Rd Sharma 2019 2020 for Class 8 Math Chapter 1 – Rational Numbers

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Rd Sharma 2019 2020 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on ExamiumClasses’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Question 1:

Verify commutativty of addition of rational numbers for each of the following pairs of rotional numbers:
(i) $\frac{- 11}{5} and \frac{4}{7}$

(ii) $\frac{4}{9} and \frac{7}{- 12}$

(iii) $\frac{- 3}{5} and \frac{- 2}{- 15}$

(iv) $\frac{2}{- 7} and \frac{12}{- 35}$

(v) $4 and \frac{- 3}{5}$

(vi) $- 4 and \frac{4}{- 7}$

Answer:

$C ommutativity of t h e addition of rational numbers means t h a t if \frac{a}{b} a n d \frac{c}{d} are two rational numbers, then \frac{a}{b} + \frac{c}{d} = \frac{c}{d} + \frac{a}{b} . (i) We have \frac{- 11}{5} and \frac{4}{7} . ∴ \frac{- 11}{5} + \frac{4}{7} = \frac{- 11 \times 7}{5 \times 7} + \frac{4 \times 5}{7 \times 5} = \frac{- 77}{35} + \frac{20}{35} = \frac{- 77 + 20}{35} = \frac{- 57}{35} \frac{4}{7} + \frac{- 11}{5} = \frac{4 \times 5}{7 \times 5} + \frac{- 11 \times 7}{5 \times 7} = \frac{20}{35} + \frac{- 77}{35} = \frac{20 - 77}{35} = \frac{- 57}{35} ∴ \frac{- 11}{5} + \frac{4}{7} = \frac{4}{7} + \frac{- 11}{5} Hence verified . (ii) We have \frac{4}{9} and \frac{7}{- 12} . ∴ \frac{4}{9} + \frac{- 7}{12} = \frac{4 \times 4}{9 \times 4} + \frac{- 7 \times 3}{12 \times 3} = \frac{16}{36} + \frac{- 21}{36} = \frac{16 - 21}{36} = \frac{- 5}{36} \frac{- 7}{12} + \frac{4}{9} = \frac{- 7 \times 3}{12 \times 3} + \frac{4 \times 4}{9 \times 4} = \frac{- 21}{36} + \frac{16}{36} = \frac{- 21 + 16}{36} = \frac{- 5}{36} ∴ \frac{4}{9} + \frac{- 7}{12} = \frac{- 7}{12} + \frac{4}{9} Hence verified . (iii) We have \frac{- 3}{5} and \frac{- 2}{- 15} or \frac{- 3}{5} and \frac{2}{15} . ∴ \frac{- 3}{5} + \frac{2}{15} = \frac{- 9}{15} + \frac{2}{15} = \frac{- 9 + 2}{15} = \frac{- 7}{15} \frac{2}{15} + \frac{- 3}{5} = \frac{2}{15} + \frac{- 9}{15} = \frac{2 - 9}{15} = \frac{- 7}{15} ∴ \frac{- 3}{5} + \frac{- 2}{- 15} = \frac{- 2}{- 15} + \frac{- 3}{5} Hence verified . (iv) We have \frac{2}{- 7} and \frac{12}{- 35} . ∴ \frac{- 2}{7} + \frac{- 12}{35} = \frac{- 2 \times 5}{7 \times 5} + \frac{- 12}{35} = \frac{- 10 - 12}{35} = \frac{- 22}{35} \frac{12}{- 35} + \frac{2}{- 7} = \frac{- 12}{35} + \frac{- 2 \times 5}{7 \times 5} = \frac{- 12 - 10}{35} = \frac{- 22}{35} ∴ \frac{2}{- 7} + \frac{12}{- 35} = \frac{12}{- 35} + \frac{2}{- 7} Hence verified . (v) We have 4 and \frac{- 3}{5} . ∴ 4 + \frac{- 3}{5} = \frac{4 \times 5}{1 \times 5} + \frac{- 3}{5} = \frac{20 - 3}{5} = \frac{17}{5} \frac{- 3}{5} + 4 = \frac{- 3}{5} + \frac{4 \times 5}{1 \times 5} = \frac{- 3 + 20}{5} = \frac{17}{5} ∴ 4 + \frac{- 3}{5} = \frac{- 3}{5} + 4 Hence verified . (vi) We have - 4 and \frac{4}{- 7} . ∴ \frac{- 4}{1} + \frac{- 4}{7} = \frac{- 4 \times 7}{1 \times 7} + \frac{- 4}{7} = \frac{- 28 - 4}{7} = \frac{- 32}{7} \frac{- 4}{7} + \frac{- 4}{1} = \frac{- 4}{7} + \frac{- 4 \times 7}{1 \times 7} = \frac{- 4 - 28}{7} = \frac{- 32}{7} ∴ - 4 + \frac{4}{- 7} = \frac{4}{- 7} - 4 Hence verified .$

Question 2:

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
(i) $x = \frac{1}{2}, y = \frac{2}{3}, z = - \frac{1}{5}$
(ii) $x = \frac{- 2}{5}, y = \frac{4}{3}, z = \frac{- 7}{10}$
(iii) $x = \frac{- 7}{11}, y = \frac{2}{- 5}, z = \frac{- 3}{22}$
(iv) $x = - 2, y = \frac{3}{5}, z = \frac{- 4}{3}$

Answer:

$We have to verify that (x + y) + z = x + (y + z) . (i) x = \frac{1}{2}, y = \frac{2}{3}, z = \frac{- 1}{5} (x + y) + z = (\frac{1}{2} + \frac{2}{3}) + \frac{- 1}{5} = (\frac{3}{6} + \frac{4}{6}) + \frac{- 1}{5} = \frac{7}{6} + \frac{- 1}{5} = \frac{35}{30} + \frac{- 6}{30} = \frac{35 - 6}{30} = \frac{29}{30} x + (y + z) = \frac{1}{2} + (\frac{2}{3} + \frac{- 1}{5}) = \frac{1}{2} + (\frac{10}{15} + \frac{- 3}{15}) = \frac{1}{2} + \frac{7}{15} = \frac{15}{30} + \frac{14}{30} = \frac{15 + 14}{30} = \frac{29}{30} ∴ (\frac{1}{2} + \frac{2}{3}) + \frac{- 1}{5} = \frac{1}{2} + (\frac{2}{3} + \frac{- 1}{5}) Hence verified . (ii) x = \frac{- 2}{5}, y = \frac{4}{3}, z = \frac{- 7}{10} (x + y) + z = (\frac{- 2}{5} + \frac{4}{3}) + \frac{- 7}{10} = (\frac{- 6}{15} + \frac{20}{15}) + \frac{- 7}{10} = \frac{14}{15} + \frac{- 7}{10} = \frac{28}{30} + \frac{- 21}{30} = \frac{28 - 21}{30} = \frac{7}{30} x + (y + z) = \frac{- 2}{5} + (\frac{4}{3} + \frac{- 7}{10}) = \frac{- 2}{5} + (\frac{40}{30} + \frac{- 21}{30}) = \frac{- 2}{5} + \frac{19}{30} = \frac{- 12}{30} + \frac{19}{30} = \frac{- 12 + 19}{30} = \frac{7}{30} ∴ \frac{- 2}{5} + \frac{4}{3}) + \frac{- 7}{10} = \frac{- 2}{5} + (\frac{4}{3} + \frac{- 7}{10}) Hence verified . (iii) x = \frac{- 7}{11}, y = \frac{2}{- 5}, z = \frac{- 3}{22} (x + y) + z = (\frac{- 7}{11} + \frac{2}{- 5}) + \frac{- 3}{22} = (\frac{- 35}{55} + \frac{- 22}{55}) + \frac{- 3}{22} = \frac{- 57}{55} + \frac{- 3}{22} = \frac{- 114}{110} + \frac{- 15}{110} = \frac{- 114 - 15}{110} = \frac{- 129}{110} x + (y + z) = \frac{- 7}{11} + (\frac{2}{- 5} + \frac{- 3}{22}) = \frac{- 7}{11} + (\frac{- 44}{110} + \frac{- 15}{110}) = \frac{- 7}{11} + \frac{- 59}{110} = \frac{- 70}{110} + \frac{- 59}{110} = \frac{- 70 - 59}{110} = \frac{- 129}{110} ∴ (\frac{- 7}{11} + \frac{2}{- 5}) + \frac{- 3}{22} = \frac{- 7}{11} + (\frac{2}{- 5} + \frac{- 3}{22}) Hence verified . (iv) x = - 2, y = \frac{3}{5}, z = \frac{- 4}{3} so, (x + y) + z = (- 2 + \frac{3}{5}) + \frac{- 4}{3} = (\frac{- 10}{5} + \frac{3}{5}) + \frac{- 4}{3} = \frac{- 7}{5} + \frac{- 4}{3} = \frac{- 21}{15} + \frac{- 20}{15} = \frac{- 21 - 20}{15} = \frac{- 41}{15} x + (y + z) = - 2 + (\frac{3}{5} + \frac{- 4}{3}) = \frac{- 2}{1} + (\frac{9}{15} + \frac{- 20}{15}) = \frac{- 2}{1} + \frac{- 11}{15} = \frac{- 30}{15} + \frac{- 11}{15} = \frac{- 30 - 11}{15} = \frac{- 41}{15} ∴ (- 2 + \frac{3}{5}) + \frac{- 4}{3} = - 2 + (\frac{3}{5} + \frac{- 4}{3}) Hence verified .$

Question 3:

Write the additive inverse of each of the following rational numbers:
(i) $\frac{- 2}{17}$
(ii) $\frac{3}{- 11}$
(iii) $\frac{- 17}{5}$
(iv) $\frac{- 11}{- 25}$

Answer:

$(i) Additive inverse is the negative of the given number . So, additive inverse of \frac{- 2}{17} is \frac{2}{17} . (ii) Additive inverse is the negative of the given number . So, additive inverse of \frac{3}{- 11} is \frac{3}{11} . (iii) Additive inverse is the negative of the given number . So, additive inverse of \frac{- 17}{5} is \frac{17}{5} . (iv) Additive inverse is the negative of the given number . So, additive inverse of \frac{- 11}{- 25} is \frac{- 11}{25} .$

Question 4:

Write the negative (additive inverse) of each of the following:
(i) $\frac{- 2}{5}$
(ii) $\frac{7}{- 9}$
(iii) $\frac{- 16}{13}$
(iv) $\frac{- 5}{1}$
(v) 0
(vi) 1
(vii) −1

Answer:

$(i) Negative of \frac{- 2}{5} is \frac{2}{5} . (ii) Negative of \frac{7}{- 9} is \frac{7}{9} . (iii) Negative of \frac{- 16}{13} is \frac{16}{13} . (iv) Negative of \frac{- 5}{1} is \frac{5}{1} . (v) Negative value of 0 is 0 . (vi) Negative of 1 is - 1 . (vi) Negative of - 1 is 1 .$

