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Chapter 2

Rational Numbers

Class - 7 Concise Mathematics Selina

Exercise 2(A)

Question 1

Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits.

Answer

Largest number of two digits = 99

Smallest number of four digits = 1000

Hence, the required rational number = 991000\dfrac{99}{1000}.

Question 2

Write the numerator of each of the following rational numbers:

(i) −125127\dfrac{-125}{127}

(ii) 37−137\dfrac{37}{-137}

(iii) −8593\dfrac{-85}{93}

(iv) 2

(v) 0

Answer

(i) Numerator of −125127\dfrac{-125}{127} is -125.

(ii) Numerator of 37−137\dfrac{37}{-137} is 37.

(iii) Numerator of −8593\dfrac{-85}{93} is -85.

(iv) 2 can be written as 21\dfrac{2}{1}, so its numerator is 2.

(v) 0 can be written as 01\dfrac{0}{1}, so its numerator is 0.

Question 3

Write the denominator of each of the following rational numbers:

(i) 7−15\dfrac{7}{-15}

(ii) −1829\dfrac{-18}{29}

(iii) −34\dfrac{-3}{4}

(iv) -7

(v) 0

Answer

(i) Denominator of 7−15\dfrac{7}{-15} is -15.

(ii) Denominator of −1829\dfrac{-18}{29} is 29.

(iii) Denominator of −34\dfrac{-3}{4} is 4.

(iv) -7 can be written as −71\dfrac{-7}{1}, so its denominator is 1.

(v) 0 can be written as 01,02,03,04,…\dfrac{0}{1}, \dfrac{0}{2}, \dfrac{0}{3}, \dfrac{0}{4}, \ldots

Zero does not have any unique denominator, so its denominator can be any non-zero number.

Question 4

Write down a rational number with numerator (-5) × (-4) and with denominator (28 - 27) × (8 - 5).

Answer

Numerator = (-5) × (-4) = 20

Denominator = (28 - 27) × (8 - 5) = 1 × 3 = 3

Hence, the required rational number = 203\dfrac{20}{3}.

Question 5

(i) −151\dfrac{-15}{1} in integer form is ............ .

(ii) 23−1\dfrac{23}{-1} in integer form is ............ .

(iii) If 18=18a18 = \dfrac{18}{a} then a = ............ .

(iv) If −57=57a-57 = \dfrac{57}{a} then a = ............ .

Answer

(i) −151\dfrac{-15}{1} in integer form is -15.

(ii) 23−1\dfrac{23}{-1} in integer form is -23.

(iii) Given,

⇒18=18a⇒18×a=18×1⇒a=1818⇒a=1\Rightarrow 18 = \dfrac{18}{a} \\[1em] \Rightarrow 18 \times a = 18 \times 1 \\[1em] \Rightarrow a = \dfrac{18}{18} \\[1em] \Rightarrow a = 1

Hence, a = 1

(iv) Given,

⇒−57=57a⇒−57×a=57×1⇒a=57−57⇒a=−1\Rightarrow -57 = \dfrac{57}{a}\\[1em] \Rightarrow -57 \times a = 57 \times 1 \\[1em] \Rightarrow a = \dfrac{57}{-57} \\[1em] \Rightarrow a = -1

Hence, a = -1

Question 6

Separate positive and negative rational numbers from the following:

−35,3−5,−3−5,35,0,−13−3,15−8,−158\dfrac{-3}{5}, \dfrac{3}{-5}, \dfrac{-3}{-5}, \dfrac{3}{5}, 0, \dfrac{-13}{-3}, \dfrac{15}{-8}, \dfrac{-15}{8}

Answer

A rational number is positive if its numerator and denominator are both of the same sign, and negative if they are of opposite signs.

Positive rational numbers: −3−5,35 and −13−3\dfrac{-3}{-5}, \dfrac{3}{5} \text{ and } \dfrac{-13}{-3}

Negative rational numbers: −35,3−5,15−8 and −158\dfrac{-3}{5}, \dfrac{3}{-5}, \dfrac{15}{-8} \text{ and } \dfrac{-15}{8}

(0 is neither positive nor negative).

Question 7

Find three rational numbers equivalent to

(i) 35\dfrac{3}{5}

(ii) 4−7\dfrac{4}{-7}

(iii) −59\dfrac{-5}{9}

(iv) 8−15\dfrac{8}{-15}

Answer

(i) Three rational numbers equivalent to 35\dfrac{3}{5} are :

3×25×2=610,3×35×3=915,3×45×4=1220\dfrac{3 \times 2}{5 \times 2} = \dfrac{6}{10}, \\[1em] \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15}, \\[1em] \dfrac{3 \times 4}{5 \times 4} = \dfrac{12}{20}

Hence, 610,915 and 1220\dfrac{6}{10}, \dfrac{9}{15} \text{ and } \dfrac{12}{20} are three rational numbers equivalent to 35\dfrac{3}{5}.

(ii) Three rational numbers equivalent to 4−7\dfrac{4}{-7} are :

4×2−7×2=8−14,4×3−7×3=12−21,4×4−7×4=16−28\dfrac{4 \times 2}{-7 \times 2} = \dfrac{8}{-14}, \\[1em] \dfrac{4 \times 3}{-7 \times 3} = \dfrac{12}{-21}, \\[1em] \dfrac{4 \times 4}{-7 \times 4} = \dfrac{16}{-28}

Hence, 8−14,12−21 and 16−28\dfrac{8}{-14}, \dfrac{12}{-21} \text{ and } \dfrac{16}{-28} are three rational numbers equivalent to 4−7\dfrac{4}{-7}.

(iii) Three rational numbers equivalent to −59\dfrac{-5}{9} are :

−5×29×2=−1018,−5×39×3=−1527,−5×49×4=−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, −1018,−1527 and −2036\dfrac{-10}{18}, \dfrac{-15}{27} \text{ and } \dfrac{-20}{36} are three rational numbers equivalent to −59\dfrac{-5}{9}.

(iv) Three rational numbers equivalent to 8−15\dfrac{8}{-15} are :

8×2−15×2=16−30,8×3−15×3=24−45,8×4−15×4=32−60\dfrac{8 \times 2}{-15 \times 2} = \dfrac{16}{-30}, \\[1em] \dfrac{8 \times 3}{-15 \times 3} = \dfrac{24}{-45}, \\[1em] \dfrac{8 \times 4}{-15 \times 4} = \dfrac{32}{-60}

Hence, 16−30,24−45 and 32−60\dfrac{16}{-30}, \dfrac{24}{-45} \text{ and } \dfrac{32}{-60} are three rational numbers equivalent to 8−15\dfrac{8}{-15}.

Question 8

Which of the following are not rational numbers:

(i) -3

(ii) 0

(iii) 04\dfrac{0}{4}

(iv) 80\dfrac{8}{0}

(v) 00\dfrac{0}{0}

Answer

A number pq\dfrac{p}{q} is a rational number only if q ≠ 0.

(i) -3 = −31\dfrac{-3}{1}, which is a rational number.

(ii) 0 = 01\dfrac{0}{1}, which is a rational number.

(iii) 04=0\dfrac{0}{4} = 0, which is a rational number.

(iv) 80\dfrac{8}{0} has denominator 0, so it is not a rational number.

(v) 00\dfrac{0}{0} has denominator 0, so it is not a rational number.

Hence, 80 and 00\dfrac{8}{0}\text{ and }\dfrac{0}{0} are not rational numbers.

Question 9

Express each of the following integers as a rational number with denominator 7:

(i) 5

(ii) -8

(iii) 0

(iv) -16

(v) 7

Answer

(i) 5=5×71×7=3575 = \dfrac{5 \times 7}{1 \times 7} = \dfrac{35}{7}

(ii) −8=−8×71×7=−567-8 = \dfrac{-8 \times 7}{1 \times 7} = \dfrac{-56}{7}

(iii) 0=0×71×7=070 = \dfrac{0 \times 7}{1 \times 7} = \dfrac{0}{7}

(iv) −16=−16×71×7=−1127-16 = \dfrac{-16 \times 7}{1 \times 7} = \dfrac{-112}{7}

(v) 7=7×71×7=4977 = \dfrac{7 \times 7}{1 \times 7} = \dfrac{49}{7}

Question 10

Express 35\dfrac{3}{5} as a rational number with denominator:

(i) 20

(ii) -20

(iii) 45

(iv) 25

(v) -35

Answer

(i) 35=3×45×4=1220\dfrac{3}{5} = \dfrac{3 \times 4}{5 \times 4} = \dfrac{12}{20}

(ii) 35=3×(−4)5×(−4)=−12−20\dfrac{3}{5} = \dfrac{3 \times (-4)}{5 \times (-4)} = \dfrac{-12}{-20}

(iii) 35=3×95×9=2745\dfrac{3}{5} = \dfrac{3 \times 9}{5 \times 9} = \dfrac{27}{45}

(iv) 35=3×55×5=1525\dfrac{3}{5} = \dfrac{3 \times 5}{5 \times 5} = \dfrac{15}{25}

(v) 35=3×(−7)5×(−7)=−21−35\dfrac{3}{5} = \dfrac{3 \times (-7)}{5 \times (-7)} = \dfrac{-21}{-35}

Question 11

Express 47\dfrac{4}{7} as a rational number with numerator:

(i) 12

(ii) -12

(iii) -16

(iv) -20

(v) 20

Answer

(i) 47=4×37×3=1221\dfrac{4}{7} = \dfrac{4 \times 3}{7 \times 3} = \dfrac{12}{21}

(ii) 47=4×(−3)7×(−3)=−12−21\dfrac{4}{7} = \dfrac{4 \times (-3)}{7 \times (-3)} = \dfrac{-12}{-21}

(iii) 47=4×(−4)7×(−4)=−16−28\dfrac{4}{7} = \dfrac{4 \times (-4)}{7 \times (-4)} = \dfrac{-16}{-28}

(iv) 47=4×(−5)7×(−5)=−20−35\dfrac{4}{7} = \dfrac{4 \times (-5)}{7 \times (-5)} = \dfrac{-20}{-35}

(v) 47=4×57×5=2035\dfrac{4}{7} = \dfrac{4 \times 5}{7 \times 5} = \dfrac{20}{35}

Question 12

Find x, such that:

(i) −23=6x-\dfrac{2}{3} = \dfrac{6}{x}

(ii) 7−4=x8\dfrac{7}{-4} = \dfrac{x}{8}

(iii) 37=x−35\dfrac{3}{7} = \dfrac{x}{-35}

(iv) −48x=6\dfrac{-48}{x} = 6

(v) 36x=3\dfrac{36}{x} = 3

(vi) −27x=9\dfrac{-27}{x} = 9

Answer

(i) Given,

⇒−23=6x⇒−2×x=3×6⇒−2x=18⇒x=18−2⇒x=−9\Rightarrow -\dfrac{2}{3} = \dfrac{6}{x} \\[1em] \Rightarrow -2 \times x = 3 \times 6 \\[1em] \Rightarrow -2x = 18 \\[1em] \Rightarrow x = \dfrac{18}{-2} \\[1em] \Rightarrow x = -9

Hence, x = -9

(ii) Given,

⇒7−4=x8⇒−4×x=7×8⇒−4x=56⇒x=56−4⇒x=−14\Rightarrow \dfrac{7}{-4} = \dfrac{x}{8}\\[1em] \Rightarrow -4 \times x = 7 \times 8\\[1em] \Rightarrow -4x = 56 \\[1em] \Rightarrow x = \dfrac{56}{-4} \\[1em] \Rightarrow x = -14

Hence, x = -14

(iii) Given,

⇒37=x−35⇒7×x=3×(−35)⇒7x=−105⇒x=−1057⇒x=−15\Rightarrow \dfrac{3}{7} = \dfrac{x}{-35}\\[1em] \Rightarrow 7 \times x = 3 \times (-35) \\[1em] \Rightarrow 7x = -105 \\[1em] \Rightarrow x = \dfrac{-105}{7} \\[1em] \Rightarrow x = -15

Hence, x = -15

(iv) Given,

⇒−48x=6⇒6x=−48⇒x=−486⇒x=−8\Rightarrow \dfrac{-48}{x} = 6\\[1em] \Rightarrow 6x = -48 \\[1em] \Rightarrow x = \dfrac{-48}{6} \\[1em] \Rightarrow x = -8

Hence, x = -8

(v) Given,

⇒36x=3⇒3x=36⇒x=363⇒x=12\Rightarrow \dfrac{36}{x} = 3\\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \dfrac{36}{3} \\[1em] \Rightarrow x = 12

Hence, x = 12

(vi) Given,

⇒−27x=9⇒9x=−27⇒x=−279⇒x=−3\Rightarrow \dfrac{-27}{x} = 9\\[1em] \Rightarrow 9x = -27 \\[1em] \Rightarrow x = \dfrac{-27}{9} \\[1em] \Rightarrow x = -3

Hence, x = -3

Question 13

Express each of the following rational numbers to the lowest terms:

(i) 1215\dfrac{12}{15}

(ii) −120144\dfrac{-120}{144}

(iii) −48−72\dfrac{-48}{-72}

(iv) 14−56\dfrac{14}{-56}

Answer

(i) By Prime Factorization,

21226331 and 315551\begin{array}{l|r} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}

HCF of 12 and 15 = 3.

1215=12÷315÷3=45\dfrac{12}{15} = \dfrac{12 ÷ 3}{15 ÷ 3} = \dfrac{4}{5}

Hence, 1215=45\dfrac{12}{15} = \dfrac{4}{5}

(ii) By Prime Factorization,

2120260230315551 and 214427223621839331\begin{array}{l|r} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

HCF of 120 and 144 = 2 × 2 × 2 × 3 = 24.

−120144=−120÷24144÷24=−56\dfrac{-120}{144} = \dfrac{-120 ÷ 24}{144 ÷ 24} = \dfrac{-5}{6}

Hence, −120144=−56\dfrac{-120}{144} = \dfrac{-5}{6}

(iii) −48−72=4872\dfrac{-48}{-72} = \dfrac{48}{72}

By Prime Factorization,

24822421226331 and 27223621839331\begin{array}{l|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

HCF of 48 and 72 = 2 × 2 × 2 × 3 = 24.

4872=48÷2472÷24=23\dfrac{48}{72} = \dfrac{48 ÷ 24}{72 ÷ 24} = \dfrac{2}{3}

Hence, −48−72=23\dfrac{-48}{-72} = \dfrac{2}{3}

(iv) By Prime Factorization,

214771 and 256228214771\begin{array}{l|r} 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}

HCF of 14 and 56 = 2 × 7 = 14.

14−56=14÷14−56÷14=1−4=−14\dfrac{14}{-56} = \dfrac{14 ÷ 14}{-56 ÷ 14} = \dfrac{1}{-4} = -\dfrac{1}{4}

Hence, 14−56=−14\dfrac{14}{-56} = -\dfrac{1}{4}

Question 14

Express each of the following rational numbers in the standard form.

(i) −7−8\dfrac{-7}{-8}

(ii) 5−12\dfrac{5}{-12}

(iii) −7−20\dfrac{-7}{-20}

(iv) 4−9\dfrac{4}{-9}

Answer

A rational number is in standard form if its denominator is positive and the rational number is in its lowest terms.

(i) −7−8=−7×(−1)−8×(−1)=78\dfrac{-7}{-8} = \dfrac{-7 \times (-1)}{-8 \times (-1)} = \dfrac{7}{8}

Hence, standard form of −7−8 is 78\dfrac{-7}{-8}\text{ is }\dfrac{7}{8}.

(ii) 5−12=5×(−1)−12×(−1)=−512\dfrac{5}{-12} = \dfrac{5 \times (-1)}{-12 \times (-1)} = \dfrac{-5}{12}

Hence, standard form of 5−12 is −512\dfrac{5}{-12}\text{ is }\dfrac{-5}{12}.

(iii) −7−20=−7×(−1)−20×(−1)=720\dfrac{-7}{-20} = \dfrac{-7 \times (-1)}{-20 \times (-1)} = \dfrac{7}{20}

Hence, standard form of −7−20 is 720\dfrac{-7}{-20}\text{ is }\dfrac{7}{20}.

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

Hence, standard form of 4−9 is −49\dfrac{4}{-9}\text{ is }\dfrac{-4}{9}.

Exercise 2(B)

Question 1

Mark the following pairs of rational numbers on the separate number lines:

(i) 34 and −14\dfrac{3}{4} \text{ and }-\dfrac{1}{4}

(ii) 25 and −35\dfrac{2}{5} \text{ and } \dfrac{-3}{5}

(iii) 56 and −23\dfrac{5}{6} \text{ and } -\dfrac{2}{3}

(iv) 25 and −45\dfrac{2}{5} \text{ and } -\dfrac{4}{5}

(v) 14 and −54\dfrac{1}{4} \text{ and } -\dfrac{5}{4}

Answer

(i) Since the denominator of each rational number is 4, divide each unit length between 0 and 1, and between 0 and -1, into four equal parts.

To represent 34\dfrac{3}{4}, move 3 parts to the right of 0; to represent −14-\dfrac{1}{4}, move 1 part to the left of 0.

Mark the following pairs of rational numbers on the separate number lines: Mathematics Solutions ICSE Class 7.

(ii) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent 25\dfrac{2}{5}, move 2 parts to the right of 0; to represent −35\dfrac{-3}{5}, move 3 parts to the left of 0.

Mark the following pairs of rational numbers on the separate number lines: Mathematics Solutions ICSE Class 7.

(iii) −23=−46-\dfrac{2}{3} = -\dfrac{4}{6}, so both numbers have denominator 6. Divide each unit length into six equal parts.

To represent 56\dfrac{5}{6}, move 5 parts to the right of 0; to represent −23=−46-\dfrac{2}{3} = -\dfrac{4}{6}, move 4 parts to the left of 0.

Mark the following pairs of rational numbers on the separate number lines: Mathematics Solutions ICSE Class 7.

(iv) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent 25\dfrac{2}{5}, move 2 parts to the right of 0; to represent −45-\dfrac{4}{5}, move 4 parts to the left of 0.

Mark the following pairs of rational numbers on the separate number lines: Mathematics Solutions ICSE Class 7.

(v) Since the denominator of each rational number is 4, divide each unit length into four equal parts.

To represent 14\dfrac{1}{4}, move 1 part to the right of 0; to represent −54-\dfrac{5}{4}, move 5 parts to the left of 0.

Mark the following pairs of rational numbers on the separate number lines: Mathematics Solutions ICSE Class 7.

Question 2

Compare:

(i) 35 and 57\dfrac{3}{5} \text{ and } \dfrac{5}{7}

(ii) −72 and 52\dfrac{-7}{2} \text{ and } \dfrac{5}{2}

(iii) -3 and 2342\dfrac{3}{4}

(iv) −112-1\dfrac{1}{2} and 0

(v) 0 and 34\dfrac{3}{4}

(vi) 3 and -1

Answer

(i) On cross-multiplying 35\dfrac{3}{5} and 57\dfrac{5}{7}:

3 × 7 = 21 and 5 × 5 = 25

Since 21 < 25,

Hence, 35<57\dfrac{3}{5} \lt \dfrac{5}{7}

(ii) −72\dfrac{-7}{2} is a negative rational number and 52\dfrac{5}{2} is a positive rational number.

Since every negative rational number is less than every positive rational number,

Hence, −72<52\dfrac{-7}{2} \lt \dfrac{5}{2}

(iii) -3 is a negative number and 2342\dfrac{3}{4} is a positive number.

Since every negative number is less than every positive number,

Hence, −3<234-3 \lt 2\dfrac{3}{4}

(iv) −112-1\dfrac{1}{2} is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, −112<0-1\dfrac{1}{2} \lt 0

(v) 0 is neither positive nor negative and 34\dfrac{3}{4} is a positive number.

Since 0 is less than every positive number,

Hence, 0<340 \lt \dfrac{3}{4}

(vi) 3 is a positive number and -1 is a negative number.

Since every positive number is greater than every negative number,

Hence, 3 > -1

Question 3

Compare:

(i) −14-\dfrac{1}{4} and 0

(ii) 14\dfrac{1}{4} and 0

(iii) −38 and 25-\dfrac{3}{8} \text{ and } \dfrac{2}{5}

(iv) −58 and 7−12\dfrac{-5}{8} \text{ and } \dfrac{7}{-12}

(v) 5−9 and −5−9\dfrac{5}{-9} \text{ and } \dfrac{-5}{-9}

(vi) −78 and 5−6\dfrac{-7}{8} \text{ and } \dfrac{5}{-6}

(vii) 27 and −3−8\dfrac{2}{7} \text{ and } \dfrac{-3}{-8}

Answer

(i) −14-\dfrac{1}{4} is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, −14<0-\dfrac{1}{4} \lt 0

(ii) 14\dfrac{1}{4} is a positive number and 0 is neither positive nor negative.

Since 0 is less than every positive number,

Hence, 14>0\dfrac{1}{4} \gt 0

(iii) −38-\dfrac{3}{8} is a negative number and 25\dfrac{2}{5} is a positive number.

Since every negative number is less than every positive number,

Hence, −38<25-\dfrac{3}{8} \lt \dfrac{2}{5}

(iv) 7−12=−712\dfrac{7}{-12} = \dfrac{-7}{12}

By Division Method,

28,1224,622,331,31,1\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

−5×38×3=−1524 and −7×212×2=−1424\dfrac{-5 \times 3}{8 \times 3} = \dfrac{-15}{24} \text{ and } \dfrac{-7 \times 2}{12 \times 2} = \dfrac{-14}{24}

Since -15 < -14,

Hence, −58<7−12\dfrac{-5}{8} \lt \dfrac{7}{-12}

(v) 5−9=−59 and −5−9=59\dfrac{5}{-9} = \dfrac{-5}{9} \text{ and } \dfrac{-5}{-9} = \dfrac{5}{9}

−59\dfrac{-5}{9} is negative and 59\dfrac{5}{9} is positive.

