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Chapter 3

Rational Numbers

Class - 7 ML Aggarwal Understanding ICSE Mathematics

Exercise 3.1

Question 1

Which of the following are positive rational numbers?

58,−311,05,7,−4,−3−13,−17−6,9−20\dfrac{5}{8}, \dfrac{-3}{11}, \dfrac{0}{5}, 7, -4, \dfrac{-3}{-13}, \dfrac{-17}{-6}, \dfrac{9}{-20}

Answer

A rational number is positive if its numerator and denominator are either both positive integers or both negative integers.

58\dfrac{5}{8} → numerator and denominator both positive → positive

−311\dfrac{-3}{11} → numerator negative, denominator positive → negative

05\dfrac{0}{5} → equal to 0 → neither positive nor negative

7=717 = \dfrac{7}{1} → both positive → positive

-4 → negative

−3−13\dfrac{-3}{-13} → both negative → positive

−17−6\dfrac{-17}{-6} → both negative → positive

9−20\dfrac{9}{-20} → numerator positive, denominator negative → negative

Hence, the positive rational numbers are 58,7,−3−13\dfrac{5}{8}, 7, \dfrac{-3}{-13} and −17−6\dfrac{-17}{-6}.

Question 2

Which of the following are negative rational numbers?

−57,4−3,−3−11,−6,9,0,−285,317\dfrac{-5}{7}, \dfrac{4}{-3}, \dfrac{-3}{-11}, -6, 9, 0, \dfrac{-28}{5}, \dfrac{31}{7}

Answer

A rational number is negative if one of its numerator and denominator is a positive integer and the other is a negative integer.

−57\dfrac{-5}{7} → numerator negative, denominator positive → negative

4−3\dfrac{4}{-3} → numerator positive, denominator negative → negative

−3−11\dfrac{-3}{-11} → both negative → positive

-6 → negative

9 → positive

0 → neither positive nor negative

−285\dfrac{-28}{5} → numerator negative, denominator positive → negative

317\dfrac{31}{7} → both positive → positive

Hence, the negative rational numbers are −57,4−3,−6\dfrac{-5}{7}, \dfrac{4}{-3}, -6 and −285\dfrac{-28}{5}.

Question 3

Find four rational numbers equivalent to each of the following rational numbers:

(i) 3−7\dfrac{3}{-7}

(ii) −5−9\dfrac{-5}{-9}

Answer

(i) Multiplying the numerator and the denominator of 3−7\dfrac{3}{-7} by 2, 3, 4 and 5, we get:

3×2−7×2=6−143×3−7×3=9−213×4−7×4=12−283×5−7×5=15−35\dfrac{3 \times 2}{-7 \times 2} = \dfrac{6}{-14} \\[1em] \dfrac{3 \times 3}{-7 \times 3} = \dfrac{9}{-21} \\[1em] \dfrac{3 \times 4}{-7 \times 4} = \dfrac{12}{-28} \\[1em] \dfrac{3 \times 5}{-7 \times 5} = \dfrac{15}{-35}

Hence, four rational numbers equivalent to 3−7\dfrac{3}{-7} are 6−14,9−21,12−28\dfrac{6}{-14}, \dfrac{9}{-21}, \dfrac{12}{-28} and 15−35\dfrac{15}{-35}.

(ii) We have −5−9=−5×(−1)−9×(−1)=59\dfrac{-5}{-9} = \dfrac{-5 \times (-1)}{-9 \times (-1)} = \dfrac{5}{9}

Multiplying the numerator and the denominator of 59\dfrac{5}{9} by 2, 3 and 4, we get:

5×29×2=10185×39×3=15275×49×4=2036\dfrac{5 \times 2}{9 \times 2} = \dfrac{10}{18} \\[1em] \dfrac{5 \times 3}{9 \times 3} = \dfrac{15}{27} \\[1em] \dfrac{5 \times 4}{9 \times 4} = \dfrac{20}{36}

Hence, four rational numbers equivalent to −5−9\dfrac{-5}{-9} are 59,1018,1527\dfrac{5}{9}, \dfrac{10}{18}, \dfrac{15}{27} and 2036\dfrac{20}{36}.

Question 4

Write each of the following rational numbers with positive denominator:

(i) 4−9\dfrac{4}{-9}

(ii) 17−33\dfrac{17}{-33}

(iii) −15−38\dfrac{-15}{-38}

Answer

To get a positive denominator, multiply the numerator and the denominator by -1.

(i) Solving,

4−9=4×(−1)−9×(−1)=−49\dfrac{4}{-9} = \dfrac{4 \times (-1)}{-9 \times (-1)} = \dfrac{-4}{9}

∴ 4−9=−49\dfrac{4}{-9} = \dfrac{-4}{9}

(ii) Solving,

17−33=17×(−1)−33×(−1)=−1733\dfrac{17}{-33} = \dfrac{17 \times (-1)}{-33 \times (-1)} = \dfrac{-17}{33}

∴ 17−33=−1733\dfrac{17}{-33} = \dfrac{-17}{33}

(iii) Solving,

−15−38=−15×(−1)−38×(−1)=1538\dfrac{-15}{-38} = \dfrac{-15 \times (-1)}{-38 \times (-1)} = \dfrac{15}{38}

∴ −15−38=1538\dfrac{-15}{-38} = \dfrac{15}{38}

Question 5

Express 5−9\dfrac{5}{-9} as a rational number with

(i) numerator = 20

(ii) numerator = -35

(iii) denominator = -54

(iv) denominator = 72

Answer

(i) To get 20 from 5, we multiply 5 by 4.

5−9=5×4−9×4=20−36\dfrac{5}{-9} = \dfrac{5 \times 4}{-9 \times 4} = \dfrac{20}{-36}

∴ 5−9=20−36\dfrac{5}{-9} = \dfrac{20}{-36}

(ii) To get -35 from 5, we multiply 5 by -7.

5−9=5×(−7)−9×(−7)=−3563\dfrac{5}{-9} = \dfrac{5 \times (-7)}{-9 \times (-7)} = \dfrac{-35}{63}

∴ 5−9=−3563\dfrac{5}{-9} = \dfrac{-35}{63}

(iii) To get -54 from -9, we multiply -9 by 6.

5−9=5×6−9×6=30−54\dfrac{5}{-9} = \dfrac{5 \times 6}{-9 \times 6} = \dfrac{30}{-54}

∴ 5−9=30−54\dfrac{5}{-9} = \dfrac{30}{-54}

(iv) To get 72 from -9, we multiply -9 by -8.

5−9=5×(−8)−9×(−8)=−4072\dfrac{5}{-9} = \dfrac{5 \times (-8)}{-9 \times (-8)} = \dfrac{-40}{72}

∴ 5−9=−4072\dfrac{5}{-9} = \dfrac{-40}{72}

Question 6

Express −80112\dfrac{-80}{112} as a rational number with

(i) numerator = -5

(ii) denominator = -14

Answer

First reduce −80112\dfrac{-80}{112} to standard form. HCF of 80 and 112 is 16.

−80112=−80÷16112÷16=−57\dfrac{-80}{112} = \dfrac{-80 \div 16}{112 \div 16} = \dfrac{-5}{7}

(i) The numerator is already -5.

∴ −80112=−57\dfrac{-80}{112} = \dfrac{-5}{7}

(ii) To get -14 from 7, we multiply 7 by -2.

−57=−5×(−2)7×(−2)=10−14\dfrac{-5}{7} = \dfrac{-5 \times (-2)}{7 \times (-2)} = \dfrac{10}{-14}

∴ −80112=10−14\dfrac{-80}{112} = \dfrac{10}{-14}

Question 7

Which of the following pairs represent the same rational number?

(i) −721,39\dfrac{-7}{21}, \dfrac{3}{9}

(ii) −1620,20−25\dfrac{-16}{20}, \dfrac{20}{-25}

(iii) −35,−1220\dfrac{-3}{5}, \dfrac{-12}{20}

(iv) 8−5,−2415\dfrac{8}{-5}, \dfrac{-24}{15}

Answer

Two rational numbers pq\dfrac{p}{q} and rs\dfrac{r}{s} are equal if and only if p×s=q×rp \times s = q \times r.

(i) For −721\dfrac{-7}{21} and 39\dfrac{3}{9}:

-7 × 9 = -63 and 21 × 3 = 63

As -63 ≠ 63, the pair does not represent the same rational number.

(ii) For −1620\dfrac{-16}{20} and 20−25\dfrac{20}{-25}:

-16 × (-25) = 400 and 20 × 20 = 400

As 400 = 400, the pair represents the same rational number.

(iii) For −35\dfrac{-3}{5} and −1220\dfrac{-12}{20}:

-3 × 20 = -60 and 5 × (-12) = -60

As -60 = -60, the pair represents the same rational number.

(iv) For 8−5\dfrac{8}{-5} and −2415\dfrac{-24}{15}:

8 × 15 = 120 and -5 × (-24) = 120

As 120 = 120, the pair represents the same rational number.

Hence, the pairs (ii), (iii) and (iv) represent the same rational number.

Question 8

Fill in the blanks:

(i) 54=...16=25...=−15...\dfrac{5}{4} = \dfrac{...}{16} = \dfrac{25}{...} = \dfrac{-15}{...}

(ii) −37=...14=9...=−6...\dfrac{-3}{7} = \dfrac{...}{14} = \dfrac{9}{...} = \dfrac{-6}{...}

Answer

(i) Solving,

54=...16\dfrac{5}{4} = \dfrac{...}{16} : To get 16 from 4, multiply by 4. So, 5 × 4 = 20.

54=25...\dfrac{5}{4} = \dfrac{25}{...} : To get 25 from 5, multiply by 5. So, 4 × 5 = 20.

54=−15...\dfrac{5}{4} = \dfrac{-15}{...} : To get -15 from 5, multiply by -3. So, 4 × (-3) = -12.

54=2016=2520=−15−12\dfrac{5}{4} = \dfrac{20}{16} = \dfrac{25}{20} = \dfrac{-15}{-12}

Hence, the blanks are 20, 20 and -12.

(ii) Solving,

−37=...14\dfrac{-3}{7} = \dfrac{...}{14} : To get 14 from 7, multiply by 2. So, -3 × 2 = -6.

−37=9...\dfrac{-3}{7} = \dfrac{9}{...} : To get 9 from -3, multiply by -3. So, 7 × (-3) = -21.

−37=−6...\dfrac{-3}{7} = \dfrac{-6}{...} : To get -6 from -3, multiply by 2. So, 7 × 2 = 14.

−37=−614=9−21=−614\dfrac{-3}{7} = \dfrac{-6}{14} = \dfrac{9}{-21} = \dfrac{-6}{14}

Hence, the blanks are -6, -21 and 14.