Question 5:

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) $\frac{2}{5} + \frac{7}{3} + \frac{- 4}{5} + \frac{- 1}{3}$
(ii) $\frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9}$
(iii) $\frac{2}{5} + \frac{8}{3} + \frac{- 11}{15} + \frac{4}{5} + \frac{- 2}{3}$
(iv) $\frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{21}$

Answer:

$(i) We have: \frac{2}{5} + \frac{7}{3} + \frac{- 4}{5} + \frac{- 1}{3} = (\frac{2}{5} + \frac{- 4}{5}) + (\frac{7}{3} + \frac{- 1}{3}) = (\frac{2 - 4}{5}) + (\frac{7 - 1}{3}) = \frac{- 2}{5} + \frac{6}{3} = \frac{- 6 + 30}{15} = \frac{24}{15} = \frac{8}{5} (ii) We have: \frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9} = (\frac{3}{7} + \frac{- 11}{7}) + (\frac{- 4}{9} + \frac{7}{9}) = (\frac{3 - 11}{7}) + (\frac{- 4 + 7}{9}) = \frac{- 8}{7} + \frac{3}{9} = \frac{- 72 + 21}{63} = \frac{- 51}{63} = \frac{- 17}{21} (iii) We have: \frac{2}{5} + \frac{8}{3} + \frac{- 11}{15} + \frac{4}{5} + \frac{- 2}{3} = (\frac{2}{5} + \frac{4}{5}) + (+ \frac{8}{3} + \frac{- 2}{3}) + \frac{- 11}{15} = (\frac{2 + 4}{5}) + (\frac{8 - 2}{3}) + \frac{- 11}{15} = \frac{6}{5} + \frac{6}{3} + \frac{- 11}{15} = \frac{18 + 30 - 11}{15} = \frac{37}{15} (iv) We have: \frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{21} = (\frac{4}{7} + \frac{- 13}{7}) + (\frac{- 8}{9}) + \frac{17}{21} = (\frac{4 - 13}{7}) + (\frac{- 8}{9}) + \frac{17}{21} = \frac{- 9}{7} + \frac{- 8}{9} + \frac{17}{21} = \frac{- 81 - 56 + 51}{63} = \frac{- 86}{63}$

Question 6:

Re-arrange suitably and find the sum in each of the following:
(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$
(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$
(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$
(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$
(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$
(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

Answer:

$(i) (\frac{11}{2} + \frac{- 25}{2}) + \frac{- 17}{3} + \frac{11}{12} = (\frac{11 - 25}{2}) + \frac{- 17}{3} + \frac{11}{12} = \frac{- 14}{2} + \frac{- 17}{3} + \frac{11}{12} = - 7 + \frac{- 17}{3} + \frac{11}{12} = \frac{- 84 - 68 + 11}{12} = \frac{- 141}{12} = \frac{- 47}{4} (ii) (\frac{- 6}{7} + \frac{- 15}{7}) + \frac{- 5}{6} + \frac{- 4}{9} = (\frac{- 6 - 15}{7}) + \frac{- 5}{6} + \frac{- 4}{9} = \frac{- 21}{7} + \frac{- 5}{6} + \frac{- 4}{9} = \frac{- 378 - 105 - 56}{126} = \frac{- 539}{126} = \frac{- 77}{18} (iii) (\frac{3}{5} + \frac{9}{5}) + (\frac{- 7}{3} + \frac{7}{3}) + \frac{- 13}{15} = (\frac{3 + 9}{5}) + (\frac{- 7 + 7}{3}) + \frac{- 13}{15} = \frac{12}{5} + \frac{- 13}{15} = \frac{36 - 13}{15} = \frac{23}{15} (iv) (\frac{4}{13} - \frac{8}{13} + \frac{9}{13}) + \frac{- 5}{8} = (\frac{4 - 8 + 9}{13}) + \frac{- 5}{8} = \frac{5}{13} + \frac{- 5}{8} = \frac{40 - 65}{104} = \frac{- 25}{104} (v) (\frac{2}{3} + \frac{1}{3}) + (\frac{- 4}{5} + \frac{2}{5}) = \frac{3}{3} + \frac{- 2}{5} = \frac{15 - 6}{15} = \frac{9}{15} = \frac{3}{5} (vi) (\frac{1}{8} + \frac{- 5}{16}) + (\frac{5}{12} + \frac{7}{12}) + (\frac{2}{7} + \frac{9}{7}) = (\frac{2 - 5}{16}) + (\frac{5 + 7}{12}) + (\frac{2 + 9}{7}) = \frac{- 3}{16} + \frac{12}{12} + \frac{11}{7} = \frac{- 63 + 336 + 528}{336} = \frac{801}{336} = \frac{267}{112}$

Question 1:

Subtract the first rational number from the second in each of the following:
(i) $\frac{3}{8}, \frac{5}{8}$
(ii) $\frac{- 7}{9}, \frac{4}{9}$
(iii) $\frac{- 2}{11}, \frac{- 9}{11}$
(iv) $\frac{11}{13}, \frac{- 4}{13}$
(v) $\frac{1}{4}, \frac{- 3}{8}$
(vi) $\frac{- 2}{3}, \frac{5}{6}$
(vii) $\frac{- 6}{7}, \frac{- 13}{14}$
(viii) $\frac{- 8}{33}, \frac{- 7}{22}$

Answer:

$(i) \frac{5}{8} - \frac{3}{8} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4} (ii) \frac{4}{9} - \frac{- 7}{9} = \frac{4 - (- 7)}{9} = \frac{4 + 7}{9} = \frac{11}{9} (iii) \frac{- 9}{11} - \frac{- 2}{11} = \frac{(- 9) - (- 2)}{11} = \frac{- 9 + 2}{11} = \frac{- 7}{11} (iv) \frac{- 4}{13} - \frac{11}{13} = \frac{- 4 - 11}{13} = \frac{- 15}{13} (v) \frac{- 3}{8} - \frac{1}{4} = \frac{- 3 - 2}{8} = \frac{- 5}{8} (vi) \frac{5}{6} - \frac{- 2}{3} = \frac{5}{6} - \frac{- 4}{6} = \frac{5 - (- 4)}{6} = \frac{5 + 4}{6} = \frac{9}{6} = \frac{3}{2} (vii) \frac{- 13}{14} - \frac{- 6}{7} = \frac{- 13 - (- 12)}{14} = \frac{- 13 + 12}{14} = \frac{- 1}{14} (viii) \frac{- 7}{22} - \frac{- 8}{33} = \frac{- 21}{66} - \frac{- 16}{66} = \frac{(- 21) - (- 16)}{66} = \frac{- 21 + 16}{66} = \frac{- 5}{66}$

Question 2:

Evaluate each of the following:
(i) $\frac{2}{3} - \frac{3}{5}$
(ii) $- \frac{4}{7} - \frac{2}{- 3}$
(iii) $\frac{4}{7} - \frac{- 5}{- 7}$
(iv) $- 2 - \frac{5}{9}$
(v) $\frac{- 3}{- 8} - \frac{- 2}{7}$
(vi) $\frac{- 4}{13} - \frac{- 5}{26}$
(vii) $\frac{- 5}{14} - \frac{- 2}{7}$
(viii) $\frac{13}{15} - \frac{12}{25}$
(ix) $\frac{- 6}{13} - \frac{- 7}{13}$
(x) $\frac{7}{24} - \frac{19}{36}$
(xi) $\frac{5}{63} - \frac{- 8}{21}$

Answer:

$(i) \frac{2}{3} - \frac{3}{5} = \frac{10}{15} - \frac{9}{15} = \frac{10 - 9}{15} = \frac{1}{15} (ii) \frac{- 4}{7} - \frac{2}{- 3} = \frac{- 12}{21} - \frac{- 14}{21} = \frac{(- 12) - (- 14)}{21} = \frac{- 12 + 14}{21} = \frac{2}{21} (iii) \frac{4}{7} - \frac{- 5}{- 7} = \frac{4 - 5}{7} = \frac{- 1}{7} (iv) - 2 - \frac{5}{9} = \frac{- 18}{9} - \frac{5}{9} = \frac{- 18 - 5}{9} = \frac{- 23}{9} (v) \frac{- 3}{- 8} - \frac{- 2}{7} = \frac{21}{56} - \frac{- 16}{56} = \frac{21 - (- 16)}{56} = \frac{21 + 16}{56} = \frac{37}{56} (vi) \frac{- 4}{13} - \frac{- 5}{26} = \frac{- 8}{26} - \frac{- 5}{26} = \frac{- 8 - (- 5)}{26} = \frac{- 8 + 5}{26} = \frac{- 3}{26} (vii) \frac{- 5}{14} - \frac{- 2}{7} = \frac{- 5}{14} - \frac{- 4}{14} = \frac{- 5 - (- 4)}{14} = \frac{- 5 + 4}{14} = \frac{- 1}{14} (viii) \frac{13}{15} - \frac{12}{25} = \frac{65}{75} - \frac{36}{75} = \frac{65 - 36}{75} = \frac{29}{75} (ix) \frac{- 6}{13} - \frac{- 7}{13} = \frac{- 6 - (- 7)}{13} = \frac{- 6 + 7}{13} = \frac{1}{13} (x) \frac{7}{24} - \frac{19}{36} = \frac{21}{72} - \frac{38}{72} = \frac{21 - 38}{72} = \frac{- 17}{72} (xi) \frac{5}{63} - \frac{- 8}{21} = \frac{5}{63} - \frac{- 24}{63} = \frac{5 - (- 24)}{63} = \frac{5 + 24}{63} = \frac{29}{63}$

Question 3:

The sum of the two numbers is $\frac{5}{9} .$ If one of the numbers is $\frac{1}{3},$ find the other.

Answer:

$It is given that t h e sum of two numbers is \frac{5}{9}, where one of the number s is \frac{1}{3} . Let the other number be x . ∴ x + \frac{1}{3} = \frac{5}{9} \Rightarrow x = \frac{5}{9} - \frac{1}{3} \Rightarrow x = \frac{5}{9} - \frac{3}{9} \Rightarrow x = \frac{5 - 3}{9} \Rightarrow x = \frac{2}{9}$

Question 4:

The sum of two numbers is $\frac{- 1}{3} .$ If one of the numbers is $\frac{- 12}{3},$ find the other.

