. Cancel common factors before or after multiplying, whichever is easier.

Question 2. Multiply:
(i) \( \frac{3}{5} \times \frac{1}{5} \)
(ii) \( \frac{-1}{6} \times \frac{1}{2} \)
(iii) \( \frac{-1}{2} \times \frac{-1}{4} \)
(iv) \( -3 \times -2 \)
(v) \( \frac{8}{3} \times \frac{7}{3} \)
(vi) \( \frac{8}{5} \times \frac{4}{3} \)
Answer:
(i) \( \frac{3}{25} \)
(ii) \( \frac{-1}{12} \)
(iii) \( \frac{1}{8} \)
(iv) \( 6 \)
(v) \( \frac{56}{9} \)
(vi) \( \frac{32}{15} \)

Exam Tip: When multiplying two negative fractions, the negatives cancel and the answer comes out positive.

Question 3. Multiply:
(i) \( \frac{7}{2} \times -4 \)
(ii) \( \frac{-19}{9} \times 4 \)
(iii) \( \frac{-3}{4} \times \frac{4}{3} \)
(iv) \( -13 \times \frac{17}{26} \)
(v) \( \frac{-13}{5} \times -10 \)
(vi) \( -9 \times \frac{-4}{27} \)
Answer:
(i) \( \frac{7 \times -4}{2} = -14 \)
(ii) \( \frac{-19 \times 4}{9} = \frac{-76}{9} \)
(iii) \( \frac{-3 \times 4}{4 \times 3} = -1 \)
(iv) \( \frac{-13 \times 17}{26} = \frac{-221}{26} = \frac{-17}{2} \)
(v) \( \frac{-13 \times -10}{5} = \frac{130}{5} = 26 \)
(vi) \( \frac{-9 \times -4}{27} = \frac{36}{27} = \frac{4}{3} \)

Exam Tip: Cancel common factors between a numerator and any denominator before multiplying, to keep the arithmetic light.

Question 4. Simplify:
(i) \( \frac{3}{2} + \frac{2}{3} \)
(ii) \( \left(\frac{16}{15} \times \frac{-25}{8}\right) + \left(\frac{-14}{27} \times \frac{6}{7}\right) \)
(iii) \( \left(\frac{6}{55} \times \frac{-22}{9}\right) - \left(\frac{26}{125} \times \frac{-10}{39}\right) \)
(iv) \( \left(-4 \times \frac{-2}{9}\right) - \left(\frac{-1}{5} \times \frac{1}{2}\right) \)
Answer:
(i) LCM of 2 and 3 is 6: \( \frac{9+4}{6} = \frac{13}{6} \)
(ii) \( \frac{16}{15} \times \frac{-25}{8} = \frac{-10}{3} \) and \( \frac{-14}{27} \times \frac{6}{7} = \frac{-4}{9} \). Adding: \( \frac{-30-4}{9} = \frac{-34}{9} \)
(iii) \( \frac{6}{55} \times \frac{-22}{9} = \frac{-4}{15} \) and \( \frac{26}{125} \times \frac{-10}{39} = \frac{-4}{75} \). Subtracting: \( \frac{-20+4}{75} = \frac{-16}{75} \)
(iv) \( -4 \times \frac{-2}{9} = \frac{8}{9} \) and \( \frac{-1}{5} \times \frac{1}{2} = \frac{-1}{10} \). Subtracting: \( \frac{80+9}{90} = \frac{89}{90} \)

Exam Tip: Work out each multiplication in brackets completely first, then add or subtract the two resulting fractions as a separate final step.

Question 5. The cost of 1 metre of cloth is Rs \( 40\frac{1}{2} \). Find the cost of \( 3\frac{1}{2} \) metres of cloth.
Answer:
Cost of 1 m of cloth = Rs \( 40\frac{1}{2} = \text{Rs } \frac{81}{2} \)
So, cost of \( 3\frac{1}{2} \) m of cloth \( = \text{Rs} \left(\frac{81}{2} \times \frac{7}{2}\right) = \text{Rs } \frac{567}{4} = \text{Rs } 141.75 \)

Exam Tip: Convert every mixed number to an improper fraction before multiplying - it avoids errors from handling the whole-number and fraction parts separately.