Question 9

Reduce each of the following rational numbers in standard form:

(i) −4530\dfrac{-45}{30}

(ii) 16−36\dfrac{16}{-36}

(iii) −3−15\dfrac{-3}{-15}

(iv) 68−119\dfrac{68}{-119}

Answer

(i) The denominator is positive. HCF of 45 and 30 is 15.

−4530=−45÷1530÷15=−32\dfrac{-45}{30} = \dfrac{-45 \div 15}{30 \div 15} = \dfrac{-3}{2}

∴ Standard form of −4530\dfrac{-45}{30} is −32\dfrac{-3}{2}.

(ii) Making the denominator positive:

16−36=16×(−1)−36×(−1)=−1636\dfrac{16}{-36} = \dfrac{16 \times (-1)}{-36 \times (-1)} = \dfrac{-16}{36}

HCF of 16 and 36 is 4.

−1636=−16÷436÷4=−49\dfrac{-16}{36} = \dfrac{-16 \div 4}{36 \div 4} = \dfrac{-4}{9}

∴ Standard form of 16−36\dfrac{16}{-36} is −49\dfrac{-4}{9}.

(iii) Making the denominator positive:

−3−15=−3×(−1)−15×(−1)=315\dfrac{-3}{-15} = \dfrac{-3 \times (-1)}{-15 \times (-1)} = \dfrac{3}{15}

HCF of 3 and 15 is 3.

315=3÷315÷3=15\dfrac{3}{15} = \dfrac{3 \div 3}{15 \div 3} = \dfrac{1}{5}

∴ Standard form of −3−15\dfrac{-3}{-15} is 15\dfrac{1}{5}.

(iv) Making the denominator positive:

68−119=68×(−1)−119×(−1)=−68119\dfrac{68}{-119} = \dfrac{68 \times (-1)}{-119 \times (-1)} = \dfrac{-68}{119}

HCF of 68 and 119 is 17.

−68119=−68÷17119÷17=−47\dfrac{-68}{119} = \dfrac{-68 \div 17}{119 \div 17} = \dfrac{-4}{7}

∴ Standard form of 68−119\dfrac{68}{-119} is −47\dfrac{-4}{7}.

Exercise 3.2

Question 1

Draw a number line and represent the following rational numbers on it:

(i) 38\dfrac{3}{8}

(ii) −34\dfrac{-3}{4}

(iii) −78\dfrac{-7}{8}

(iv) −178\dfrac{-17}{8}

Answer

(i) To represent 38\dfrac{3}{8}, divide the gap between 0 and 1 into 8 equal parts and take the 3rd point to the right of 0.

Draw a number line and represent the following rational numbers on it. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

(ii) To represent −34\dfrac{-3}{4}, divide the gap between 0 and -1 into 4 equal parts and take the 3rd point to the left of 0.

Draw a number line and represent the following rational numbers on it. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

(iii) To represent −78\dfrac{-7}{8}, divide the gap between 0 and -1 into 8 equal parts and take the 7th point to the left of 0.

Draw a number line and represent the following rational numbers on it. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

(iv) Since −178=−218\dfrac{-17}{8} = -2\dfrac{1}{8}, it lies between -2 and -3. Divide the gap between -2 and -3 into 8 equal parts and take the 1st point to the left of -2.

Draw a number line and represent the following rational numbers on it. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

Question 2

The points P, Q, R, S, T, U, A and B on the number line are such that TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S respectively.

The points P, Q, R, S, T, U, A and B on the number line are such that TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S respectively. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

Answer

From the number line, A and B are at 2 and 3, and the gap AB is divided into 3 equal parts by P and Q, so AP = PQ = QB = 13\dfrac{1}{3}.

P=2+13=73Q=2+23=83P = 2 + \dfrac{1}{3} = \dfrac{7}{3} \\[1em] Q = 2 + \dfrac{2}{3} = \dfrac{8}{3}

Also, T and U are at -1 and -2, and the gap TU is divided into 3 equal parts by R and S, so TR = RS = SU = 13\dfrac{1}{3}.

R=−1−13=−43S=−1−23=−53R = -1 - \dfrac{1}{3} = \dfrac{-4}{3} \\[1em] S = -1 - \dfrac{2}{3} = \dfrac{-5}{3}

Hence, P, Q, R and S represent 73,83,−43\dfrac{7}{3}, \dfrac{8}{3}, \dfrac{-4}{3} and −53\dfrac{-5}{3} respectively.

Question 3

State whether the following statements are true or false:

(i) The rational number −13−5\dfrac{-13}{-5} lies to the right of zero on the number line.

(ii) The rational numbers −5−7\dfrac{-5}{-7} and 7−9\dfrac{7}{-9} lie on opposite sides of zero on the number line.

(iii) The rational numbers −176\dfrac{-17}{6} and 8−15\dfrac{8}{-15} lie on opposite sides of zero on the number line.

Answer

(i) −13−5=135\dfrac{-13}{-5} = \dfrac{13}{5}, which is a positive rational number, so it lies to the right of zero.

Hence, the statement is True.

(ii) −5−7=57\dfrac{-5}{-7} = \dfrac{5}{7} is positive (right of zero) and 7−9=−79\dfrac{7}{-9} = \dfrac{-7}{9} is negative (left of zero). So, they lie on opposite sides of zero.

Hence, the statement is True.

(iii) −176\dfrac{-17}{6} is negative (left of zero) and 8−15=−815\dfrac{8}{-15} = \dfrac{-8}{15} is also negative (left of zero). So, they lie on the same side of zero.

Hence, the statement is False.

Question 4

Which of the two rational numbers is greater in each of the following pairs?

(i) −47,0\dfrac{-4}{7}, 0

(ii) 5−9,37\dfrac{5}{-9}, \dfrac{3}{7}

(iii) −9−5,0\dfrac{-9}{-5}, 0

(iv) 7−5,−21−23\dfrac{7}{-5}, \dfrac{-21}{-23}

Answer

(i) −47\dfrac{-4}{7} is a negative rational number and every negative rational number is less than zero.

Hence, the greater rational number is 0.

(ii) 5−9=−59\dfrac{5}{-9} = \dfrac{-5}{9} is negative and 37\dfrac{3}{7} is positive. Every positive rational number is greater than every negative rational number.

Hence, the greater rational number is 37\dfrac{3}{7}.

(iii) −9−5=95\dfrac{-9}{-5} = \dfrac{9}{5} is a positive rational number and every positive rational number is greater than zero.

Hence, the greater rational number is −9−5\dfrac{-9}{-5}.

(iv) 7−5=−75\dfrac{7}{-5} = \dfrac{-7}{5} is negative and −21−23=2123\dfrac{-21}{-23} = \dfrac{21}{23} is positive. Every positive rational number is greater than every negative rational number.

Hence, the greater rational number is −21−23\dfrac{-21}{-23}.

Question 5

Fill in the boxes with the correct symbol out of >, < and =:

(i) −45653−57\dfrac{-4}{5} \boxed{\phantom{653}} \dfrac{-5}{7}

(ii) −85653−74\dfrac{-8}{5} \boxed{\phantom{653}} \dfrac{-7}{4}

(iii) −7865342−48\dfrac{-7}{8} \boxed{\phantom{653}} \dfrac{42}{-48}

(iv) 1−3653−14\dfrac{1}{-3} \boxed{\phantom{653}} \dfrac{-1}{4}

(v) −38653−27-\dfrac{3}{8} \boxed{\phantom{653}} -\dfrac{2}{7}

(vi) −43653−32\dfrac{-4}{3} \boxed{\phantom{653}} -\dfrac{3}{2}

Answer

(i) LCM of 5 and 7 is 35.

−4×75×7=−2835,−5×57×5=−2535\dfrac{-4 \times 7}{5 \times 7} = \dfrac{-28}{35}, \quad \dfrac{-5 \times 5}{7 \times 5} = \dfrac{-25}{35}

As -28 < -25, so −45\dfrac{-4}{5} < −57\dfrac{-5}{7}.

∴ −45\dfrac{-4}{5} < −57\dfrac{-5}{7}

(ii) LCM of 5 and 4 is 20.

−8×45×4=−3220,−7×54×5=−3520\dfrac{-8 \times 4}{5 \times 4} = \dfrac{-32}{20}, \quad \dfrac{-7 \times 5}{4 \times 5} = \dfrac{-35}{20}

As -32 > -35, so −85\dfrac{-8}{5} > −74\dfrac{-7}{4}.

∴ −85\dfrac{-8}{5} > −74\dfrac{-7}{4}

(iii) 42−48=−4248=−78\dfrac{42}{-48} = \dfrac{-42}{48} = \dfrac{-7}{8}

∴ −78\dfrac{-7}{8} = 42−48\dfrac{42}{-48}

(iv) 1−3=−13\dfrac{1}{-3} = \dfrac{-1}{3}. LCM of 3 and 4 is 12.

−1×43×4=−412,−1×34×3=−312\dfrac{-1 \times 4}{3 \times 4} = \dfrac{-4}{12}, \quad \dfrac{-1 \times 3}{4 \times 3} = \dfrac{-3}{12}

As -4 < -3, so 1−3\dfrac{1}{-3} < −14\dfrac{-1}{4}.

∴ 1−3\dfrac{1}{-3} < −14\dfrac{-1}{4}

(v) LCM of 8 and 7 is 56.

−3×78×7=−2156,−2×87×8=−1656\dfrac{-3 \times 7}{8 \times 7} = \dfrac{-21}{56}, \quad \dfrac{-2 \times 8}{7 \times 8} = \dfrac{-16}{56}

As -21 < -16, so −38-\dfrac{3}{8} < −27-\dfrac{2}{7}.

∴ −38-\dfrac{3}{8} < −27-\dfrac{2}{7}

(vi) LCM of 3 and 2 is 6.

−4×23×2=−86,−3×32×3=−96\dfrac{-4 \times 2}{3 \times 2} = \dfrac{-8}{6}, \quad \dfrac{-3 \times 3}{2 \times 3} = \dfrac{-9}{6}

As -8 > -9, so −43\dfrac{-4}{3} > −32-\dfrac{3}{2}.

∴ −43\dfrac{-4}{3} > −32-\dfrac{3}{2}

Question 6

Arrange the following rational numbers in ascending order:

(i) −37,−32,−34\dfrac{-3}{7}, \dfrac{-3}{2}, \dfrac{-3}{4}

(ii) −34,5−12,9−24,−716\dfrac{-3}{4}, \dfrac{5}{-12}, \dfrac{9}{-24}, \dfrac{-7}{16}

Answer

(i) LCM of 7, 2 and 4 is 28.

−3×47×4=−1228−3×142×14=−4228−3×74×7=−2128\dfrac{-3 \times 4}{7 \times 4} = \dfrac{-12}{28} \\[1em] \dfrac{-3 \times 14}{2 \times 14} = \dfrac{-42}{28} \\[1em] \dfrac{-3 \times 7}{4 \times 7} = \dfrac{-21}{28}

As -42 < -21 < -12,

−4228<−2128<−1228\dfrac{-42}{28} \lt \dfrac{-21}{28} \lt \dfrac{-12}{28}

Hence, the ascending order is −32,−34,−37\dfrac{-3}{2}, \dfrac{-3}{4}, \dfrac{-3}{7}.