Answer:

$It is given that t h e sum of two numbers is \frac{- 1}{3}, where one of the number s is \frac{- 12}{3} . Let the other number b e x . ∴ x + \frac{- 12}{3} = \frac{- 1}{3} \Rightarrow x = \frac{- 1}{3} - \frac{- 12}{3} \Rightarrow x = \frac{- 1 - (- 12)}{3} = \frac{- 1 + 12}{3} = \frac{11}{3}$

Question 5:

The sum of two numbers is $\frac{- 4}{3} .$ If one of the numbers is −5, find the other.

Answer:

$It is given that the sum of two numbers is \frac{- 4}{3}, where one of the number s is - 5 . Let the other number be x . ∴ x + (- 5) = \frac{- 4}{3} \Rightarrow x = \frac{- 4}{3} - (\frac{- 5}{1}) \Rightarrow x = \frac{- 4}{3} - \frac{- 15}{3} \Rightarrow x = \frac{- 4 - (- 15)}{3} \Rightarrow x = \frac{- 4 + 15}{3} = \frac{11}{3}$

Question 6:

The sum of two rational numbers is −8. If one of the numbers is $\frac{- 15}{7},$ find the other.

Answer:

$It is given that t h e sum of two rational numbers is - 8, where one of the number s is \frac{- 15}{7} . Le t the other rational number be x . ∴ x + (\frac{- 15}{7}) = - 8 \Rightarrow x = \frac{- 8}{1} - \frac{- 15}{7} \Rightarrow x = \frac{- 56}{7} - \frac{- 15}{7} = \frac{- 56 - (- 15)}{7} = \frac{- 56 + 15}{7} = \frac{- 41}{7} Therefore, the other rational number is \frac{- 41}{7} .$

Question 7:

What should be added to $\frac{- 7}{8}$ so as to get $\frac{5}{9} ?$

Answer:

$Let x be added to \frac{- 7}{8} so as to get \frac{5}{9} . ∴ x + \frac{- 7}{8} = \frac{5}{9} \Rightarrow x = \frac{5}{9} - \frac{- 7}{8} \Rightarrow x = \frac{40}{72} - \frac{- 63}{72} \Rightarrow x = \frac{40 - (- 63)}{72} \Rightarrow x = \frac{40 + 63}{72} = \frac{103}{72}$

Question 8:

What number should be added to $\frac{- 5}{11}$ so as to get $\frac{26}{33} ?$

Answer:

$Let x be added . ∴ x + \frac{- 5}{11} = \frac{26}{33} \Rightarrow x = \frac{26}{33} - \frac{- 5}{11} \Rightarrow x = \frac{26}{33} - \frac{- 15}{33} \Rightarrow x = \frac{26 - (- 15)}{33} \Rightarrow x = \frac{26 + 15}{33} \Rightarrow x = \frac{41}{33}$

Question 9:

What number should be added to $\frac{- 5}{7}$ to get $\frac{- 2}{3} ?$

Answer:

$Let x be added . ∴ x + \frac{- 5}{7} = \frac{- 2}{3} \Rightarrow x = \frac{- 2}{3} - \frac{- 5}{7} \Rightarrow x = \frac{- 14}{21} - \frac{- 15}{21} \Rightarrow x = \frac{- 14 - (- 15)}{21} \Rightarrow x = \frac{- 14 + 15}{21} \Rightarrow x = \frac{1}{21}$

Question 10:

What number should be subtracted from $\frac{- 5}{3}$ to get $\frac{5}{6} ?$

Answer:

$Let x be subtracted . ∴ \frac{- 5}{3} - x = \frac{5}{6} \Rightarrow x = \frac{- 5}{3} - \frac{5}{6} \Rightarrow x = \frac{- 10}{6} - \frac{5}{6} \Rightarrow x = \frac{- 10 - 5}{6} = \frac{- 15}{6} = \frac{- 5}{2}$

Question 11:

What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4} ?$

Answer:

$Let, x be subtracted . ∴ \frac{3}{7} - x = \frac{5}{4} \Rightarrow x = \frac{3}{7} - \frac{5}{4} \Rightarrow x = \frac{12}{28} - \frac{35}{28} \Rightarrow x = \frac{12 - 35}{28} = \frac{- 23}{28}$

Question 12:

What should be added to $(\frac{2}{3} + \frac{3}{5})$ to get $\frac{- 2}{15} ?$

Answer:

$Let x be added . ∴ x + (\frac{2}{3} + \frac{3}{5}) = \frac{- 2}{15} \Rightarrow x + (\frac{10}{15} + \frac{9}{15}) = \frac{- 2}{15} \Rightarrow x + (\frac{10 + 9}{15}) = \frac{- 2}{15} \Rightarrow x = \frac{- 2}{15} - \frac{19}{15} \Rightarrow x = \frac{- 2 - 19}{15} \Rightarrow x = \frac{- 21}{15} \Rightarrow x = \frac{- 7}{5}$

Question 13:

What should be added to $(\frac{1}{2} + \frac{1}{3} + \frac{1}{5})$ to get 3?

Answer:

$Let x be added . ∴ x + (\frac{1}{2} + \frac{1}{3} + \frac{1}{5}) = 3 \Rightarrow x + (\frac{15}{30} + \frac{10}{30} + \frac{6}{30}) = 3 \Rightarrow x + (\frac{15 + 10 + 6}{30}) = 3 \Rightarrow x + \frac{31}{30} = 3 \Rightarrow x = \frac{3}{1} - \frac{31}{30} \Rightarrow x = \frac{90}{30} - \frac{31}{30} \Rightarrow x = \frac{90 - 31}{30} \Rightarrow x = \frac{59}{30}$

Question 14:

What should be subtracted from $(\frac{3}{4} - \frac{2}{3})$ to get $\frac{- 1}{6} ?$

Answer:

$Let x be subtracted . ∴ (\frac{3}{4} - \frac{2}{3}) - x = \frac{- 1}{6} \Rightarrow (\frac{9}{12} - \frac{8}{12}) - x = \frac{- 1}{6} \Rightarrow (\frac{9 - 8}{12}) - x = \frac{- 1}{6} \Rightarrow x = \frac{1}{12} - \frac{- 1}{6} \Rightarrow x = \frac{1}{12} - \frac{- 2}{12} \Rightarrow x = \frac{1 - (- 2)}{12} \Rightarrow x = \frac{1 + 2}{12} \Rightarrow x = \frac{3}{12} \Rightarrow x = \frac{1}{4}$

Question 15:

Fill in the blanks:
(i) $\frac{- 4}{13} - \frac{- 3}{26} = . . .$
(ii) $\frac{- 9}{14} + . . . = - 1$
(iii) $\frac{- 7}{9} + . . . = 3$
(iv) $. . . + \frac{15}{23} = 4$

Answer:

$(i) \frac{- 4}{13} - \frac{- 3}{26} = \frac{- 8}{26} - \frac{- 3}{26} = \frac{- 8 - (- 3)}{26} = \frac{- 8 + 3}{26} = \frac{- 5}{26} (ii) \frac{- 9}{14} + x = - 1 \Rightarrow x = - 1 - \frac{- 9}{14} \Rightarrow x = \frac{- 14}{14} - \frac{- 9}{14} \Rightarrow x = \frac{- 14 - (- 9)}{14} \Rightarrow x = \frac{- 14 + 9}{14} \Rightarrow x = \frac{- 5}{14} (iii) \frac{- 7}{9} + x = 3 \Rightarrow x = 3 - \frac{- 7}{9} \Rightarrow x = \frac{27}{9} - \frac{- 7}{9} \Rightarrow x = \frac{27 - (- 7)}{9} \Rightarrow x = \frac{27 + 7}{9} \Rightarrow x = \frac{34}{9} (iv) x + \frac{15}{23} = 4 \Rightarrow x = 4 - \frac{15}{23} \Rightarrow x = \frac{92}{23} - \frac{15}{23} \Rightarrow x = \frac{92 - 15}{23} \Rightarrow x = \frac{77}{23}$

Question 1:

Simplify each of the following and write as a rational number of the form $\frac{p}{q} :$
(i) $\frac{3}{4} + \frac{5}{6} + \frac{- 7}{8}$
(ii) $\frac{2}{3} + \frac{- 5}{6} + \frac{- 7}{9}$
(iii) $\frac{- 11}{2} + \frac{7}{6} + \frac{- 5}{8}$
(iv) $\frac{- 4}{5} + \frac{- 7}{10} + \frac{- 8}{15}$
(v) $\frac{- 9}{10} + \frac{22}{15} + \frac{13}{- 20}$
(vi) $\frac{5}{3} + \frac{3}{- 2} + \frac{- 7}{3} + 3$

Answer:

$(i) \frac{3}{4} + \frac{5}{6} + \frac{- 7}{8} = \frac{18}{24} + \frac{20}{24} + \frac{- 21}{24} = \frac{18 + 20 + (- 21)}{24} = \frac{18 + 20 - 21}{24} = \frac{17}{24} (ii) \frac{2}{3} + \frac{- 5}{6} + \frac{- 7}{9} = \frac{12}{18} + \frac{- 15}{18} + \frac{- 14}{18} = \frac{12 + (- 15) + (- 14)}{18} = \frac{12 - 15 - 14}{18} = \frac{- 17}{18} (iii) \frac{- 11}{2} + \frac{7}{6} + \frac{- 5}{8} = \frac{- 132}{24} + \frac{28}{24} + \frac{- 15}{24} = \frac{(- 132) + 28 + (- 15)}{24} = \frac{- 132 + 28 - 15}{24} = \frac{- 119}{24} (iv) \frac{- 4}{5} + \frac{- 7}{10} + \frac{- 8}{15} = \frac{- 24}{30} + \frac{- 21}{30} + \frac{- 16}{30} = \frac{(- 24) + (- 21) + (- 16)}{30} = \frac{- 24 - 21 - 16}{30} = \frac{- 61}{30} (v) \frac{- 9}{10} + \frac{22}{15} + \frac{13}{- 20} = \frac{- 54}{60} + \frac{88}{60} + \frac{- 39}{60} = \frac{(- 54) + 88 + (- 39)}{60} = \frac{- 54 + 88 - 39}{60} = \frac{- 5}{60} = \frac{- 1}{12} (vi) \frac{5}{3} + \frac{3}{- 2} + \frac{- 7}{3} + 3 = \frac{10}{6} + \frac{- 9}{6} + \frac{- 14}{6} + \frac{18}{6} = \frac{10 + (- 9) + (- 14) + 18}{6} = \frac{10 - 9 - 14 + 18}{6} = \frac{5}{6}$

Question 2:

Express each of the following as a rational number of the form $\frac{p}{q} :$
(i) $\frac{- 8}{3} + \frac{- 1}{4} + \frac{- 11}{6} + \frac{3}{8} - 3$
(ii) $\frac{6}{7} + 1 + \frac{- 7}{9} + \frac{19}{21} + \frac{- 12}{7}$
(iii) $\frac{15}{2} + \frac{9}{8} + \frac{- 11}{3} + 6 + \frac{- 7}{6}$
(iv) $\frac{- 7}{4} + 0 + \frac{- 9}{5} + \frac{19}{10} + \frac{11}{14}$
(v) $\frac{- 7}{4} + \frac{5}{3} + \frac{- 1}{2} + \frac{- 5}{6} + 2$

Answer:

$(i) \frac{- 8}{3} + \frac{- 1}{4} + \frac{- 11}{6} + \frac{3}{8} - 3 = \frac{- 64}{24} + \frac{- 6}{24} + \frac{- 44}{24} + \frac{9}{24} - \frac{72}{24} = \frac{(- 64) + (- 6) + (- 44) + 9 + (- 72)}{24} = \frac{- 64 - 6 - 44 + 9 - 72}{24} = \frac{- 177}{24} = \frac{- 59}{8} (ii) \frac{6}{7} + 1 + \frac{- 7}{9} + \frac{19}{21} + \frac{- 12}{7} = \frac{54}{63} + \frac{63}{63} + \frac{- 49}{63} + \frac{57}{63} + \frac{- 108}{63} = \frac{54 + 63 + (- 49) + 57 + (- 108)}{63} = \frac{54 + 63 - 49 + 57 - 108}{63} = \frac{17}{63} (iii) \frac{15}{2} + \frac{9}{8} + \frac{- 11}{3} + 6 + \frac{- 7}{6} = \frac{180}{24} + \frac{27}{24} + \frac{- 88}{24} + \frac{144}{24} + \frac{- 28}{24} = \frac{180 + 27 + (- 88) + 144 + (- 28)}{24} = \frac{180 + 27 - 88 + 144 - 28}{24} = \frac{235}{24} (iv) \frac{- 7}{4} + 0 + \frac{- 9}{5} + \frac{19}{10} + \frac{11}{14} = \frac{- 245}{140} + \frac{- 252}{140} + \frac{266}{140} + \frac{110}{140} = \frac{(- 245) + (- 252) + 266 + 110}{140} = \frac{- 245 - 252 + 266 + 110}{140} = \frac{- 121}{140} (v) \frac{- 7}{4} + \frac{5}{3} + \frac{- 1}{2} + \frac{- 5}{6} + 2 = \frac{- 21}{12} + \frac{20}{12} + \frac{- 6}{12} + \frac{- 10}{12} + \frac{24}{12} = \frac{(- 21) + 20 + (- 6) + (- 10) + 24}{12} = \frac{- 21 + 20 - 6 - 10 + 24}{12} = \frac{7}{12}$

Question 3:

Simplify:
(i) $\frac{- 3}{2} + \frac{5}{4} - \frac{7}{4}$

(ii) $\frac{5}{3} - \frac{7}{6} + \frac{- 2}{3}$

(iii) $\frac{5}{4} - \frac{7}{6} - \frac{- 2}{3}$

(iv) $\frac{- 2}{5} - \frac{- 3}{10} - \frac{- 4}{7}$

(v) $\frac{5}{6} + \frac{- 2}{5} - \frac{- 2}{15}$

(vi) $\frac{3}{8} - \frac{- 2}{9} + \frac{- 5}{36}$

Answer:

$(i) \frac{- 3}{2} + \frac{5}{4} - \frac{7}{4} Taking t h e L . C . M . of the denominators : \frac{- 6}{4} + \frac{5}{4} - \frac{7}{4} = \frac{- 6 + 5 - 7}{4} = \frac{- 8}{4} = - 2 (ii) \frac{5}{3} - \frac{7}{6} + \frac{- 2}{3} Taking t h e L . C . M . of the denominators : \frac{10}{6} - \frac{7}{6} + \frac{- 4}{6} = \frac{10 - 7 + (- 4)}{6} = \frac{10 - 7 - 4}{6} = \frac{- 1}{6} (iii) \frac{5}{4} - \frac{7}{6} - \frac{- 2}{3} Taking t h e L . C . M . of the denominators : \frac{15}{12} - \frac{14}{12} - \frac{- 8}{12} = \frac{15 - 14 - (- 8)}{12} = \frac{15 - 14 + 8}{12} = \frac{9}{12} = \frac{3}{4} (iv) \frac{- 2}{5} - \frac{- 3}{10} - \frac{- 4}{7} Taking t h e L . C . M . of the denominators : \frac{- 28}{70} - \frac{- 21}{70} - \frac{- 40}{70} = \frac{(- 28) - (- 21) - (- 40)}{70} = \frac{- 28 + 21 + 40}{70} = \frac{33}{70} (v) \frac{5}{6} + \frac{- 2}{5} - \frac{- 2}{15} Taking t h e L . C . M . of the denominators : \frac{25}{30} + \frac{- 12}{30} - \frac{- 4}{30} = \frac{25 + (- 12) - (- 4)}{30} = \frac{25 - 12 + 4}{30} = \frac{17}{30} (vi) \frac{3}{8} - \frac{- 2}{9} + \frac{- 5}{36} Taking t h e L . C . M . of the denominators : \frac{27}{72} - \frac{- 16}{72} + \frac{- 10}{72} = \frac{27 - (- 16) + (- 10)}{72} = \frac{27 + 16 - 10}{72} = \frac{33}{72} = \frac{11}{24}$

Question 1:

Multiply:
(i) $\frac{7}{11} by \frac{5}{4}$
(ii) $\frac{5}{7} by \frac{- 3}{4}$
(iii) $\frac{- 2}{9} by \frac{5}{11}$
(iv) $\frac{- 3}{17} by \frac{- 5}{- 4}$
(v) $\frac{9}{- 7} by \frac{36}{- 11}$
(vi) $\frac{- 11}{13} by \frac{- 21}{7}$
(vii) $- \frac{3}{5} by - \frac{4}{7}$
(viii) $- \frac{15}{11} by 7$

Answer:

$(i) \frac{7}{11} \times \frac{5}{4} = \frac{7 \times 5}{11 \times 4} = \frac{35}{44} (ii) \frac{5}{7} \times \frac{- 3}{4} = \frac{5 \times (- 3)}{7 \times 4} = \frac{- 15}{28} (iii) \frac{- 2}{9} \times \frac{5}{11} = \frac{(- 2) \times 5}{9 \times 11} = \frac{- 10}{99} (iv) \frac{- 3}{17} \times \frac{- 5}{- 4} = \frac{(- 3) \times (- 5)}{17 \times (- 4)} = \frac{15}{- 68} (v) \frac{9}{- 7} \times \frac{36}{- 11} = \frac{9 \times 36}{(- 7) \times (- 11)} = \frac{324}{77} (vi) \frac{- 11}{13} \times \frac{- 21}{7} = \frac{(- 11) \times (- 21)}{13 \times 7} = \frac{231}{91} = \frac{33}{13} (vii) \frac{- 3}{5} \times \frac{- 4}{7} = \frac{(- 3) \times (- 4)}{5 \times 7} = \frac{12}{35} (viii) \frac{- 15}{11} \times 7 = \frac{- 15}{11} \times \frac{7}{1} = \frac{(- 15) \times 7}{11 \times 1} = \frac{- 105}{11}$

Question 2:

Multiply:
(i) $\frac{- 5}{17} by \frac{51}{- 60}$

(ii) $\frac{- 6}{11} by \frac{- 55}{36}$

(iii) $\frac{- 8}{25} by \frac{- 5}{16}$

(iv) $\frac{6}{7} by \frac{- 49}{36}$

(v) $\frac{8}{- 9} by \frac{- 7}{- 16}$

(vi) $\frac{- 8}{9} by \frac{3}{64}$

Answer:

$(i) \frac{- 5}{17} \times \frac{51}{- 60} = \frac{- 5}{17} \times \frac{3 \times 17}{- 5 \times 3 \times 4} = \frac{- 5 \times 3 \times 17}{- 17 \times 5 \times 3 \times 4} = \frac{1}{4} (ii) \frac{- 6}{11} \times \frac{- 55}{36} = \frac{- 6}{11} \times \frac{- 5 \times 11}{6 \times 6} = \frac{5}{6} (iii) \frac{- 8}{25} \times \frac{- 5}{16} = \frac{- 1}{5 \times 5} \times \frac{- 5}{2} = \frac{1}{10} (iv) \frac{6}{7} \times \frac{- 49}{36} = \frac{6}{7} \times \frac{- 7 \times 7}{6 \times 6} = \frac{- 7}{6} (v) \frac{8}{- 9} \times \frac{- 7}{- 16} = \frac{1}{- 9} \times \frac{- 7}{- 2} = \frac{- 7}{18} (vi) \frac{- 8}{9} \times \frac{3}{64} = \frac{- 8}{3 \times 3} \times \frac{3}{8 \times 8} = \frac{- 1}{24}$

Question 3:

Simplify each of the following and express the result as a rational number in standard form:
(i) $\frac{- 16}{21} \times \frac{14}{5}$

(ii) $\frac{7}{6} \times \frac{- 3}{28}$

(iii) $\frac{- 19}{36} \times 16$

(iv) $\frac{- 13}{9} \times \frac{27}{- 26}$

(v) $\frac{- 9}{16} \times \frac{- 64}{- 27}$

(vi) $\frac{- 50}{7} \times \frac{14}{3}$

(vii) $\frac{- 11}{9} \times \frac{- 81}{- 88}$

(viii) $\frac{- 5}{9} \times \frac{72}{- 25}$

Answer:

$(i) \frac{- 16}{21} \times \frac{14}{5} = \frac{- 2 \times 2 \times 2 \times 2}{3 \times 7} \times \frac{2 \times 7}{5} = \frac{- 32}{15} (ii) \frac{7}{6} \times \frac{- 3}{28} = \frac{7}{2 \times 3} \times \frac{- 3}{2 \times 2 \times 7} = \frac{- 1}{8} (iii) \frac{- 19}{36} \times 16 = \frac{- 19}{2 \times 2 \times 3 \times 3} \times 2 \times 2 \times 2 \times 2 = \frac{- 76}{9} (iv) \frac{- 13}{9} \times \frac{27}{- 26} = \frac{- 13}{3 \times 3} \times \frac{3 \times 3 \times 3}{- 2 \times 13} = \frac{3}{2} (v) \frac{- 9}{16} \times \frac{- 64}{- 27} = \frac{- 9}{16} \times \frac{- 4 \times 16}{- 3 \times 9} = \frac{- 4}{3} (vi) \frac{- 50}{7} \times \frac{14}{3} = \frac{- 50}{7} \times \frac{2 \times 7}{3} = \frac{- 100}{3} (vii) \frac{- 11}{9} \times \frac{- 81}{- 88} = \frac{- 11}{9} \times \frac{- 9 \times 9}{- 8 \times 11} = \frac{- 9}{8} (viii) \frac{- 5}{9} \times \frac{72}{- 25} = \frac{- 5}{9} \times \frac{8 \times 9}{- 5 \times 5} = \frac{8}{5}$