Since every negative number is less than every positive number,

Hence, 5−9<−5−9\dfrac{5}{-9} \lt \dfrac{-5}{-9}

(vi) 5−6=−56\dfrac{5}{-6} = \dfrac{-5}{6}

By prime factorization,

2824221 and 26331\begin{array}{l|r} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

−7×38×3=−2124 and −5×46×4=−2024\dfrac{-7 \times 3}{8 \times 3} = \dfrac{-21}{24} \text{ and } \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24}

Since -21 < -20,

Hence, −78<5−6\dfrac{-7}{8} \lt \dfrac{5}{-6}

(vii) −3−8=38\dfrac{-3}{-8} = \dfrac{3}{8}

On cross-multiplying 27 and 38\dfrac{2}{7} \text{ and } \dfrac{3}{8}:

2×8=16 and 7×3=212 \times 8 = 16 \text{ and } 7 \times 3 = 21

Since 16 < 21,

Hence, 27<−3−8\dfrac{2}{7} \lt \dfrac{-3}{-8}

Question 4

Arrange the given rational numbers in ascending order:

(i) 710,−11−30 and 5−15\dfrac{7}{10}, \dfrac{-11}{-30} \text{ and } \dfrac{5}{-15}

(ii) 4−9,−512 and 2−3\dfrac{4}{-9}, \dfrac{-5}{12} \text{ and } \dfrac{2}{-3}

Answer

(i) Writing each number with a positive denominator:

−11−30=1130 and 5−15=−13\dfrac{-11}{-30} = \dfrac{11}{30} \text{ and } \dfrac{5}{-15} = \dfrac{-1}{3}

So, the numbers are 710,1130 and −13\dfrac{7}{10}, \dfrac{11}{30} \text{ and } \dfrac{-1}{3}.

By Division Method,

210,30,335,15,355,5,11,1,1\begin{array}{l|r} 2 & 10, 30, 3 \\ \hline 3 & 5, 15, 3 \\ \hline 5 & 5, 5, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 10, 30 and 3 is 2 × 3 × 5 = 30

7×310×3=2130,11×130×1=1130,−1×103×10=−1030\dfrac{7 \times 3}{10 \times 3} = \dfrac{21}{30}, \\[1em] \dfrac{11 \times 1}{30 \times 1} = \dfrac{11}{30}, \\[1em] \dfrac{-1 \times 10}{3 \times 10} = \dfrac{-10}{30}

Since -10 < 11 < 21,

−1030<1130<2130\dfrac{-10}{30} \lt \dfrac{11}{30} \lt \dfrac{21}{30}

Hence, 5−15<−11−30<710\dfrac{5}{-15} \lt \dfrac{-11}{-30} \lt \dfrac{7}{10}.

(ii) Writing each number with a positive denominator:

4−9=−49 and 2−3=−23\dfrac{4}{-9} = \dfrac{-4}{9} \text{ and } \dfrac{2}{-3} = \dfrac{-2}{3}

So, the numbers are −49,−512 and −23\dfrac{-4}{9}, \dfrac{-5}{12} \text{ and } \dfrac{-2}{3}.

By Division Method,

29,12,329,6,339,3,333,1,11,1,1\begin{array}{l|r} 2 & 9, 12, 3 \\ \hline 2 & 9, 6, 3 \\ \hline 3 & 9, 3, 3 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 9, 12 and 3 is 2 × 2 × 3 × 3 = 36

−4×49×4=−1636,−5×312×3=−1536,−2×123×12=−2436\dfrac{-4 \times 4}{9 \times 4} = \dfrac{-16}{36}, \\[1em] \dfrac{-5 \times 3}{12 \times 3} = \dfrac{-15}{36}, \\[1em] \dfrac{-2 \times 12}{3 \times 12} = \dfrac{-24}{36}

Since -24 < -16 < -15,

−2436<−1636<−1536\dfrac{-24}{36} \lt \dfrac{-16}{36} \lt \dfrac{-15}{36}

Hence, 2−3<4−9<−512\dfrac{2}{-3} \lt \dfrac{4}{-9} \lt \dfrac{-5}{12}.

Question 5

Arrange the given rational numbers in descending order:

(i) 58,13−16 and −712\dfrac{5}{8}, \dfrac{13}{-16} \text{ and } \dfrac{-7}{12}

(ii) 3−10,−1330 and 8−20\dfrac{3}{-10}, \dfrac{-13}{30} \text{ and } \dfrac{8}{-20}

Answer

(i) Writing each number with a positive denominator:

13−16=−1316\dfrac{13}{-16} = \dfrac{-13}{16}

So, the numbers are 58,−1316 and −712\dfrac{5}{8}, \dfrac{-13}{16} \text{ and } \dfrac{-7}{12}.

By Division Method,

28,16,1224,8,622,4,321,2,331,1,31,1,1\begin{array}{l|r} 2 & 8, 16, 12 \\ \hline 2 & 4, 8, 6 \\ \hline 2 & 2, 4, 3 \\ \hline 2 & 1, 2, 3 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 8, 16 and 12 is 2 × 2 × 2 × 2 × 3 = 48

5×68×6=3048,−13×316×3=−3948,−7×412×4=−2848\dfrac{5 \times 6}{8 \times 6} = \dfrac{30}{48}, \\[1em] \dfrac{-13 \times 3}{16 \times 3} = \dfrac{-39}{48}, \\[1em] \dfrac{-7 \times 4}{12 \times 4} = \dfrac{-28}{48}

Since 30 > -28 > -39,

3048>−2848>−3948\dfrac{30}{48} \gt \dfrac{-28}{48} \gt \dfrac{-39}{48}

Hence, 58>−712>13−16\dfrac{5}{8} \gt \dfrac{-7}{12} \gt \dfrac{13}{-16}.

(ii) Writing each number with a positive denominator:

3−10=−310 and 8−20=−25\dfrac{3}{-10} = \dfrac{-3}{10} \text{ and } \dfrac{8}{-20} = \dfrac{-2}{5}

So, the numbers are −310,−1330 and −25\dfrac{-3}{10}, \dfrac{-13}{30} \text{ and } \dfrac{-2}{5}.

By Division Method,

210,30,535,15,555,5,51,1,1\begin{array}{l|r} 2 & 10, 30, 5 \\ \hline 3 & 5, 15, 5 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1, 1 \end{array}

LCM of 10, 30 and 5 is 2 × 3 × 5 = 30

−3×310×3=−930,−13×130×1=−1330,−2×65×6=−1230\dfrac{-3 \times 3}{10 \times 3} = \dfrac{-9}{30}, \\[1em] \dfrac{-13 \times 1}{30 \times 1} = \dfrac{-13}{30}, \\[1em] \dfrac{-2 \times 6}{5 \times 6} = \dfrac{-12}{30}

Since -9 > -12 > -13,

−930>−1230>−1330\dfrac{-9}{30} \gt \dfrac{-12}{30} \gt \dfrac{-13}{30}

Hence, 3−10>8−20>−1330\dfrac{3}{-10} \gt \dfrac{8}{-20} \gt \dfrac{-13}{30}.

Question 6

Fill in the blanks:

(i) 58 and 310\dfrac{5}{8} \text{ and } \dfrac{3}{10} are on the .......... side of zero.

(ii) −58 and 310-\dfrac{5}{8} \text{ and } \dfrac{3}{10} are on the .......... sides of zero.

(iii) −58 and −310-\dfrac{5}{8} \text{ and } -\dfrac{3}{10} are on the .......... side of zero.

(iv) 58 and −310\dfrac{5}{8} \text{ and } -\dfrac{3}{10} are on the .......... sides of zero.

Answer

(i) 58 and 310\dfrac{5}{8} \text{ and } \dfrac{3}{10} are both positive, so they are on the same side of zero.

(ii) −58-\dfrac{5}{8} is negative and 310\dfrac{3}{10} is positive, so they are on the opposite sides of zero.

(iii) −58 and −310-\dfrac{5}{8} \text{ and } -\dfrac{3}{10} are both negative, so they are on the same side of zero.

(iv) 58\dfrac{5}{8} is positive and −310-\dfrac{3}{10} is negative, so they are on the opposite sides of zero.

Question 7

Insert three rational numbers between:

(i) −23 and 34-\dfrac{2}{3} \text{ and } \dfrac{3}{4}

(ii) 57 and 79\dfrac{5}{7} \text{ and } \dfrac{7}{9}

(iii) −58 and −16-\dfrac{5}{8} \text{ and } -\dfrac{1}{6}

Answer

(i) By Division Method,

23,423,233,11,1\begin{array}{l|r} 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 3 and 4 = 2 × 2 × 3 = 12

−23=−2×43×4=−812 and 34=3×34×3=912-\dfrac{2}{3} = \dfrac{-2 \times 4}{3 \times 4} = \dfrac{-8}{12} \text{ and } \dfrac{3}{4} = \dfrac{3 \times 3}{4 \times 3} = \dfrac{9}{12}

Now, −812<−712<−612<−512<912\dfrac{-8}{12} \lt \dfrac{-7}{12} \lt \dfrac{-6}{12} \lt \dfrac{-5}{12} \lt \dfrac{9}{12}

Hence, −712,−612 and −512\dfrac{-7}{12}, \dfrac{-6}{12} \text{ and } \dfrac{-5}{12} are three rational numbers between −23 and 34-\dfrac{2}{3} \text{ and } \dfrac{3}{4}.

(ii) By Division Method,

37,937,377,11,1\begin{array}{l|r} 3 & 7, 9 \\ \hline 3 & 7, 3 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}

LCM of 7 and 9 = 3 × 3 × 7 = 63

57=5×97×9=456379=7×79×7=4963\dfrac{5}{7} = \dfrac{5 \times 9}{7 \times 9} = \dfrac{45}{63} \\[1em] \dfrac{7}{9} = \dfrac{7 \times 7}{9 \times 7} = \dfrac{49}{63}

Now, 4563<4663<4763<4863<4963\dfrac{45}{63} \lt \dfrac{46}{63} \lt \dfrac{47}{63} \lt \dfrac{48}{63} \lt \dfrac{49}{63}

Hence, 4663,4763 and 4863\dfrac{46}{63}, \dfrac{47}{63} \text{ and } \dfrac{48}{63} are three rational numbers between 57 and 79\dfrac{5}{7} \text{ and } \dfrac{7}{9}.

(iii) By Division Method,

28,624,322,331,31,1\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

−58=−5×38×3=−1524−16=−1×46×4=−424-\dfrac{5}{8} = \dfrac{-5 \times 3}{8 \times 3} = \dfrac{-15}{24} \\[1em] -\dfrac{1}{6} = \dfrac{-1 \times 4}{6 \times 4} = \dfrac{-4}{24}

Now, −1524<−1424<−1324<−1224<−424\dfrac{-15}{24} \lt \dfrac{-14}{24} \lt \dfrac{-13}{24} \lt \dfrac{-12}{24} \lt \dfrac{-4}{24}

Hence, −1424,−1324 and −1224\dfrac{-14}{24}, \dfrac{-13}{24} \text{ and } \dfrac{-12}{24} are three rational numbers between −58 and −16-\dfrac{5}{8} \text{ and } -\dfrac{1}{6}.

Question 8

Insert four rational numbers between:

(i) −38 and −56-\dfrac{3}{8} \text{ and } -\dfrac{5}{6}

(ii) −45 and 23-\dfrac{4}{5} \text{ and } \dfrac{2}{3}

(iii) -3 and 6

(iv) 0 and 6

Answer

(i) 5−6=−56\dfrac{5}{-6} = -\dfrac{5}{6}

By Division Method,

28,624,322,331,31,1\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 8 and 6 = 2 × 2 × 2 × 3 = 24

−56=−5×46×4=−2024−38=−3×38×3=−924-\dfrac{5}{6} = \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24} \\[1em] -\dfrac{3}{8} = \dfrac{-3 \times 3}{8 \times 3} = \dfrac{-9}{24}

Now, −2024<−1924<−1824<−1724<−1624<−924\dfrac{-20}{24} \lt \dfrac{-19}{24} \lt \dfrac{-18}{24} \lt \dfrac{-17}{24} \lt \dfrac{-16}{24} \lt \dfrac{-9}{24}

Hence, −1924,−1824,−1724 and −1624\dfrac{-19}{24}, \dfrac{-18}{24}, \dfrac{-17}{24} \text{ and } \dfrac{-16}{24} are four rational numbers between −38 and −56-\dfrac{3}{8} \text{ and } -\dfrac{5}{6}.

(ii) By Division Method,

35,355,11,1\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}

LCM of 5 and 3 = 3 × 5 = 15

−45=−4×35×3=−121523=2×53×5=1015-\dfrac{4}{5} = \dfrac{-4 \times 3}{5 \times 3} = \dfrac{-12}{15} \\[1em] \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}

Now, −1215<−1115<−1015<−915<−815<1015\dfrac{-12}{15} \lt \dfrac{-11}{15} \lt \dfrac{-10}{15} \lt \dfrac{-9}{15} \lt \dfrac{-8}{15} \lt \dfrac{10}{15}

Hence, −1115,−1015,−915 and −815\dfrac{-11}{15}, \dfrac{-10}{15}, \dfrac{-9}{15} \text{ and } \dfrac{-8}{15} are four rational numbers between −45 and 23-\dfrac{4}{5} \text{ and } \dfrac{2}{3}.

(iii) The integers between -3 and 6 are -2, -1, 0, 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, -2, -1, 0 and 1 are four rational numbers between -3 and 6.

(iv) The integers between 0 and 6 are 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, 1, 2, 3 and 4 are four rational numbers between 0 and 6.

Exercise 2(C)

Question 1(i)

Add:

75 and 25\dfrac{7}{5} \text{ and } \dfrac{2}{5}

Answer

Solving,

⇒75+25=7+25=95=145\Rightarrow \dfrac{7}{5} + \dfrac{2}{5} \\[1em] = \dfrac{7 + 2}{5} \\[1em] = \dfrac{9}{5} \\[1em] = 1\dfrac{4}{5}

Hence, 75+25=145\dfrac{7}{5} + \dfrac{2}{5} = 1\dfrac{4}{5}

Question 1(ii)

Add:

−49 and 29\dfrac{-4}{9} \text{ and } \dfrac{2}{9}

Answer

Solving,

⇒−49+29=−4+29=−29\Rightarrow \dfrac{-4}{9} + \dfrac{2}{9} \\[1em] = \dfrac{-4 + 2}{9} \\[1em] = \dfrac{-2}{9}

Hence, −49+29=−29\dfrac{-4}{9} + \dfrac{2}{9} = \dfrac{-2}{9}

Question 1(iii)

Add:

5−12 and 112\dfrac{5}{-12} \text{ and } \dfrac{1}{12}

Answer

5−12=−512\dfrac{5}{-12} = \dfrac{-5}{12}

Solving,

⇒−512+112=−5+112=−412=−13\Rightarrow \dfrac{-5}{12} + \dfrac{1}{12} \\[1em] = \dfrac{-5 + 1}{12} \\[1em] = \dfrac{-4}{12} \\[1em] = \dfrac{-1}{3}

Hence, 5−12+112=−13\dfrac{5}{-12} + \dfrac{1}{12} = \dfrac{-1}{3}

Question 1(iv)

Add:

4−15 and −7−15\dfrac{4}{-15} \text{ and } \dfrac{-7}{-15}

Answer

4−15=−415 and −7−15=715\dfrac{4}{-15} = \dfrac{-4}{15} \text{ and } \dfrac{-7}{-15} = \dfrac{7}{15}

Solving,

⇒−415+715=−4+715=315=15\Rightarrow \dfrac{-4}{15} + \dfrac{7}{15} \\[1em] = \dfrac{-4 + 7}{15} \\[1em] = \dfrac{3}{15} \\[1em] = \dfrac{1}{5}

Hence, 4−15+−7−15=15\dfrac{4}{-15} + \dfrac{-7}{-15} = \dfrac{1}{5}

Question 1(v)

Add:

−725 and 9−25\dfrac{-7}{25} \text{ and } \dfrac{9}{-25}

Answer

9−25=−925\dfrac{9}{-25} = \dfrac{-9}{25}

Solving,

⇒−725+−925=−7+(−9)25=−1625\Rightarrow \dfrac{-7}{25} + \dfrac{-9}{25} \\[1em] = \dfrac{-7 + (-9)}{25} \\[1em] = \dfrac{-16}{25}

Hence, −725+9−25=−1625\dfrac{-7}{25} + \dfrac{9}{-25} = \dfrac{-16}{25}

Question 1(vi)

Add:

−726 and 7−26\dfrac{-7}{26} \text{ and } \dfrac{7}{-26}

Answer

7−26=−726\dfrac{7}{-26} = \dfrac{-7}{26}

Solving,

⇒−726+−726=−7+(−7)26=−1426=−713\Rightarrow \dfrac{-7}{26} + \dfrac{-7}{26} \\[1em] = \dfrac{-7 + (-7)}{26} \\[1em] = \dfrac{-14}{26} \\[1em] = \dfrac{-7}{13}

Hence, −726+7−26=−713\dfrac{-7}{26} + \dfrac{7}{-26} = \dfrac{-7}{13}

Question 2(i)

Add:

−25 and 37\dfrac{-2}{5} \text{ and } \dfrac{3}{7}

Answer

By Division Method,

55,771,71,1\begin{array}{l|r} 5 & 5, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}

LCM of 5 and 7 = 5 × 7 = 35

⇒−2×75×7+3×57×5=−1435+1535=−14+1535=135\Rightarrow \dfrac{-2 \times 7}{5 \times 7} + \dfrac{3 \times 5}{7 \times 5} \\[1em] = \dfrac{-14}{35} + \dfrac{15}{35} \\[1em] = \dfrac{-14 + 15}{35} \\[1em] = \dfrac{1}{35}

Hence, −25+37=135\dfrac{-2}{5} + \dfrac{3}{7} = \dfrac{1}{35}

Question 2(ii)

Add:

−56 and 49\dfrac{-5}{6} \text{ and } \dfrac{4}{9}

Answer

By Division Method,

26,933,931,31,1\begin{array}{l|r} 2 & 6, 9 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 6 and 9 = 2 × 3 × 3 = 18

⇒−5×36×3+4×29×2=−1518+818=−15+818=−718\Rightarrow \dfrac{-5 \times 3}{6 \times 3} + \dfrac{4 \times 2}{9 \times 2} \\[1em] = \dfrac{-15}{18} + \dfrac{8}{18} \\[1em] = \dfrac{-15 + 8}{18} \\[1em] = \dfrac{-7}{18}

Hence, −56+49=−718\dfrac{-5}{6} + \dfrac{4}{9} = \dfrac{-7}{18}

Question 2(iii)

Add:

-3 and 23\dfrac{2}{3}

Answer

Solving,

−31+23\dfrac{-3}{1} + \dfrac{2}{3}

LCM of 1 and 3 = 3

⇒−3×31×3+2×13×1=−93+23=−9+23=−73=−213\Rightarrow \dfrac{-3 \times 3}{1 \times 3} + \dfrac{2 \times 1}{3 \times 1} \\[1em] = \dfrac{-9}{3} + \dfrac{2}{3} \\[1em] = \dfrac{-9 + 2}{3} \\[1em] = \dfrac{-7}{3} \\[1em] = -2\dfrac{1}{3}

Hence, −3+23=−213-3 + \dfrac{2}{3} = -2\dfrac{1}{3}

Question 2(iv)

Add:

−59 and 718\dfrac{-5}{9} \text{ and } \dfrac{7}{18}

Answer

By Division Method,

29,1839,933,31,1\begin{array}{l|r} 2 & 9, 18 \\ \hline 3 & 9, 9 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 9 and 18 = 2 × 3 × 3 = 18

⇒−5×29×2+7×118×1=−1018+718=−10+718=−318=−16\Rightarrow \dfrac{-5 \times 2}{9 \times 2} + \dfrac{7 \times 1}{18 \times 1} \\[1em] = \dfrac{-10}{18} + \dfrac{7}{18} \\[1em] = \dfrac{-10 + 7}{18} \\[1em] = \dfrac{-3}{18} \\[1em] = \dfrac{-1}{6}

Hence, −59+718=−16\dfrac{-5}{9} + \dfrac{7}{18} = \dfrac{-1}{6}

Question 2(v)

Add:

−724 and −548\dfrac{-7}{24} \text{ and } \dfrac{-5}{48}

Answer

By Division Method,

224,48212,2426,1223,633,31,1\begin{array}{l|r} 2 & 24, 48 \\ \hline 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 24 and 48 = 2 × 2 × 2 × 2 × 3 = 48

⇒−7×224×2+−5×148×1=−1448+−548=−14+(−5)48=−1948\Rightarrow \dfrac{-7 \times 2}{24 \times 2} + \dfrac{-5 \times 1}{48 \times 1} \\[1em] = \dfrac{-14}{48} + \dfrac{-5}{48} \\[1em] = \dfrac{-14 + (-5)}{48} \\[1em] = \dfrac{-19}{48}

Hence, −724+−548=−1948\dfrac{-7}{24} + \dfrac{-5}{48} = \dfrac{-19}{48}

Question 2(vi)

Add:

1−18 and 5−27\dfrac{1}{-18} \text{ and } \dfrac{5}{-27}

Answer

1−18=−118 and 5−27=−527\dfrac{1}{-18} = \dfrac{-1}{18} \text{ and } \dfrac{5}{-27} = \dfrac{-5}{27}

By Division Method,

218,2739,2733,931,31,1\begin{array}{l|r} 2 & 18, 27 \\ \hline 3 & 9, 27 \\ \hline 3 & 3, 9 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 18 and 27 = 2 × 3 × 3 × 3 = 54