(ii) Writing each with a positive denominator:

5−12=−512,9−24=−924=−38\dfrac{5}{-12} = \dfrac{-5}{12}, \quad \dfrac{9}{-24} = \dfrac{-9}{24} = \dfrac{-3}{8}

So the numbers are −34,−512,−38,−716\dfrac{-3}{4}, \dfrac{-5}{12}, \dfrac{-3}{8}, \dfrac{-7}{16}. LCM of 4, 12, 8 and 16 is 48.

−3×124×12=−3648−5×412×4=−2048−3×68×6=−1848−7×316×3=−2148\dfrac{-3 \times 12}{4 \times 12} = \dfrac{-36}{48} \\[1em] \dfrac{-5 \times 4}{12 \times 4} = \dfrac{-20}{48} \\[1em] \dfrac{-3 \times 6}{8 \times 6} = \dfrac{-18}{48} \\[1em] \dfrac{-7 \times 3}{16 \times 3} = \dfrac{-21}{48}

As -36 < -21 < -20 < -18,

−3648<−2148<−2048<−1848\dfrac{-36}{48} \lt \dfrac{-21}{48} \lt \dfrac{-20}{48} \lt \dfrac{-18}{48}

Hence, the ascending order is −34,−716,5−12,9−24\dfrac{-3}{4}, \dfrac{-7}{16}, \dfrac{5}{-12}, \dfrac{9}{-24}.

Question 7

Arrange the following rational numbers in descending order:

(i) −310,−1120,−715,17−30\dfrac{-3}{10}, \dfrac{-11}{20}, \dfrac{-7}{15}, \dfrac{17}{-30}

(ii) −710,−1115,2−5,19−30\dfrac{-7}{10}, \dfrac{-11}{15}, \dfrac{2}{-5}, \dfrac{19}{-30}

Answer

(i) Writing each with a positive denominator, 17−30=−1730\dfrac{17}{-30} = \dfrac{-17}{30}.

So the numbers are −310,−1120,−715,−1730\dfrac{-3}{10}, \dfrac{-11}{20}, \dfrac{-7}{15}, \dfrac{-17}{30}. LCM of 10, 20, 15 and 30 is 60.

−3×610×6=−1860−11×320×3=−3360−7×415×4=−2860−17×230×2=−3460\dfrac{-3 \times 6}{10 \times 6} = \dfrac{-18}{60} \\[1em] \dfrac{-11 \times 3}{20 \times 3} = \dfrac{-33}{60} \\[1em] \dfrac{-7 \times 4}{15 \times 4} = \dfrac{-28}{60} \\[1em] \dfrac{-17 \times 2}{30 \times 2} = \dfrac{-34}{60}

As -18 > -28 > -33 > -34,

−1860>−2860>−3360>−3460\dfrac{-18}{60} \gt \dfrac{-28}{60} \gt \dfrac{-33}{60} \gt \dfrac{-34}{60}

Hence, the descending order is −310,−715,−1120,17−30\dfrac{-3}{10}, \dfrac{-7}{15}, \dfrac{-11}{20}, \dfrac{17}{-30}.

(ii) Writing each with a positive denominator, 2−5=−25\dfrac{2}{-5} = \dfrac{-2}{5} and 19−30=−1930\dfrac{19}{-30} = \dfrac{-19}{30}.

So the numbers are −710,−1115,−25,−1930\dfrac{-7}{10}, \dfrac{-11}{15}, \dfrac{-2}{5}, \dfrac{-19}{30}. LCM of 10, 15, 5 and 30 is 30.

−7×310×3=−2130−11×215×2=−2230−2×65×6=−1230−19×130×1=−1930\dfrac{-7 \times 3}{10 \times 3} = \dfrac{-21}{30} \\[1em] \dfrac{-11 \times 2}{15 \times 2} = \dfrac{-22}{30} \\[1em] \dfrac{-2 \times 6}{5 \times 6} = \dfrac{-12}{30} \\[1em] \dfrac{-19 \times 1}{30 \times 1} = \dfrac{-19}{30}

As -12 > -19 > -21 > -22,

−1230>−1930>−2130>−2230\dfrac{-12}{30} \gt \dfrac{-19}{30} \gt \dfrac{-21}{30} \gt \dfrac{-22}{30}

Hence, the descending order is 2−5,19−30,−710,−1115\dfrac{2}{-5}, \dfrac{19}{-30}, \dfrac{-7}{10}, \dfrac{-11}{15}.

Question 8

Express the following rational numbers as decimals:

(i) 3583\dfrac{5}{8}

(ii) −21125\dfrac{-21}{125}

(iii) −332-\dfrac{3}{32}

(iv) −71180-7\dfrac{11}{80}

Answer

(i) Solving,

358=3+58=3+5×1258×125=3+6251000=3+0.625=3.6253\dfrac{5}{8} = 3 + \dfrac{5}{8} = 3 + \dfrac{5 \times 125}{8 \times 125} = 3 + \dfrac{625}{1000} = 3 + 0.625 = 3.625

∴ 358=3.6253\dfrac{5}{8} = 3.625

(ii) Solving,

−21125=−21×8125×8=−1681000=−0.168\dfrac{-21}{125} = \dfrac{-21 \times 8}{125 \times 8} = \dfrac{-168}{1000} = -0.168

∴ −21125=−0.168\dfrac{-21}{125} = -0.168

(iii) Solving,

−332=−3×312532×3125=−9375100000=−0.09375-\dfrac{3}{32} = -\dfrac{3 \times 3125}{32 \times 3125} = -\dfrac{9375}{100000} = -0.09375

∴ −332=−0.09375-\dfrac{3}{32} = -0.09375

(iv) Solving,

−71180=−(7+1180)=−(7+11×12580×125)=−(7+137510000)=−(7+0.1375)=−7.1375-7\dfrac{11}{80} = -\left(7 + \dfrac{11}{80}\right) = -\left(7 + \dfrac{11 \times 125}{80 \times 125}\right) = -\left(7 + \dfrac{1375}{10000}\right) = -(7 + 0.1375) = -7.1375

∴ −71180=−7.1375-7\dfrac{11}{80} = -7.1375

Exercise 3.3

Question 1

Add the following pairs of rational numbers:

(i) 311,−511\dfrac{3}{11}, \dfrac{-5}{11}

(ii) 49,5−9\dfrac{4}{9}, \dfrac{5}{-9}

(iii) 5−7,−2−7\dfrac{5}{-7}, \dfrac{-2}{-7}

(iv) −25,34\dfrac{-2}{5}, \dfrac{3}{4}

Answer

(i) Solving,

311+−511=3+(−5)11=−211\dfrac{3}{11} + \dfrac{-5}{11} = \dfrac{3 + (-5)}{11} = \dfrac{-2}{11}

∴ 311+−511=−211\dfrac{3}{11} + \dfrac{-5}{11} = \dfrac{-2}{11}

(ii) 5−9=−59\dfrac{5}{-9} = \dfrac{-5}{9}

49+−59=4+(−5)9=−19\dfrac{4}{9} + \dfrac{-5}{9} = \dfrac{4 + (-5)}{9} = \dfrac{-1}{9}

∴ 49+5−9=−19\dfrac{4}{9} + \dfrac{5}{-9} = \dfrac{-1}{9}

(iii) 5−7=−57\dfrac{5}{-7} = \dfrac{-5}{7} and −2−7=27\dfrac{-2}{-7} = \dfrac{2}{7}

−57+27=−5+27=−37\dfrac{-5}{7} + \dfrac{2}{7} = \dfrac{-5 + 2}{7} = \dfrac{-3}{7}

∴ 5−7+−2−7=−37\dfrac{5}{-7} + \dfrac{-2}{-7} = \dfrac{-3}{7}

(iv) LCM of 5 and 4 is 20.

−2×45×4+3×54×5=−820+1520=−8+1520=720\dfrac{-2 \times 4}{5 \times 4} + \dfrac{3 \times 5}{4 \times 5} \\[1em] = \dfrac{-8}{20} + \dfrac{15}{20} \\[1em] = \dfrac{-8 + 15}{20} \\[1em] = \dfrac{7}{20}

∴ −25+34=720\dfrac{-2}{5} + \dfrac{3}{4} = \dfrac{7}{20}

Question 2

Find the sum:

(i) 27−4+−158\dfrac{27}{-4} + \dfrac{-15}{8}

(ii) −118+−38\dfrac{-1}{18} + \dfrac{-3}{8}

(iii) −316+238-3\dfrac{1}{6} + 2\dfrac{3}{8}

(iv) −245+4310-2\dfrac{4}{5} + 4\dfrac{3}{10}

Answer

(i) 27−4=−274\dfrac{27}{-4} = \dfrac{-27}{4}. LCM of 4 and 8 is 8.

⇒−27×24×2+−15×18×1=−548+−158=−54+(−15)8=−698\Rightarrow \dfrac{-27 \times 2}{4 \times 2} + \dfrac{-15 \times 1}{8 \times 1} \\[1em] = \dfrac{-54}{8} + \dfrac{-15}{8} \\[1em] = \dfrac{-54 + (-15)}{8} \\[1em] = \dfrac{-69}{8}

∴ 27−4+−158=−698\dfrac{27}{-4} + \dfrac{-15}{8} = \dfrac{-69}{8}

(ii) LCM of 18 and 8 is 72.

⇒−1×418×4+−3×98×9=−472+−2772=−4+(−27)72=−3172\Rightarrow \dfrac{-1 \times 4}{18 \times 4} + \dfrac{-3 \times 9}{8 \times 9} \\[1em] = \dfrac{-4}{72} + \dfrac{-27}{72} \\[1em] = \dfrac{-4 + (-27)}{72} \\[1em] = \dfrac{-31}{72}

∴ −118+−38=−3172\dfrac{-1}{18} + \dfrac{-3}{8} = \dfrac{-31}{72}

(iii) −316=−196-3\dfrac{1}{6} = \dfrac{-19}{6} and 238=1982\dfrac{3}{8} = \dfrac{19}{8}. LCM of 6 and 8 is 24.

⇒−19×46×4+19×38×3=−7624+5724=−76+5724=−1924\Rightarrow \dfrac{-19 \times 4}{6 \times 4} + \dfrac{19 \times 3}{8 \times 3} \\[1em] = \dfrac{-76}{24} + \dfrac{57}{24} \\[1em] = \dfrac{-76 + 57}{24} \\[1em] = \dfrac{-19}{24}

∴ −316+238=−1924-3\dfrac{1}{6} + 2\dfrac{3}{8} = \dfrac{-19}{24}

(iv) −245=−145-2\dfrac{4}{5} = \dfrac{-14}{5} and 4310=43104\dfrac{3}{10} = \dfrac{43}{10}.