Question 4:

Simplify:
(i) $(\frac{25}{8} \times \frac{2}{5}) - (\frac{3}{5} \times \frac{- 10}{9})$

(ii) $(\frac{1}{2} \times \frac{1}{4}) + (\frac{1}{2} \times 6)$

(iii) $(- 5 \times \frac{2}{15}) - (- 6 \times \frac{2}{9})$

(iv) $(\frac{- 9}{4} \times \frac{5}{3}) + (\frac{13}{2} \times \frac{5}{6})$

(v) $(\frac{- 4}{3} \times \frac{12}{- 5}) + (\frac{3}{7} \times \frac{21}{15})$

(vi) $(\frac{13}{5} \times \frac{8}{3}) - (\frac{- 5}{2} \times \frac{11}{3})$

(vii) $(\frac{13}{7} \times \frac{11}{26}) - (\frac{- 4}{3} \times \frac{5}{6})$

(viii) $(\frac{8}{5} \times \frac{- 3}{2}) + (\frac{- 3}{10} \times \frac{11}{16})$

Answer:

$(i) (\frac{25}{8} \times \frac{2}{5}) - (\frac{3}{5} \times \frac{- 10}{9}) = \frac{5}{4} - \frac{- 2}{3} = \frac{5 \times 3 + 2 \times 4}{12} = \frac{23}{12} (ii) (\frac{1}{2} \times \frac{1}{4}) + (\frac{1}{2} \times 6) = \frac{1}{8} + 3 = \frac{1 + 3 \times 8}{8} = \frac{25}{8} (iii) (- 5 \times \frac{2}{15}) - (- 6 \times \frac{2}{9}) = \frac{- 2}{3} - \frac{- 4}{3} = \frac{- 2 + 4}{3} = \frac{2}{3} (iv) (\frac{- 9}{4} \times \frac{5}{3}) + (\frac{13}{2} \times \frac{5}{6}) = \frac{- 3 \times 5}{4} + \frac{13 \times 5}{12} = \frac{- 15}{4} + \frac{65}{12} = \frac{- 15 \times 3 + 65}{12} = \frac{20}{12} = \frac{5}{3} (v) (\frac{- 4}{3} \times \frac{12}{- 5}) + (\frac{3}{7} \times \frac{21}{15}) = \frac{4 \times 4}{5} + \frac{1 \times 3}{5} = \frac{16}{5} + \frac{3}{5} = \frac{16 + 3}{5} = \frac{19}{5} (vi) (\frac{13}{5} \times \frac{8}{3}) - (\frac{- 5}{2} \times \frac{11}{3}) = \frac{13 \times 8}{15} - \frac{- 5 \times 11}{6} = \frac{104}{15} - \frac{- 55}{6} = \frac{104 \times 2 + 55 \times 5}{30} = \frac{483}{30} (vii) (\frac{13}{7} \times \frac{11}{26}) - (\frac{- 4}{3} \times \frac{5}{6}) = \frac{1 \times 11}{7 \times 2} - \frac{- 2 \times 5}{3 \times 3} = \frac{11}{14} - \frac{- 10}{9} = \frac{11 \times 9 + 10 \times 14}{126} = \frac{239}{126} (viii) (\frac{8}{5} \times \frac{- 3}{2}) + (\frac{- 3}{10} \times \frac{11}{16}) = \frac{4 \times (- 3)}{5} + \frac{- 3 \times 11}{10 \times 16} = \frac{- 12}{5} + \frac{- 33}{160} = \frac{- 12 \times 32 - 33}{160} = \frac{- 384 - 33}{160} = \frac{- 417}{160}$

Question 5:

Simplify:
(i) $(\frac{3}{2} \times \frac{1}{6}) + (\frac{5}{3} \times \frac{7}{2}) - (\frac{13}{8} \times \frac{4}{3})$

(ii) $(\frac{1}{4} \times \frac{2}{7}) - (\frac{5}{14} \times \frac{- 2}{3}) + (\frac{3}{7} \times \frac{9}{2})$

(iii) $(\frac{13}{9} \times \frac{- 15}{2}) + (\frac{7}{3} \times \frac{8}{5}) + (\frac{3}{5} \times \frac{1}{2})$

(iv) $(\frac{3}{11} \times \frac{5}{6}) - (\frac{9}{12} \times \frac{4}{3}) + (\frac{5}{13} \times \frac{6}{15})$

Answer:

$(i) (\frac{3}{2} \times \frac{1}{6}) + (\frac{5}{3} \times \frac{7}{2}) - (\frac{13}{8} \times \frac{4}{3}) = \frac{1}{4} + \frac{35}{6} - \frac{13}{6} = \frac{1 \times 3 + 35 \times 2 - 13 \times 2}{12} = \frac{3 + 70 - 26}{12} = \frac{47}{12} (ii) (\frac{1}{4} \times \frac{2}{7}) - (\frac{5}{14} \times \frac{- 2}{3}) + (\frac{3}{7} \times \frac{9}{2}) = \frac{1}{14} - \frac{- 5}{21} + \frac{27}{14} = \frac{1 \times 3 + 5 \times 2 + 27 \times 3}{42} = \frac{94}{42} = \frac{47}{21} (iii) (\frac{13}{9} \times \frac{- 15}{2}) + (\frac{7}{3} \times \frac{8}{5}) + (\frac{3}{5} \times \frac{1}{2}) = \frac{- 13 \times 5}{6} + \frac{7 \times 8}{15} + \frac{3}{10} = \frac{- 65}{6} + \frac{56}{15} + \frac{3}{10} = \frac{- 65 \times 5 + 56 \times 2 + 3 \times 3}{30} = \frac{- 204}{30} = \frac{- 34}{5} (iv) (\frac{3}{11} \times \frac{5}{6}) - (\frac{9}{12} \times \frac{4}{3}) + (\frac{5}{13} \times \frac{6}{15}) = \frac{5}{22} - 1 + \frac{2}{13} = \frac{5 \times 13 - 286 + 2 \times 22}{286} = \frac{65 - 286 + 44}{286} = \frac{- 177}{286}$

Question 1:

Verify the property: x × y = y × x by taking:
(i) $x = - \frac{1}{3}, y = \frac{2}{7}$
(ii) $x = \frac{- 3}{5}, y = \frac{- 11}{13}$
(iii) $x = 2, y = \frac{7}{- 8}$
(iv) $x = 0, y = \frac{- 15}{8}$

Answer:

$We have to verify that x \times y = y \times x . (i) x = \frac{- 1}{3}, y = \frac{2}{7} x \times y = \frac{- 1}{3} \times \frac{2}{7} = \frac{- 2}{21} y \times x = \frac{2}{7} \times \frac{- 1}{3} = \frac{- 2}{21} ∴ \frac{- 1}{3} \times \frac{2}{7} = \frac{2}{7} \times \frac{- 1}{3} Hence verified . (ii) x = \frac{- 3}{5}, y = \frac{- 11}{13} x \times y = \frac{- 3}{5} \times \frac{- 11}{13} = \frac{33}{65} y \times x = \frac{- 11}{13} \times \frac{- 3}{5} = \frac{33}{65} ∴ \frac{- 3}{5} \times \frac{- 11}{13} = \frac{- 11}{13} \times \frac{- 3}{5} Hence verified . (iii) x = 2, y = \frac{7}{- 8} x \times y = 2 \times \frac{7}{- 8} = \frac{7}{- 4} y \times x = \frac{7}{- 8} \times 2 = \frac{7}{- 4} ∴ 2 \times \frac{7}{- 8} = \frac{7}{- 8} \times 2 Hence verified . (iv) x = 0, y = \frac{- 15}{8} x \times y = 0 \times \frac{- 15}{8} = 0 y \times x = \frac{- 15}{8} \times 0 = 0 ∴ 0 \times \frac{- 15}{8} = \frac{- 15}{8} \times 0 Hence verified .$

Question 2:

Verify the property: x × (y × z) = (x × y) × z by taking:

(i) $x = \frac{- 7}{3}, y = \frac{12}{5}, z = \frac{4}{9}$

(ii) $x = 0, y = \frac{- 3}{5}, z = \frac{- 9}{4}$

(iii) $x = \frac{1}{2}, y = \frac{5}{- 4}, z = \frac{- 7}{5}$

(iv) $x = \frac{5}{7}, y = \frac{- 12}{13}, z = \frac{- 7}{18}$

Answer:

$We have to verify that x \times (y \times z) = (x \times y) \times z . (i) x = \frac{- 7}{3}, y = \frac{12}{5}, z = \frac{4}{9} x \times (y \times z) = \frac{- 7}{3} \times (\frac{12}{5} \times \frac{4}{9}) = \frac{- 7}{3} \times \frac{16}{15} = \frac{- 112}{45} (x \times y) \times z = (\frac{- 7}{3} \times \frac{12}{5}) \times \frac{4}{9} = \frac{- 28}{5} \times \frac{4}{9} = \frac{- 112}{45} ∴ \frac{- 7}{8} \times (\frac{15}{5} \times \frac{4}{9}) = (\frac{- 7}{8} \times \frac{15}{5}) \times \frac{4}{9} Hence verified . (ii) x = 0, y = \frac{- 3}{5}, z = \frac{- 9}{4} x \times (y \times z) = 0 \times (\frac{- 3}{5} \times \frac{- 9}{4}) = 0 \times \frac{27}{20} = 0 (x \times y) \times z = (0 \times \frac{- 3}{5}) \times \frac{- 9}{4} = 0 \times \frac{- 9}{4} = 0 ∴ 0 \times (\frac{- 3}{5} \times \frac{- 9}{4}) = (0 \times \frac{- 3}{5}) \times \frac{- 9}{4} Hence verified . (iii) x = \frac{1}{2}, y = \frac{5}{- 4}, z = \frac{- 7}{4} x \times (y \times z) = \frac{1}{2} \times (\frac{5}{- 4} \times \frac{- 7}{4}) = \frac{1}{2} \times \frac{35}{16} = \frac{35}{32} (x \times y) \times z = (\frac{1}{2} \times \frac{5}{- 4}) \times \frac{- 7}{4} = \frac{5}{- 8} \times \frac{- 7}{4} = \frac{35}{32} ∴ \frac{1}{2} \times (\frac{5}{- 4} \times \frac{- 7}{4}) = (\frac{1}{2} \times \frac{5}{- 4}) \times \frac{- 7}{4} Hence verified . (iv) x = \frac{5}{7}, y = \frac{- 12}{13}, z = \frac{- 7}{18} x \times (y \times z) = \frac{5}{7} \times (\frac{- 12}{13} \times \frac{- 7}{18}) = \frac{5}{7} \times \frac{14}{39} = \frac{10}{39} (x \times y) \times z = \frac{5}{7} \times \frac{- 12}{13}) \times \frac{- 7}{18} = \frac{- 60}{91} \times \frac{- 7}{18} = \frac{10}{39} ∴ \frac{5}{7} \times (\frac{- 12}{13} \times \frac{- 7}{18}) = (\frac{5}{7} \times \frac{- 12}{13}) \times \frac{- 7}{18} Hence verified .$