⇒−1×318×3+−5×227×2=−354+−1054=−3+(−10)54=−1354\Rightarrow \dfrac{-1 \times 3}{18 \times 3} + \dfrac{-5 \times 2}{27 \times 2} \\[1em] = \dfrac{-3}{54} + \dfrac{-10}{54} \\[1em] = \dfrac{-3 + (-10)}{54} \\[1em] = \dfrac{-13}{54}

Hence, 1−18+5−27=−1354\dfrac{1}{-18} + \dfrac{5}{-27} = \dfrac{-13}{54}

Question 2(vii)

Add:

−925 and 1−75\dfrac{-9}{25} \text{ and } \dfrac{1}{-75}

Answer

1−75=−175\dfrac{1}{-75} = \dfrac{-1}{75}

By Division Method,

325,75525,2555,51,1\begin{array}{l|r} 3 & 25, 75 \\ \hline 5 & 25, 25 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}

LCM of 25 and 75 = 3 × 5 × 5 = 75

⇒−9×325×3+−1×175×1=−2775+−175=−27+(−1)75=−2875\Rightarrow \dfrac{-9 \times 3}{25 \times 3} + \dfrac{-1 \times 1}{75 \times 1} \\[1em] = \dfrac{-27}{75} + \dfrac{-1}{75} \\[1em] = \dfrac{-27 + (-1)}{75} \\[1em] = \dfrac{-28}{75}

Hence, −925+1−75=−2875\dfrac{-9}{25} + \dfrac{1}{-75} = \dfrac{-28}{75}

Question 2(viii)

Add:

13−16 and −1124\dfrac{13}{-16} \text{ and } \dfrac{-11}{24}

Answer

13−16=−1316\dfrac{13}{-16} = \dfrac{-13}{16}

By Division Method,

216,2428,1224,622,331,31,1\begin{array}{l|r} 2 & 16, 24 \\ \hline 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 16 and 24 = 2 × 2 × 2 × 2 × 3 = 48

⇒−13×316×3+−11×224×2=−3948+−2248=−39+(−22)48=−6148=−11348\Rightarrow \dfrac{-13 \times 3}{16 \times 3} + \dfrac{-11 \times 2}{24 \times 2} \\[1em] = \dfrac{-39}{48} + \dfrac{-22}{48} \\[1em] = \dfrac{-39 + (-22)}{48} \\[1em] = \dfrac{-61}{48} \\[1em] = -1\dfrac{13}{48}

Hence, 13−16+−1124=−11348\dfrac{13}{-16} + \dfrac{-11}{24} = -1\dfrac{13}{48}

Question 2(ix)

Add:

−9−16 and −118\dfrac{-9}{-16} \text{ and } \dfrac{-11}{8}

Answer

−9−16=916\dfrac{-9}{-16} = \dfrac{9}{16}

By Division Method,

216,828,424,222,11,1\begin{array}{l|r} 2 & 16, 8 \\ \hline 2 & 8, 4 \\ \hline 2 & 4, 2 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}

LCM of 16 and 8 = 2 × 2 × 2 × 2 = 16

⇒9×116×1+−11×28×2=916+−2216=9+(−22)16=−1316\Rightarrow \dfrac{9 \times 1}{16 \times 1} + \dfrac{-11 \times 2}{8 \times 2} \\[1em] = \dfrac{9}{16} + \dfrac{-22}{16} \\[1em] = \dfrac{9 + (-22)}{16} \\[1em] = \dfrac{-13}{16}

Hence, −9−16+−118=−1316\dfrac{-9}{-16} + \dfrac{-11}{8} = \dfrac{-13}{16}

Question 3(i)

Evaluate:

−25+35+−15\dfrac{-2}{5} + \dfrac{3}{5} + \dfrac{-1}{5}

Answer

Solving,

⇒−25+35+−15=−2+3+(−1)5=05=0\Rightarrow \dfrac{-2}{5} + \dfrac{3}{5} + \dfrac{-1}{5} \\[1em] = \dfrac{-2 + 3 + (-1)}{5} \\[1em] = \dfrac{0}{5} \\[1em] = 0

Hence, −25+35+−15=0\dfrac{-2}{5} + \dfrac{3}{5} + \dfrac{-1}{5} = 0

Question 3(ii)

Evaluate:

−89+49+−29\dfrac{-8}{9} + \dfrac{4}{9} + \dfrac{-2}{9}

Answer

Solving,

⇒−89+49+−29=−8+4+(−2)9=−69=−23\Rightarrow \dfrac{-8}{9} + \dfrac{4}{9} + \dfrac{-2}{9} \\[1em] = \dfrac{-8 + 4 + (-2)}{9} \\[1em] = \dfrac{-6}{9} \\[1em] = \dfrac{-2}{3}

Hence, −89+49+−29=−23\dfrac{-8}{9} + \dfrac{4}{9} + \dfrac{-2}{9} = \dfrac{-2}{3}

Question 3(iii)

Evaluate:

5−24+−18+316\dfrac{5}{-24} + \dfrac{-1}{8} + \dfrac{3}{16}

Answer

Solving,

5−24+−18+316\dfrac{5}{-24} + \dfrac{-1}{8} + \dfrac{3}{16}

By division method:

224,8,16212,4,826,2,423,1,233,1,11,1,1\begin{array}{l|r} 2 & 24, 8, 16 \\ \hline 2 & 12, 4, 8 \\ \hline 2 & 6, 2, 4 \\ \hline 2 & 3, 1, 2 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 24, 8 and 16 = 2 × 2 × 2 × 2 × 3 = 48

−5×224×2+−1×68×6+3×316×3=−1048+−648+948=−10+(−6)+948=−748\dfrac{-5 \times 2}{24 \times 2} + \dfrac{-1 \times 6}{8 \times 6} + \dfrac{3 \times 3}{16 \times 3} \\[1em] = \dfrac{-10}{48} + \dfrac{-6}{48} + \dfrac{9}{48} \\[1em] = \dfrac{-10 + (-6) + 9}{48} \\[1em] = \dfrac{-7}{48}

Hence, 5−24+−18+316=−748\dfrac{5}{-24} + \dfrac{-1}{8} + \dfrac{3}{16} = \dfrac{-7}{48}

Question 3(iv)

Evaluate:

−76+4−15+−4−30\dfrac{-7}{6} + \dfrac{4}{-15} + \dfrac{-4}{-30}

Answer

4−15=−415 and −4−30=430\dfrac{4}{-15} = \dfrac{-4}{15} \text{ and } \dfrac{-4}{-30} = \dfrac{4}{30}

LCM of 6, 15 and 30 = 30

⇒−7×56×5+−4×215×2+4×130×1=−3530+−830+430=−35+(−8)+430=−3930=−1310=−1310\Rightarrow \dfrac{-7 \times 5}{6 \times 5} + \dfrac{-4 \times 2}{15 \times 2} + \dfrac{4 \times 1}{30 \times 1} \\[1em] = \dfrac{-35}{30} + \dfrac{-8}{30} + \dfrac{4}{30} \\[1em] = \dfrac{-35 + (-8) + 4}{30} \\[1em] = \dfrac{-39}{30} \\[1em] = \dfrac{-13}{10} \\[1em] = -1\dfrac{3}{10}

Hence, −76+4−15+−4−30=−1310\dfrac{-7}{6} + \dfrac{4}{-15} + \dfrac{-4}{-30} = -1\dfrac{3}{10}

Question 3(v)

Evaluate:

−2+25+−215-2 + \dfrac{2}{5} + \dfrac{-2}{15}

Answer

−21+25+−215\dfrac{-2}{1} + \dfrac{2}{5} + \dfrac{-2}{15}

LCM of 1, 5 and 15 = 15

⇒−2×151×15+2×35×3+−2×115×1=−3015+615+−215=−30+6+(−2)15=−2615=−11115\Rightarrow \dfrac{-2 \times 15}{1 \times 15} + \dfrac{2 \times 3}{5 \times 3} + \dfrac{-2 \times 1}{15 \times 1} \\[1em] = \dfrac{-30}{15} + \dfrac{6}{15} + \dfrac{-2}{15} \\[1em] = \dfrac{-30 + 6 + (-2)}{15} \\[1em] = \dfrac{-26}{15} \\[1em] = -1\dfrac{11}{15}

Hence, −2+25+−215=−11115-2 + \dfrac{2}{5} + \dfrac{-2}{15} = -1\dfrac{11}{15}

Question 3(vi)

Evaluate:

−1112+516+−38\dfrac{-11}{12} + \dfrac{5}{16} + \dfrac{-3}{8}

Answer

Solving,

By division method,

212,16,826,8,423,4,223,2,133,1,11,1,1\begin{array}{l|r} 2 & 12, 16, 8 \\ \hline 2 & 6, 8, 4 \\ \hline 2 & 3, 4, 2 \\ \hline 2 & 3, 2, 1 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 12, 16 and 8 = 2 × 2 × 2 × 2 × 3 = 48

⇒−11×412×4+5×316×3+−3×68×6=−4448+1548+−1848=−44+15+(−18)48=−4748\Rightarrow \dfrac{-11 \times 4}{12 \times 4} + \dfrac{5 \times 3}{16 \times 3} + \dfrac{-3 \times 6}{8 \times 6} \\[1em] = \dfrac{-44}{48} + \dfrac{15}{48} + \dfrac{-18}{48} \\[1em] = \dfrac{-44 + 15 + (-18)}{48} \\[1em] = \dfrac{-47}{48}

Hence, −1112+516+−38=−4748\dfrac{-11}{12} + \dfrac{5}{16} + \dfrac{-3}{8} = \dfrac{-47}{48}

Question 4(i)

Evaluate:

−1118+−39+2−3-\dfrac{11}{18} + \dfrac{-3}{9} + \dfrac{2}{-3}

Answer

2−3=−23\dfrac{2}{-3} = \dfrac{-2}{3}

By Division Method,

218,9,339,9,333,3,11,1,1\begin{array}{l|r} 2 & 18, 9, 3 \\ \hline 3 & 9, 9, 3 \\ \hline 3 & 3, 3, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 18, 9 and 3 = 2 × 3 × 3 = 18

⇒−11×118×1+−3×29×2+−2×63×6=−1118+−618+−1218=−11+(−6)+(−12)18=−2918=−11118\Rightarrow \dfrac{-11 \times 1}{18 \times 1} + \dfrac{-3 \times 2}{9 \times 2} + \dfrac{-2 \times 6}{3 \times 6} \\[1em] = \dfrac{-11}{18} + \dfrac{-6}{18} + \dfrac{-12}{18} \\[1em] = \dfrac{-11 + (-6) + (-12)}{18} \\[1em] = \dfrac{-29}{18} \\[1em] = -1\dfrac{11}{18}

Hence, −1118+−39+2−3=−11118-\dfrac{11}{18} + \dfrac{-3}{9} + \dfrac{2}{-3} = -1\dfrac{11}{18}

Question 4(ii)

Evaluate:

−94+133+256\dfrac{-9}{4} + \dfrac{13}{3} + \dfrac{25}{6}

Answer

By Division Method,

24,3,622,3,331,3,31,1,1\begin{array}{l|r} 2 & 4, 3, 6 \\ \hline 2 & 2, 3, 3 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 4, 3 and 6 = 2 × 2 × 3 = 12

⇒−9×34×3+13×43×4+25×26×2=−2712+5212+5012=−27+52+5012=7512=254=614\Rightarrow \dfrac{-9 \times 3}{4 \times 3} + \dfrac{13 \times 4}{3 \times 4} + \dfrac{25 \times 2}{6 \times 2} \\[1em] = \dfrac{-27}{12} + \dfrac{52}{12} + \dfrac{50}{12} \\[1em] = \dfrac{-27 + 52 + 50}{12} \\[1em] = \dfrac{75}{12} \\[1em] = \dfrac{25}{4} \\[1em] = 6\dfrac{1}{4}

Hence, −94+133+256=614\dfrac{-9}{4} + \dfrac{13}{3} + \dfrac{25}{6} = 6\dfrac{1}{4}

Question 4(iii)

Evaluate:

−5+5−8+−5−12-5 + \dfrac{5}{-8} + \dfrac{-5}{-12}

Answer

5−8=−58 and −5−12=512\dfrac{5}{-8} = \dfrac{-5}{8}\text{ and }\dfrac{-5}{-12} = \dfrac{5}{12}

By Division Method,

21,8,1221,4,621,2,331,1,31,1,1\begin{array}{l|r} 2 & 1, 8, 12 \\ \hline 2 & 1, 4, 6 \\ \hline 2 & 1, 2, 3 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 1, 8 and 12 = 2 × 2 × 2 × 3 = 24

⇒−5×241×24+−5×38×3+5×212×2=−12024+−1524+1024=−120+(−15)+1024=−12524=−5524\Rightarrow \dfrac{-5 \times 24}{1 \times 24} + \dfrac{-5 \times 3}{8 \times 3} + \dfrac{5 \times 2}{12 \times 2} \\[1em] = \dfrac{-120}{24} + \dfrac{-15}{24} + \dfrac{10}{24} \\[1em] = \dfrac{-120 + (-15) + 10}{24} \\[1em] = \dfrac{-125}{24} \\[1em] = -5\dfrac{5}{24}

Hence, −5+5−8+−5−12=−5524-5 + \dfrac{5}{-8} + \dfrac{-5}{-12} = -5\dfrac{5}{24}

Question 4(iv)

Evaluate:

−23+52+2-\dfrac{2}{3} + \dfrac{5}{2} + 2

Answer

By Division Method,

23,2,133,1,11,1,1\begin{array}{l|r} 2 & 3, 2, 1 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 3, 2 and 1 = 2 × 3 = 6

⇒−2×23×2+5×32×3+2×61×6=−46+156+126=−4+15+126=236=356\Rightarrow \dfrac{-2 \times 2}{3 \times 2} + \dfrac{5 \times 3}{2 \times 3} + \dfrac{2 \times 6}{1 \times 6} \\[1em] = \dfrac{-4}{6} + \dfrac{15}{6} + \dfrac{12}{6} \\[1em] = \dfrac{-4 + 15 + 12}{6} \\[1em] = \dfrac{23}{6} \\[1em] = 3\dfrac{5}{6}

Hence, −23+52+2=356-\dfrac{2}{3} + \dfrac{5}{2} + 2 = 3\dfrac{5}{6}

Question 4(v)

Evaluate:

5+−34+−585 + \dfrac{-3}{4} + \dfrac{-5}{8}

Answer

By Division Method,

21,4,821,2,421,1,21,1,1\begin{array}{l|r} 2 & 1, 4, 8 \\ \hline 2 & 1, 2, 4 \\ \hline 2 & 1, 1, 2 \\ \hline & 1, 1, 1 \end{array}

LCM of 1, 4 and 8 = 2 × 2 × 2 = 8

⇒5×81×8+−3×24×2+−5×18×1=408+−68+−58=40+(−6)+(−5)8=298=358\Rightarrow \dfrac{5 \times 8}{1 \times 8} + \dfrac{-3 \times 2}{4 \times 2} + \dfrac{-5 \times 1}{8 \times 1} \\[1em] = \dfrac{40}{8} + \dfrac{-6}{8} + \dfrac{-5}{8} \\[1em] = \dfrac{40 + (-6) + (-5)}{8} \\[1em] = \dfrac{29}{8} \\[1em] = 3\dfrac{5}{8}

Hence, 5+−34+−58=3585 + \dfrac{-3}{4} + \dfrac{-5}{8} = 3\dfrac{5}{8}

Question 5(i)

Subtract:

29 from 59\dfrac{2}{9} \text{ from } \dfrac{5}{9}

Answer

Solving,

⇒59−29=5−29=39=13\Rightarrow \dfrac{5}{9} - \dfrac{2}{9} \\[1em] = \dfrac{5 - 2}{9} \\[1em] = \dfrac{3}{9} \\[1em] = \dfrac{1}{3}

Hence, subtracting 29\dfrac{2}{9} from 59\dfrac{5}{9} gives 13\dfrac{1}{3}.

Question 5(ii)

Subtract:

−611 from −3−11\dfrac{-6}{11} \text{ from } \dfrac{-3}{-11}

Answer

−3−11=311\dfrac{-3}{-11} = \dfrac{3}{11}

Solving,

⇒311−−611=3−(−6)11=3+611=911\Rightarrow \dfrac{3}{11} - \dfrac{-6}{11} \\[1em] = \dfrac{3 - (-6)}{11} \\[1em] = \dfrac{3 + 6}{11} \\[1em] = \dfrac{9}{11}

Hence, subtracting −611\dfrac{-6}{11} from −3−11\dfrac{-3}{-11} gives 911\dfrac{9}{11}.

Question 5(iii)

Subtract:

−215 from −815\dfrac{-2}{15} \text{ from } \dfrac{-8}{15}

Answer

Solving,

⇒−815−−215=−8−(−2)15=−8+215=−615=−25\Rightarrow \dfrac{-8}{15} - \dfrac{-2}{15} \\[1em] = \dfrac{-8 - (-2)}{15} \\[1em] = \dfrac{-8 + 2}{15} \\[1em] = \dfrac{-6}{15} \\[1em] = \dfrac{-2}{5}

Hence, subtracting −215\dfrac{-2}{15} from −815\dfrac{-8}{15} gives −25\dfrac{-2}{5}.

Question 5(iv)

Subtract:

1118 from −518\dfrac{11}{18} \text{ from } \dfrac{-5}{18}

Answer

Solving,

⇒−518−1118=−5−1118=−1618=−89\Rightarrow \dfrac{-5}{18} - \dfrac{11}{18} \\[1em] = \dfrac{-5 - 11}{18} \\[1em] = \dfrac{-16}{18} \\[1em] = \dfrac{-8}{9}

Hence, subtracting 1118\dfrac{11}{18} from −518\dfrac{-5}{18} gives −89\dfrac{-8}{9}.

Question 5(v)

Subtract:

−411\dfrac{-4}{11} from -2.

Answer

Solving,

⇒−2−−411=−21+411\Rightarrow -2 - \dfrac{-4}{11} \\[1em] = \dfrac{-2}{1} + \dfrac{4}{11}

LCM of 1 and 11 = 11

=−2×111×11+4×111×1=−2211+411=−22+411=−1811=−1711= \dfrac{-2 \times 11}{1 \times 11} + \dfrac{4 \times 1}{11 \times 1} \\[1em] = \dfrac{-22}{11} + \dfrac{4}{11} \\[1em] = \dfrac{-22 + 4}{11} \\[1em] = \dfrac{-18}{11} \\[1em] = -1\dfrac{7}{11}

Hence, subtracting −411\dfrac{-4}{11} from −2-2 gives −1711-1\dfrac{7}{11}.

Question 6(i)

Subtract:

−310 from 15-\dfrac{3}{10} \text{ from } \dfrac{1}{5}

Answer

Solving,

⇒15−(−310)=15+310\Rightarrow \dfrac{1}{5} - \left(-\dfrac{3}{10}\right) \\[1em] = \dfrac{1}{5} + \dfrac{3}{10}

By Division Method,

25,1055,51,1\begin{array}{l|r} 2 & 5, 10 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}

LCM of 5 and 10 = 2 × 5 = 10

=1×25×2+3×110×1=210+310=2+310=510=12= \dfrac{1 \times 2}{5 \times 2} + \dfrac{3 \times 1}{10 \times 1} \\[1em] = \dfrac{2}{10} + \dfrac{3}{10} \\[1em] = \dfrac{2 + 3}{10} \\[1em] = \dfrac{5}{10} \\[1em] = \dfrac{1}{2}

Hence, the difference = 12\dfrac{1}{2}.

Question 6(ii)

Subtract:

−625 from −85\dfrac{-6}{25} \text{ from } \dfrac{-8}{5}

Answer

Solving,

⇒−85−−625=−85+625\Rightarrow \dfrac{-8}{5} - \dfrac{-6}{25} \\[1em] = \dfrac{-8}{5} + \dfrac{6}{25}

By Division Method,

55,2551,51,1\begin{array}{l|r} 5 & 5, 25 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 5 and 25 = 5 × 5 = 25

=−8×55×5+6×125×1=−4025+625=−40+625=−3425=−1925= \dfrac{-8 \times 5}{5 \times 5} + \dfrac{6 \times 1}{25 \times 1} \\[1em] = \dfrac{-40}{25} + \dfrac{6}{25} \\[1em] = \dfrac{-40 + 6}{25} \\[1em] = \dfrac{-34}{25} \\[1em] = -1\dfrac{9}{25}

Hence, the difference = −1925-1\dfrac{9}{25}.

Question 6(iii)

Subtract:

−74\dfrac{-7}{4} from -2

Answer

Solving,

⇒−2−−74=−21+74\Rightarrow -2 - \dfrac{-7}{4} \\[1em] = \dfrac{-2}{1} + \dfrac{7}{4}

By Division Method,

21,421,21,1\begin{array}{l|r} 2 & 1, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}

LCM of 1 and 4 = 2 × 2 = 4

=−2×41×4+7×14×1=−84+74=−8+74=−14= \dfrac{-2 \times 4}{1 \times 4} + \dfrac{7 \times 1}{4 \times 1} \\[1em] = \dfrac{-8}{4} + \dfrac{7}{4} \\[1em] = \dfrac{-8 + 7}{4} \\[1em] = \dfrac{-1}{4}

Hence, the difference = −14\dfrac{-1}{4}.