LCM of 5 and 10 is 10.

⇒−14×25×2+43×110×1=−2810+4310=−28+4310=1510=32=112\Rightarrow \dfrac{-14 \times 2}{5 \times 2} + \dfrac{43 \times 1}{10 \times 1} \\[1em] = \dfrac{-28}{10} + \dfrac{43}{10} \\[1em] = \dfrac{-28 + 43}{10} \\[1em] = \dfrac{15}{10} = \dfrac{3}{2} = 1\dfrac{1}{2}

∴ −245+4310=112-2\dfrac{4}{5} + 4\dfrac{3}{10} = 1\dfrac{1}{2}

Question 3

Subtract:

(i) −613\dfrac{-6}{13} from 413\dfrac{4}{13}

(ii) −12\dfrac{-1}{2} from −23\dfrac{-2}{3}

(iii) 59\dfrac{5}{9} from −23\dfrac{-2}{3}

Answer

(i) Solving,

413−−613=413+613=4+613=1013\dfrac{4}{13} - \dfrac{-6}{13} = \dfrac{4}{13} + \dfrac{6}{13} = \dfrac{4 + 6}{13} = \dfrac{10}{13}

∴ The result is 1013\dfrac{10}{13}.

(ii) LCM of 3 and 2 is 6.

⇒−23−−12=−23+12=−2×23×2+1×32×3=−46+36=−4+36=−16\Rightarrow \dfrac{-2}{3} - \dfrac{-1}{2} = \dfrac{-2}{3} + \dfrac{1}{2} \\[1em] = \dfrac{-2 \times 2}{3 \times 2} + \dfrac{1 \times 3}{2 \times 3} \\[1em] = \dfrac{-4}{6} + \dfrac{3}{6} \\[1em] = \dfrac{-4 + 3}{6} \\[1em] = \dfrac{-1}{6}

∴ The result is −16\dfrac{-1}{6}.

(iii) LCM of 3 and 9 is 9.

⇒−23−59=−2×33×3−5×19×1=−69−59=−6−59=−119\Rightarrow \dfrac{-2}{3} - \dfrac{5}{9} = \dfrac{-2 \times 3}{3 \times 3} - \dfrac{5 \times 1}{9 \times 1} \\[1em] = \dfrac{-6}{9} - \dfrac{5}{9} \\[1em] = \dfrac{-6 - 5}{9} \\[1em] = \dfrac{-11}{9}

∴ The result is −119\dfrac{-11}{9}.

Question 4

Find:

(i) 563−(−621)\dfrac{5}{63} - \left(\dfrac{-6}{21}\right)

(ii) −613−(−715)\dfrac{-6}{13} - \left(\dfrac{-7}{15}\right)

(iii) 318−(−156)3\dfrac{1}{8} - \left(-1\dfrac{5}{6}\right)

Answer

(i)

563−(−621)=563+621\dfrac{5}{63} - \left(\dfrac{-6}{21}\right) = \dfrac{5}{63} + \dfrac{6}{21}

LCM of 63 and 21 is 63.

⇒5×163×1+6×321×3=563+1863=5+1863=2363\Rightarrow \dfrac{5 \times 1}{63 \times 1} + \dfrac{6 \times 3}{21 \times 3} \\[1em] = \dfrac{5}{63} + \dfrac{18}{63} \\[1em] = \dfrac{5 + 18}{63} \\[1em] = \dfrac{23}{63}

∴ 563−(−621)=2363\dfrac{5}{63} - \left(\dfrac{-6}{21}\right) = \dfrac{23}{63}

(ii)

−613−(−715)=−613+715\dfrac{-6}{13} - \left(\dfrac{-7}{15}\right) = \dfrac{-6}{13} + \dfrac{7}{15}

LCM of 13 and 15 is 195.

⇒−6×1513×15+7×1315×13=−90195+91195=−90+91195=1195\Rightarrow \dfrac{-6 \times 15}{13 \times 15} + \dfrac{7 \times 13}{15 \times 13} \\[1em] = \dfrac{-90}{195} + \dfrac{91}{195} \\[1em] = \dfrac{-90 + 91}{195} \\[1em] = \dfrac{1}{195}

∴ −613−(−715)=1195\dfrac{-6}{13} - \left(\dfrac{-7}{15}\right) = \dfrac{1}{195}

(iii) 318=2583\dfrac{1}{8} = \dfrac{25}{8} and −156=−116-1\dfrac{5}{6} = \dfrac{-11}{6}.

258−(−116)=258+116\dfrac{25}{8} - \left(\dfrac{-11}{6}\right) = \dfrac{25}{8} + \dfrac{11}{6}

LCM of 8 and 6 is 24.

⇒25×38×3+11×46×4=7524+4424=75+4424=11924=42324\Rightarrow \dfrac{25 \times 3}{8 \times 3} + \dfrac{11 \times 4}{6 \times 4} \\[1em] = \dfrac{75}{24} + \dfrac{44}{24} \\[1em] = \dfrac{75 + 44}{24} \\[1em] = \dfrac{119}{24} = 4\dfrac{23}{24}

∴ 318−(−156)=423243\dfrac{1}{8} - \left(-1\dfrac{5}{6}\right) = 4\dfrac{23}{24}

Question 5

The sum of two rational numbers is 25\dfrac{2}{5}. If one of them is −47\dfrac{-4}{7}, find the other number.

Answer

Let the other number be xx.

x+−47=25⇒x=25−−47⇒x=25+47x + \dfrac{-4}{7} = \dfrac{2}{5} \\[1em] \Rightarrow x = \dfrac{2}{5} - \dfrac{-4}{7} \\[1em] \Rightarrow x = \dfrac{2}{5} + \dfrac{4}{7}

LCM of 5 and 7 is 35.

⇒x=2×75×7+4×57×5⇒x=1435+2035⇒x=14+2035⇒x=3435\Rightarrow x = \dfrac{2 \times 7}{5 \times 7} + \dfrac{4 \times 5}{7 \times 5} \\[1em] \Rightarrow x = \dfrac{14}{35} + \dfrac{20}{35} \\[1em] \Rightarrow x = \dfrac{14 + 20}{35} \\[1em] \Rightarrow x = \dfrac{34}{35}

Hence, the other number is 3435\dfrac{34}{35}.

Question 6

What rational number should be added to −512\dfrac{-5}{12} to get −78\dfrac{-7}{8}?

Answer

Let the required number be xx.

−512+x=−78⇒x=−78−−512⇒x=−78+512\dfrac{-5}{12} + x = \dfrac{-7}{8} \\[1em] \Rightarrow x = \dfrac{-7}{8} - \dfrac{-5}{12} \\[1em] \Rightarrow x = \dfrac{-7}{8} + \dfrac{5}{12}

LCM of 8 and 12 is 24.

⇒x=−7×38×3+5×212×2⇒x=−2124+1024⇒x=−21+1024⇒x=−1124\Rightarrow x = \dfrac{-7 \times 3}{8 \times 3} + \dfrac{5 \times 2}{12 \times 2} \\[1em] \Rightarrow x = \dfrac{-21}{24} + \dfrac{10}{24} \\[1em] \Rightarrow x = \dfrac{-21 + 10}{24} \\[1em] \Rightarrow x = \dfrac{-11}{24}

Hence, the required number is −1124\dfrac{-11}{24}.

Question 7

What rational number should be subtracted from −23\dfrac{-2}{3} to get −56\dfrac{-5}{6}?

Answer

Let the required number be xx.

−23−x=−56⇒x=−23−−56⇒x=−23+56\dfrac{-2}{3} - x = \dfrac{-5}{6} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \dfrac{-5}{6} \\[1em] \Rightarrow x = \dfrac{-2}{3} + \dfrac{5}{6}

LCM of 3 and 6 is 6.

⇒x=−2×23×2+5×16×1⇒x=−46+56⇒x=−4+56⇒x=16\Rightarrow x = \dfrac{-2 \times 2}{3 \times 2} + \dfrac{5 \times 1}{6 \times 1} \\[1em] \Rightarrow x = \dfrac{-4}{6} + \dfrac{5}{6} \\[1em] \Rightarrow x = \dfrac{-4 + 5}{6} \\[1em] \Rightarrow x = \dfrac{1}{6}

Hence, the required number is 16\dfrac{1}{6}.

Question 8

Find the product:

(i) 23×−78\dfrac{2}{3} \times \dfrac{-7}{8}

(ii) −67×57\dfrac{-6}{7} \times \dfrac{5}{7}

(iii) −29×(−5)\dfrac{-2}{9} \times (-5)

(iv) −511×(11−5)\dfrac{-5}{11} \times \left(\dfrac{11}{-5}\right)

(v) 835×21−32\dfrac{8}{35} \times \dfrac{21}{-32}

(vi) −105128×(−12935)\dfrac{-105}{128} \times \left(-1\dfrac{29}{35}\right)

Answer

(i) Solving,

23×−78=2×(−7)3×8=−1424=−712\dfrac{2}{3} \times \dfrac{-7}{8} = \dfrac{2 \times (-7)}{3 \times 8} = \dfrac{-14}{24} = \dfrac{-7}{12}

∴ 23×−78=−712\dfrac{2}{3} \times \dfrac{-7}{8} = \dfrac{-7}{12}

(ii) Solving,

−67×57=−6×57×7=−3049\dfrac{-6}{7} \times \dfrac{5}{7} = \dfrac{-6 \times 5}{7 \times 7} = \dfrac{-30}{49}

∴ −67×57=−3049\dfrac{-6}{7} \times \dfrac{5}{7} = \dfrac{-30}{49}

(iii) Solving,

−29×(−5)=−29×−51=−2×(−5)9×1=109\dfrac{-2}{9} \times (-5) = \dfrac{-2}{9} \times \dfrac{-5}{1} = \dfrac{-2 \times (-5)}{9 \times 1} = \dfrac{10}{9}

∴ −29×(−5)=109\dfrac{-2}{9} \times (-5) = \dfrac{10}{9}

(iv) Solving,

−511×11−5=−5×1111×(−5)=−55−55=1\dfrac{-5}{11} \times \dfrac{11}{-5} = \dfrac{-5 \times 11}{11 \times (-5)} = \dfrac{-55}{-55} = 1

∴ −511×(11−5)=1\dfrac{-5}{11} \times \left(\dfrac{11}{-5}\right) = 1

(v) Solving,

835×21−32=8×2135×(−32)=168−1120=−320\dfrac{8}{35} \times \dfrac{21}{-32} = \dfrac{8 \times 21}{35 \times (-32)} = \dfrac{168}{-1120} = \dfrac{-3}{20}