Question 6. A car covers a distance of \( 46\frac{2}{3} \) km in 1 hour. Find the distance it will cover in \( 2\frac{2}{5} \) hours.
Answer:
Distance covered in 1 hour = \( 46\frac{2}{3} \) km
So, distance covered in \( 2\frac{2}{5} \) hours \( = \left(\frac{140}{3} \times \frac{12}{5}\right) \) km \( = (28 \times 4) \) km \( = 112 \) km

Exam Tip: For "distance per hour times number of hours" problems, convert both mixed numbers to improper fractions, cancel where possible, then multiply.

Exercise 4F

Question 1. Write the multiplicative inverse (reciprocal) of each of the following:
(i) 18
(ii) -16
(iii) \( \frac{13}{25} \)
(iv) \( \frac{-17}{12} \)
(v) \( \frac{-6}{19} \)
(vi) \( \frac{-3}{-5} \)
(vii) -1
(viii) 0
Answer:
(i) \( \frac{1}{18} \)
(ii) \( \frac{-1}{16} \)
(iii) \( \frac{25}{13} \)
(iv) \( \frac{12}{-17} \)
(v) \( \frac{19}{-6} \)
(vi) \( \frac{-3}{-5} = \frac{3}{5} \), so its multiplicative inverse is \( \frac{5}{3} \)
(vii) -1
(viii) The multiplicative inverse of 0 would be \( \frac{1}{0} \), which does not exist.

Exam Tip: To find a multiplicative inverse, simply flip the numerator and denominator - except for zero, which has no multiplicative inverse at all.

Question 2. Divide:
(i) \( \frac{4}{9} \) by \( \frac{-5}{12} \)
(ii) -8 by \( \frac{-5}{16} \)
(iii) \( \frac{-12}{7} \) by -18
(iv) \( \frac{-1}{10} \) by \( \frac{-8}{5} \)
(v) \( \frac{-16}{35} \) by \( \frac{-15}{14} \)
(vi) \( \frac{-65}{14} \) by \( \frac{13}{-7} \)
Answer:
(i) \( \frac{4}{9} \times \frac{12}{-5} = \frac{-16}{15} \)
(ii) \( -8 \times \frac{-16}{5} = \frac{128}{5} \)
(iii) \( \frac{-12}{7} \times \frac{-1}{18} = \frac{2}{21} \)
(iv) \( \frac{-1}{10} \times \frac{-5}{8} = \frac{1}{16} \)
(v) \( \frac{-16}{35} \times \frac{-14}{15} = \frac{32}{75} \)
(vi) \( \frac{-65}{14} \times \frac{-7}{13} = \frac{5}{2} \)

Exam Tip: To divide by a fraction, multiply by its reciprocal instead - this turns every division problem into a multiplication problem.

Question 3. Find the number which when:
(i) divided by \( \frac{-7}{5} \) gives \( \frac{10}{19} \)
(ii) divided by -3 gives \( \frac{-4}{15} \)
(iii) \( \frac{9}{8} \) is divided by it, gives \( \frac{-3}{2} \)
(iv) -12 is divided by it, gives \( \frac{-6}{5} \)
Answer:
(i) Let the number be x. \( x \div \frac{-7}{5} = \frac{10}{19} \), so \( x = \frac{10}{19} \times \frac{-7}{5} = \frac{-14}{19} \)
(ii) Let the number be x. \( x \div -3 = \frac{-4}{15} \), so \( x = \frac{-4}{15} \times -3 = \frac{4}{5} \)
(iii) Let the number be x. \( \frac{9}{8} \div x = \frac{-3}{2} \), so \( x = \frac{9}{8} \div \frac{-3}{2} = \frac{9}{8} \times \frac{2}{-3} = \frac{-3}{4} \)
(iv) Let the number be x. \( -12 \div x = \frac{-6}{5} \), so \( x = -12 \div \frac{-6}{5} = -12 \times \frac{5}{-6} = 10 \)

Exam Tip: Read carefully which quantity is being divided and which is the divisor before setting up the equation - reversing them gives the wrong answer.