Question 6(iv)

Subtract:

−1621\dfrac{-16}{21} from 1

Answer

Solving,

⇒1−−1621=11+1621\Rightarrow 1 - \dfrac{-16}{21} \\[1em] = \dfrac{1}{1} + \dfrac{16}{21}

By Division Method,

31,2171,71,1\begin{array}{l|r} 3 & 1, 21 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}

LCM of 1 and 21 = 3 × 7 = 21

=1×211×21+16×121×1=2121+1621=21+1621=3721=11621= \dfrac{1 \times 21}{1 \times 21} + \dfrac{16 \times 1}{21 \times 1} \\[1em] = \dfrac{21}{21} + \dfrac{16}{21} \\[1em] = \dfrac{21 + 16}{21} \\[1em] = \dfrac{37}{21} \\[1em] = 1\dfrac{16}{21}

Hence, the difference = 116211\dfrac{16}{21}.

Question 6(v)

Subtract:

−815\dfrac{-8}{15} from 0

Answer

Solving,

⇒0−−815=0+815=815\Rightarrow 0 - \dfrac{-8}{15} \\[1em] = 0 + \dfrac{8}{15} \\[1em] = \dfrac{8}{15}

Hence, the difference = 815\dfrac{8}{15}.

Question 6(vi)

Subtract:

0 from −38\dfrac{-3}{8}

Answer

Solving,

⇒−38−0=−38\Rightarrow \dfrac{-3}{8} - 0 \\[1em] = \dfrac{-3}{8}

Hence, the difference = −38\dfrac{-3}{8}.

Question 6(vii)

Subtract:

-2 from −310\dfrac{-3}{10}

Answer

Solving,

⇒−310−(−2)=−310+21\Rightarrow \dfrac{-3}{10} - (-2) \\[1em] = \dfrac{-3}{10} + \dfrac{2}{1}

By Division Method,

210,155,11,1\begin{array}{l|r} 2 & 10, 1 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}

LCM of 10 and 1 = 2 × 5 = 10

=−3×110×1+2×101×10=−310+2010=−3+2010=1710=1710= \dfrac{-3 \times 1}{10 \times 1} + \dfrac{2 \times 10}{1 \times 10} \\[1em] = \dfrac{-3}{10} + \dfrac{20}{10} \\[1em] = \dfrac{-3 + 20}{10} \\[1em] = \dfrac{17}{10} \\[1em] = 1\dfrac{7}{10}

Hence, the difference = 17101\dfrac{7}{10}.

Question 6(viii)

Subtract:

58 from −516\dfrac{5}{8} \text{ from } \dfrac{-5}{16}

Answer

By Division Method,

216,828,424,222,11,1\begin{array}{l|r} 2 & 16, 8 \\ \hline 2 & 8, 4 \\ \hline 2 & 4, 2 \\ \hline 2 & 2, 1 \\ \hline & 1, 1 \end{array}

LCM of 16 and 8 = 2 × 2 × 2 × 2 = 16

Solving,

⇒−516−58=−5×116×1−5×28×2=−516−1016=−5−1016=−1516\Rightarrow \dfrac{-5}{16} - \dfrac{5}{8} \\[1em] = \dfrac{-5 \times 1}{16 \times 1} - \dfrac{5 \times 2}{8 \times 2} \\[1em] = \dfrac{-5}{16} - \dfrac{10}{16} \\[1em] = \dfrac{-5 - 10}{16} \\[1em] = \dfrac{-15}{16}

Hence, the difference = −1516\dfrac{-15}{16}.

Question 6(ix)

Subtract:

4 from −313-\dfrac{3}{13}

Answer

Solving,

⇒−313−4=−313−41\Rightarrow -\dfrac{3}{13} - 4 \\[1em] = \dfrac{-3}{13} - \dfrac{4}{1}

LCM of 13 and 1 = 13

=−3×113×1−4×131×13=−313−5213=−3−5213=−5513=−4313= \dfrac{-3 \times 1}{13 \times 1} - \dfrac{4 \times 13}{1 \times 13} \\[1em] = \dfrac{-3}{13} - \dfrac{52}{13} \\[1em] = \dfrac{-3 - 52}{13} \\[1em] = \dfrac{-55}{13} \\[1em] = -4\dfrac{3}{13}

Hence, the difference = −4313-4\dfrac{3}{13}.

Question 7

The sum of two rational numbers is 1124\dfrac{11}{24}. If one of them is 38\dfrac{3}{8}, find the other.

Answer

Let the other number be x.

According to the question,

x+38=1124\text{x} + \dfrac{3}{8} = \dfrac{11}{24}

x=1124−38\text{x} = \dfrac{11}{24} - \dfrac{3}{8}

By Division Method,

224,8212,426,233,11,1\begin{array}{l|r} 2 & 24, 8 \\ \hline 2 & 12, 4 \\ \hline 2 & 6, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 24 and 8 = 2 × 2 × 2 × 3 = 24

x=11×124×1−3×38×3=1124−924=11−924=224=112\text{x} = \dfrac{11 \times 1}{24 \times 1} - \dfrac{3 \times 3}{8 \times 3} \\[1em] = \dfrac{11}{24} - \dfrac{9}{24} \\[1em] = \dfrac{11 - 9}{24} \\[1em] = \dfrac{2}{24} \\[1em] = \dfrac{1}{12}

Hence, the other rational number is 112\dfrac{1}{12}.

Question 8

The sum of two rational numbers is −712\dfrac{-7}{12}. If one of them is 1324\dfrac{13}{24}, find the other.

Answer

Let the other number be x.

According to the question,

x+1324=−712\text{x} + \dfrac{13}{24} = \dfrac{-7}{12}

x=−712−1324\text{x} = \dfrac{-7}{12} - \dfrac{13}{24}

By Division Method,

212,2426,1223,633,31,1\begin{array}{l|r} 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 12 and 24 = 2 × 2 × 2 × 3 = 24

x=−7×212×2−13×124×1=−1424−1324=−14−1324=−2724=−98=−118\text{x} = \dfrac{-7 \times 2}{12 \times 2} - \dfrac{13 \times 1}{24 \times 1} \\[1em] = \dfrac{-14}{24} - \dfrac{13}{24} \\[1em] = \dfrac{-14 - 13}{24} \\[1em] = \dfrac{-27}{24} \\[1em] = \dfrac{-9}{8} \\[1em] = -1\dfrac{1}{8}

Hence, the other rational number is −118-1\dfrac{1}{8}.

Question 9

The sum of two rational numbers is -4. If one of them is −1312-\dfrac{13}{12}, find the other.

Answer

Let the other number be x.

According to the question,

x+−1312=−4x=−4−(−1312)=−41+1312\text{x} + \dfrac{-13}{12} = -4 \\[1em] \text{x} = -4 - \left(-\dfrac{13}{12}\right) \\[1em] = \dfrac{-4}{1} + \dfrac{13}{12}

By Division Method,

21,1221,631,31,1\begin{array}{l|r} 2 & 1, 12 \\ \hline 2 & 1, 6 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 1 and 12 = 2 × 2 × 3 = 12

x=−4×121×12+13×112×1=−4812+1312=−48+1312=−3512=−21112\text{x} = \dfrac{-4 \times 12}{1 \times 12} + \dfrac{13 \times 1}{12 \times 1} \\[1em] = \dfrac{-48}{12} + \dfrac{13}{12} \\[1em] = \dfrac{-48 + 13}{12} \\[1em] = \dfrac{-35}{12} \\[1em] = -2\dfrac{11}{12}

Hence, the other rational number is −21112-2\dfrac{11}{12}.

Question 10

What should be added to −332-\dfrac{3}{32} to get 5396\dfrac{53}{96}?

Answer

Let x be added to −332-\dfrac{3}{32}.

⇒−332+x=5396⇒x=5396−(−332)⇒x=5396+332\Rightarrow -\dfrac{3}{32} + x = \dfrac{53}{96} \\[1em] \Rightarrow x = \dfrac{53}{96} - \left(-\dfrac{3}{32}\right) \\[1em] \Rightarrow x = \dfrac{53}{96} + \dfrac{3}{32}

By Division Method,

296,32248,16224,8212,426,233,11,1\begin{array}{l|r} 2 & 96, 32 \\ \hline 2 & 48, 16 \\ \hline 2 & 24, 8 \\ \hline 2 & 12, 4 \\ \hline 2 & 6, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 96 and 32 = 2 × 2 × 2 × 2 × 2 × 3 = 96

⇒x=53×196×1+3×332×3⇒x=5396+996⇒x=53+996⇒x=6296⇒x=3148\Rightarrow x = \dfrac{53 \times 1}{96 \times 1} + \dfrac{3 \times 3}{32 \times 3} \\[1em] \Rightarrow x = \dfrac{53}{96} + \dfrac{9}{96} \\[1em] \Rightarrow x = \dfrac{53 + 9}{96} \\[1em] \Rightarrow x = \dfrac{62}{96} \\[1em] \Rightarrow x = \dfrac{31}{48}

Hence, 3148\dfrac{31}{48} should be added to −332-\dfrac{3}{32} to get 5396\dfrac{53}{96}.

Question 11

What should be added to −320\dfrac{-3}{20} to get 29202\dfrac{9}{20}?

Answer

2920=49202\dfrac{9}{20} = \dfrac{49}{20}

Let x be added to −320\dfrac{-3}{20} to get 29202\dfrac{9}{20}.

⇒−320+x=4920⇒x=4920−−320⇒x=49−(−3)20⇒x=49+320⇒x=5220⇒x=135⇒x=235\Rightarrow \dfrac{-3}{20} + x = \dfrac{49}{20} \\[1em] \Rightarrow x = \dfrac{49}{20} - \dfrac{-3}{20} \\[1em] \Rightarrow x = \dfrac{49 - (-3)}{20} \\[1em] \Rightarrow x = \dfrac{49 + 3}{20} \\[1em] \Rightarrow x = \dfrac{52}{20} \\[1em] \Rightarrow x = \dfrac{13}{5} \\[1em] \Rightarrow x = 2\dfrac{3}{5}

Hence, 2352\dfrac{3}{5} should be added to −320\dfrac{-3}{20} to get 29202\dfrac{9}{20}.

Question 12

What should be subtracted from −45\dfrac{-4}{5} to get 1?

Answer

Let x be subtracted from −45\dfrac{-4}{5}.

⇒−45−x=1⇒x=−45−1⇒x=−45−11\Rightarrow \dfrac{-4}{5} - x = 1 \\[1em] \Rightarrow x = \dfrac{-4}{5} - 1 \\[1em] \Rightarrow x = \dfrac{-4}{5} - \dfrac{1}{1}

LCM of 5 and 1 = 5

⇒x=−4×15×1−1×51×5⇒x=−45−55⇒x=−4−55⇒x=−95⇒x=−145\Rightarrow x = \dfrac{-4 \times 1}{5 \times 1} - \dfrac{1 \times 5}{1 \times 5} \\[1em] \Rightarrow x = \dfrac{-4}{5} - \dfrac{5}{5} \\[1em] \Rightarrow x = \dfrac{-4 - 5}{5} \\[1em] \Rightarrow x = \dfrac{-9}{5} \\[1em] \Rightarrow x = -1\dfrac{4}{5}

Hence, −145-1\dfrac{4}{5} should be subtracted from −45\dfrac{-4}{5} to get 1.

Question 13

The sum of two numbers is −65-\dfrac{6}{5}. If one of them is -2, find the other.

Answer

Let the other number be x.

According to the question,

⇒x+(−2)=−65⇒x=−65−(−2)=−65+21\Rightarrow \text{x} + (-2) = \dfrac{-6}{5} \\[1em] \Rightarrow \text{x} = -\dfrac{6}{5} - (-2) \\[1em] = \dfrac{-6}{5} + \dfrac{2}{1}

LCM of 5 and 1 = 5

⇒x=−6×15×1+2×51×5=−65+105=−6+105=45\Rightarrow \text{x} = \dfrac{-6 \times 1}{5 \times 1} + \dfrac{2 \times 5}{1 \times 5} \\[1em] = \dfrac{-6}{5} + \dfrac{10}{5} \\[1em] = \dfrac{-6 + 10}{5} \\[1em] = \dfrac{4}{5}

Hence, the other number is 45\dfrac{4}{5}.

Question 14

What should be added to −712\dfrac{-7}{12} to get 38\dfrac{3}{8}?

Answer

Let x be added to −712\dfrac{-7}{12} to get 38\dfrac{3}{8}.

⇒−712+x=38⇒x=38−−712⇒x=38+712\Rightarrow \dfrac{-7}{12} + x = \dfrac{3}{8} \\[1em] \Rightarrow x = \dfrac{3}{8} - \dfrac{-7}{12} \\[1em] \Rightarrow x = \dfrac{3}{8} + \dfrac{7}{12}

By Division Method,

28,1224,622,331,31,1\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

⇒x=3×38×3+7×212×2⇒x=924+1424⇒x=9+1424⇒x=2324\Rightarrow x = \dfrac{3 \times 3}{8 \times 3} + \dfrac{7 \times 2}{12 \times 2} \\[1em] \Rightarrow x = \dfrac{9}{24} + \dfrac{14}{24} \\[1em] \Rightarrow x = \dfrac{9 + 14}{24} \\[1em] \Rightarrow x = \dfrac{23}{24}

Hence, 2324\dfrac{23}{24} should be added to −712\dfrac{-7}{12} to get 38\dfrac{3}{8}.

Question 15

What should be subtracted from 59\dfrac{5}{9} to get 95\dfrac{9}{5}?

Answer

Let x be subtracted from 59\dfrac{5}{9} to get 95\dfrac{9}{5}.

59−x=95⇒x=59−95\dfrac{5}{9} - x = \dfrac{9}{5} \\[1em] \Rightarrow x = \dfrac{5}{9} - \dfrac{9}{5}

By Division Method,

39,533,551,51,1\begin{array}{l|r} 3 & 9, 5 \\ \hline 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 9 and 5 = 3 × 3 × 5 = 45

⇒x=5×59×5−9×95×9⇒x=2545−8145⇒x=25−8145⇒x=−5645⇒x=−11145\Rightarrow x = \dfrac{5 \times 5}{9 \times 5} - \dfrac{9 \times 9}{5 \times 9} \\[1em] \Rightarrow x = \dfrac{25}{45} - \dfrac{81}{45} \\[1em] \Rightarrow x = \dfrac{25 - 81}{45} \\[1em] \Rightarrow x = \dfrac{-56}{45} \\[1em] \Rightarrow x = -1\dfrac{11}{45}

Hence, −11145-1\dfrac{11}{45} should be subtracted from 59\dfrac{5}{9} to get 95\dfrac{9}{5}.

Exercise 2(D)

Question 1(i)

Evaluate:

54×37\dfrac{5}{4} \times \dfrac{3}{7}

Answer

Solving,

⇒54×37=5×34×7=1528\Rightarrow \dfrac{5}{4} \times \dfrac{3}{7} \\[1em] = \dfrac{5 \times 3}{4 \times 7} \\[1em] = \dfrac{15}{28}

Hence, 54×37=1528\dfrac{5}{4} \times \dfrac{3}{7} = \dfrac{15}{28}

Question 1(ii)

Evaluate:

23×−67\dfrac{2}{3} \times -\dfrac{6}{7}

Answer

Solving,

⇒23×−67=2×(−6)3×7=−1221=−47\Rightarrow \dfrac{2}{3} \times -\dfrac{6}{7} \\[1em] = \dfrac{2 \times (-6)}{3 \times 7} \\[1em] = \dfrac{-12}{21} \\[1em] = \dfrac{-4}{7}

Hence, 23×−67=−47\dfrac{2}{3} \times -\dfrac{6}{7} = \dfrac{-4}{7}

Question 1(iii)

Evaluate:

(−125)×(10−3)\left(\dfrac{-12}{5}\right) \times \left(\dfrac{10}{-3}\right)

Answer

Solving,

⇒(−125)×(10−3)=−12×105×(−3)=−120−15=8\Rightarrow \left(\dfrac{-12}{5}\right) \times \left(\dfrac{10}{-3}\right) \\[1em] = \dfrac{-12 \times 10}{5 \times (-3)} \\[1em] = \dfrac{-120}{-15} \\[1em] = 8

Hence, (−125)×(10−3)=8\left(\dfrac{-12}{5}\right) \times \left(\dfrac{10}{-3}\right) = 8

Question 1(iv)

Evaluate:

−4539×−1315-\dfrac{45}{39} \times \dfrac{-13}{15}

Answer

Solving,

⇒−4539×−1315=−45×(−13)39×15=585585=1\Rightarrow -\dfrac{45}{39} \times \dfrac{-13}{15} \\[1em] = \dfrac{-45 \times (-13)}{39 \times 15} \\[1em] = \dfrac{585}{585} \\[1em] = 1

Hence, −4539×−1315=1-\dfrac{45}{39} \times \dfrac{-13}{15} = 1

Question 1(v)

Evaluate:

318×(−225)3\dfrac{1}{8} \times \left(-2\dfrac{2}{5}\right)

Answer

Solving,

⇒318×(−225)=258×(−125)=25×(−12)8×5=−30040=−152=−712\Rightarrow 3\dfrac{1}{8} \times \left(-2\dfrac{2}{5}\right) \\[1em] = \dfrac{25}{8} \times \left(-\dfrac{12}{5}\right) \\[1em] = \dfrac{25 \times (-12)}{8 \times 5} \\[1em] = \dfrac{-300}{40} \\[1em] = \dfrac{-15}{2} \\[1em] = -7\dfrac{1}{2}

Hence, 318×(−225)=−7123\dfrac{1}{8} \times \left(-2\dfrac{2}{5}\right) = -7\dfrac{1}{2}

Question 1(vi)

Evaluate:

21425×(−516)2\dfrac{14}{25} \times \left(\dfrac{-5}{16}\right)

Answer

Solving,

⇒21425×(−516)=6425×(−516)=64×(−5)25×16=−320400=−45\Rightarrow 2\dfrac{14}{25} \times \left(\dfrac{-5}{16}\right) \\[1em] = \dfrac{64}{25} \times \left(\dfrac{-5}{16}\right) \\[1em] = \dfrac{64 \times (-5)}{25 \times 16} \\[1em] = \dfrac{-320}{400} \\[1em] = \dfrac{-4}{5}

Hence, 21425×(−516)=−452\dfrac{14}{25} \times \left(\dfrac{-5}{16}\right) = \dfrac{-4}{5}

Question 1(vii)

Evaluate:

(−89)×(−316)\left(\dfrac{-8}{9}\right) \times \left(\dfrac{-3}{16}\right)

Answer

Solving,

⇒(−89)×(−316)=−8×(−3)9×16=24144=16\Rightarrow \left(\dfrac{-8}{9}\right) \times \left(\dfrac{-3}{16}\right) \\[1em] = \dfrac{-8 \times (-3)}{9 \times 16} \\[1em] = \dfrac{24}{144} \\[1em] = \dfrac{1}{6}

Hence, (−89)×(−316)=16\left(\dfrac{-8}{9}\right) \times \left(\dfrac{-3}{16}\right) = \dfrac{1}{6}

Question 1(viii)

Evaluate:

(5−27)×(−920)\left(\dfrac{5}{-27}\right) \times \left(\dfrac{-9}{20}\right)

Answer

Solving,

⇒(5−27)×(−920)=(−527)×(−920)=−5×(−9)27×20=45540=112\Rightarrow \left(\dfrac{5}{-27}\right) \times \left(\dfrac{-9}{20}\right) \\[1em] = \left(\dfrac{-5}{27}\right) \times \left(\dfrac{-9}{20}\right) \\[1em] = \dfrac{-5 \times (-9)}{27 \times 20} \\[1em] = \dfrac{45}{540} \\[1em] = \dfrac{1}{12}

Hence, (5−27)×(−920)=112\left(\dfrac{5}{-27}\right) \times \left(\dfrac{-9}{20}\right) = \dfrac{1}{12}

Question 2(i)

Multiply:

325 and 45\dfrac{3}{25} \text{ and } \dfrac{4}{5}

Answer

Solving,

⇒325×45=3×425×5=12125\Rightarrow \dfrac{3}{25} \times \dfrac{4}{5} \\[1em] = \dfrac{3 \times 4}{25 \times 5} \\[1em] = \dfrac{12}{125}

Hence, 325×45=12125\dfrac{3}{25} \times \dfrac{4}{5} = \dfrac{12}{125}

Question 2(ii)

Multiply:

118 and 10231\dfrac{1}{8} \text{ and } 10\dfrac{2}{3}

Answer

Solving,

⇒118×1023=98×323=9×328×3=28824=12\Rightarrow 1\dfrac{1}{8} \times 10\dfrac{2}{3} \\[1em] = \dfrac{9}{8} \times \dfrac{32}{3} \\[1em] = \dfrac{9 \times 32}{8 \times 3} \\[1em] = \dfrac{288}{24} \\[1em] = 12

Hence, 118×1023=121\dfrac{1}{8} \times 10\dfrac{2}{3} = 12

Question 2(iii)

Multiply:

623 and −386\dfrac{2}{3} \text{ and } \dfrac{-3}{8}

Answer

Solving,

⇒623×−38=203×−38=20×(−3)3×8=−6024=−52=−212\Rightarrow 6\dfrac{2}{3} \times \dfrac{-3}{8} \\[1em] = \dfrac{20}{3} \times \dfrac{-3}{8} \\[1em] = \dfrac{20 \times (-3)}{3 \times 8} \\[1em] = \dfrac{-60}{24} \\[1em] = \dfrac{-5}{2} \\[1em] = -2\dfrac{1}{2}

Hence, 623×−38=−2126\dfrac{2}{3} \times \dfrac{-3}{8} = -2\dfrac{1}{2}

Question 2(iv)

Multiply:

−1315 and −2526\dfrac{-13}{15} \text{ and } \dfrac{-25}{26}

Answer

Solving,

⇒−1315×−2526=−13×(−25)15×26=325390=56\Rightarrow \dfrac{-13}{15} \times \dfrac{-25}{26} \\[1em] = \dfrac{-13 \times (-25)}{15 \times 26} \\[1em] = \dfrac{325}{390} \\[1em] = \dfrac{5}{6}