∴ 835×21−32=−320\dfrac{8}{35} \times \dfrac{21}{-32} = \dfrac{-3}{20}

(vi) −12935=−6435-1\dfrac{29}{35} = \dfrac{-64}{35}

−105128×−6435=−105×(−64)128×35=67204480=32=112\dfrac{-105}{128} \times \dfrac{-64}{35} = \dfrac{-105 \times (-64)}{128 \times 35} = \dfrac{6720}{4480} = \dfrac{3}{2} = 1\dfrac{1}{2}

∴ −105128×(−12935)=112\dfrac{-105}{128} \times \left(-1\dfrac{29}{35}\right) = 1\dfrac{1}{2}

Question 9

Find the value of:

(i) (−6)÷25(-6) \div \dfrac{2}{5}

(ii) −110÷−85\dfrac{-1}{10} \div \dfrac{-8}{5}

(iii) −6514÷13−7\dfrac{-65}{14} \div \dfrac{13}{-7}

(iv) (−6)÷335(-6) \div 3\dfrac{3}{5}

(v) −4849÷72−35\dfrac{-48}{49} \div \dfrac{72}{-35}

(vi) 317÷(−3334)3\dfrac{1}{7} \div \left(\dfrac{-33}{34}\right)

Answer

(i) Solving,

(−6)÷25=−6×52=−6×52=−302=−15(-6) \div \dfrac{2}{5} = -6 \times \dfrac{5}{2} = \dfrac{-6 \times 5}{2} = \dfrac{-30}{2} = -15

∴ (−6)÷25=−15(-6) \div \dfrac{2}{5} = -15

(ii) Solving,

−110÷−85=−110×5−8=−1×510×(−8)=−5−80=116\dfrac{-1}{10} \div \dfrac{-8}{5} = \dfrac{-1}{10} \times \dfrac{5}{-8} = \dfrac{-1 \times 5}{10 \times (-8)} = \dfrac{-5}{-80} = \dfrac{1}{16}

∴ −110÷−85=116\dfrac{-1}{10} \div \dfrac{-8}{5} = \dfrac{1}{16}

(iii) Solving,

−6514÷13−7=−6514×−713=−65×(−7)14×13=455182=52=212\dfrac{-65}{14} \div \dfrac{13}{-7} = \dfrac{-65}{14} \times \dfrac{-7}{13} = \dfrac{-65 \times (-7)}{14 \times 13} = \dfrac{455}{182} = \dfrac{5}{2} = 2\dfrac{1}{2}

∴ −6514÷13−7=212\dfrac{-65}{14} \div \dfrac{13}{-7} = 2\dfrac{1}{2}

(iv) 335=1853\dfrac{3}{5} = \dfrac{18}{5}

(−6)÷185=−6×518=−6×518=−3018=−53=−123(-6) \div \dfrac{18}{5} = -6 \times \dfrac{5}{18} = \dfrac{-6 \times 5}{18} = \dfrac{-30}{18} = \dfrac{-5}{3} = -1\dfrac{2}{3}

∴ (−6)÷335=−123(-6) \div 3\dfrac{3}{5} = -1\dfrac{2}{3}

(v) Solving,

−4849÷72−35=−4849×−3572=−48×(−35)49×72=16803528=1021\dfrac{-48}{49} \div \dfrac{72}{-35} = \dfrac{-48}{49} \times \dfrac{-35}{72} = \dfrac{-48 \times (-35)}{49 \times 72} = \dfrac{1680}{3528} = \dfrac{10}{21}

∴ −4849÷72−35=1021\dfrac{-48}{49} \div \dfrac{72}{-35} = \dfrac{10}{21}

(vi) 317=2273\dfrac{1}{7} = \dfrac{22}{7}

227÷−3334=227×34−33=22×347×(−33)=748−231=−6821=−3521\dfrac{22}{7} \div \dfrac{-33}{34} = \dfrac{22}{7} \times \dfrac{34}{-33} = \dfrac{22 \times 34}{7 \times (-33)} = \dfrac{748}{-231} = \dfrac{-68}{21} = -3\dfrac{5}{21}

∴ 317÷(−3334)=−35213\dfrac{1}{7} \div \left(\dfrac{-33}{34}\right) = -3\dfrac{5}{21}

Question 10

The product of two rational numbers is 1835\dfrac{18}{35}. If one of them is −25\dfrac{-2}{5}, find the other number.

Answer

Let the other number be xx.

x×−25=1835⇒x=1835÷−25⇒x=1835×5−2⇒x=18×535×(−2)⇒x=90−70⇒x=−97=−127x \times \dfrac{-2}{5} = \dfrac{18}{35} \\[1em] \Rightarrow x = \dfrac{18}{35} \div \dfrac{-2}{5} \\[1em] \Rightarrow x = \dfrac{18}{35} \times \dfrac{5}{-2} \\[1em] \Rightarrow x = \dfrac{18 \times 5}{35 \times (-2)} \\[1em] \Rightarrow x = \dfrac{90}{-70} \\[1em] \Rightarrow x = \dfrac{-9}{7} = -1\dfrac{2}{7}

Hence, the other number is −127-1\dfrac{2}{7}.

Question 11

Find the value of:

(i) (1321÷3942)×(−35)\left(\dfrac{13}{21} \div \dfrac{39}{42}\right) \times \left(\dfrac{-3}{5}\right)

(ii) (−5521)÷(711×512)\left(-5\dfrac{5}{21}\right) \div \left(\dfrac{7}{11} \times \dfrac{5}{12}\right)

Answer

(i) First solve the bracket:

1321÷3942=1321×4239=13×4221×39=546819=23\dfrac{13}{21} \div \dfrac{39}{42} = \dfrac{13}{21} \times \dfrac{42}{39} = \dfrac{13 \times 42}{21 \times 39} = \dfrac{546}{819} = \dfrac{2}{3}

Now,

23×−35=2×(−3)3×5=−615=−25\dfrac{2}{3} \times \dfrac{-3}{5} = \dfrac{2 \times (-3)}{3 \times 5} = \dfrac{-6}{15} = \dfrac{-2}{5}

∴ (1321÷3942)×(−35)=−25\left(\dfrac{13}{21} \div \dfrac{39}{42}\right) \times \left(\dfrac{-3}{5}\right) = \dfrac{-2}{5}

(ii) First solve the bracket:

711×512=7×511×12=35132\dfrac{7}{11} \times \dfrac{5}{12} = \dfrac{7 \times 5}{11 \times 12} = \dfrac{35}{132}

Also, −5521=−11021-5\dfrac{5}{21} = \dfrac{-110}{21}. Now,

−11021÷35132=−11021×13235=−110×13221×35=−14520735=−96849=−193749\dfrac{-110}{21} \div \dfrac{35}{132} = \dfrac{-110}{21} \times \dfrac{132}{35} = \dfrac{-110 \times 132}{21 \times 35} = \dfrac{-14520}{735} = \dfrac{-968}{49} = -19\dfrac{37}{49}

∴ (−5521)÷(711×512)=−193749\left(-5\dfrac{5}{21}\right) \div \left(\dfrac{7}{11} \times \dfrac{5}{12}\right) = -19\dfrac{37}{49}

Question 12

Find the reciprocal of the following:

(i) 313÷−465\dfrac{3}{13} \div \dfrac{-4}{65}

(ii) (−5×1215)−(−3×29)\left(-5 \times \dfrac{12}{15}\right) - \left(-3 \times \dfrac{2}{9}\right)

Answer

(i) First find the value:

313÷−465=313×65−4=3×6513×(−4)=195−52=−154\dfrac{3}{13} \div \dfrac{-4}{65} = \dfrac{3}{13} \times \dfrac{65}{-4} = \dfrac{3 \times 65}{13 \times (-4)} = \dfrac{195}{-52} = \dfrac{-15}{4}

Reciprocal of −154\dfrac{-15}{4} is −415\dfrac{-4}{15}.

Hence, the reciprocal is −415\dfrac{-4}{15}.

(ii) First find the value:

(−5×1215)=−6015=−4(−3×29)=−69=−23\left(-5 \times \dfrac{12}{15}\right) = \dfrac{-60}{15} = -4 \\[1em] \left(-3 \times \dfrac{2}{9}\right) = \dfrac{-6}{9} = \dfrac{-2}{3}

So,

−4−(−23)=−4+23=−123+23=−12+23=−103-4 - \left(\dfrac{-2}{3}\right) = -4 + \dfrac{2}{3} = \dfrac{-12}{3} + \dfrac{2}{3} = \dfrac{-12 + 2}{3} = \dfrac{-10}{3}

Reciprocal of −103\dfrac{-10}{3} is −310\dfrac{-3}{10}.

Hence, the reciprocal is −310\dfrac{-3}{10}.

Mental Maths

Question 1

Fill in the blanks:

(i) Two rational numbers are called equivalent if they have .... value.

(ii) The rational number −5−11\dfrac{-5}{-11} lies to the .... of zero on the number line.

(iii) The standard form of the rational number 14−12\dfrac{14}{-12} is ....

(iv) The multiplicative inverse of −315-3\dfrac{1}{5} is ....

(v) If p and q are positive integers, then pq\dfrac{p}{q} is a .... rational number and p−q\dfrac{p}{-q} is a .... rational number.

Answer

(i) Two rational numbers are called equivalent if they have the same value.

(ii) −5−11=511\dfrac{-5}{-11} = \dfrac{5}{11}, which is positive, so it lies to the right of zero on the number line.

(iii) Making the denominator positive, 14−12=−1412\dfrac{14}{-12} = \dfrac{-14}{12}. HCF of 14 and 12 is 2, so −1412=−76\dfrac{-14}{12} = \dfrac{-7}{6}.

The standard form is −76\dfrac{-7}{6}.

(iv) −315=−165-3\dfrac{1}{5} = \dfrac{-16}{5}. The multiplicative inverse is −516\dfrac{-5}{16}.

(v) If p and q are positive integers, then pq\dfrac{p}{q} is a positive rational number and p−q\dfrac{p}{-q} is a negative rational number.

Question 2

State whether the following statements are true (T) or false (F):

(i) Zero is the smallest rational number.

(ii) Every integer is a rational number.

(iii) Every rational number is an integer.

(iv) Every fraction is a rational number.

(v) Every rational number is a fraction.

(vi) The reciprocal of -1 is -1.

(vii) The difference of two rational numbers is always a rational number.

(viii) The quotient of two integers is always a rational number.

(ix) The value of a rational number remains the same if both its numerator and denominator are multiplied (or divided) by the same (non-zero) integer.

Answer

(i) False. Negative rational numbers are smaller than zero, so zero is not the smallest rational number.

(ii) True. Every integer can be written in the form pq\dfrac{p}{q} with q=1q = 1.

(iii) False. For example, 23\dfrac{2}{3} is a rational number but not an integer.

(iv) True. Every fraction is a rational number.

(v) False. For example, −23\dfrac{-2}{3} is a rational number but not a fraction.

(vi) True. The reciprocal of -1 is -1.