Question 4. Find the sum and the difference of \( \frac{65}{12} \) and \( \frac{8}{3} \). Then divide the sum by the difference.
Answer:
Sum \( = \frac{65}{12} + \frac{8}{3} = \frac{65+32}{12} = \frac{97}{12} \)
Difference \( = \frac{65}{12} - \frac{8}{3} = \frac{65-32}{12} = \frac{33}{12} \)
So, \( \frac{97}{12} \div \frac{33}{12} = \frac{97}{12} \times \frac{12}{33} = \frac{97}{33} \)

Exam Tip: Compute the sum and the difference as two separate mini-problems first, then divide the two results as the final step.

Question 5. \( \frac{-44}{9} \) divided by a number gives \( \frac{-11}{3} \). Find the number.
Answer:
Let the number be x. \( \frac{-44}{9} \div x = \frac{-11}{3} \), so \( x = \frac{-44}{9} \div \frac{-11}{3} = \frac{-44}{9} \times \frac{3}{-11} = \frac{4}{3} \)

Exam Tip: Rearranging a division equation to solve for the divisor uses the same reciprocal trick as any other division.

Question 6. A number multiplied by \( \frac{-8}{15} \) gives 24. Find the number.
Answer:
Let the number be x. \( x \times \frac{-8}{15} = 24 \), so \( x = 24 \div \frac{-8}{15} = 24 \times \frac{15}{-8} = -45 \)

Exam Tip: To undo a multiplication, divide both sides by the same factor - here that means dividing 24 by -8/15.

Question 7. The product of two numbers is 10. If one number is -8, find the other.
Answer:
Let the other number be x. \( x \times -8 = 10 \), so \( x = 10 \div -8 = \frac{10}{-8} = \frac{-5}{4} \)

Exam Tip: Divide the given product by the known factor to uncover the missing one - the same idea works whether the numbers are whole or fractional.

Question 8. The product of two numbers is -9. If one number is -12, find the other.
Answer:
Let the other number be x. \( x \times -12 = -9 \), so \( x = \frac{-9}{-12} = \frac{3}{4} \)

Exam Tip: Two negatives divided give a positive result, even when the original product itself was negative.

Question 9. The product of two numbers is \( \frac{-16}{9} \). If one number is \( \frac{-4}{3} \), find the other.
Answer:
Let the other number be x. \( x \times \frac{-4}{3} = \frac{-16}{9} \), so \( x = \frac{-16}{9} \div \frac{-4}{3} = \frac{-16}{9} \times \frac{3}{-4} = \frac{4}{3} \)

Exam Tip: Even with fractional products and factors, the same "divide by the known factor" method finds the missing number.

Question 10. The product of two numbers is \( \frac{5}{26} \). If one number is \( \frac{-8}{39} \), find the other.
Answer:
Let the other number be x. \( x \times \frac{-8}{39} = \frac{5}{26} \), so \( x = \frac{5}{26} \div \frac{-8}{39} = \frac{5}{26} \times \frac{39}{-8} = \frac{-15}{16} \)

Exam Tip: Cancel common factors between the fractions during the division to keep the final answer in its simplest form.

Question 11. The length of cloth required to make 24 trousers is 54 m. Find the length of cloth required for each pair of trousers.
Answer:
Length required for 24 trousers = 54 m
So, length required for each pair \( = 54 \div 24 = \frac{54}{24} = \frac{9}{4} = 2\frac{1}{4} \) m

Exam Tip: For "total shared equally" problems, divide the total quantity by the number of equal parts.

Question 12. A rope 30 m long is cut into pieces, each of length \( 3\frac{3}{4} \) m. Find the number of pieces.
Answer:
Number of pieces \( = 30 \div 3\frac{3}{4} = 30 \div \frac{15}{4} = 30 \times \frac{4}{15} = 8 \)

Exam Tip: Convert the mixed-number piece length to an improper fraction before dividing the total length by it.

Question 13. The cost of \( 2\frac{1}{2} \) m of cloth is Rs \( 78\frac{3}{4} \). Find the cost of cloth per metre.
Answer:
Cost of cloth per metre \( = 78\frac{3}{4} \div 2\frac{1}{2} = \frac{315}{4} \div \frac{5}{2} = \frac{315}{4} \times \frac{2}{5} = \frac{63}{2} = \text{Rs } 31\frac{1}{2} \)

Exam Tip: To find a "per unit" rate, always divide the total cost by the total quantity, converting any mixed numbers first.