Hence, −1315×−2526=56\dfrac{-13}{15} \times \dfrac{-25}{26} = \dfrac{5}{6}

Question 2(v)

Multiply:

1161\dfrac{1}{6} and 18

Answer

Solving,

⇒116×18=76×181=7×186×1=1266=21\Rightarrow 1\dfrac{1}{6} \times 18 \\[1em] = \dfrac{7}{6} \times \dfrac{18}{1} \\[1em] = \dfrac{7 \times 18}{6 \times 1} \\[1em] = \dfrac{126}{6} \\[1em] = 21

Hence, 116×18=211\dfrac{1}{6} \times 18 = 21

Question 2(vi)

Multiply:

21142\dfrac{1}{14} and -7

Answer

Solving,

⇒2114×(−7)=2914×−71=29×(−7)14×1=−20314=−292=−1412\Rightarrow 2\dfrac{1}{14} \times (-7) \\[1em] = \dfrac{29}{14} \times \dfrac{-7}{1} \\[1em] = \dfrac{29 \times (-7)}{14 \times 1} \\[1em] = \dfrac{-203}{14} \\[1em] = \dfrac{-29}{2} \\[1em] = -14\dfrac{1}{2}

Hence, 2114×(−7)=−14122\dfrac{1}{14} \times (-7) = -14\dfrac{1}{2}

Question 2(vii)

Multiply:

5185\dfrac{1}{8} and -16

Answer

Solving,

⇒518×(−16)=418×−161=41×(−16)8×1=−6568=−82\Rightarrow 5\dfrac{1}{8} \times (-16) \\[1em] = \dfrac{41}{8} \times \dfrac{-16}{1} \\[1em] = \dfrac{41 \times (-16)}{8 \times 1} \\[1em] = \dfrac{-656}{8} \\[1em] = -82

Hence, 518×(−16)=−825\dfrac{1}{8} \times (-16) = -82

Question 2(viii)

Multiply:

35 and −1825\dfrac{-18}{25}

Answer

Solving,

⇒35×−1825=351×−1825=35×(−18)1×25=−63025=−1265=−2515\Rightarrow 35 \times \dfrac{-18}{25} \\[1em] = \dfrac{35}{1} \times \dfrac{-18}{25} \\[1em] = \dfrac{35 \times (-18)}{1 \times 25} \\[1em] = \dfrac{-630}{25} \\[1em] = \dfrac{-126}{5} \\[1em] = -25\dfrac{1}{5}

Hence, 35×−1825=−251535 \times \dfrac{-18}{25} = -25\dfrac{1}{5}

Question 2(ix)

Multiply:

623 and −386\dfrac{2}{3} \text{ and } -\dfrac{3}{8}

Answer

Solving,

⇒623×−38=203×−38=20×(−3)3×8=−6024=−52=−212\Rightarrow 6\dfrac{2}{3} \times -\dfrac{3}{8} \\[1em] = \dfrac{20}{3} \times \dfrac{-3}{8} \\[1em] = \dfrac{20 \times (-3)}{3 \times 8} \\[1em] = \dfrac{-60}{24} \\[1em] = \dfrac{-5}{2} \\[1em] = -2\dfrac{1}{2}

Hence, 623×−38=−2126\dfrac{2}{3} \times -\dfrac{3}{8} = -2\dfrac{1}{2}

Question 2(x)

Multiply:

3353\dfrac{3}{5} and -10

Answer

Solving,

⇒335×(−10)=185×−101=18×(−10)5×1=−1805=−36\Rightarrow 3\dfrac{3}{5} \times (-10) \\[1em] = \dfrac{18}{5} \times \dfrac{-10}{1} \\[1em] = \dfrac{18 \times (-10)}{5 \times 1} \\[1em] = \dfrac{-180}{5} \\[1em] = -36

Hence, 335×(−10)=−363\dfrac{3}{5} \times (-10) = -36

Question 2(xi)

Multiply:

2728\dfrac{27}{28} and -14

Answer

Solving,

⇒2728×(−14)=2728×−141=27×(−14)28×1=−37828=−272=−1312\Rightarrow \dfrac{27}{28} \times (-14) \\[1em] = \dfrac{27}{28} \times \dfrac{-14}{1} \\[1em] = \dfrac{27 \times (-14)}{28 \times 1} \\[1em] = \dfrac{-378}{28} \\[1em] = \dfrac{-27}{2} \\[1em] = -13\dfrac{1}{2}

Hence, 2728×(−14)=−1312\dfrac{27}{28} \times (-14) = -13\dfrac{1}{2}

Question 2(xii)

Multiply:

-24 and 516\dfrac{5}{16}

Answer

Solving,

⇒−24×516=−241×516=−24×51×16=−12016=−152=−712\Rightarrow -24 \times \dfrac{5}{16} \\[1em] = \dfrac{-24}{1} \times \dfrac{5}{16} \\[1em] = \dfrac{-24 \times 5}{1 \times 16} \\[1em] = \dfrac{-120}{16} \\[1em] = \dfrac{-15}{2} \\[1em] = -7\dfrac{1}{2}

Hence, −24×516=−712-24 \times \dfrac{5}{16} = -7\dfrac{1}{2}

Question 3(i)

Evaluate:

(−6×518)−(−429)\left(-6 \times \dfrac{5}{18}\right) - \left(-4\dfrac{2}{9}\right)

Answer

Solving,

⇒(−6×518)−(−429)=(−6×51×18)−(−389)=−3018+389=−53+389\Rightarrow \left(-6 \times \dfrac{5}{18}\right) - \left(-4\dfrac{2}{9}\right) \\[1em] = \left(\dfrac{-6 \times 5}{1 \times 18}\right) - \left(\dfrac{-38}{9}\right) \\[1em] = \dfrac{-30}{18} + \dfrac{38}{9} \\[1em] = \dfrac{-5}{3} + \dfrac{38}{9}

LCM of 3 and 9 = 9

=−5×33×3+38×19×1=−159+389=−15+389=239=259= \dfrac{-5 \times 3}{3 \times 3} + \dfrac{38 \times 1}{9 \times 1} \\[1em] = \dfrac{-15}{9} + \dfrac{38}{9} \\[1em] = \dfrac{-15 + 38}{9} \\[1em] = \dfrac{23}{9} \\[1em] = 2\dfrac{5}{9}

Hence, (−6×518)−(−429)=259\left(-6 \times \dfrac{5}{18}\right) - \left(-4\dfrac{2}{9}\right) = 2\dfrac{5}{9}

Question 3(ii)

Evaluate:

(78×87)+(−59)×(6−25)\left(\dfrac{7}{8} \times \dfrac{8}{7}\right) + \left(\dfrac{-5}{9}\right) \times \left(\dfrac{6}{-25}\right)

Answer

Solving,

⇒(78×87)+(−59×6−25)=7×88×7+−5×69×(−25)=5656+−30−225=1+215\Rightarrow \left(\dfrac{7}{8} \times \dfrac{8}{7}\right) + \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) \\[1em] = \dfrac{7 \times 8}{8 \times 7} + \dfrac{-5 \times 6}{9 \times (-25)} \\[1em] = \dfrac{56}{56} + \dfrac{-30}{-225} \\[1em] = 1 + \dfrac{2}{15}

LCM of 1 and 15 = 15

=1×151×15+2×115×1=1515+215=15+215=1715=1215= \dfrac{1 \times 15}{1 \times 15} + \dfrac{2 \times 1}{15 \times 1} \\[1em] = \dfrac{15}{15} + \dfrac{2}{15} \\[1em] = \dfrac{15 + 2}{15} \\[1em] = \dfrac{17}{15} \\[1em] = 1\dfrac{2}{15}

Hence, (78×87)+(−59)×(6−25)=1215\left(\dfrac{7}{8} \times \dfrac{8}{7}\right) + \left(\dfrac{-5}{9}\right) \times \left(\dfrac{6}{-25}\right) = 1\dfrac{2}{15}

Question 3(iii)

Evaluate:

(11−9×2144)+(−59)×(63−100)\left(\dfrac{11}{-9} \times \dfrac{21}{44}\right) + \left(\dfrac{-5}{9}\right) \times \left(\dfrac{63}{-100}\right)

Answer

Solving,

⇒(11−9×2144)+(−59×63−100)=11×21−9×44+−5×639×(−100)=231−396+−315−900=−712+720\Rightarrow \left(\dfrac{11}{-9} \times \dfrac{21}{44}\right) + \left(\dfrac{-5}{9} \times \dfrac{63}{-100}\right) \\[1em] = \dfrac{11 \times 21}{-9 \times 44} + \dfrac{-5 \times 63}{9 \times (-100)} \\[1em] = \dfrac{231}{-396} + \dfrac{-315}{-900} \\[1em] = \dfrac{-7}{12} + \dfrac{7}{20}

LCM of 12 and 20 = 60

=−7×512×5+7×320×3=−3560+2160=−35+2160=−1460=−730= \dfrac{-7 \times 5}{12 \times 5} + \dfrac{7 \times 3}{20 \times 3} \\[1em] = \dfrac{-35}{60} + \dfrac{21}{60} \\[1em] = \dfrac{-35 + 21}{60} \\[1em] = \dfrac{-14}{60} \\[1em] = \dfrac{-7}{30}

Hence, (11−9×2144)+(−59)×(63−100)=−730\left(\dfrac{11}{-9} \times \dfrac{21}{44}\right) + \left(\dfrac{-5}{9}\right) \times \left(\dfrac{63}{-100}\right) = \dfrac{-7}{30}

Question 3(iv)

Evaluate:

(−59×6−25)+(2421×78)\left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) + \left(\dfrac{24}{21} \times \dfrac{7}{8}\right)

Answer

Solving,

⇒(−59×6−25)+(2421×78)=−5×69×(−25)+24×721×8=−30−225+168168=215+1\Rightarrow \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) + \left(\dfrac{24}{21} \times \dfrac{7}{8}\right) \\[1em] = \dfrac{-5 \times 6}{9 \times (-25)} + \dfrac{24 \times 7}{21 \times 8} \\[1em] = \dfrac{-30}{-225} + \dfrac{168}{168} \\[1em] = \dfrac{2}{15} + 1

LCM of 15 and 1 = 15

=2×115×1+1×151×15=215+1515=2+1515=1715=1215= \dfrac{2 \times 1}{15 \times 1} + \dfrac{1 \times 15}{1 \times 15} \\[1em] = \dfrac{2}{15} + \dfrac{15}{15} \\[1em] = \dfrac{2 + 15}{15} \\[1em] = \dfrac{17}{15} \\[1em] = 1\dfrac{2}{15}

Hence, (−59×6−25)+(2421×78)=1215\left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) + \left(\dfrac{24}{21} \times \dfrac{7}{8}\right) = 1\dfrac{2}{15}

Question 3(v)

Evaluate:

(−3539×−137)−(790×−1814)\left(\dfrac{-35}{39} \times \dfrac{-13}{7}\right) - \left(\dfrac{7}{90} \times \dfrac{-18}{14}\right)

Answer

Solving,

⇒(−3539×−137)−(790×−1814)=−35×(−13)39×7−7×(−18)90×14=455273−−1261260=53−(−110)=53+110\Rightarrow \left(\dfrac{-35}{39} \times \dfrac{-13}{7}\right) - \left(\dfrac{7}{90} \times \dfrac{-18}{14}\right) \\[1em] = \dfrac{-35 \times (-13)}{39 \times 7} - \dfrac{7 \times (-18)}{90 \times 14} \\[1em] = \dfrac{455}{273} - \dfrac{-126}{1260} \\[1em] = \dfrac{5}{3} - \left(\dfrac{-1}{10}\right) \\[1em] = \dfrac{5}{3} + \dfrac{1}{10}

LCM of 3 and 10 = 30

=5×103×10+1×310×3=5030+330=50+330=5330=12330= \dfrac{5 \times 10}{3 \times 10} + \dfrac{1 \times 3}{10 \times 3} \\[1em] = \dfrac{50}{30} + \dfrac{3}{30} \\[1em] = \dfrac{50 + 3}{30} \\[1em] = \dfrac{53}{30} \\[1em] = 1\dfrac{23}{30}

Hence, (−3539×−137)−(790×−1814)=12330\left(\dfrac{-35}{39} \times \dfrac{-13}{7}\right) - \left(\dfrac{7}{90} \times \dfrac{-18}{14}\right) = 1\dfrac{23}{30}

Question 3(vi)

Evaluate:

(−45×32)+(9−5×103)−(−32×−14)\left(\dfrac{-4}{5} \times \dfrac{3}{2}\right) + \left(\dfrac{9}{-5} \times \dfrac{10}{3}\right) - \left(\dfrac{-3}{2} \times \dfrac{-1}{4}\right)

Answer

Solving,

⇒(−45×32)+(9−5×103)−(−32×−14)=−4×35×2+9×10−5×3−−3×(−1)2×4=−1210+90−15−38=−65+(−6)−38\Rightarrow \left(\dfrac{-4}{5} \times \dfrac{3}{2}\right) + \left(\dfrac{9}{-5} \times \dfrac{10}{3}\right) - \left(\dfrac{-3}{2} \times \dfrac{-1}{4}\right) \\[1em] = \dfrac{-4 \times 3}{5 \times 2} + \dfrac{9 \times 10}{-5 \times 3} - \dfrac{-3 \times (-1)}{2 \times 4} \\[1em] = \dfrac{-12}{10} + \dfrac{90}{-15} - \dfrac{3}{8} \\[1em] = \dfrac{-6}{5} + (-6) - \dfrac{3}{8}

LCM of 5, 1 and 8 = 40

=−6×85×8+−6×401×40−3×58×5=−4840+−24040−1540=−48+(−240)−1540=−30340=−72340= \dfrac{-6 \times 8}{5 \times 8} + \dfrac{-6 \times 40}{1 \times 40} - \dfrac{3 \times 5}{8 \times 5} \\[1em] = \dfrac{-48}{40} + \dfrac{-240}{40} - \dfrac{15}{40} \\[1em] = \dfrac{-48 + (-240) - 15}{40} \\[1em] = \dfrac{-303}{40} \\[1em] = -7\dfrac{23}{40}

Hence, (−45×32)+(9−5×103)−(−32×−14)=−72340\left(\dfrac{-4}{5} \times \dfrac{3}{2}\right) + \left(\dfrac{9}{-5} \times \dfrac{10}{3}\right) - \left(\dfrac{-3}{2} \times \dfrac{-1}{4}\right) = -7\dfrac{23}{40}

Question 4

Find the cost of 3123\dfrac{1}{2} m cloth, if one metre cloth costs ₹ 32512325\dfrac{1}{2}.

Answer

Cost of 1 m cloth = ₹ 32512325\dfrac{1}{2} = ₹ 6512\dfrac{651}{2}

Cost of 3123\dfrac{1}{2} m cloth = 6512×312\dfrac{651}{2} \times 3\dfrac{1}{2}

=6512×72=651×72×2=45574=113914= \dfrac{651}{2} \times \dfrac{7}{2} \\[1em] = \dfrac{651 \times 7}{2 \times 2} \\[1em] = \dfrac{4557}{4} \\[1em] = 1139\dfrac{1}{4}

Hence, the cost of 3123\dfrac{1}{2} m cloth is ₹ 1139141139\dfrac{1}{4}.

Question 5

A bus is moving with a speed of 651265\dfrac{1}{2} km per hour. How much distance will it cover in 1131\dfrac{1}{3} hours?

Answer

Speed = 651265\dfrac{1}{2} km per hour = 1312\dfrac{131}{2} km per hour

Time = 1131\dfrac{1}{3} hours = 43\dfrac{4}{3} hours

Distance covered = Speed × Time

=1312×43=131×42×3=5246=2623=8713= \dfrac{131}{2} \times \dfrac{4}{3} \\[1em] = \dfrac{131 \times 4}{2 \times 3} \\[1em] = \dfrac{524}{6} \\[1em] = \dfrac{262}{3} \\[1em] = 87\dfrac{1}{3}

Hence, the bus will cover 871387\dfrac{1}{3} km in 1131\dfrac{1}{3} hours.

Question 6(i)

Divide:

1528 by 34\dfrac{15}{28} \text{ by } \dfrac{3}{4}

Answer

Solving,

⇒1528÷34=1528×43=15×428×3=6084=57\Rightarrow \dfrac{15}{28} ÷ \dfrac{3}{4} \\[1em] = \dfrac{15}{28} \times \dfrac{4}{3} \\[1em] = \dfrac{15 \times 4}{28 \times 3} \\[1em] = \dfrac{60}{84} \\[1em] = \dfrac{5}{7}

Hence, 1528÷34=57\dfrac{15}{28} ÷ \dfrac{3}{4} = \dfrac{5}{7}

Question 6(ii)

Divide:

−209 by −59\dfrac{-20}{9} \text{ by } \dfrac{-5}{9}

Answer

Solving,

⇒−209÷−59=−209×9−5=−20×99×(−5)=−180−45=4\Rightarrow \dfrac{-20}{9} ÷ \dfrac{-5}{9} \\[1em] = \dfrac{-20}{9} \times \dfrac{9}{-5} \\[1em] = \dfrac{-20 \times 9}{9 \times (-5)} \\[1em] = \dfrac{-180}{-45} \\[1em] = 4

Hence, −209÷−59=4\dfrac{-20}{9} ÷ \dfrac{-5}{9} = 4

Question 6(iii)

Divide:

16−5 by −87\dfrac{16}{-5} \text{ by } \dfrac{-8}{7}

Answer

Solving,

⇒16−5÷−87=−165×7−8=−16×75×(−8)=−112−40=145=245\Rightarrow \dfrac{16}{-5} ÷ \dfrac{-8}{7} \\[1em] = \dfrac{-16}{5} \times \dfrac{7}{-8} \\[1em] = \dfrac{-16 \times 7}{5 \times (-8)} \\[1em] = \dfrac{-112}{-40} \\[1em] = \dfrac{14}{5} \\[1em] = 2\dfrac{4}{5}

Hence, 16−5÷−87=245\dfrac{16}{-5} ÷ \dfrac{-8}{7} = 2\dfrac{4}{5}

Question 6(iv)

Divide:

-7 by −145\dfrac{-14}{5}

Answer

Solving,

⇒−7÷−145=−71×5−14=−7×51×(−14)=−35−14=52=212\Rightarrow -7 ÷ \dfrac{-14}{5} \\[1em] = \dfrac{-7}{1} \times \dfrac{5}{-14} \\[1em] = \dfrac{-7 \times 5}{1 \times (-14)} \\[1em] = \dfrac{-35}{-14} \\[1em] = \dfrac{5}{2} \\[1em] = 2\dfrac{1}{2}

Hence, −7÷−145=212-7 ÷ \dfrac{-14}{5} = 2\dfrac{1}{2}

Question 6(v)

Divide:

-14 by 7−2\dfrac{7}{-2}

Answer

Solving,

⇒−14÷7−2=−141×−27=−14×(−2)1×7=287=4\Rightarrow -14 ÷ \dfrac{7}{-2} \\[1em] = \dfrac{-14}{1} \times \dfrac{-2}{7} \\[1em] = \dfrac{-14 \times (-2)}{1 \times 7} \\[1em] = \dfrac{28}{7} \\[1em] = 4

Hence, −14÷7−2=4-14 ÷ \dfrac{7}{-2} = 4

Question 6(vi)

Divide:

−229 by 1118\dfrac{-22}{9} \text{ by } \dfrac{11}{18}

Answer

Solving,

⇒−229÷1118=−229×1811=−22×189×11=−39699=−4\Rightarrow \dfrac{-22}{9} ÷ \dfrac{11}{18} \\[1em] = \dfrac{-22}{9} \times \dfrac{18}{11} \\[1em] = \dfrac{-22 \times 18}{9 \times 11} \\[1em] = \dfrac{-396}{99} \\[1em] = -4

Hence, −229÷1118=−4\dfrac{-22}{9} ÷ \dfrac{11}{18} = -4

Question 6(vii)

Divide:

35 by −79\dfrac{-7}{9}

Answer

Solving,

⇒35÷−79=351×9−7=35×91×(−7)=315−7=−45\Rightarrow 35 ÷ \dfrac{-7}{9} \\[1em] = \dfrac{35}{1} \times \dfrac{9}{-7} \\[1em] = \dfrac{35 \times 9}{1 \times (-7)} \\[1em] = \dfrac{315}{-7} \\[1em] = -45

Hence, 35÷−79=−4535 ÷ \dfrac{-7}{9} = -45

Question 6(viii)

Divide:

2144 by −119\dfrac{21}{44} \text{ by } -\dfrac{11}{9}

Answer

Solving,

⇒2144÷−119=2144×9−11=21×944×(−11)=189−484=−189484\Rightarrow \dfrac{21}{44} ÷ -\dfrac{11}{9} \\[1em] = \dfrac{21}{44} \times \dfrac{9}{-11} \\[1em] = \dfrac{21 \times 9}{44 \times (-11)} \\[1em] = \dfrac{189}{-484} \\[1em] = \dfrac{-189}{484}

Hence, 2144÷−119=−189484\dfrac{21}{44} ÷ -\dfrac{11}{9} = \dfrac{-189}{484}

Question 7(i)

Evaluate:

3512+1233\dfrac{5}{12} + 1\dfrac{2}{3}

Answer

Solving,

⇒3512+123=4112+53\Rightarrow 3\dfrac{5}{12} + 1\dfrac{2}{3} \\[1em] = \dfrac{41}{12} + \dfrac{5}{3}

LCM of 12 and 3 is 2 × 2 × 3 = 12

=41×112×1+5×43×4=4112+2012=41+2012=6112=5112= \dfrac{41 \times 1}{12 \times 1} + \dfrac{5 \times 4}{3 \times 4} \\[1em] = \dfrac{41}{12} + \dfrac{20}{12} \\[1em] = \dfrac{41 + 20}{12} \\[1em] = \dfrac{61}{12} \\[1em] = 5\dfrac{1}{12}

Hence, 3512+123=51123\dfrac{5}{12} + 1\dfrac{2}{3} = 5\dfrac{1}{12}

Question 7(ii)

Evaluate:

3512−1233\dfrac{5}{12} - 1\dfrac{2}{3}

Answer

Solving,

⇒3512−123=4112−53\Rightarrow 3\dfrac{5}{12} - 1\dfrac{2}{3} \\[1em] = \dfrac{41}{12} - \dfrac{5}{3}

LCM of 12 and 3 = 12

=41×112×1−5×43×4=4112−2012=41−2012=2112=74=134= \dfrac{41 \times 1}{12 \times 1} - \dfrac{5 \times 4}{3 \times 4} \\[1em] = \dfrac{41}{12} - \dfrac{20}{12} \\[1em] = \dfrac{41 - 20}{12} \\[1em] = \dfrac{21}{12} \\[1em] = \dfrac{7}{4} \\[1em] = 1\dfrac{3}{4}

Hence, 3512−123=1343\dfrac{5}{12} - 1\dfrac{2}{3} = 1\dfrac{3}{4}

Question 7(iii)

Evaluate:

(3512+123)÷(3512−123)\left(3\dfrac{5}{12} + 1\dfrac{2}{3}\right) \div \left(3\dfrac{5}{12} - 1\dfrac{2}{3}\right)

Answer

Solving,

From parts (i) and (ii):

3512+123=6112 and 3512−123=21123\dfrac{5}{12} + 1\dfrac{2}{3} = \dfrac{61}{12} \text{ and } 3\dfrac{5}{12} - 1\dfrac{2}{3} = \dfrac{21}{12}

Solving,

⇒(3512+123)÷(3512−123)=6112÷2112=6112×1221=61×1212×21=6121=21921\Rightarrow \left(3\dfrac{5}{12} + 1\dfrac{2}{3}\right) \div \left(3\dfrac{5}{12} - 1\dfrac{2}{3}\right) \\[1em] = \dfrac{61}{12} ÷ \dfrac{21}{12} \\[1em] = \dfrac{61}{12} \times \dfrac{12}{21} \\[1em] = \dfrac{61 \times 12}{12 \times 21} \\[1em] = \dfrac{61}{21} \\[1em] = 2\dfrac{19}{21}

Hence, (3512+123)÷(3512−123)=21921\left(3\dfrac{5}{12} + 1\dfrac{2}{3}\right) \div \left(3\dfrac{5}{12} - 1\dfrac{2}{3}\right) = 2\dfrac{19}{21}

Question 8

The product of two numbers is 14. If one of the numbers is −87\dfrac{-8}{7}, find the other.