(vii) True. The difference of two rational numbers is always a rational number.

(viii) False. Division by zero is not defined, so the quotient of two integers is not always a rational number.

(ix) True. The value of a rational number remains the same if both numerator and denominator are multiplied (or divided) by the same non-zero integer.

Multiple Choice Questions

Question 3

The rational number 110−132\dfrac{110}{-132} when reduced to standard form is

  1. 10−12\dfrac{10}{-12}

  2. 5−6\dfrac{5}{-6}

  3. −56\dfrac{-5}{6}

  4. 110−132\dfrac{110}{-132}

Answer

Making the denominator positive, 110−132=−110132\dfrac{110}{-132} = \dfrac{-110}{132}. HCF of 110 and 132 is 22.

−110132=−110÷22132÷22=−56\dfrac{-110}{132} = \dfrac{-110 \div 22}{132 \div 22} = \dfrac{-5}{6}

Hence, Option 3 is the correct option.

Question 4

Which of the following is not equal to the others?

  1. 21−56\dfrac{21}{-56}

  2. −1540\dfrac{-15}{40}

  3. −616\dfrac{-6}{16}

  4. 1848\dfrac{18}{48}

Answer

Reducing each to standard form:

21−56=−2156=−38\dfrac{21}{-56} = \dfrac{-21}{56} = \dfrac{-3}{8}

−1540=−38\dfrac{-15}{40} = \dfrac{-3}{8}

−616=−38\dfrac{-6}{16} = \dfrac{-3}{8}

1848=38\dfrac{18}{48} = \dfrac{3}{8}

The first three equal −38\dfrac{-3}{8}, but 1848=38\dfrac{18}{48} = \dfrac{3}{8} is different.

Hence, Option 4 is the correct option.

Question 5

The multiplicative inverse of −49\dfrac{-4}{9} is

  1. 49\dfrac{4}{9}

  2. −94\dfrac{-9}{4}

  3. 94\dfrac{9}{4}

  4. none of these

Answer

The multiplicative inverse of −49\dfrac{-4}{9} is 9−4=−94\dfrac{9}{-4} = \dfrac{-9}{4}.

Hence, Option 2 is the correct option.

Question 6

The reciprocal of the rational number −237-2\dfrac{3}{7} is

  1. −177-\dfrac{17}{7}

  2. 717\dfrac{7}{17}

  3. −717-\dfrac{7}{17}

  4. none of these

Answer

−237=−177-2\dfrac{3}{7} = \dfrac{-17}{7}. Its reciprocal is 7−17=−717\dfrac{7}{-17} = -\dfrac{7}{17}.

Hence, Option 3 is the correct option.

Question 7

The product of rational number −25\dfrac{-2}{5} and its multiplicative inverse is

  1. 1

  2. 0

  3. 425\dfrac{4}{25}

  4. 25\dfrac{2}{5}

Answer

The product of any non-zero rational number and its multiplicative inverse is always 1.

Hence, Option 1 is the correct option.

Question 8

The product of rational number −23\dfrac{-2}{3} and its additive inverse is

  1. 1

  2. 23\dfrac{2}{3}

  3. 49\dfrac{4}{9}

  4. −49\dfrac{-4}{9}

Answer

The additive inverse of −23\dfrac{-2}{3} is 23\dfrac{2}{3}.

−23×23=−2×23×3=−49\dfrac{-2}{3} \times \dfrac{2}{3} = \dfrac{-2 \times 2}{3 \times 3} = \dfrac{-4}{9}

Hence, Option 4 is the correct option.

Question 9

The sum of rational number −13\dfrac{-1}{3} and its reciprocal is

  1. 0

  2. 1

  3. −103\dfrac{-10}{3}

  4. −310\dfrac{-3}{10}

Answer

The reciprocal of −13\dfrac{-1}{3} is -3.

−13+(−3)=−13+−93=−1+(−9)3=−103\dfrac{-1}{3} + (-3) = \dfrac{-1}{3} + \dfrac{-9}{3} = \dfrac{-1 + (-9)}{3} = \dfrac{-10}{3}

Hence, Option 3 is the correct option.

Question 10

−35−(−215)\dfrac{-3}{5} - \left(\dfrac{-2}{15}\right) is equal to

  1. −115\dfrac{-11}{5}

  2. −115\dfrac{-1}{15}

  3. −715\dfrac{-7}{15}

  4. 715\dfrac{7}{15}

Answer

−35−(−215)=−35+215\dfrac{-3}{5} - \left(\dfrac{-2}{15}\right) = \dfrac{-3}{5} + \dfrac{2}{15}

LCM of 5 and 15 is 15.

=−3×35×3+2×115×1=−915+215=−9+215=−715= \dfrac{-3 \times 3}{5 \times 3} + \dfrac{2 \times 1}{15 \times 1} \\[1em] = \dfrac{-9}{15} + \dfrac{2}{15} \\[1em] = \dfrac{-9 + 2}{15} \\[1em] = \dfrac{-7}{15}

Hence, Option 3 is the correct option.

Question 11

(−513)×(−178)\left(-5\dfrac{1}{3}\right) \times \left(-1\dfrac{7}{8}\right) is equal to

  1. 10

  2. -10

  3. 57245\dfrac{7}{24}

  4. −5724-5\dfrac{7}{24}

Answer

−513=−163-5\dfrac{1}{3} = \dfrac{-16}{3} and −178=−158-1\dfrac{7}{8} = \dfrac{-15}{8}.

−163×−158=−16×(−15)3×8=24024=10\dfrac{-16}{3} \times \dfrac{-15}{8} = \dfrac{-16 \times (-15)}{3 \times 8} = \dfrac{240}{24} = 10

Hence, Option 1 is the correct option.

Question 12

(−213)÷21112\left(-2\dfrac{1}{3}\right) \div 2\dfrac{11}{12} is equal to

  1. −45-\dfrac{4}{5}

  2. 45\dfrac{4}{5}

  3. 411\dfrac{4}{11}

  4. −411-\dfrac{4}{11}

Answer

−213=−73-2\dfrac{1}{3} = \dfrac{-7}{3} and 21112=35122\dfrac{11}{12} = \dfrac{35}{12}.

−73÷3512=−73×1235=−7×123×35=−84105=−45\dfrac{-7}{3} \div \dfrac{35}{12} = \dfrac{-7}{3} \times \dfrac{12}{35} = \dfrac{-7 \times 12}{3 \times 35} = \dfrac{-84}{105} = -\dfrac{4}{5}

Hence, Option 1 is the correct option.

Question 13

In the standard form of a rational number, the denominator is always

  1. 0

  2. a negative integer

  3. 1

  4. a positive integer

Answer

In the standard form of a rational number, the denominator is always a positive integer.

Hence, Option 4 is the correct option.

Question 14

The sum of two rational numbers is -1. If one of them is −57\dfrac{-5}{7}, then the other is

  1. 57\dfrac{5}{7}

  2. −27\dfrac{-2}{7}

  3. 127\dfrac{12}{7}

  4. −127\dfrac{-12}{7}

Answer

Let the other number be xx.

x+−57=−1⇒x=−1−−57⇒x=−1+57⇒x=−77+57⇒x=−7+57⇒x=−27x + \dfrac{-5}{7} = -1 \\[1em] \Rightarrow x = -1 - \dfrac{-5}{7} \\[1em] \Rightarrow x = -1 + \dfrac{5}{7} \\[1em] \Rightarrow x = \dfrac{-7}{7} + \dfrac{5}{7} \\[1em] \Rightarrow x = \dfrac{-7 + 5}{7} \\[1em] \Rightarrow x = \dfrac{-2}{7}

Hence, Option 2 is the correct option.

Statement I-II Type Questions

Question 15

Statement I: −2−3=23=46\dfrac{-2}{-3} = \dfrac{2}{3} = \dfrac{4}{6}

Statement II: The rational number −2−3\dfrac{-2}{-3} is in the standard form.

  1. Statement I is true but statement II is false.

  2. Statement I is false but statement II is true.

  3. Both Statement I and statement II are true.

  4. Both Statement I and statement II are false.

Answer

Statement I: −2−3=23=46\dfrac{-2}{-3} = \dfrac{2}{3} = \dfrac{4}{6}, which is correct. So Statement I is true.

Statement II: A rational number is in standard form only if its denominator is positive. Since −2−3\dfrac{-2}{-3} has a negative denominator, it is not in standard form. So Statement II is false.

Hence, Option 1 is the correct option.

Question 16

Statement I: 30412589>4782−844921\dfrac{304}{12589} \gt \dfrac{4782}{-844921}

Statement II: The two numbers 30412589\dfrac{304}{12589} and 4782−844921\dfrac{4782}{-844921} are rational numbers.

  1. Statement I is true but statement II is false.

  2. Statement I is false but statement II is true.

  3. Both Statement I and statement II are true.

  4. Both Statement I and statement II are false.

Answer

Statement I: 30412589\dfrac{304}{12589} is a positive rational number, while 4782−844921\dfrac{4782}{-844921} is a negative rational number. Every positive rational number is greater than every negative rational number, so 30412589>4782−844921\dfrac{304}{12589} \gt \dfrac{4782}{-844921}. So Statement I is true.

Statement II: Both numbers are of the form pq\dfrac{p}{q} where p and q are integers and q≠0q \neq 0, so both are rational numbers. So Statement II is true.

Hence, Option 3 is the correct option.

Question 17

Statement I: The multiplicative inverse of 0 is 0.

Statement II: The additive inverse of 0 is 0.

  1. Statement I is true but statement II is false.

  2. Statement I is false but statement II is true.

  3. Both Statement I and statement II are true.

  4. Both Statement I and statement II are false.

Answer

Statement I: Since there is no rational number which when multiplied by 0 gives 1, the multiplicative inverse (reciprocal) of 0 is not defined. So Statement I is false.

Statement II: 0 + 0 = 0, so the additive inverse of 0 is 0. So Statement II is true.

Hence, Option 2 is the correct option.

Question 18

Statement I: All positive rational numbers are greater than 0.

Statement II: 0 is the smallest rational number.

  1. Statement I is true but statement II is false.

  2. Statement I is false but statement II is true.

  3. Both Statement I and statement II are true.

  4. Both Statement I and statement II are false.

Answer

Statement I: Every positive rational number is greater than zero. So Statement I is true.

Statement II: Negative rational numbers are smaller than 0, so 0 is not the smallest rational number. So Statement II is false.

Hence, Option 1 is the correct option.

Check Your Progress

Question 1

Write five rational numbers equivalent to 5−11\dfrac{5}{-11}.

Answer

First write 5−11\dfrac{5}{-11} with a positive denominator: 5−11=−511\dfrac{5}{-11} = \dfrac{-5}{11}.