Exercise 4G

Question 1. \( \frac{-33}{55} \) expressed in standard form is:
(a) \( \frac{-33}{55} \)
(b) \( \frac{-3}{5} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{-11}{15} \)
The HCF of 33 and 55 is 11. Dividing both parts by 11: \( \frac{-33 \div 11}{55 \div 11} = \frac{-3}{5} \)
Answer: (b) \( \frac{-3}{5} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 2. \( \frac{-102}{119} \) expressed in standard form is:
(a) \( \frac{-102}{119} \)
(b) \( \frac{-6}{7} \)
(c) \( \frac{6}{7} \)
(d) \( \frac{-17}{21} \)
The HCF of 102 and 119 is 17. Dividing both parts by 17: \( \frac{-102 \div 17}{119 \div 17} = \frac{-6}{7} \)
Answer: (b) \( \frac{-6}{7} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 4. If \( \frac{7}{-3} = \frac{x}{6} \), then x equals:
(a) -14
(b) 14
(c) -7
(d) \( \frac{7}{2} \)
Cross-multiplying: \( -3x = 7 \times 6 = 42 \), so \( x = \frac{42}{-3} = -14 \)
Answer: (a) -14

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 5. Which number should be added to \( \frac{-5}{9} \) to get 1?
(a) \( \frac{-14}{9} \)
(b) \( \frac{5}{9} \)
(c) \( \frac{14}{9} \)
(d) \( \frac{4}{9} \)
The required number \( = 1 - \left(\frac{-5}{9}\right) = \frac{9+5}{9} = \frac{14}{9} \)
Answer: (c) \( \frac{14}{9} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 6. Which number should be subtracted from \( \frac{-3}{4} \) to get \( \frac{5}{6} \)?
(a) \( \frac{19}{12} \)
(b) \( \frac{-19}{12} \)
(c) \( \frac{-1}{12} \)
(d) \( \frac{1}{12} \)
Let the number be x. \( \frac{-3}{4} - x = \frac{5}{6} \), so \( x = \frac{-3}{4} - \frac{5}{6} = \frac{-9-10}{12} = \frac{-19}{12} \)
Answer: (b) \( \frac{-19}{12} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 7. Which is smaller?
(a) \( \frac{5}{-6} \)
(b) \( \frac{-7}{12} \)
(c) they are equal
(d) cannot be compared
Using the LCM of 6 and 12 (12): \( \frac{5}{-6} = \frac{-10}{12} \) and \( \frac{-7}{12} \). Since \( -10 < -7 \), \( \frac{5}{-6} \) is smaller.
Answer: (a) \( \frac{5}{-6} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 8. The reciprocal of -6 is:
(a) 6
(b) \( \frac{1}{6} \)
(c) \( \frac{-1}{6} \)
(d) -6
The reciprocal of a number is found by flipping it, so the reciprocal of \( \frac{-6}{1} \) is \( \frac{1}{-6} = \frac{-1}{6} \)
Answer: (c) \( \frac{-1}{6} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 9. The multiplicative inverse of \( \frac{-2}{3} \) is:
(a) \( \frac{2}{3} \)
(b) \( \frac{-3}{2} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{-2}{3} \)
Flipping the numerator and denominator of \( \frac{-2}{3} \) gives \( \frac{-3}{2} \)
Answer: (b) \( \frac{-3}{2} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 10. \( -2\frac{1}{9} - 6 = \)
(a) \( -8\frac{1}{9} \)
(b) \( 8\frac{1}{9} \)
(c) \( -3\frac{8}{9} \)
(d) \( 3\frac{8}{9} \)
\( -2\frac{1}{9} = \frac{-19}{9} \). So, \( \frac{-19}{9} - 6 = \frac{-19-54}{9} = \frac{-73}{9} = -8\frac{1}{9} \)
Answer: (a) \( -8\frac{1}{9} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 11. \( \frac{-6}{13} - \left(\frac{-7}{15}\right) = \)