Answer

Product of two numbers = 14

Let the other number = x

One number = −87\dfrac{-8}{7}

According to question,

⇒x×−87=14⇒x=14÷−87=141×7−8=14×71×(−8)=98−8=−494=−1214\Rightarrow x \times \dfrac{-8}{7} = 14\\[1em] \Rightarrow x = 14 ÷ \dfrac{-8}{7}\\[1em] = \dfrac{14}{1} \times \dfrac{7}{-8} \\[1em] = \dfrac{14 \times 7}{1 \times (-8)} \\[1em] = \dfrac{98}{-8} \\[1em] = \dfrac{-49}{4} \\[1em] = -12\dfrac{1}{4}

Hence, the other number is −1214-12\dfrac{1}{4}.

Question 9

The cost of 11 pens is ₹ 243424\dfrac{3}{4}. Find the cost of one pen.

Answer

Cost of 11 pens = ₹ 2434=₹99424\dfrac{3}{4} = ₹ \dfrac{99}{4}

Cost of one pen = 994÷11\dfrac{99}{4} ÷ 11

=994×111=99×14×11=9944=94=214= \dfrac{99}{4} \times \dfrac{1}{11} \\[1em] = \dfrac{99 \times 1}{4 \times 11} \\[1em] = \dfrac{99}{44} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4}

Hence, the cost of one pen is ₹ 2142\dfrac{1}{4}.

Question 10

If 6 identical articles can be bought for ₹ 26172\dfrac{6}{17}. Find the cost of each article.

Answer

Cost of 6 articles = ₹ 2617=₹40172\dfrac{6}{17} = ₹ \dfrac{40}{17}

Cost of each article = 4017÷6\dfrac{40}{17} ÷ 6

=4017×16=40×117×6=40102=2051= \dfrac{40}{17} \times \dfrac{1}{6} \\[1em] = \dfrac{40 \times 1}{17 \times 6} \\[1em] = \dfrac{40}{102} \\[1em] = \dfrac{20}{51}

Hence, the cost of each article is ₹ 2051\dfrac{20}{51}.

Question 11

By what number should −38\dfrac{-3}{8} be multiplied so that the product is −916\dfrac{-9}{16}?

Answer

Let the required number be x.

⇒−38×x=−916⇒x=−916÷−38⇒x=−916×8−3⇒x=−9×816×(−3)⇒x=−72−48⇒x=32⇒x=112\Rightarrow \dfrac{-3}{8} \times x = \dfrac{-9}{16} \\[1em] \Rightarrow x = \dfrac{-9}{16} ÷ \dfrac{-3}{8} \\[1em] \Rightarrow x = \dfrac{-9}{16} \times \dfrac{8}{-3} \\[1em] \Rightarrow x = \dfrac{-9 \times 8}{16 \times (-3)} \\[1em] \Rightarrow x = \dfrac{-72}{-48} \\[1em] \Rightarrow x = \dfrac{3}{2} \\[1em] \Rightarrow x = 1\dfrac{1}{2}

Hence, −38\dfrac{-3}{8} should be multiplied by 112 to get −9161\dfrac{1}{2} \text{ to get } \dfrac{-9}{16}.

Question 12

By what number should −57\dfrac{-5}{7} be divided so that the result is −1528\dfrac{-15}{28}?

Answer

Let the required number be x.

⇒−57÷x=−1528⇒−57×1x=−1528⇒1x=−1528÷−57⇒1x=−1528×7−5⇒1x=−15×728×(−5)⇒1x=−105−140⇒1x=34⇒x=43⇒x=113\Rightarrow \dfrac{-5}{7} ÷ x = \dfrac{-15}{28} \\[1em] \Rightarrow \dfrac{-5}{7} \times \dfrac{1}{x} = \dfrac{-15}{28} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-15}{28} ÷ \dfrac{-5}{7} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-15}{28} \times \dfrac{7}{-5} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-15 \times 7}{28 \times (-5)} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{-105}{-140} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{3}{4} \\[1em] \Rightarrow x = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}

Hence, −57\dfrac{-5}{7} should be divided by 1131\dfrac{1}{3} to get −1528\dfrac{-15}{28}.

Question 13

Evaluate: (3215+85)÷(3215−85)\left(\dfrac{32}{15} + \dfrac{8}{5}\right) \div \left(\dfrac{32}{15} - \dfrac{8}{5}\right).

Answer

By Division Method,

315,555,51,1\begin{array}{l|r} 3 & 15, 5 \\ \hline 5 & 5, 5 \\ \hline & 1, 1 \end{array}

LCM of 15 and 5 = 3 × 5 = 15

⇒3215+85=32×115×1+8×35×3=3215+2415=5615⇒3215−85=32×115×1−8×35×3=3215−2415=815\Rightarrow \dfrac{32}{15} + \dfrac{8}{5} = \dfrac{32 \times 1}{15 \times 1} + \dfrac{8 \times 3}{5 \times 3} = \dfrac{32}{15} + \dfrac{24}{15} = \dfrac{56}{15} \\[1em] \Rightarrow \dfrac{32}{15} - \dfrac{8}{5} = \dfrac{32 \times 1}{15 \times 1} - \dfrac{8 \times 3}{5 \times 3} = \dfrac{32}{15} - \dfrac{24}{15} = \dfrac{8}{15}

Solving,

⇒(3215+85)÷(3215−85)=5615÷815=5615×158=56×1515×8=568=7\Rightarrow\left(\dfrac{32}{15} + \dfrac{8}{5}\right) \div \left(\dfrac{32}{15} - \dfrac{8}{5}\right) \\[1em] = \dfrac{56}{15} ÷ \dfrac{8}{15} \\[1em] = \dfrac{56}{15} \times \dfrac{15}{8} \\[1em] = \dfrac{56 \times 15}{15 \times 8} \\[1em] = \dfrac{56}{8} \\[1em] = 7

Hence, (3215+85)÷(3215−85)=7\left(\dfrac{32}{15} + \dfrac{8}{5}\right) \div \left(\dfrac{32}{15} - \dfrac{8}{5}\right) = 7

Question 14

Seven equal pieces are made out of a rope of 215721\dfrac{5}{7} m. Find the length of each piece.

Answer

Length of rope = 2157 m=152721\dfrac{5}{7} \text{ m} = \dfrac{152}{7} m

Length of each piece = 1527÷7\dfrac{152}{7} ÷ 7

=1527×17=152×17×7=15249=3549= \dfrac{152}{7} \times \dfrac{1}{7} \\[1em] = \dfrac{152 \times 1}{7 \times 7} \\[1em] = \dfrac{152}{49} \\[1em] = 3\dfrac{5}{49}

Hence, the length of each piece is 35493\dfrac{5}{49} m.

Exercise 2(E)

Question 1(i)

Evaluate:

−23+34\dfrac{-2}{3} + \dfrac{3}{4}

Answer

By Division Method,

23,423,233,11,1\begin{array}{l|r} 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 3 and 4 = 2 × 2 × 3 = 12

Solving,

⇒−2×43×4+3×34×3=−812+912=−8+912=112\Rightarrow \dfrac{-2 \times 4}{3 \times 4} + \dfrac{3 \times 3}{4 \times 3} \\[1em] = \dfrac{-8}{12} + \dfrac{9}{12} \\[1em] = \dfrac{-8 + 9}{12} \\[1em] = \dfrac{1}{12}

Hence, −23+34=112\dfrac{-2}{3} + \dfrac{3}{4} = \dfrac{1}{12}

Question 1(ii)

Evaluate:

7−27+1118\dfrac{7}{-27} + \dfrac{11}{18}

Answer

By Division Method,

227,18327,939,333,11,1\begin{array}{l|r} 2 & 27, 18 \\ \hline 3 & 27, 9 \\ \hline 3 & 9, 3 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 27 and 18 is 2 × 3 × 3 × 3 = 54

Solving,

⇒7−27+1118=−727+1118=−7×227×2+11×318×3=−1454+3354=−14+3354=1954\Rightarrow \dfrac{7}{-27} + \dfrac{11}{18} \\[1em] = \dfrac{-7}{27} + \dfrac{11}{18} \\[1em] = \dfrac{-7 \times 2}{27 \times 2} + \dfrac{11 \times 3}{18 \times 3} \\[1em] = \dfrac{-14}{54} + \dfrac{33}{54} \\[1em] = \dfrac{-14 + 33}{54} \\[1em] = \dfrac{19}{54}

Hence, 7−27+1118=1954\dfrac{7}{-27} + \dfrac{11}{18} = \dfrac{19}{54}

Question 1(iii)

Evaluate:

−38+−512\dfrac{-3}{8} + \dfrac{-5}{12}

Answer

By Division Method,

28,1224,622,331,31,1\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 8 and 12 is 2 × 2 × 2 × 3 = 24

Solving,

⇒−3×38×3+−5×212×2=−924+−1024=−9+(−10)24=−1924\Rightarrow \dfrac{-3 \times 3}{8 \times 3} + \dfrac{-5 \times 2}{12 \times 2} \\[1em] = \dfrac{-9}{24} + \dfrac{-10}{24} \\[1em] = \dfrac{-9 + (-10)}{24} \\[1em] = \dfrac{-19}{24}

Hence, −38+−512=−1924\dfrac{-3}{8} + \dfrac{-5}{12} = \dfrac{-19}{24}

Question 1(iv)

Evaluate:

9−16+−5−12\dfrac{9}{-16} + \dfrac{-5}{-12}

Answer

By Division Method,

216,1228,624,322,331,31,1\begin{array}{l|r} 2 & 16, 12 \\ \hline 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 16 and 12 is 2 × 2 × 2 × 2 × 3 = 48

Solving,

⇒9−16+−5−12=−916+512=−9×316×3+5×412×4=−2748+2048=−27+2048=−748\Rightarrow \dfrac{9}{-16} + \dfrac{-5}{-12} \\[1em] = \dfrac{-9}{16} + \dfrac{5}{12} \\[1em] = \dfrac{-9 \times 3}{16 \times 3} + \dfrac{5 \times 4}{12 \times 4} \\[1em] = \dfrac{-27}{48} + \dfrac{20}{48} \\[1em] = \dfrac{-27 + 20}{48} \\[1em] = \dfrac{-7}{48}

Hence, 9−16+−5−12=−748\dfrac{9}{-16} + \dfrac{-5}{-12} = \dfrac{-7}{48}

Question 1(v)

Evaluate:

−59+−712+1118\dfrac{-5}{9} + \dfrac{-7}{12} + \dfrac{11}{18}

Answer

Solving,

By Division Method,

29,12,1829,6,939,3,933,1,31,1,1\begin{array}{l|r} 2 & 9, 12, 18 \\ \hline 2 & 9, 6, 9 \\ \hline 3 & 9, 3, 9 \\ \hline 3 & 3, 1, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 9, 12 and 18 is 2 × 2 × 3 × 3 = 36

⇒−5×49×4+−7×312×3+11×218×2=−2036+−2136+2236=−20+(−21)+2236=−1936\Rightarrow \dfrac{-5 \times 4}{9 \times 4} + \dfrac{-7 \times 3}{12 \times 3} + \dfrac{11 \times 2}{18 \times 2} \\[1em] = \dfrac{-20}{36} + \dfrac{-21}{36} + \dfrac{22}{36} \\[1em] = \dfrac{-20 + (-21) + 22}{36} \\[1em] = \dfrac{-19}{36}

Hence, −59+−712+1118=−1936\dfrac{-5}{9} + \dfrac{-7}{12} + \dfrac{11}{18} = \dfrac{-19}{36}

Question 1(vi)

Evaluate:

7−26+1639\dfrac{7}{-26} + \dfrac{16}{39}

Answer

By Division Method,

226,39313,391313,131,1\begin{array}{l|r} 2 & 26, 39 \\ \hline 3 & 13, 39 \\ \hline 13 & 13, 13 \\ \hline & 1, 1 \end{array}

LCM of 26 and 39 is 2 × 3 × 13 = 78

Solving,

⇒7−26+1639=−726+1639=−7×326×3+16×239×2=−2178+3278=−21+3278=1178\Rightarrow \dfrac{7}{-26} + \dfrac{16}{39} \\[1em] = \dfrac{-7}{26} + \dfrac{16}{39} \\[1em] = \dfrac{-7 \times 3}{26 \times 3} + \dfrac{16 \times 2}{39 \times 2} \\[1em] = \dfrac{-21}{78} + \dfrac{32}{78} \\[1em] = \dfrac{-21 + 32}{78} \\[1em] = \dfrac{11}{78}

Hence, 7−26+1639=1178\dfrac{7}{-26} + \dfrac{16}{39} = \dfrac{11}{78}

Question 1(vii)

Evaluate:

−23−(−57)-\dfrac{2}{3} - \left(\dfrac{-5}{7}\right)

Answer

By Division Method,

33,771,71,1\begin{array}{l|r} 3 & 3, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}

LCM of 3 and 7 is 3 × 7 = 21

Solving,

⇒−23−(−57)=−23+57=−2×73×7+5×37×3=−1421+1521=−14+1521=121\Rightarrow -\dfrac{2}{3} - \left(\dfrac{-5}{7}\right) \\[1em] = -\dfrac{2}{3} + \dfrac{5}{7} \\[1em] = \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{1}{21}

Hence, −23−(−57)=121-\dfrac{2}{3} - \left(\dfrac{-5}{7}\right) = \dfrac{1}{21}

Question 1(viii)

Evaluate:

−57−(−38)-\dfrac{5}{7} - \left(-\dfrac{3}{8}\right)

Answer

By Division Method,

27,827,427,277,11,1\begin{array}{l|r} 2 & 7, 8 \\ \hline 2 & 7, 4 \\ \hline 2 & 7, 2 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}

LCM of 7 and 8 is 2 × 2 × 2 × 7 = 56

Solving,

⇒−57−(−38)=−57+38=−5×87×8+3×78×7=−4056+2156=−40+2156=−1956\Rightarrow -\dfrac{5}{7} - \left(-\dfrac{3}{8}\right) \\[1em] = -\dfrac{5}{7} + \dfrac{3}{8} \\[1em] = \dfrac{-5 \times 8}{7 \times 8} + \dfrac{3 \times 7}{8 \times 7} \\[1em] = \dfrac{-40}{56} + \dfrac{21}{56} \\[1em] = \dfrac{-40 + 21}{56} \\[1em] = \dfrac{-19}{56}

Hence, −57−(−38)=−1956-\dfrac{5}{7} - \left(-\dfrac{3}{8}\right) = \dfrac{-19}{56}

Question 1(ix)

Evaluate:

726+2+−1113\dfrac{7}{26} + 2 + \dfrac{-11}{13}

Answer

By Division Method,

226,1,131313,1,131,1,1\begin{array}{l|r} 2 & 26, 1, 13 \\ \hline 13 & 13, 1, 13 \\ \hline & 1, 1, 1 \end{array}

LCM of 26, 1 and 13 is 2 × 13 = 26

Solving,

⇒726+2+−1113=726+21+−1113=7×126×1+2×261×26+−11×213×2=726+5226+−2226=7+52+(−22)26=3726=11126\Rightarrow \dfrac{7}{26} + 2 + \dfrac{-11}{13} \\[1em] = \dfrac{7}{26} + \dfrac{2}{1} + \dfrac{-11}{13} \\[1em] = \dfrac{7 \times 1}{26 \times 1} + \dfrac{2 \times 26}{1 \times 26} + \dfrac{-11 \times 2}{13 \times 2} \\[1em] = \dfrac{7}{26} + \dfrac{52}{26} + \dfrac{-22}{26} \\[1em] = \dfrac{7 + 52 + (-22)}{26} \\[1em] = \dfrac{37}{26} \\[1em] = 1\dfrac{11}{26}

Hence, 726+2+−1113=11126\dfrac{7}{26} + 2 + \dfrac{-11}{13} = 1\dfrac{11}{26}

Question 1(x)

Evaluate:

−1+2−3+56-1 + \dfrac{2}{-3} + \dfrac{5}{6}

Answer

By Division Method,

21,3,631,3,31,1,1\begin{array}{l|r} 2 & 1, 3, 6 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 1, 3 and 6 is 2 × 3 = 6

Solving,

⇒−1+2−3+56=−11+−23+56=−1×61×6+−2×23×2+5×16×1=−66+−46+56=−6+(−4)+56=−56\Rightarrow -1 + \dfrac{2}{-3} + \dfrac{5}{6} \\[1em] = \dfrac{-1}{1} + \dfrac{-2}{3} + \dfrac{5}{6} \\[1em] = \dfrac{-1 \times 6}{1 \times 6} + \dfrac{-2 \times 2}{3 \times 2} + \dfrac{5 \times 1}{6 \times 1} \\[1em] = \dfrac{-6}{6} + \dfrac{-4}{6} + \dfrac{5}{6} \\[1em] = \dfrac{-6 + (-4) + 5}{6} \\[1em] = \dfrac{-5}{6}

Hence, −1+2−3+56=−56-1 + \dfrac{2}{-3} + \dfrac{5}{6} = \dfrac{-5}{6}

Question 2

The sum of two rational numbers is −38\dfrac{-3}{8}. If one of them is 316\dfrac{3}{16}, find the other.

Answer

Let x be the other number.

⇒316+x=−38⇒x=−38−316\Rightarrow \dfrac{3}{16} + x = \dfrac{-3}{8} \\[1em] \Rightarrow x = \dfrac{-3}{8} - \dfrac{3}{16}

By Division Method,

28,1624,822,421,21,1\begin{array}{l|r} 2 & 8, 16 \\ \hline 2 & 4, 8 \\ \hline 2 & 2, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}

LCM of 8 and 16 is 2 × 2 × 2 × 2 = 16

⇒x=−3×28×2−3×116×1⇒x=−616−316⇒x=−6−316⇒x=−916\Rightarrow x = \dfrac{-3 \times 2}{8 \times 2} - \dfrac{3 \times 1}{16 \times 1} \\[1em] \Rightarrow x = \dfrac{-6}{16} - \dfrac{3}{16} \\[1em] \Rightarrow x = \dfrac{-6 - 3}{16} \\[1em] \Rightarrow x = \dfrac{-9}{16}

Hence, the other rational number is −916\dfrac{-9}{16}.

Question 3

The sum of two rational numbers is -5. If one of them is −5225\dfrac{-52}{25}, find the other.

Answer

Let x be the other number.