Multiplying the numerator and the denominator by 2, 3, 4 and 5:

−5×211×2=−1022−5×311×3=−1533−5×411×4=−2044−5×511×5=−2555\dfrac{-5 \times 2}{11 \times 2} = \dfrac{-10}{22} \\[1em] \dfrac{-5 \times 3}{11 \times 3} = \dfrac{-15}{33} \\[1em] \dfrac{-5 \times 4}{11 \times 4} = \dfrac{-20}{44} \\[1em] \dfrac{-5 \times 5}{11 \times 5} = \dfrac{-25}{55}

Hence, five rational numbers equivalent to 5−11\dfrac{5}{-11} are −511,−1022,−1533,−2044\dfrac{-5}{11}, \dfrac{-10}{22}, \dfrac{-15}{33}, \dfrac{-20}{44} and −2555\dfrac{-25}{55}.

Question 2

Express 9−15\dfrac{9}{-15} as a rational number with:

(i) denominator 5

(ii) numerator -12

(iii) denominator 30

Answer

First reduce 9−15\dfrac{9}{-15} to standard form. 9−15=−915=−35\dfrac{9}{-15} = \dfrac{-9}{15} = \dfrac{-3}{5} (dividing by HCF 3).

(i) The denominator is already 5.

9−15=−35\dfrac{9}{-15} = \dfrac{-3}{5}

∴ 9−15=−35\dfrac{9}{-15} = \dfrac{-3}{5}

(ii) To get -12 from -3, multiply by 4.

−35=−3×45×4=−1220\dfrac{-3}{5} = \dfrac{-3 \times 4}{5 \times 4} = \dfrac{-12}{20}

∴ 9−15=−1220\dfrac{9}{-15} = \dfrac{-12}{20}

(iii) To get 30 from 5, multiply by 6.

−35=−3×65×6=−1830\dfrac{-3}{5} = \dfrac{-3 \times 6}{5 \times 6} = \dfrac{-18}{30}

∴ 9−15=−1830\dfrac{9}{-15} = \dfrac{-18}{30}

Question 3

Write each of the following numbers in standard form:

(i) 78−91\dfrac{78}{-91}

(ii) −216162\dfrac{-216}{162}

(iii) −195−520\dfrac{-195}{-520}

Answer

(i) Making the denominator positive:

78−91=−7891\dfrac{78}{-91} = \dfrac{-78}{91}

HCF of 78 and 91 is 13.

−7891=−78÷1391÷13=−67\dfrac{-78}{91} = \dfrac{-78 \div 13}{91 \div 13} = \dfrac{-6}{7}

∴ Standard form of 78−91\dfrac{78}{-91} is −67\dfrac{-6}{7}.

(ii) The denominator is positive. HCF of 216 and 162 is 54.

−216162=−216÷54162÷54=−43\dfrac{-216}{162} = \dfrac{-216 \div 54}{162 \div 54} = \dfrac{-4}{3}

∴ Standard form of −216162\dfrac{-216}{162} is −43\dfrac{-4}{3}.

(iii) Making the denominator positive:

−195−520=195520\dfrac{-195}{-520} = \dfrac{195}{520}

HCF of 195 and 520 is 65.

195520=195÷65520÷65=38\dfrac{195}{520} = \dfrac{195 \div 65}{520 \div 65} = \dfrac{3}{8}

∴ Standard form of −195−520\dfrac{-195}{-520} is 38\dfrac{3}{8}.

Question 4

Which of the following are pairs of equivalent rational numbers?

(i) −413,60−195\dfrac{-4}{13}, \dfrac{60}{-195}

(ii) 7−15,−35−75\dfrac{7}{-15}, \dfrac{-35}{-75}

(iii) 16−20,−5670\dfrac{16}{-20}, \dfrac{-56}{70}

Answer

(i) 60−195=−60195=−413\dfrac{60}{-195} = \dfrac{-60}{195} = \dfrac{-4}{13} (dividing by HCF 15). So −413=60−195\dfrac{-4}{13} = \dfrac{60}{-195}. Equivalent.

(ii) 7−15=−715\dfrac{7}{-15} = \dfrac{-7}{15} and −35−75=3575=715\dfrac{-35}{-75} = \dfrac{35}{75} = \dfrac{7}{15}. Since −715≠715\dfrac{-7}{15} \neq \dfrac{7}{15}, they are not equivalent.

(iii) 16−20=−1620=−45\dfrac{16}{-20} = \dfrac{-16}{20} = \dfrac{-4}{5} and −5670=−45\dfrac{-56}{70} = \dfrac{-4}{5} (dividing by HCF 14). So they are equivalent.

Hence, the pairs (i) and (iii) are pairs of equivalent rational numbers.

Question 5

Arrange the following rational numbers in ascending order:

(i) −56,−1718,23−24,−11−13\dfrac{-5}{6}, \dfrac{-17}{18}, \dfrac{23}{-24}, \dfrac{-11}{-13}

(ii) −256,15−4,−178,−5312\dfrac{-25}{6}, \dfrac{15}{-4}, \dfrac{-17}{8}, \dfrac{-53}{12}

Answer

(i) Writing each with a positive denominator: 23−24=−2324\dfrac{23}{-24} = \dfrac{-23}{24} and −11−13=1113\dfrac{-11}{-13} = \dfrac{11}{13}.

So the numbers are −56,−1718,−2324,1113\dfrac{-5}{6}, \dfrac{-17}{18}, \dfrac{-23}{24}, \dfrac{11}{13}. The first three are negative and the last is positive.

LCM of 6, 18 and 24 is 72.

−5×126×12=−6072−17×418×4=−6872−23×324×3=−6972\dfrac{-5 \times 12}{6 \times 12} = \dfrac{-60}{72} \\[1em] \dfrac{-17 \times 4}{18 \times 4} = \dfrac{-68}{72} \\[1em] \dfrac{-23 \times 3}{24 \times 3} = \dfrac{-69}{72}

As -69 < -68 < -60, the negatives in ascending order are −2324,−1718,−56\dfrac{-23}{24}, \dfrac{-17}{18}, \dfrac{-5}{6}, and the positive 1113\dfrac{11}{13} is the greatest.

Hence, the ascending order is 23−24,−1718,−56,−11−13\dfrac{23}{-24}, \dfrac{-17}{18}, \dfrac{-5}{6}, \dfrac{-11}{-13}.

(ii) Writing each with a positive denominator: 15−4=−154\dfrac{15}{-4} = \dfrac{-15}{4}.

So the numbers are −256,−154,−178,−5312\dfrac{-25}{6}, \dfrac{-15}{4}, \dfrac{-17}{8}, \dfrac{-53}{12}. LCM of 6, 4, 8 and 12 is 24.

−25×46×4=−10024−15×64×6=−9024−17×38×3=−5124−53×212×2=−10624\dfrac{-25 \times 4}{6 \times 4} = \dfrac{-100}{24} \\[1em] \dfrac{-15 \times 6}{4 \times 6} = \dfrac{-90}{24} \\[1em] \dfrac{-17 \times 3}{8 \times 3} = \dfrac{-51}{24} \\[1em] \dfrac{-53 \times 2}{12 \times 2} = \dfrac{-106}{24}

As -106 < -100 < -90 < -51,

−10624<−10024<−9024<−5124\dfrac{-106}{24} \lt \dfrac{-100}{24} \lt \dfrac{-90}{24} \lt \dfrac{-51}{24}

Hence, the ascending order is −5312,−256,15−4,−178\dfrac{-53}{12}, \dfrac{-25}{6}, \dfrac{15}{-4}, \dfrac{-17}{8}.

Question 6

Arrange the rational numbers −710,5−8,2−3,−14,−35\dfrac{-7}{10}, \dfrac{5}{-8}, \dfrac{2}{-3}, \dfrac{-1}{4}, \dfrac{-3}{5} in descending order.

Answer

Writing each with a positive denominator: 5−8=−58\dfrac{5}{-8} = \dfrac{-5}{8} and 2−3=−23\dfrac{2}{-3} = \dfrac{-2}{3}.

So the numbers are −710,−58,−23,−14,−35\dfrac{-7}{10}, \dfrac{-5}{8}, \dfrac{-2}{3}, \dfrac{-1}{4}, \dfrac{-3}{5}. LCM of 10, 8, 3, 4 and 5 is 120.

−7×1210×12=−84120−5×158×15=−75120−2×403×40=−80120−1×304×30=−30120−3×245×24=−72120\dfrac{-7 \times 12}{10 \times 12} = \dfrac{-84}{120} \\[1em] \dfrac{-5 \times 15}{8 \times 15} = \dfrac{-75}{120} \\[1em] \dfrac{-2 \times 40}{3 \times 40} = \dfrac{-80}{120} \\[1em] \dfrac{-1 \times 30}{4 \times 30} = \dfrac{-30}{120} \\[1em] \dfrac{-3 \times 24}{5 \times 24} = \dfrac{-72}{120}

As -30 > -72 > -75 > -80 > -84,

−30120>−72120>−75120>−80120>−84120\dfrac{-30}{120} \gt \dfrac{-72}{120} \gt \dfrac{-75}{120} \gt \dfrac{-80}{120} \gt \dfrac{-84}{120}

Hence, the descending order is −14,−35,5−8,2−3,−710\dfrac{-1}{4}, \dfrac{-3}{5}, \dfrac{5}{-8}, \dfrac{2}{-3}, \dfrac{-7}{10}.

Question 7

Find the sum:

(i) −23+(5−7)\dfrac{-2}{3} + \left(\dfrac{5}{-7}\right)

(ii) −1112+−59-1\dfrac{1}{12} + \dfrac{-5}{9}

(iii) 225+(−4310)2\dfrac{2}{5} + \left(-4\dfrac{3}{10}\right)

Answer

(i) 5−7=−57\dfrac{5}{-7} = \dfrac{-5}{7}. LCM of 3 and 7 is 21.

−2×73×7+−5×37×3=−1421+−1521=−14+(−15)21=−2921\dfrac{-2 \times 7}{3 \times 7} + \dfrac{-5 \times 3}{7 \times 3} \\[1em] = \dfrac{-14}{21} + \dfrac{-15}{21} \\[1em] = \dfrac{-14 + (-15)}{21} \\[1em] = \dfrac{-29}{21}

∴ −23+(5−7)=−2921\dfrac{-2}{3} + \left(\dfrac{5}{-7}\right) = \dfrac{-29}{21}

(ii) −1112=−1312-1\dfrac{1}{12} = \dfrac{-13}{12}. LCM of 12 and 9 is 36.

−13×312×3+−5×49×4=−3936+−2036=−39+(−20)36=−5936=−12336\dfrac{-13 \times 3}{12 \times 3} + \dfrac{-5 \times 4}{9 \times 4} \\[1em] = \dfrac{-39}{36} + \dfrac{-20}{36} \\[1em] = \dfrac{-39 + (-20)}{36} \\[1em] = \dfrac{-59}{36} = -1\dfrac{23}{36}

∴ −1112+−59=−12336-1\dfrac{1}{12} + \dfrac{-5}{9} = -1\dfrac{23}{36}

(iii) 225=1252\dfrac{2}{5} = \dfrac{12}{5} and −4310=−4310-4\dfrac{3}{10} = \dfrac{-43}{10}. LCM of 5 and 10 is 10.