Question 3:

Verify the property: x × (y + z) = x × y + x × z by taking:

(i) $x = \frac{- 3}{7}, y = \frac{12}{13}, z = \frac{- 5}{6}$

(ii) $x = \frac{- 12}{5}, y = \frac{- 15}{4}, z = \frac{8}{3}$

(iii) $x = \frac{- 8}{3}, y = \frac{5}{6}, z = \frac{- 13}{12}$

(iv) $x = \frac{- 3}{4}, y = \frac{- 5}{2}, z = \frac{7}{6}$

Answer:

$We have to verify that x \times (y + z) = x \times y + x \times z . (i) x = \frac{- 3}{7}, y = \frac{12}{13}, z = \frac{- 5}{6} x \times (y + z) = \frac{- 3}{7} \times (\frac{12}{13} + \frac{- 5}{6}) = \frac{- 3}{7} \times \frac{72 - 65}{78} = \frac{- 3}{7} \times \frac{7}{78} = \frac{- 1}{26} x \times y + x \times z = \frac{- 3}{7} \times \frac{12}{13} + \frac{- 3}{7} \times \frac{- 5}{6} = \frac{- 36}{91} + \frac{5}{14} = \frac{- 36 \times 2 + 5 \times 13}{182} = \frac{- 72 + 65}{182} = \frac{- 1}{26} ∴ \frac{- 3}{7} \times (\frac{12}{13} + \frac{- 5}{6}) = \frac{- 3}{7} \times \frac{12}{13} + \frac{- 3}{7} \times \frac{- 5}{6} Hence verified . (ii) x = \frac{- 12}{5}, y = \frac{- 15}{4}, z = \frac{8}{3} x \times (y + z) = \frac{- 12}{5} \times (\frac{- 15}{4} + \frac{8}{3}) = \frac{- 12}{5} \times \frac{- 45 + 32}{12} = \frac{- 12}{5} \times \frac{- 13}{12} = \frac{13}{5} x \times y + x \times z = \frac{- 12}{5} \times \frac{- 15}{4} + \frac{- 12}{5} \times \frac{8}{3} = \frac{9}{1} + \frac{- 32}{5} = \frac{45 - 32}{5} = \frac{13}{5} ∴ \frac{- 12}{5} \times (\frac{- 15}{4} + \frac{8}{3}) = \frac{- 12}{5} \times \frac{- 15}{4} + \frac{- 12}{5} \times \frac{8}{3} Hence verified . (iii) x = \frac{- 8}{3}, y = \frac{5}{6}, z = \frac{- 13}{12} x \times (y + z) = \frac{- 8}{3} \times (\frac{5}{6} + \frac{- 13}{12}) = \frac{- 8}{3} \times \frac{10 - 13}{12} = \frac{- 8}{3} \times \frac{- 3}{12} = \frac{2}{3} x \times y + x \times z = \frac{- 8}{3} \times \frac{5}{6} + \frac{- 8}{3} \times \frac{- 13}{12} = \frac{- 20}{9} + \frac{26}{9} = \frac{- 20 + 26}{9} = \frac{2}{3} ∴ \frac{- 8}{3} \times (\frac{5}{6} + \frac{- 13}{12}) = \frac{- 8}{3} \times \frac{5}{6} + \frac{- 8}{3} \times \frac{- 13}{12} Hence verified . (iv) x = \frac{- 3}{4}, y = \frac{- 5}{2}, z = \frac{7}{6} x \times (y + z) = \frac{- 3}{4} \times (\frac{- 5}{2} + \frac{7}{6}) = \frac{- 3}{4} \times \frac{- 15 + 7}{6} = \frac{- 3}{4} \times \frac{- 8}{6} = 1 x \times y + x \times z = \frac{- 3}{4} \times \frac{- 5}{2} + \frac{- 3}{4} \times \frac{7}{6} = \frac{15}{8} + \frac{- 7}{8} = \frac{15 - 7}{8} = 1 ∴ \frac{- 3}{4} \times (\frac{- 5}{2} + \frac{7}{6}) = \frac{- 3}{4} \times \frac{- 5}{2} + \frac{- 3}{4} \times \frac{7}{6} Hence verified .$

Question 4:

Use the distributivity of multiplication of rational numbers over their addition to simplify:
(i) $\frac{3}{5} \times (\frac{35}{24} + \frac{10}{1})$

(ii) $\frac{- 5}{4} \times (\frac{8}{5} + \frac{16}{5})$

(iii) $\frac{2}{7} \times (\frac{7}{16} - \frac{21}{4})$

(iv) $\frac{3}{4} \times (\frac{8}{9} - 40)$

Answer:

$(i) \frac{3}{5} \times (\frac{35}{24} + \frac{10}{1}) = \frac{3}{5} \times \frac{35}{24} + \frac{3}{5} \times \frac{10}{1} = \frac{7}{8} + \frac{6}{1} = \frac{7 + 48}{8} = \frac{55}{8} (ii) \frac{- 5}{4} \times (\frac{8}{5} + \frac{16}{5}) = \frac{- 5}{4} \times \frac{8}{5} + \frac{- 5}{4} \times \frac{16}{5} = \frac{- 2}{1} + \frac{- 4}{1} = - 6 (iii) \frac{2}{7} \times (\frac{7}{16} - \frac{21}{4}) = \frac{2}{7} \times \frac{7}{16} - \frac{2}{7} \times \frac{21}{4} = \frac{1}{8} - \frac{3}{2} = \frac{1 - 12}{8} = \frac{- 11}{8} (iv) \frac{3}{4} \times (\frac{8}{9} - 40) = \frac{3}{4} \times \frac{8}{9} - \frac{3}{4} \times 40 = \frac{2}{3} - 30 = \frac{2 - 90}{3} = \frac{- 88}{3}$

Question 5:

Find the multiplicative inverse (reciprocal) of each of the following rational numbers:
(i) 9
(ii) −7
(iii) $\frac{12}{5}$
(iv) $\frac{- 7}{9}$
(v) $\frac{- 3}{- 5}$
(vi) $\frac{2}{3} \times \frac{9}{4}$
(vii) $\frac{- 5}{8} \times \frac{16}{15}$
(viii) $- 2 \times \frac{- 3}{5}$
(ix) −1
(x) $\frac{0}{3}$
(xi) 1

Answer:

$(i) Multiplicative inverse (reciprocal) of 9 = \frac{1}{9} (ii) Multiplicative inverse (reciprocal) of - 7 = \frac{- 1}{7} (iii) Multiplicative inverse (reciprocal) of \frac{12}{5} = \frac{5}{12} (iv) Multiplicative inverse (reciprocal) of \frac{- 7}{9} = \frac{- 9}{7} (v) Multiplicative inverse (reciprocal) of \frac{- 3}{- 5} = \frac{- 5}{- 3} o r \frac{5}{3} (vi) Multiplicative inverse (reciprocal) of \frac{2}{3} \times \frac{9}{4} = \frac{3}{2} \times \frac{4}{9} = \frac{2}{3} (vii) Multiplicative inverse (reciprocal) of \frac{- 5}{8} \times \frac{16}{15} = \frac{8}{- 5} \times \frac{15}{16} = \frac{- 3}{2} (viii) Multiplicative inverse (reciprocal) of - 2 \times \frac{- 3}{5} = \frac{1}{- 2} \times \frac{5}{- 3} = \frac{5}{6} (ix) Multiplicative inverse (reciprocal) of - 1 = \frac{1}{- 1} = - 1 (x) Multiplicative inverse (reciprocal) of \frac{0}{3} = \frac{3}{0} = undefined (ix) Multiplicative inverse (reciprocal) of 1 = \frac{1}{1} = 1$

Question 6:

Name the property of multiplication of rational numbers illustrated by the following statements:
(i) $\frac{- 5}{16} \times \frac{8}{15} = \frac{8}{15} \times \frac{- 5}{16}$

(ii) $\frac{- 17}{5} \times 9 = 9 \times \frac{- 17}{5}$

(iii) $\frac{7}{4} \times (\frac{- 8}{3} + \frac{- 13}{12}) = \frac{7}{4} \times \frac{- 8}{3} + \frac{7}{4} \times \frac{- 13}{12}$

(iv) $\frac{- 5}{9} \times (\frac{4}{15} \times \frac{- 9}{8}) = (\frac{- 5}{9} \times \frac{4}{15}) \times \frac{- 9}{8}$

(v) $\frac{13}{- 17} \times 1 = \frac{13}{- 17} = 1 \times \frac{13}{- 17}$

(vi) $\frac{- 11}{16} \times \frac{16}{- 11} = 1$

(vii) $\frac{2}{13} \times 0 = 0 = 0 \times \frac{2}{13}$

(viii) $\frac{- 3}{2} \times \frac{5}{4} + \frac{- 3}{2} \times \frac{- 7}{6} = \frac{- 3}{2} \times (\frac{5}{4} + \frac{- 7}{6})$

Answer:

(i) Commutative property
(ii) Commutative property
(iii) Distributivity of multiplication over addition
(iv) Associativity of multiplication
(v) Existence of identity for multiplication
(vi) Existence of multiplicative inverse
(vii) Multiplication by 0
(viii) Distributive property

Question 7:

Fill in the blanks:
(i) The product of two positive rational numbers is always …..
(ii) The product of a positive rational number and a negative rational number is always …..
(iii) The product of two negative rational numbers is always …..
(iv) The reciprocal of a positive rational number is …..
(v) The reciprocal of a negative rational number is …..
(vi) Zero has ….. reciprocal.
(vii) The product of a rational number and its reciprocal is …..
(viii) The numbers ….. and ….. are their own reciprocals.
(ix) If a is reciprocal of b, then the reciprocal of b is …..
(x) The number 0 is ….. the reciprocal of any number.
(xi) Reciprocal of $\frac{1}{a}, a \neq 0$ is …..
(xii) (17 × 12)⁻¹ = 17⁻¹ × …..