According to question,

⇒−5225+x=−5⇒x=−51−−5225\Rightarrow \dfrac{-52}{25} + x = -5 \\[1em] \Rightarrow x = \dfrac{-5}{1} - \dfrac{-52}{25}

By Division Method,

51,2551,51,1\begin{array}{l|r} 5 & 1, 25 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 1 and 25 is 5 × 5 = 25

⇒x=−5×251×25−−52×125×1⇒x=−12525−−5225⇒x=−125−(−52)25⇒x=−125+5225⇒x=−7325⇒x=−22325\Rightarrow x = \dfrac{-5 \times 25}{1 \times 25} - \dfrac{-52 \times 1}{25 \times 1} \\[1em] \Rightarrow x = \dfrac{-125}{25} - \dfrac{-52}{25} \\[1em] \Rightarrow x = \dfrac{-125 - (-52)}{25} \\[1em] \Rightarrow x = \dfrac{-125 + 52}{25} \\[1em] \Rightarrow x = \dfrac{-73}{25} \\[1em] \Rightarrow x = -2\dfrac{23}{25}

Hence, the other rational number is −22325-2\dfrac{23}{25}.

Question 4

What rational number should be added to −316-\dfrac{3}{16} to get 1124\dfrac{11}{24}?

Answer

Let x be added to −316-\dfrac{3}{16}.

⇒−316+x=1124⇒x=1124−(−316)⇒x=1124+316\Rightarrow -\dfrac{3}{16} + x = \dfrac{11}{24} \\[1em] \Rightarrow x = \dfrac{11}{24} - \left(-\dfrac{3}{16}\right) \\[1em] \Rightarrow x = \dfrac{11}{24} + \dfrac{3}{16}

By Division Method,

224,16212,826,423,233,11,1\begin{array}{l|r} 2 & 24, 16 \\ \hline 2 & 12, 8 \\ \hline 2 & 6, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 24 and 16 is 2 × 2 × 2 × 2 × 3 = 48

⇒x=11×224×2+3×316×3⇒x=2248+948⇒x=22+948⇒x=3148\Rightarrow x = \dfrac{11 \times 2}{24 \times 2} + \dfrac{3 \times 3}{16 \times 3} \\[1em] \Rightarrow x = \dfrac{22}{48} + \dfrac{9}{48} \\[1em] \Rightarrow x = \dfrac{22 + 9}{48} \\[1em] \Rightarrow x = \dfrac{31}{48}

Hence, 3148\dfrac{31}{48} should be added to −316-\dfrac{3}{16} to get 1124\dfrac{11}{24}.

Question 5

What rational number should be added to −35-\dfrac{3}{5} to get 2?

Answer

Let x be added to −35-\dfrac{3}{5}.

⇒−35+x=2⇒x=21−(−35)⇒x=21+35\Rightarrow -\dfrac{3}{5} + x = 2 \\[1em] \Rightarrow x = \dfrac{2}{1} - \left(-\dfrac{3}{5}\right) \\[1em] \Rightarrow x = \dfrac{2}{1} + \dfrac{3}{5}

LCM of 1 and 5 is 5.

⇒x=2×51×5+3×15×1⇒x=105+35⇒x=10+35⇒x=135⇒x=235\Rightarrow x = \dfrac{2 \times 5}{1 \times 5} + \dfrac{3 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \dfrac{10}{5} + \dfrac{3}{5} \\[1em] \Rightarrow x = \dfrac{10 + 3}{5} \\[1em] \Rightarrow x = \dfrac{13}{5} \\[1em] \Rightarrow x = 2\dfrac{3}{5}

Hence, 2352\dfrac{3}{5} should be added to −35-\dfrac{3}{5} to get 2.

Question 6

What rational number should be subtracted from −512-\dfrac{5}{12} to get 524\dfrac{5}{24}?

Answer

Let xx be subtracted from −512-\dfrac{5}{12}.

⇒−512−x=524⇒x=−512−524\Rightarrow -\dfrac{5}{12} - x = \dfrac{5}{24} \\[1em] \Rightarrow x = -\dfrac{5}{12} - \dfrac{5}{24}

By Division Method,

212,2426,1223,633,31,1\begin{array}{l|r} 2 & 12, 24 \\ \hline 2 & 6, 12 \\ \hline 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 12 and 24 is 2 × 2 × 2 × 3 = 24

⇒x=−5×212×2−5×124×1⇒x=−1024−524⇒x=−10−524⇒x=−1524⇒x=−58\Rightarrow x = \dfrac{-5 \times 2}{12 \times 2} - \dfrac{5 \times 1}{24 \times 1} \\[1em] \Rightarrow x = \dfrac{-10}{24} - \dfrac{5}{24} \\[1em] \Rightarrow x = \dfrac{-10 - 5}{24} \\[1em] \Rightarrow x = \dfrac{-15}{24} \\[1em] \Rightarrow x = \dfrac{-5}{8}

Hence, −58\dfrac{-5}{8} should be subtracted from −512-\dfrac{5}{12} to get 524\dfrac{5}{24}.

Question 7

What rational number should be subtracted from 58\dfrac{5}{8} to get 85\dfrac{8}{5}?

Answer

Let x be subtracted from 58\dfrac{5}{8}.

⇒58−x=85⇒x=58−85\Rightarrow \dfrac{5}{8} - x = \dfrac{8}{5} \\[1em] \Rightarrow x = \dfrac{5}{8} - \dfrac{8}{5}

By Division Method,

28,524,522,551,51,1\begin{array}{l|r} 2 & 8, 5 \\ \hline 2 & 4, 5 \\ \hline 2 & 2, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 8 and 5 is 2 × 2 × 2 × 5 = 40

⇒x=5×58×5−8×85×8⇒x=2540−6440⇒x=25−6440⇒x=−3940\Rightarrow x = \dfrac{5 \times 5}{8 \times 5} - \dfrac{8 \times 8}{5 \times 8} \\[1em] \Rightarrow x = \dfrac{25}{40} - \dfrac{64}{40} \\[1em] \Rightarrow x = \dfrac{25 - 64}{40} \\[1em] \Rightarrow x = \dfrac{-39}{40}

Hence, −3940\dfrac{-39}{40} should be subtracted from 58\dfrac{5}{8} to get 85\dfrac{8}{5}.

Question 8(i)

Evaluate:

(78×2421)+(−59×6−25)\left(\dfrac{7}{8} \times \dfrac{24}{21}\right) + \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right)

Answer

Solving,

⇒(78×2421)+(−59×6−25)=7×248×21+−5×69×(−25)=168168+−30−225=1+215\Rightarrow \left(\dfrac{7}{8} \times \dfrac{24}{21}\right) + \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) \\[1em] = \dfrac{7 \times 24}{8 \times 21} + \dfrac{-5 \times 6}{9 \times (-25)} \\[1em] = \dfrac{168}{168} + \dfrac{-30}{-225} \\[1em] = 1 + \dfrac{2}{15}

By Division Method,

31,1551,51,1\begin{array}{l|r} 3 & 1, 15 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 1 and 15 is 3 × 5 = 15

=1×151×15+2×115×1=1515+215=15+215=1715=1215= \dfrac{1 \times 15}{1 \times 15} + \dfrac{2 \times 1}{15 \times 1} \\[1em] = \dfrac{15}{15} + \dfrac{2}{15} \\[1em] = \dfrac{15 + 2}{15} \\[1em] = \dfrac{17}{15} \\[1em] = 1\dfrac{2}{15}

Hence, (78×2421)+(−59×6−25)=1215\left(\dfrac{7}{8} \times \dfrac{24}{21}\right) + \left(\dfrac{-5}{9} \times \dfrac{6}{-25}\right) = 1\dfrac{2}{15}

Question 8(ii)

Evaluate:

(815×−2516)+(−1835×56)\left(\dfrac{8}{15} \times \dfrac{-25}{16}\right) + \left(\dfrac{-18}{35} \times \dfrac{5}{6}\right)

Answer

Solving,

⇒(815×−2516)+(−1835×56)=8×(−25)15×16+−18×535×6=−200240+−90210=−56+−37\Rightarrow \left(\dfrac{8}{15} \times \dfrac{-25}{16}\right) + \left(\dfrac{-18}{35} \times \dfrac{5}{6}\right) \\[1em] = \dfrac{8 \times (-25)}{15 \times 16} + \dfrac{-18 \times 5}{35 \times 6} \\[1em] = \dfrac{-200}{240} + \dfrac{-90}{210} \\[1em] = \dfrac{-5}{6} + \dfrac{-3}{7}

By Division Method,

26,733,771,71,1\begin{array}{l|r} 2 & 6, 7 \\ \hline 3 & 3, 7 \\ \hline 7 & 1, 7 \\ \hline & 1, 1 \end{array}

LCM of 6 and 7 is 2 × 3 × 7 = 42

=−5×76×7+−3×67×6=−3542+−1842=−35+(−18)42=−5342=−11142= \dfrac{-5 \times 7}{6 \times 7} + \dfrac{-3 \times 6}{7 \times 6} \\[1em] = \dfrac{-35}{42} + \dfrac{-18}{42} \\[1em] = \dfrac{-35 + (-18)}{42} \\[1em] = \dfrac{-53}{42} \\[1em] = -1\dfrac{11}{42}

Hence, (815×−2516)+(−1835×56)=−11142\left(\dfrac{8}{15} \times \dfrac{-25}{16}\right) + \left(\dfrac{-18}{35} \times \dfrac{5}{6}\right) = -1\dfrac{11}{42}

Question 8(iii)

Evaluate:

(1833×−2227)−(1325×−7526)\left(\dfrac{18}{33} \times \dfrac{-22}{27}\right) - \left(\dfrac{13}{25} \times \dfrac{-75}{26}\right)

Answer

Solving,

⇒(1833×−2227)−(1325×−7526)=18×(−22)33×27−13×(−75)25×26=−396891−−975650=−49−(−32)=−49+32\Rightarrow \left(\dfrac{18}{33} \times \dfrac{-22}{27}\right) - \left(\dfrac{13}{25} \times \dfrac{-75}{26}\right) \\[1em] = \dfrac{18 \times (-22)}{33 \times 27} - \dfrac{13 \times (-75)}{25 \times 26} \\[1em] = \dfrac{-396}{891} - \dfrac{-975}{650} \\[1em] = \dfrac{-4}{9} - \left(\dfrac{-3}{2}\right) \\[1em] = \dfrac{-4}{9} + \dfrac{3}{2}

By Division Method,

29,239,133,11,1\begin{array}{l|r} 2 & 9, 2 \\ \hline 3 & 9, 1 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 9 and 2 is 2 × 3 × 3 = 18

=−4×29×2+3×92×9=−818+2718=−8+2718=1918=1118= \dfrac{-4 \times 2}{9 \times 2} + \dfrac{3 \times 9}{2 \times 9} \\[1em] = \dfrac{-8}{18} + \dfrac{27}{18} \\[1em] = \dfrac{-8 + 27}{18} \\[1em] = \dfrac{19}{18} \\[1em] = 1\dfrac{1}{18}

Hence, (1833×−2227)−(1325×−7526)=1118\left(\dfrac{18}{33} \times \dfrac{-22}{27}\right) - \left(\dfrac{13}{25} \times \dfrac{-75}{26}\right) = 1\dfrac{1}{18}

Question 8(iv)

Evaluate:

(−137×−3539)−(−745×914)\left(\dfrac{-13}{7} \times \dfrac{-35}{39}\right) - \left(\dfrac{-7}{45} \times \dfrac{9}{14}\right)

Answer

Solving,

⇒(−137×−3539)−(−745×914)=−13×(−35)7×39−−7×945×14=455273−−63630=53−(−110)=53+110\Rightarrow \left(\dfrac{-13}{7} \times \dfrac{-35}{39}\right) - \left(\dfrac{-7}{45} \times \dfrac{9}{14}\right) \\[1em] = \dfrac{-13 \times (-35)}{7 \times 39} - \dfrac{-7 \times 9}{45 \times 14} \\[1em] = \dfrac{455}{273} - \dfrac{-63}{630} \\[1em] = \dfrac{5}{3} - \left(\dfrac{-1}{10}\right) \\[1em] = \dfrac{5}{3} + \dfrac{1}{10}

By Division Method,

23,1033,551,51,1\begin{array}{l|r} 2 & 3, 10 \\ \hline 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}

LCM of 3 and 10 is 2 × 3 × 5 = 30

=5×103×10+1×310×3=5030+330=50+330=5330=12330= \dfrac{5 \times 10}{3 \times 10} + \dfrac{1 \times 3}{10 \times 3} \\[1em] = \dfrac{50}{30} + \dfrac{3}{30} \\[1em] = \dfrac{50 + 3}{30} \\[1em] = \dfrac{53}{30} \\[1em] = 1\dfrac{23}{30}

Hence, (−137×−3539)−(−745×914)=12330\left(\dfrac{-13}{7} \times \dfrac{-35}{39}\right) - \left(\dfrac{-7}{45} \times \dfrac{9}{14}\right) = 1\dfrac{23}{30}

Question 9

The product of two rational numbers is 24. If one of them is −3611\dfrac{-36}{11}, find the other.

Answer

Product of two rational numbers = 24

One of them = −3611\dfrac{-36}{11}

Let the other number = x

According to question,

x = 24÷−361124 ÷ \dfrac{-36}{11}

=241×11−36=24×111×(−36)=264−36=−223=−713= \dfrac{24}{1} \times \dfrac{11}{-36} \\[1em] = \dfrac{24 \times 11}{1 \times (-36)} \\[1em] = \dfrac{264}{-36} \\[1em] = \dfrac{-22}{3} \\[1em] = -7\dfrac{1}{3}

Hence, the other rational number is −713-7\dfrac{1}{3}.

Question 10

By what rational number should we multiply 20−9\dfrac{20}{-9}, so that the product is −59\dfrac{-5}{9}?

Answer

Let the required rational number be xx.

⇒20−9×x=−59⇒x=−59÷20−9⇒x=−59÷−209⇒x=−59×9−20⇒x=−5×99×(−20)⇒x=−45−180⇒x=14\Rightarrow \dfrac{20}{-9} \times x = \dfrac{-5}{9} \\[1em] \Rightarrow x = \dfrac{-5}{9} ÷ \dfrac{20}{-9} \\[1em] \Rightarrow x = \dfrac{-5}{9} ÷ \dfrac{-20}{9} \\[1em] \Rightarrow x = \dfrac{-5}{9} \times \dfrac{9}{-20} \\[1em] \Rightarrow x = \dfrac{-5 \times 9}{9 \times (-20)} \\[1em] \Rightarrow x = \dfrac{-45}{-180} \\[1em] \Rightarrow x = \dfrac{1}{4}

Hence, 20−9\dfrac{20}{-9} should be multiplied by 14\dfrac{1}{4} to get −59\dfrac{-5}{9}.

Question 11

State true or false:

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

(ii) −611-\dfrac{6}{11} is greater than 411\dfrac{4}{11}

(iii) −932+523=−9+532+23=−455-\dfrac{9}{32} + \dfrac{5}{23} = \dfrac{-9+5}{32+23} = \dfrac{-4}{55}

(iv) 1−315=−2151 - \dfrac{3}{15} = \dfrac{-2}{15}

Answer

(i) False.

The quotient of two integers is not always a rational number, because if the divisor (denominator) is 0, then the quotient is not defined and so it is not a rational number.

(ii) False.

−611-\dfrac{6}{11} is a negative rational number and 411\dfrac{4}{11} is a positive rational number. A negative rational number is always less than a positive rational number. So, −611<411-\dfrac{6}{11} \lt \dfrac{4}{11}.

(iii) False.

While adding rational numbers, we do not add the numerators and the denominators separately. Taking LCM of 32 and 23 as 736:

⇒−932+523=−9×2332×23+5×3223×32=−207736+160736=−47736.\Rightarrow -\dfrac{9}{32} + \dfrac{5}{23} \\[1em] = \dfrac{-9 \times 23}{32 \times 23} + \dfrac{5 \times 32}{23 \times 32} \\[1em] = \dfrac{-207}{736} + \dfrac{160}{736} \\[1em] = \dfrac{-47}{736}.

Thus, sum is not equal to −455\dfrac{-4}{55}.

(iv) False.

1−315=1515−315=15−315=1215=451 - \dfrac{3}{15} = \dfrac{15}{15} - \dfrac{3}{15} = \dfrac{15 - 3}{15} \\[1em] = \dfrac{12}{15} = \dfrac{4}{5}

which is not equal to −215\dfrac{-2}{15}.

Question 12

Find x, if:

(i) −58+x=712-\dfrac{5}{8} + x = \dfrac{7}{12}

(ii) 25+x=−2\dfrac{2}{5} + x = -2

(iii) 2+x=−232 + x = -\dfrac{2}{3}

(iv) 212x=33132\dfrac{1}{2}x = 33\dfrac{1}{3}

(v) −935x=35-\dfrac{9}{35}x = \dfrac{3}{5}

Answer

(i) Solving,

⇒−58+x=712⇒x=712+58\Rightarrow -\dfrac{5}{8} + x = \dfrac{7}{12} \\[1em] \Rightarrow x = \dfrac{7}{12} + \dfrac{5}{8}

By Division Method,

212,826,423,233,11,1\begin{array}{l|r} 2 & 12, 8 \\ \hline 2 & 6, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 12 and 8 = 2 × 2 × 2 × 3 = 24

⇒x=7×212×2+5×38×3⇒x=1424+1524⇒x=14+1524⇒x=2924⇒x=1524\Rightarrow x = \dfrac{7 \times 2}{12 \times 2} + \dfrac{5 \times 3}{8 \times 3} \\[1em] \Rightarrow x = \dfrac{14}{24} + \dfrac{15}{24} \\[1em] \Rightarrow x = \dfrac{14 + 15}{24} \\[1em] \Rightarrow x = \dfrac{29}{24} \\[1em] \Rightarrow x = 1\dfrac{5}{24}

Hence, x = 15241\dfrac{5}{24}

(ii) Solving,

⇒25+x=−2⇒x=−21−25\Rightarrow \dfrac{2}{5} + x = -2 \\[1em] \Rightarrow x = \dfrac{-2}{1} - \dfrac{2}{5}

LCM of 1 and 5 is 5.

⇒x=−2×51×5−2×15×1⇒x=−105−25⇒x=−10−25⇒x=−125⇒x=−225\Rightarrow x = \dfrac{-2 \times 5}{1 \times 5} - \dfrac{2 \times 1}{5 \times 1} \\[1em] \Rightarrow x = \dfrac{-10}{5} - \dfrac{2}{5} \\[1em] \Rightarrow x = \dfrac{-10 - 2}{5} \\[1em] \Rightarrow x = \dfrac{-12}{5} \\[1em] \Rightarrow x = -2\dfrac{2}{5}

Hence, x = −225-2\dfrac{2}{5}

(iii) Solving,

⇒2+x=−23⇒x=−23−21\Rightarrow 2 + x = -\dfrac{2}{3} \\[1em] \Rightarrow x = -\dfrac{2}{3} - \dfrac{2}{1}

LCM of 3 and 1 is 3.

⇒x=−2×13×1−2×31×3⇒x=−23−63⇒x=−2−63⇒x=−83⇒x=−223\Rightarrow x = \dfrac{-2 \times 1}{3 \times 1} - \dfrac{2 \times 3}{1 \times 3} \\[1em] \Rightarrow x = \dfrac{-2}{3} - \dfrac{6}{3} \\[1em] \Rightarrow x = \dfrac{-2 - 6}{3} \\[1em] \Rightarrow x = \dfrac{-8}{3} \\[1em] \Rightarrow x = -2\dfrac{2}{3}

Hence, x = −223-2\dfrac{2}{3}

(iv) Solving,

⇒212x=3313⇒52x=1003⇒x=1003÷52⇒x=1003×25⇒x=100×23×5⇒x=20015⇒x=403⇒x=1313\Rightarrow 2\dfrac{1}{2}x = 33\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{5}{2}x = \dfrac{100}{3} \\[1em] \Rightarrow x = \dfrac{100}{3} ÷ \dfrac{5}{2} \\[1em] \Rightarrow x = \dfrac{100}{3} \times \dfrac{2}{5} \\[1em] \Rightarrow x = \dfrac{100 \times 2}{3 \times 5} \\[1em] \Rightarrow x = \dfrac{200}{15} \\[1em] \Rightarrow x = \dfrac{40}{3} \\[1em] \Rightarrow x = 13\dfrac{1}{3}

Hence, x = 131313\dfrac{1}{3}

(v) Solving,

⇒−935x=35⇒x=35÷(−935)⇒x=35×35−9⇒x=3×355×(−9)⇒x=105−45⇒x=−73⇒x=−213\Rightarrow -\dfrac{9}{35}x = \dfrac{3}{5} \\[1em] \Rightarrow x = \dfrac{3}{5} ÷ \left(-\dfrac{9}{35}\right) \\[1em] \Rightarrow x = \dfrac{3}{5} \times \dfrac{35}{-9} \\[1em] \Rightarrow x = \dfrac{3 \times 35}{5 \times (-9)} \\[1em] \Rightarrow x = \dfrac{105}{-45} \\[1em] \Rightarrow x = \dfrac{-7}{3} \\[1em] \Rightarrow x = -2\dfrac{1}{3}

Hence, x = −213-2\dfrac{1}{3}

Question 13

Manish walks 89\dfrac{8}{9} km from a place P towards East. From there, he walks 2122\dfrac{1}{2} km towards West. Find his final position from the place P.

Answer

Let the distance covered towards East be positive and towards West be negative.

Distance walked towards East = 89\dfrac{8}{9} km

Distance walked towards West = 2122\dfrac{1}{2} km = 52\dfrac{5}{2} km

Final position from P = 89−52\dfrac{8}{9} - \dfrac{5}{2}

By Division Method,

29,239,133,11,1\begin{array}{l|r} 2 & 9, 2 \\ \hline 3 & 9, 1 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 9 and 2 is 2 × 3 × 3 = 18

=8×29×2−5×92×9=1618−4518=16−4518=−2918=−11118= \dfrac{8 \times 2}{9 \times 2} - \dfrac{5 \times 9}{2 \times 9} \\[1em] = \dfrac{16}{18} - \dfrac{45}{18} \\[1em] = \dfrac{16 - 45}{18} \\[1em] = \dfrac{-29}{18} \\[1em] = -1\dfrac{11}{18}

The negative sign shows the final position is towards West.