12×25×2+−43×110×1=2410+−4310=24+(−43)10=−1910=−1910\dfrac{12 \times 2}{5 \times 2} + \dfrac{-43 \times 1}{10 \times 1} \\[1em] = \dfrac{24}{10} + \dfrac{-43}{10} \\[1em] = \dfrac{24 + (-43)}{10} \\[1em] = \dfrac{-19}{10} = -1\dfrac{9}{10}

∴ 225+(−4310)=−19102\dfrac{2}{5} + \left(-4\dfrac{3}{10}\right) = -1\dfrac{9}{10}

Question 8

Subtract:

(i) −1124\dfrac{-11}{24} from −536\dfrac{-5}{36}

(ii) −815\dfrac{-8}{15} from −125-1\dfrac{2}{5}

(iii) −229-2\dfrac{2}{9} from −3512-3\dfrac{5}{12}

Answer

(i)

−536−−1124=−536+1124\dfrac{-5}{36} - \dfrac{-11}{24} = \dfrac{-5}{36} + \dfrac{11}{24}

LCM of 36 and 24 is 72.

=−5×236×2+11×324×3=−1072+3372=−10+3372=2372= \dfrac{-5 \times 2}{36 \times 2} + \dfrac{11 \times 3}{24 \times 3} \\[1em] = \dfrac{-10}{72} + \dfrac{33}{72} \\[1em] = \dfrac{-10 + 33}{72} \\[1em] = \dfrac{23}{72}

∴ The result is 2372\dfrac{23}{72}.

(ii) −125=−75-1\dfrac{2}{5} = \dfrac{-7}{5}.

−75−−815=−75+815\dfrac{-7}{5} - \dfrac{-8}{15} = \dfrac{-7}{5} + \dfrac{8}{15}

LCM of 5 and 15 is 15.

=−7×35×3+8×115×1=−2115+815=−21+815=−1315= \dfrac{-7 \times 3}{5 \times 3} + \dfrac{8 \times 1}{15 \times 1} \\[1em] = \dfrac{-21}{15} + \dfrac{8}{15} \\[1em] = \dfrac{-21 + 8}{15} \\[1em] = \dfrac{-13}{15}

∴ The result is −1315\dfrac{-13}{15}.

(iii) −3512=−4112-3\dfrac{5}{12} = \dfrac{-41}{12} and −229=−209-2\dfrac{2}{9} = \dfrac{-20}{9}.

−4112−−209=−4112+209\dfrac{-41}{12} - \dfrac{-20}{9} = \dfrac{-41}{12} + \dfrac{20}{9}

LCM of 12 and 9 is 36.

=−41×312×3+20×49×4=−12336+8036=−123+8036=−4336=−1736= \dfrac{-41 \times 3}{12 \times 3} + \dfrac{20 \times 4}{9 \times 4} \\[1em] = \dfrac{-123}{36} + \dfrac{80}{36} \\[1em] = \dfrac{-123 + 80}{36} \\[1em] = \dfrac{-43}{36} = -1\dfrac{7}{36}

∴ The result is −1736-1\dfrac{7}{36}.

Question 9

What should be subtracted from −34-\dfrac{3}{4} to get −58\dfrac{-5}{8}?

Answer

Let the required number be xx.

−34−x=−58⇒x=−34−−58⇒x=−34+58-\dfrac{3}{4} - x = \dfrac{-5}{8} \\[1em] \Rightarrow x = -\dfrac{3}{4} - \dfrac{-5}{8} \\[1em] \Rightarrow x = -\dfrac{3}{4} + \dfrac{5}{8}

LCM of 4 and 8 is 8.

⇒x=−3×24×2+5×18×1⇒x=−68+58⇒x=−6+58⇒x=−18\Rightarrow x = \dfrac{-3 \times 2}{4 \times 2} + \dfrac{5 \times 1}{8 \times 1} \\[1em] \Rightarrow x = \dfrac{-6}{8} + \dfrac{5}{8} \\[1em] \Rightarrow x = \dfrac{-6 + 5}{8} \\[1em] \Rightarrow x = \dfrac{-1}{8}

Hence, −18\dfrac{-1}{8} should be subtracted.

Question 10

Find the following products:

(i) −311×(−7)\dfrac{-3}{11} \times (-7)

(ii) −512×1625\dfrac{-5}{12} \times \dfrac{16}{25}

(iii) (−258)×(137)\left(-2\dfrac{5}{8}\right) \times \left(1\dfrac{3}{7}\right)

Answer

(i)

−311×(−7)=−311×−71=−3×(−7)11×1=2111=11011\dfrac{-3}{11} \times (-7) = \dfrac{-3}{11} \times \dfrac{-7}{1} = \dfrac{-3 \times (-7)}{11 \times 1} = \dfrac{21}{11} = 1\dfrac{10}{11}

∴ −311×(−7)=2111\dfrac{-3}{11} \times (-7) = \dfrac{21}{11}

(ii)

−512×1625=−5×1612×25=−80300=−415\dfrac{-5}{12} \times \dfrac{16}{25} = \dfrac{-5 \times 16}{12 \times 25} = \dfrac{-80}{300} = \dfrac{-4}{15}

∴ −512×1625=−415\dfrac{-5}{12} \times \dfrac{16}{25} = \dfrac{-4}{15}

(iii) −258=−218-2\dfrac{5}{8} = \dfrac{-21}{8} and 137=1071\dfrac{3}{7} = \dfrac{10}{7}.

−218×107=−21×108×7=−21056=−154=−334\dfrac{-21}{8} \times \dfrac{10}{7} = \dfrac{-21 \times 10}{8 \times 7} = \dfrac{-210}{56} = \dfrac{-15}{4} = -3\dfrac{3}{4}

∴ (−258)×(137)=−334\left(-2\dfrac{5}{8}\right) \times \left(1\dfrac{3}{7}\right) = -3\dfrac{3}{4}

Question 11

Find the value of:

(i) −813÷3−26\dfrac{-8}{13} \div \dfrac{3}{-26}

(ii) 317÷11−123\dfrac{1}{7} \div \dfrac{11}{-12}

(iii) (−37×−23)÷16−21\left(\dfrac{-3}{7} \times \dfrac{-2}{3}\right) \div \dfrac{16}{-21}

Answer

(i)

−813÷3−26=−813×−263=−8×(−26)13×3=20839=163=513\dfrac{-8}{13} \div \dfrac{3}{-26} = \dfrac{-8}{13} \times \dfrac{-26}{3} = \dfrac{-8 \times (-26)}{13 \times 3} = \dfrac{208}{39} = \dfrac{16}{3} = 5\dfrac{1}{3}

∴ −813÷3−26=163\dfrac{-8}{13} \div \dfrac{3}{-26} = \dfrac{16}{3}

(ii) 317=2273\dfrac{1}{7} = \dfrac{22}{7}.

227÷11−12=227×−1211=22×(−12)7×11=−26477=−247=−337\dfrac{22}{7} \div \dfrac{11}{-12} = \dfrac{22}{7} \times \dfrac{-12}{11} = \dfrac{22 \times (-12)}{7 \times 11} = \dfrac{-264}{77} = \dfrac{-24}{7} = -3\dfrac{3}{7}

∴ 317÷11−12=−3373\dfrac{1}{7} \div \dfrac{11}{-12} = -3\dfrac{3}{7}

(iii) First solve the bracket:

−37×−23=−3×(−2)7×3=621=27\dfrac{-3}{7} \times \dfrac{-2}{3} = \dfrac{-3 \times (-2)}{7 \times 3} = \dfrac{6}{21} = \dfrac{2}{7}

Now,

27÷16−21=27×−2116=2×(−21)7×16=−42112=−38\dfrac{2}{7} \div \dfrac{16}{-21} = \dfrac{2}{7} \times \dfrac{-21}{16} = \dfrac{2 \times (-21)}{7 \times 16} = \dfrac{-42}{112} = \dfrac{-3}{8}

∴ (−37×−23)÷16−21=−38\left(\dfrac{-3}{7} \times \dfrac{-2}{3}\right) \div \dfrac{16}{-21} = \dfrac{-3}{8}

Question 12

From a rope 15 m long, 4134\dfrac{1}{3} m is cut off and 35\dfrac{3}{5} of the remaining is cut off again. Find the length of the remaining part of the rope.

Answer

Length of rope cut off first =413= 4\dfrac{1}{3} m =133= \dfrac{13}{3} m.

Remaining length after first cut:

15−133=453−133=45−133=323 m15 - \dfrac{13}{3} = \dfrac{45}{3} - \dfrac{13}{3} = \dfrac{45 - 13}{3} = \dfrac{32}{3} \text{ m}

Now 35\dfrac{3}{5} of the remaining is cut off, so the part left is (1−35)=25\left(1 - \dfrac{3}{5}\right) = \dfrac{2}{5} of 323\dfrac{32}{3}.

25×323=2×325×3=6415=4415 m\dfrac{2}{5} \times \dfrac{32}{3} = \dfrac{2 \times 32}{5 \times 3} = \dfrac{64}{15} = 4\dfrac{4}{15} \text{ m}

Hence, the length of the remaining part of the rope is 44154\dfrac{4}{15} m.

Question 13

Perimeter of a rectangle is 2 m less than 25\dfrac{2}{5} of the perimeter of a square. If the perimeter of the square is 40 m, find the length and breadth of the rectangle given that the breadth is 13\dfrac{1}{3} of the length.

Answer

Perimeter of the square = 40 m.

25 of perimeter of square=25×40=16 m\dfrac{2}{5} \text{ of perimeter of square} = \dfrac{2}{5} \times 40 = 16 \text{ m}

Perimeter of rectangle = 16 - 2 = 14 m.

Let the length be ll. Then breadth =13l= \dfrac{1}{3}l.

2(l+b)=14⇒l+13l=7⇒3l+l3=7⇒4l3=7⇒l=7×34=214 m2(l + b) = 14 \\[1em] \Rightarrow l + \dfrac{1}{3}l = 7 \\[1em] \Rightarrow \dfrac{3l + l}{3} = 7 \\[1em] \Rightarrow \dfrac{4l}{3} = 7 \\[1em] \Rightarrow l = \dfrac{7 \times 3}{4} = \dfrac{21}{4} \text{ m}

Breadth:

b=13×214=2112=74 mb = \dfrac{1}{3} \times \dfrac{21}{4} = \dfrac{21}{12} = \dfrac{7}{4} \text{ m}

Hence, the length of the rectangle is 214\dfrac{21}{4} m and the breadth is 74\dfrac{7}{4} m.

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