Answer:

(i) Positive
(ii) Negative
(iii) Positive
(iv) Positive
(v) Negative
(vi) No
(vii) 1
(viii) -1 and 1
(ix) a
(x) not
(xi) a
(xii) $12^{- 1}$

Question 8:

Fill in the blanks:
(i) $- 4 \times \frac{7}{9} = \frac{7}{9} \times . . . . . .$
(ii) $\frac{5}{11} \times \frac{- 3}{8} = \frac{- 3}{8} \times . . . . . .$
(iii) $\frac{1}{2} \times (\frac{3}{4} + \frac{- 5}{12}) = \frac{1}{2} \times . . . . . . + . . . . . . \times \frac{- 5}{12}$
(iv) $\frac{- 4}{5} \times (\frac{5}{7} + \frac{- 8}{9}) = (\frac{- 4}{5} \times . . . . .) \times \frac{- 8}{9}$

Answer:

$(i) - 4 x \times y = y \times x (commutativity) (ii) \frac{5}{11} x \times y = y \times x (commutativity) (iii) \frac{3}{4}; \frac{1}{2} x \times (y + z) = x \times y + x \times z (d istributivity of multiplication over addition) (iv) \frac{5}{7} x \times (y \times z) = (x \times y) \times z (a ssociativity of multiplication)$

Question 1:

Divide:
(i) $1 by \frac{1}{2}$

(ii) $5 by \frac{- 5}{7}$

(iii) $\frac{- 3}{4} by \frac{9}{- 16}$

(iv) $\frac{- 7}{8} by \frac{- 21}{16}$

(v) $\frac{7}{- 4} by \frac{63}{64}$

(vi) $0 by \frac{- 7}{5}$

(vii) $\frac{- 3}{4} by - 6$

(viii) $\frac{2}{3} by \frac{- 7}{12}$

(ix) $- 4 by \frac{- 3}{5}$

(x) $\frac{- 3}{13} by \frac{- 4}{65}$

Answer:

$(i) 1 \div \frac{1}{2} = 1 \times \frac{2}{1} = 2 (ii) 5 \div \frac{- 5}{7} = 5 \times \frac{7}{- 5} = - 7 (iii) \frac{- 3}{4} \div \frac{9}{- 16} = \frac{- 3}{4} \times \frac{- 16}{9} = \frac{4}{3} (iv) \frac{- 7}{8} \div \frac{- 21}{16} = \frac{- 7}{8} \times \frac{16}{- 21} = \frac{2}{3} (v) \frac{7}{- 4} \div \frac{63}{64} = \frac{7}{- 4} \times \frac{64}{63} = \frac{- 16}{9} (vi) 0 \div \frac{- 7}{5} = 0 \times \frac{5}{- 7} = 0 (vii) \frac{- 3}{4} \div - 6 = \frac{- 3}{4} \times \frac{1}{- 6} = \frac{1}{8} (viii) \frac{2}{3} \div \frac{- 7}{12} = \frac{2}{3} \times \frac{12}{- 7} = \frac{- 8}{7} (ix) - 4 \div \frac{- 3}{5} = - 4 \times \frac{5}{- 3} = \frac{20}{3} (x) \frac{- 3}{13} \div \frac{- 4}{65} = \frac{- 3}{13} \times \frac{65}{- 4} = \frac{15}{4}$

Question 2:

Find the value and express as a rational number in standard form:
(i) $\frac{2}{5} \div \frac{26}{15}$

(ii) $\frac{10}{3} \div \frac{- 35}{12}$

(iii) $- 6 \div (\frac{- 8}{17})$

(iv) $\frac{- 40}{99} \div (- 20)$

(v) $\frac{- 22}{27} \div \frac{- 110}{18}$

(vi) $\frac{- 36}{125} \div \frac{- 3}{75}$

Answer:

$(i) \frac{2}{5} \div \frac{26}{15} = \frac{2}{5} \times \frac{15}{26} = \frac{3}{13} (ii) \frac{10}{3} \div \frac{- 35}{12} = \frac{10}{3} \times \frac{12}{- 35} = \frac{- 8}{7} (iii) - 6 \div \frac{- 8}{17} = - 6 \times \frac{17}{- 8} = \frac{51}{4} (iv) \frac{- 40}{99} \div (- 20) = \frac{- 40}{99} \times \frac{1}{- 20} = \frac{2}{99} (v) \frac{- 22}{27} \div \frac{- 110}{18} = \frac{- 22}{27} \times \frac{18}{- 110} = \frac{2}{15} (vi) \frac{- 36}{125} \div \frac{- 3}{75} = \frac{- 36}{125} \times \frac{75}{- 3} = \frac{36}{5}$

Question 3:

The product of two rational numbers is 15. If one of the numbers is −10, find the other.

Answer:

$Let the other number b e x . ∴ x \times (- 10) = 15 or x = \frac{15}{- 10} = \frac{3}{- 2} So, the other number is \frac{- 3}{2} .$

Question 4:

The product of two rational numbers is $\frac{- 8}{9} .$ If one of the numbers is $\frac{- 4}{15},$ find the other.

Answer:

$Let the other number b e x . ∴ x \times \frac{- 4}{15} = \frac{- 8}{9} or x = \frac{- 8}{9} \div \frac{- 4}{15} or x = \frac{- 8}{9} \times \frac{15}{- 4} or x = \frac{10}{3} T h u s, the other number is \frac{10}{3} .$

Question 5:

By what number should we multiply $\frac{- 1}{6}$ so that the product may be $\frac{- 23}{9} ?$

Answer:

$Let the number b e x . ∴ x \times \frac{- 1}{6} = \frac{- 23}{9} x = \frac{- 23}{9} \div \frac{- 1}{6} x = \frac{- 23}{9} \times \frac{6}{- 1} = \frac{46}{3} Therefore, the other number is \frac{46}{3} .$

Question 6:

By what number should we multiply $\frac{- 15}{28}$ so that the product may be $\frac{- 5}{7} ?$

Answer:

$Let the other number b e x . ∴ x \times \frac{- 15}{28} = \frac{- 5}{7} or x = \frac{- 5}{7} \div \frac{- 15}{28} or x = \frac{- 5}{7} \times \frac{28}{- 15} or x = \frac{4}{3} T h u s, the other number is \frac{4}{3} .$

Question 7:

By what number should we multiply $\frac{- 8}{13}$ so that the product may be 24?

Answer:

$Let the number b e x . ∴ x \times \frac{- 8}{13} = 24 or x = 24 \div \frac{- 8}{13} or x = 24 \times \frac{13}{- 8} or x = - 39 T h u s, the number is - 39 .$

Question 8:

By what number should $\frac{- 3}{4}$ be multiplied in order to produce $\frac{2}{3} ?$

Answer:

$Let the other number that should be multiplied with \frac{- 3}{4} to produce \frac{2}{3} b e x . ∴ x \times \frac{- 3}{4} = \frac{2}{3} or x = \frac{2}{3} \div \frac{- 3}{4} or x = \frac{2}{3} \times \frac{4}{- 3} or x = \frac{- 8}{9} T h u s, the number is \frac{- 8}{9} .$

Question 9:

Find (x + y) ÷ (x − y), if
(i) $x = \frac{2}{3}, y = \frac{3}{2}$
(ii) $x = \frac{2}{5}, y = \frac{1}{2}$
(iii) $x = \frac{5}{4}, y = \frac{- 1}{3}$
(iv) $x = \frac{2}{7}, y = \frac{4}{3}$
(v) $x = \frac{1}{4}, y = \frac{3}{2}$

Answer:

Question 10:

The cost of $7 \frac{2}{3}$ metres of rope is Rs $12 \frac{3}{4} .$ Find its cost per metre.

Answer:

$The cost of 7 \frac{2}{3} metres of rope is Rs 12 \frac{3}{4} . ∴ C ost per metre = 12 \frac{3}{4} \div 7 \frac{2}{3} = \frac{51}{4} \div \frac{23}{3} = \frac{51}{4} \times \frac{3}{23} = \frac{153}{92} = Rs 1 \frac{61}{92}$

Question 11:

The cost of $2 \frac{1}{3}$ metres of cloth is Rs $75 \frac{1}{4} .$ Find the cost of cloth per metre.

Answer:

$The cost of 2 \frac{1}{3} metres of cloth is Rs 75 \frac{1}{4} . ∴ C ost per metre = 75 \frac{1}{4} \div 2 \frac{1}{3} = \frac{301}{4} \div \frac{7}{3} = \frac{301}{4} \times \frac{3}{7} = \frac{129}{4} = Rs 32 \frac{1}{4} T h u s, Rs 32 \frac{1}{4} or Rs 32.25 is the cost of cloth per metre .$

Question 12:

By what number should $\frac{- 33}{16}$ be divided to get $\frac{- 11}{4} ?$

Answer:

$Let the number b e x . ∴ \frac{- 33}{16} \div x = \frac{- 11}{4} or \frac{- 33}{16} \times \frac{1}{x} = \frac{- 11}{4} or \frac{1}{x} = \frac{- 11}{4} \times \frac{16}{- 33} or \frac{1}{x} = \frac{4}{3} or x = \frac{3}{4} T h u s, the number is \frac{3}{4} .$

Question 13:

Divide the sum of $\frac{- 13}{5}$ and $\frac{12}{7}$ by the product of $\frac{- 31}{7} and \frac{- 1}{2} .$

Answer:

$(\frac{- 13}{5} + \frac{12}{7}) \div (\frac{- 31}{7} \times \frac{- 1}{2}) = \frac{- 13 \times 7 + 12 \times 5}{35} \div \frac{31}{14} = \frac{- 91 + 60}{35} \div \frac{31}{14} = \frac{- 31}{35} \times \frac{14}{31} = \frac{- 2}{5}$

Question 14:

Divide the sum of $\frac{65}{12} and \frac{12}{7}$ by their difference.

Answer:

$(\frac{65}{12} + \frac{12}{7}) \div (\frac{65}{12} - \frac{12}{7}) = \frac{65 \times 7 + 12 \times 12}{84} \div \frac{65 \times 7 - 12 \times 12}{84} = \frac{455 + 144}{84} \div \frac{455 - 144}{84} = \frac{599}{84} \div \frac{311}{84} = \frac{599}{84} \times \frac{84}{311} = \frac{599}{311}$

Question 15:

If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Answer:

$Cloth needed to prepare 24 trousers = 54 m ∴ L ength of t h e cloth required for each trousers = 54 \div 24 = \frac{54}{24} = \frac{9}{4} m = 2 \frac{1}{4} metres$

Question 1:

Find a rational number between −3 and 1.

Answer:

$R ational number between - 3 and 1 = \frac{- 3 + 1}{2} = \frac{- 2}{2} = - 1$

Question 2:

Find any five rational numbers less than 2.