Hence, Manish is 111181\dfrac{11}{18} km towards West from the place P.

Question 14

State true or false:

(i) If pq\dfrac{p}{q} is a rational number and m is an integer, then pq=p×mq×m\dfrac{p}{q} = \dfrac{p \times m}{q \times m}.

(ii) If q is a positive integer and p and q are co-prime numbers, then pq\dfrac{p}{q} is a rational number.

Answer

(i) False.

The statement pq=p×mq×m\dfrac{p}{q} = \dfrac{p \times m}{q \times m} holds only when m is a non-zero integer. If m = 0, then p×mq×m=00\dfrac{p \times m}{q \times m} = \dfrac{0}{0}, which is not defined. So, the statement is not true for every integer m.

(ii) True.

Since q is a positive integer, q ≠ 0, and p is an integer. So pq\dfrac{p}{q} is of the form pq\dfrac{p}{q} where p and q are integers and q ≠ 0. Hence, pq\dfrac{p}{q} is a rational number.

Question 15

Find x such that −38 and x16-\dfrac{3}{8} \text { and }\dfrac{x}{16} are equivalent rational numbers.

Answer

Since −38 and x16-\dfrac{3}{8} \text { and }\dfrac{x}{16} are equivalent rational numbers,

⇒−38=x16⇒−3×16=8×x⇒−48=8x⇒x=−488⇒x=−6\Rightarrow -\dfrac{3}{8} = \dfrac{x}{16} \\[1em] \Rightarrow -3 \times 16 = 8 \times x \\[1em] \Rightarrow -48 = 8x \\[1em] \Rightarrow x = \dfrac{-48}{8} \\[1em] \Rightarrow x = -6

Hence, x = -6

Question 16

What should be added to (−176+−78)\left(-\dfrac{17}{6} + \dfrac{-7}{8}\right) to get -2?

Answer

First, find −176+−78-\dfrac{17}{6} + \dfrac{-7}{8}.

By Division Method,

26,823,423,233,11,1\begin{array}{l|r} 2 & 6, 8 \\ \hline 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}

LCM of 6 and 8 is 2 × 2 × 2 × 3 = 24

⇒−176+−78=−17×46×4+−7×38×3=−6824+−2124=−68+(−21)24=−8924\Rightarrow -\dfrac{17}{6} + \dfrac{-7}{8}\\[1em] = \dfrac{-17 \times 4}{6 \times 4} + \dfrac{-7 \times 3}{8 \times 3} \\[1em] = \dfrac{-68}{24} + \dfrac{-21}{24}\\[1em] = \dfrac{-68 + (-21)}{24} \\[1em] = \dfrac{-89}{24}

Let x be added to −8924\dfrac{-89}{24}.

⇒−8924+x=−2⇒x=−21−−8924\Rightarrow \dfrac{-89}{24} + x = -2 \\[1em] \Rightarrow x = \dfrac{-2}{1} - \dfrac{-89}{24}

By Division Method,

21,2421,1221,631,31,1\begin{array}{l|r} 2 & 1, 24 \\ \hline 2 & 1, 12 \\ \hline 2 & 1, 6 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}

LCM of 1 and 24 is 2 × 2 × 2 × 3 = 24

⇒x=−2×241×24−−8924⇒x=−4824−−8924⇒x=−48−(−89)24⇒x=−48+8924⇒x=4124⇒x=11724\Rightarrow x = \dfrac{-2 \times 24}{1 \times 24} - \dfrac{-89}{24} \\[1em] \Rightarrow x = \dfrac{-48}{24} - \dfrac{-89}{24} \\[1em] \Rightarrow x = \dfrac{-48 - (-89)}{24} \\[1em] \Rightarrow x = \dfrac{-48 + 89}{24} \\[1em] \Rightarrow x = \dfrac{41}{24} \\[1em] \Rightarrow x = 1\dfrac{17}{24}

Hence, 117241\dfrac{17}{24} should be added to (−176+−78)\left(-\dfrac{17}{6} + \dfrac{-7}{8}\right) to get -2.

Multiple Choice Questions

Question 1

The rational number equivalent to −23\dfrac{-2}{3} and with numerator -10 is:

  1. −103\dfrac{-10}{3}

  2. −1030\dfrac{-10}{30}

  3. −1015\dfrac{-10}{15}

  4. 1015\dfrac{10}{15}

Answer

To get numerator -10 from -2, we multiply by 5.

−23=−2×53×5=−1015\dfrac{-2}{3} = \dfrac{-2 \times 5}{3 \times 5} = \dfrac{-10}{15}

Hence, Option 3 is the correct option.

Question 2

The largest rational number out of −56,−1924 and 37−12\dfrac{-5}{6}, \dfrac{-19}{24} \text{ and } \dfrac{37}{-12} is:

  1. −56-\dfrac{5}{6}

  2. −1924\dfrac{-19}{24}

  3. 37−12\dfrac{37}{-12}

  4. none of these

Answer

By Division Method,

26,24,1223,12,623,6,333,3,31,1,1\begin{array}{l|r} 2 & 6, 24, 12 \\ \hline 2 & 3, 12, 6 \\ \hline 2 & 3, 6, 3 \\ \hline 3 & 3, 3, 3 \\ \hline & 1, 1, 1 \end{array}

LCM of 6, 24 and 12 is 2 × 2 × 2 × 3 = 24

Converting each rational number to denominator 24:

⇒−56=−5×46×4=−2024−1924=−192437−12=−3712=−37×212×2=−7424\Rightarrow \dfrac{-5}{6} = \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24} \\[1em] \dfrac{-19}{24} = \dfrac{-19}{24} \\[1em] \dfrac{37}{-12} = \dfrac{-37}{12} = \dfrac{-37 \times 2}{12 \times 2} = \dfrac{-74}{24}

Since −74<−20<−19-74 \lt -20 \lt -19, the largest rational number is −1924\dfrac{-19}{24}.

Hence, Option 2 is the correct option.

Question 3

−88−136\dfrac{-88}{-136} in standard form is:

  1. 88−13\dfrac{88}{-13}

  2. −88136\dfrac{-88}{136}

  3. 1117\dfrac{11}{17}

  4. −1117-\dfrac{11}{17}

Answer

−88−136=88136\dfrac{-88}{-136} = \dfrac{88}{136}

HCF of 88 and 136 is 8.

88136=88÷8136÷8=1117\dfrac{88}{136} = \dfrac{88 \div 8}{136 \div 8} = \dfrac{11}{17}

Hence, Option 3 is the correct option.

Question 4

Which of the following is true for the rational numbers −35-\dfrac{3}{5} and −13-\dfrac{1}{3}?

  1. −35>−13-\dfrac{3}{5} \gt -\dfrac{1}{3}

  2. −13>−35-\dfrac{1}{3} \gt -\dfrac{3}{5}

  3. none of these

Answer

By Division Method,

35,355,11,1\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}

LCM of 5 and 3 is 3 × 5 = 15

⇒−35=−3×35×3=−915⇒−13=−1×53×5=−515\Rightarrow -\dfrac{3}{5} = \dfrac{-3 \times 3}{5 \times 3} = \dfrac{-9}{15} \\[1em] \Rightarrow -\dfrac{1}{3} = \dfrac{-1 \times 5}{3 \times 5} = \dfrac{-5}{15}

Since -5 > -9, we have −515>−915\dfrac{-5}{15} \gt \dfrac{-9}{15}.

Hence, −13>−35-\dfrac{1}{3} \gt -\dfrac{3}{5}

Hence, Option 2 is the correct option.

Question 5

According to the number line given above, the values of A and B are: Concise Mathematics Solutions ICSE Class 7

According to the number line given above, the values of A and B are:

  1. A=−54 and B=73A = -\dfrac{5}{4} \text{ and } B = \dfrac{7}{3}

  2. A=−54 and B=−73A = -\dfrac{5}{4} \text{ and } B = -\dfrac{7}{3}

  3. A=−74 and B=73A = -\dfrac{7}{4} \text{ and } B = \dfrac{7}{3}

  4. A=74 and B=−73A = \dfrac{7}{4} \text{ and } B = -\dfrac{7}{3}

Answer

From the number line, A lies between -2 and -1 at the point −74-\dfrac{7}{4}, and B lies between 2 and 3 at the point 73\dfrac{7}{3}.

Hence, A=−74 and B=73A = -\dfrac{7}{4} \text{ and } B = \dfrac{7}{3}

Hence, Option 3 is the correct option.

Question 6

The rational number between −13 and 13-\dfrac{1}{3} \text { and }\dfrac{1}{3} is:

  1. 1

  2. -1

  3. 0

  4. −23-\dfrac{2}{3}

Answer

A rational number between −13 and 13-\dfrac{1}{3} \text { and } \dfrac{1}{3} is their mean.

12(−13+13)=12×0=0\dfrac{1}{2}\left(-\dfrac{1}{3} + \dfrac{1}{3}\right) = \dfrac{1}{2} \times 0 = 0

Hence, 0 lies between −13 and 13-\dfrac{1}{3} \text { and }\dfrac{1}{3}.

Hence, Option 3 is the correct option.

Question 7

A boy walks 23\dfrac{2}{3} km from a place P towards north and then from there 56\dfrac{5}{6} km towards south. The position of the boy from the place P is:

  1. 16\dfrac{1}{6} km towards north

  2. 16\dfrac{1}{6} km towards south

  3. 96\dfrac{9}{6} km towards north

  4. 96\dfrac{9}{6} km towards south

Answer

Taking the direction towards north as positive and south as negative:

By Division Method,

23,633,31,1\begin{array}{l|r} 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 3 and 6 is 2 × 3 = 6

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

The negative sign shows the boy is 16\dfrac{1}{6} km towards south of P.

Hence, Option 2 is the correct option.

Question 8

The rational number that should be added to −37 to get −57-\dfrac{3}{7} \text { to get } -\dfrac{5}{7} is:

  1. 87\dfrac{8}{7}

  2. −87-\dfrac{8}{7}

  3. −27-\dfrac{2}{7}

  4. 27\dfrac{2}{7}

Answer

Let x be added to −37-\dfrac{3}{7}.

⇒−37+x=−57⇒x=−57−(−37)⇒x=−57+37⇒x=−5+37⇒x=−27\Rightarrow -\dfrac{3}{7} + x = -\dfrac{5}{7} \\[1em] \Rightarrow x = -\dfrac{5}{7} - \left(-\dfrac{3}{7}\right) \\[1em] \Rightarrow x = \dfrac{-5}{7} + \dfrac{3}{7} \\[1em] \Rightarrow x = \dfrac{-5 + 3}{7} \\[1em] \Rightarrow x = \dfrac{-2}{7}

Hence, Option 3 is the correct option.

Question 9

The rational number that should be subtracted from −27 to get −57-\dfrac{2}{7}\text { to get }-\dfrac{5}{7} is:

  1. 37\dfrac{3}{7}

  2. −37-\dfrac{3}{7}

  3. 27\dfrac{2}{7}

  4. −27-\dfrac{2}{7}

Answer

Let x be subtracted from −27-\dfrac{2}{7}.

⇒−27−x=−57⇒x=−27−(−57)⇒x=−27+57⇒x=−2+57⇒x=37\Rightarrow -\dfrac{2}{7} - x = -\dfrac{5}{7} \\[1em] \Rightarrow x = -\dfrac{2}{7} - \left(-\dfrac{5}{7}\right) \\[1em] \Rightarrow x = \dfrac{-2}{7} + \dfrac{5}{7} \\[1em] \Rightarrow x = \dfrac{-2 + 5}{7} \\[1em] \Rightarrow x = \dfrac{3}{7}

Hence, Option 1 is the correct option.

Question 10

The sum of two rational numbers is 2. If one of them is −12-\dfrac{1}{2}; the other is:

  1. 52\dfrac{5}{2}

  2. −52-\dfrac{5}{2}

  3. 32\dfrac{3}{2}

  4. −32-\dfrac{3}{2}

Answer

Let x be the other number.

⇒−12+x=2⇒x=2−(−12)⇒x=21+12⇒x=2×21×2+12⇒x=42+12⇒x=52\Rightarrow -\dfrac{1}{2} + x = 2 \\[1em] \Rightarrow x = 2 - \left(-\dfrac{1}{2}\right) \\[1em] \Rightarrow x = \dfrac{2}{1} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{2 \times 2}{1 \times 2} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{4}{2} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{5}{2}

Hence, Option 1 is the correct option.

Question 11

If the cost of 1 m cloth is ₹ 205820\dfrac{5}{8}, then the cost of 2272\dfrac{2}{7} m is:

  1. ₹ (2058+227)\left(20\dfrac{5}{8} + 2\dfrac{2}{7}\right)

  2. ₹ (2058÷227)\left(20\dfrac{5}{8} \div 2\dfrac{2}{7}\right)

  3. ₹ (2058×227)\left(20\dfrac{5}{8} \times 2\dfrac{2}{7}\right)

  4. none of these

Answer

The cost of 2272\dfrac{2}{7} m of cloth is obtained by multiplying the cost of 1 m by the length.

Cost=2058×227\text{Cost} = 20\dfrac{5}{8} \times 2\dfrac{2}{7}

Hence, Option 3 is the correct option.

Question 12

The product of two rational numbers is -1. If one of them is 56\dfrac{5}{6}, then the other rational number is:

  1. 65\dfrac{6}{5}

  2. −65-\dfrac{6}{5}

  3. −56-\dfrac{5}{6}

  4. none of these

Answer

Let x be the other rational number.

⇒56×x=−1⇒x=−1÷56⇒x=−1×65⇒x=−65\Rightarrow \dfrac{5}{6} \times x = -1 \\[1em] \Rightarrow x = -1 \div \dfrac{5}{6} \\[1em] \Rightarrow x = -1 \times \dfrac{6}{5} \\[1em] \Rightarrow x = -\dfrac{6}{5}

Hence, Option 2 is the correct option.

Question 13

The distance covered in 2292\dfrac{2}{9} hours at the speed of 5255\dfrac{2}{5} km per hour is:

  1. (275+209)\left(\dfrac{27}{5} + \dfrac{20}{9}\right) km

  2. (275÷209)\left(\dfrac{27}{5} \div \dfrac{20}{9}\right) km

  3. (209÷275)\left(\dfrac{20}{9} \div \dfrac{27}{5}\right) km

  4. none of these

Answer

Distance = Speed × Time

Distance=525×229=275×209\text{Distance} = 5\dfrac{2}{5} \times 2\dfrac{2}{9} = \dfrac{27}{5} \times \dfrac{20}{9}

The distance is obtained by multiplication, which is not given in options 1, 2 or 3.

Hence, Option 4 is the correct option.

Question 14

2582\dfrac{5}{8} divided by 1161\dfrac{1}{6} gives:

  1. 49\dfrac{4}{9}

  2. 319243\dfrac{19}{24}

  3. 2142\dfrac{1}{4}

  4. none of these

Answer

⇒258÷116=218÷76=218×67=21×68×7=12656=94=214\Rightarrow 2\dfrac{5}{8} \div 1\dfrac{1}{6} = \dfrac{21}{8} \div \dfrac{7}{6} \\[1em] = \dfrac{21}{8} \times \dfrac{6}{7} \\[1em] = \dfrac{21 \times 6}{8 \times 7} \\[1em] = \dfrac{126}{56} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4}

Hence, Option 3 is the correct option.

Question 15

−13,−59 and −43-\dfrac{1}{3}, \dfrac{-5}{9} \text{ and }\dfrac{-4}{3} in ascending order are:

  1. −13<−59<−43-\dfrac{1}{3} \lt \dfrac{-5}{9} \lt \dfrac{-4}{3}

  2. −13<−43<−59-\dfrac{1}{3} \lt \dfrac{-4}{3} \lt \dfrac{-5}{9}

  3. −43<−59<−13\dfrac{-4}{3} \lt \dfrac{-5}{9} \lt -\dfrac{1}{3}

  4. −59<−43<−13\dfrac{-5}{9} \lt \dfrac{-4}{3} \lt -\dfrac{1}{3}

Answer

By Division Method,

33,9,331,3,11,1,1\begin{array}{l|r} 3 & 3, 9, 3 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}

LCM of 3 and 9 = 3 × 3 = 9

Converting each rational number to denominator 9:

⇒−13=−1×33×3=−39⇒−59=−59⇒−43=−4×33×3=−129\Rightarrow -\dfrac{1}{3} = \dfrac{-1 \times 3}{3 \times 3} = \dfrac{-3}{9} \\[1em] \Rightarrow \dfrac{-5}{9} = \dfrac{-5}{9} \\[1em] \Rightarrow \dfrac{-4}{3} = \dfrac{-4 \times 3}{3 \times 3} = \dfrac{-12}{9}

Comparing the numerators: -12 < -5 < -3

Hence, −43<−59<−13\dfrac{-4}{3} \lt \dfrac{-5}{9} \lt -\dfrac{1}{3}

Hence, Option 3 is the correct option.

Statement I-II Type Questions

Question 16

Statement 1: 1.6666.... can be expressed as a rational number.

Statement 2: A decimal number is said to be non-terminating but recurring decimal if in the decimal part a digit or a group of digits repeats itself again and again.

Which of the following options is correct?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Statement 1 is true, since 1.6666.... is a non-terminating recurring decimal which equals 53\dfrac{5}{3}, a rational number.

Statement 2 is true, as it correctly defines a non-terminating recurring decimal.

Hence, both the statements are true.

Hence, Option 1 is the correct option.

Assertion-Reason Type Questions

Question 17

Assertion (A): The product (multiplication) of two rational numbers is 38\dfrac{3}{8}. If one of them is −320-\dfrac{3}{20}, then the other rational number is −52-\dfrac{5}{2}.

Reason (R): The product of two rational numbers ab\dfrac{a}{b} and cd\dfrac{c}{d} is a×cb×d\dfrac{a \times c}{b \times d}.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Let x be the other rational number.

⇒−320×x=38⇒x=38÷(−320)⇒x=38×(−203)⇒x=3×(−20)8×3⇒x=−6024⇒x=−52\Rightarrow -\dfrac{3}{20} \times x = \dfrac{3}{8} \\[1em] \Rightarrow x = \dfrac{3}{8} \div \left(-\dfrac{3}{20}\right) \\[1em] \Rightarrow x = \dfrac{3}{8} \times \left(-\dfrac{20}{3}\right) \\[1em] \Rightarrow x = \dfrac{3 \times (-20)}{8 \times 3} \\[1em] \Rightarrow x = \dfrac{-60}{24} \\[1em] \Rightarrow x = -\dfrac{5}{2}

So, Assertion (A) is true. Reason (R) correctly states the rule for multiplication of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

Question 18

Assertion (A): On dividing the sum of 6512 and 83\dfrac{65}{12}\text { and }\dfrac{8}{3} by their difference, you get 9733\dfrac{97}{33}.

Reason (R): If pq and rs\dfrac{p}{q} \text{ and } \dfrac{r}{s} are two rational numbers such that rs≠0\dfrac{r}{s} \neq 0, then pq÷rs=pq×sr\dfrac{p}{q} \div \dfrac{r}{s} = \dfrac{p}{q} \times \dfrac{s}{r}.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

By Division Method,

212,326,333,31,1\begin{array}{l|r} 2 & 12, 3 \\ \hline 2 & 6, 3 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}

LCM of 12 and 3 is 2 × 2 × 3 = 12

Sum:

6512+83=6512+8×43×4=6512+3212=9712\dfrac{65}{12} + \dfrac{8}{3} = \dfrac{65}{12} + \dfrac{8 \times 4}{3 \times 4} = \dfrac{65}{12} + \dfrac{32}{12} = \dfrac{97}{12}

Difference:

6512−83=6512−3212=3312\dfrac{65}{12} - \dfrac{8}{3} = \dfrac{65}{12} - \dfrac{32}{12} = \dfrac{33}{12}

Dividing the sum by the difference:

9712÷3312=9712×1233=9733\dfrac{97}{12} \div \dfrac{33}{12} = \dfrac{97}{12} \times \dfrac{12}{33} = \dfrac{97}{33}

So, Assertion (A) is true. Reason (R) correctly states the rule for division of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

Question 19

Assertion (A): 1 and 0 are two co-prime integers, hence 10\dfrac{1}{0} is rational.

Reason (R): Every integer is a rational number. Its converse is also true.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

A rational number is of the form pq\dfrac{p}{q} where q≠0q \neq 0. Since 10\dfrac{1}{0} has denominator 0, it is not a rational number. So, Assertion (A) is false.

Every integer is a rational number, but its converse "every rational number is an integer" is not true (for example, 12\dfrac{1}{2} is rational but not an integer). So, Reason (R) is also false.

Hence, both A and R are false.

Hence, Option 4 is the correct option.

Question 20

Assertion (A): 34−51 and −23\dfrac{34}{-51} \text { and }\dfrac{-2}{3} are equivalent rational numbers.

Reason (R): If pq\dfrac{p}{q} is a rational number and n is a non-zero integer, then p×nq×n=pq\dfrac{p \times n}{q \times n} = \dfrac{p}{q}

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Reducing 34−51\dfrac{34}{-51} to standard form:

HCF of 34 and 51 is 17.

34−51=34÷17−51÷17=2−3=−23\dfrac{34}{-51} = \dfrac{34 \div 17}{-51 \div 17} = \dfrac{2}{-3} = \dfrac{-2}{3}

So, 34−51 and −23\dfrac{34}{-51} \text { and } \dfrac{-2}{3} are equivalent rational numbers and Assertion (A) is true. Reason (R) correctly states the property of equivalent rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

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