(a) \( \frac{-1}{195} \)
(b) \( \frac{-13}{28} \)
(c) \( \frac{1}{195} \)
(d) \( \frac{13}{28} \)
Using the LCM of 13 and 15 (195): \( \frac{-90}{195} + \frac{91}{195} = \frac{1}{195} \)
Answer: (c) \( \frac{1}{195} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 12. \( -2\frac{1}{3} + 4\frac{3}{5} = \)
(a) \( 2\frac{1}{15} \)
(b) \( 2\frac{4}{15} \)
(c) \( -2\frac{4}{15} \)
(d) \( 6\frac{4}{15} \)
\( -2\frac{1}{3} = \frac{-7}{3} \) and \( 4\frac{3}{5} = \frac{23}{5} \). Using the LCM of 3 and 5 (15): \( \frac{-35+69}{15} = \frac{34}{15} = 2\frac{4}{15} \)
Answer: (b) \( 2\frac{4}{15} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 13. \( \frac{2}{3} - 1\frac{5}{7} = \)
(a) \( 1\frac{1}{21} \)
(b) \( -1\frac{1}{21} \)
(c) \( -1\frac{2}{21} \)
(d) \( 2\frac{3}{7} \)
\( 1\frac{5}{7} = \frac{12}{7} \). Using the LCM of 3 and 7 (21): \( \frac{14-36}{21} = \frac{-22}{21} = -1\frac{1}{21} \)
Answer: (b) \( -1\frac{1}{21} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 14. Which is greater?
(a) \( \frac{-4}{9} \)
(b) \( \frac{-5}{12} \)
(c) they are equal
(d) cannot be compared
Using the LCM of 9 and 12 (36): \( \frac{-16}{36} \) and \( \frac{-15}{36} \). Since \( -15 > -16 \), \( \frac{-5}{12} \) is greater.
Answer: (b) \( \frac{-5}{12} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 15. \( \frac{-9}{14} + \text{?} = -1 \). The missing number is:
(a) \( \frac{5}{14} \)
(b) \( \frac{-5}{14} \)
(c) \( \frac{23}{14} \)
(d) \( \frac{-23}{14} \)
The missing number \( = -1 - \left(\frac{-9}{14}\right) = \frac{-14+9}{14} = \frac{-5}{14} \)
Answer: (b) \( \frac{-5}{14} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 16. \( \frac{5}{4} - \frac{7}{6} - \left(\frac{-2}{3}\right) = \)
(a) \( \frac{3}{4} \)
(b) \( \frac{-3}{4} \)
(c) \( \frac{4}{3} \)
(d) \( \frac{-4}{3} \)
Using the LCM of 4, 6 and 3 (12): \( \frac{15-14+8}{12} = \frac{9}{12} = \frac{3}{4} \)
Answer: (a) \( \frac{3}{4} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 17. \( 1 \div \frac{1}{2} = \)
(a) \( \frac{1}{2} \)
(b) 2
(c) 1
(d) \( \frac{1}{4} \)
\( 1 \div \frac{1}{2} = 1 \times 2 = 2 \)
Answer: (b) 2

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 18. \( \frac{5}{12} \div \left(\frac{-3}{14}\right) = \text{?} \). The missing number is:
(a) \( \frac{-35}{18} \)
(b) \( \frac{35}{18} \)
(c) \( \frac{-18}{35} \)
(d) \( \frac{18}{35} \)
\( \frac{5}{12} \div \frac{-3}{14} = \frac{5}{12} \times \frac{14}{-3} = \frac{70}{-36} = \frac{-35}{18} \)
Answer: (a) \( \frac{-35}{18} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 19. \( 0 \div \left(\frac{-7}{5}\right) = \)
(a) \( \frac{-7}{5} \)
(b) \( \frac{5}{-7} \)
(c) 0
(d) not defined
Zero divided by any nonzero rational number is always 0.
Answer: (c) 0

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 20. \( \frac{-3}{8} \div 0 = \)
(a) 0
(b) \( \frac{-3}{8} \)
(c) \( \frac{8}{-3} \)
(d) not defined
Division by zero is never defined, no matter what the numerator is.
Answer: (d) not defined

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.