Answer:

$We can write : 2 = \frac{2}{1} = \frac{2 \times 5}{1 \times 5} = \frac{10}{5} Integers less than 10 are 0, 1, 2, 3, 4, 5 . . . 9 . Hence, five rational numbers less than 2 are \frac{0}{5}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5} and \frac{4}{5} .$

Question 3:

Find two rational numbers between $\frac{- 2}{9} and \frac{5}{9} .$

Answer:

$S i n c e b o t h t h e f r a c t i o n s (\frac{- 2}{9} a n d \frac{5}{9}) h a v e t h e s a m e d e n o m i n a t o r, t h e i n t e g e r s b e t w e e n t h e n u m e r a t o r s (- 2 a n d 5) a r e - 1, 0, 1, 2, 3, 4 . H e n c e, t w o r a t i o n a l n u m b e r s b e t w e e n \frac{- 2}{9} a n d \frac{5}{9} a r e \frac{0}{9} o r 0 a n d \frac{1}{9} .$

Question 4:

Find two rational numbers between $\frac{1}{5} and \frac{1}{2} .$

Answer:

$R ational number between \frac{1}{5} and \frac{1}{2} = \frac{(\frac{1}{5} + \frac{1}{2})}{2} = \frac{\frac{2 + 5}{10}}{2} = \frac{7}{20} R ational number between \frac{1}{5} and \frac{7}{20} = \frac{(\frac{1}{5} + \frac{7}{20})}{2} = \frac{\frac{4 + 7}{20}}{2} = \frac{11}{40} Therefore, two rational numbers between \frac{1}{5} and \frac{1}{2} are \frac{7}{20} and \frac{11}{40} .$

Question 5:

Find ten rational numbers between $\frac{1}{4} and \frac{1}{2} .$

Answer:

$T h e L . C . M o f t h e d e n o m i n a t o r s (2 a n d 4) i s 4 . S o, w e c a n w r i t e \frac{1}{4} a s i t i s . A l s o, \frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4} A s t h e i n t e g e r s b e t w e e n t h e n u m e r a t o r s 1 a n d 2 o f b o t h t h e f r a c t i o n s a r e n o t s u f f i c i e n t, w e w i l l m u l t i p l y t h e f r a c t i o n s b y 20 . ∴ \frac{1}{4} = \frac{1 \times 20}{4 \times 20} = \frac{20}{80} \frac{2}{4} = \frac{2 \times 20}{4 \times 20} = \frac{40}{80} B e t w e e n 20 a n d 40, t h e r e a r e 19 i n t e g e r s . T h e y a r e 21, 22, 23, 24, 25, 26, 27 . . . . 39, 40 . T h u s, \frac{21}{40}, \frac{22}{40}, \frac{23}{40}, \frac{24}{40}, \frac{25}{40}, . . . . . . . . . . . . . . . . . . . \frac{38}{40} a n d \frac{39}{40} a r e t h e f r a c t i o n s . W e c a n t a k e a n y 10 o f t h e s e .$

Question 6:

Find ten rational numbers between $\frac{- 2}{5} and \frac{1}{2} .$

Answer:

$L . C . M of the denominator s (2 and 5) is 10 . W e can write : \frac{- 2}{5} = \frac{- 2 \times 2}{5 \times 2} = \frac{- 4}{10} and \frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10} S i n c e t h e integers between the numerators (- 4 and 5) of both t h e fractions a r e n o t s u f f i c i e n t, we w i l l multiply the fractions by 2 . ∴ \frac{- 4}{10} = \frac{- 4 \times 2}{10 \times 2} = \frac{- 8}{20} \frac{5}{10} = \frac{5 \times 2}{10 \times 2} = \frac{10}{20} T h e r e a r e 17 i n t e g e r s b etween - 8 and 10, w h i c h a r e - 7, - 6, - 5, - 4 . . . . . . . . . . . . . . . . . . . 8, 9 . T h e s e c a n b e w r i t t e n a s : \frac{- 7}{20}, \frac{- 6}{20}, \frac{- 5}{20}, \frac{- 4}{20}, \frac{- 3}{20}, . . . . . . . . . . . . . . . . . . . \frac{8}{20} and \frac{9}{20} We can take any 10 of these .$

Question 7:

Find ten rational numbers between $\frac{3}{5} and \frac{3}{4} .$

Answer:

$T h e L . C . M o f t h e d e n o m i n a t o r s 5 a n d 4 o f b o t h t h e f r a c t i o n s i s 20 . W e c a n w r i t e : \frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20} \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} S i n c e t h e i n t e g e r s b e t w e e n t h e n u m e r a t o r s 12 a n d 15 a r e n o t s u f f i c i e n t, w e w i l l m u l t i p l y b o t h t h e f r a c t i o n s b y 5 . \frac{12}{20} = \frac{12 \times 5}{20 \times 5} = \frac{60}{100} \frac{15}{20} = \frac{15 \times 5}{20 \times 5} = \frac{75}{100} T h e r e a r e 14 i n t e g e r s b e t w e e n 60 a n d 75 . T h e y a r e 61, 62, 63 . . . . . . . 73 a n d 74 . T h e r e f o r e, \frac{60}{100}, \frac{61}{100}, \frac{62}{100} . . . . . . . . . . \frac{73}{100} a n d \frac{74}{100} a r e t h e 14 f r a c t i o n s . W e c a n t a k e a n y 10 o f t h e s e .$

Question 1:

Add the following rational numbers.
(i) $\frac{- 5}{7} and \frac{3}{7}$
(ii) $\frac{- 15}{4} and \frac{7}{4}$
(iii) $\frac{- 8}{11} and \frac{- 4}{11}$
(iv) $\frac{6}{13} and \frac{- 9}{13}$

Answer:

$(i) \frac{- 5}{7} + \frac{3}{7} = \frac{- 5 + 3}{7} = \frac{- 2}{7} (ii) \frac{- 15}{4} + \frac{7}{4} = \frac{- 15 + 7}{4} = \frac{- 8}{4} = - 2 (iii) \frac{- 8}{11} + \frac{- 4}{11} = \frac{- 8 - 4}{11} = \frac{- 12}{11} (iv) \frac{6}{13} + \frac{- 9}{13} = \frac{6 - 9}{13} = \frac{- 3}{13}$

Question 2:

Add the following rational numbers:
(i) $\frac{3}{4} and \frac{- 5}{8}$
(ii) $\frac{5}{- 9} and \frac{7}{3}$
(iii) $- 3 and \frac{3}{5}$
(iv) $\frac{- 7}{27} and \frac{11}{18}$
(v) $\frac{31}{- 4} and \frac{- 5}{8}$
(vi) $\frac{5}{36} and \frac{- 7}{12}$
(vii) $\frac{- 5}{16} and \frac{7}{24}$
(viii) $\frac{7}{- 18} and \frac{8}{27}$

Answer:

$(i) Clearly, denominators of the given numbers are positive. The L.C.M. of denominators 4 and 8 is 8. Now, we will express \frac{3}{4} in the form in which it takes the denominator is 8 . \frac{3 \times 2}{4 \times 2} = \frac{6}{8} \frac{3}{4} + \frac{- 5}{8} = \frac{6}{8} + \frac{- 5}{8} = \frac{6 + (- 5)}{8} = \frac{6 - 5}{8} = \frac{1}{8} (ii) \frac{5}{- 9} + \frac{7}{3} = \frac{- 5}{9} + \frac{7}{3} The L.C.M. of denominators 9 and 3 is 9 . Now, we will express \frac{7}{3} in the form in which it takes the denominator is 9 . \frac{7 \times 3}{3 \times 3} = \frac{21}{9} \frac{- 5}{9} + \frac{7}{3} = \frac{- 5}{9} + \frac{21}{9} = \frac{(- 5) + 21}{9} = \frac{- 5 + 21}{9} = \frac{16}{9} (iii) - 3 + \frac{3}{5} = \frac{- 3}{1} + \frac{3}{5} The L.C.M. of denominators 1 and 5 is 5 . Now, we will express \frac{- 3}{1} in the form in which it takes the denominator 5 . \frac{- 3 \times 5}{1 \times 5} = \frac{- 15}{5} So \frac{- 3}{1} + \frac{3}{5} = \frac{- 15}{5} + \frac{3}{5} = \frac{- 15 + 3}{5} = \frac{- 12}{5} (iv) The L.C.M. of denominators 27 and 18 is 54 . Now, we will express \frac{- 7}{27} and \frac{11}{18} in the form in which they take the denominator 54 . \frac{- 7 \times 2}{27 \times 2} = \frac{- 14}{54} \frac{11 \times 3}{18 \times 3} = \frac{33}{54} \frac{- 7}{27} + \frac{11}{18} = \frac{- 14}{54} + \frac{33}{54} = \frac{- 14 + 33}{54} = \frac{19}{54} (v) We have \frac{31}{- 4} + \frac{- 5}{8} = \frac{- 31}{4} + \frac{- 5}{8} The L.C.M. of denominators 4 and 8 is 8 . Now, we will express \frac{- 31}{4} in the form in which it takes the denominator 8 . \frac{- 31 \times 2}{4 \times 2} = \frac{- 62}{8} So \frac{- 31}{4} + \frac{- 5}{8} = \frac{- 62}{8} + \frac{- 5}{8} = \frac{(- 62) + (- 5)}{8} = \frac{- 62 - 5}{8} = \frac{- 67}{8} (vi) The L.C.M. of denominators 36 and 12 is 36 . Now, we will express \frac{- 7}{12} in the form in whic h it takes the denominator 36 . \frac{- 7 \times 3}{12 \times 3} = \frac{- 21}{36} So \frac{5}{36} + \frac{- 7}{12} = \frac{5}{36} + \frac{- 21}{36} = \frac{5 + (- 21)}{36} = \frac{5 - 21}{36} = \frac{- 16}{36} = \frac{- 4}{9} (vii) The L.C.M. of denominators 16 and 24 is 48 . Now, we will express \frac{- 5}{16} and \frac{7}{24} in the form in which they take the denominator 48 . \frac{- 5 \times 3}{16 \times 3} = \frac{- 15}{48} \frac{7 \times 2}{24 \times 2} = \frac{14}{48} So \frac{- 5}{16} + \frac{7}{24} = \frac{- 15}{48} + \frac{14}{48} = \frac{(- 15) + 14}{48} = \frac{- 15 + 14}{48} = \frac{- 1}{48} (viii) \frac{7}{- 18} + \frac{8}{27} = \frac{- 7}{18} + \frac{8}{27} The L.C.M. of denominators 18 and 27 is 54 . Now, we will express \frac{- 7}{18} and \frac{8}{27} in the form in which they take the denominator 54 . \frac{- 7 \times 3}{18 \times 3} = \frac{- 21}{54} \frac{8 \times 2}{27 \times 2} = \frac{16}{54} So \frac{- 7}{18} + \frac{8}{27} = \frac{- 21}{54} + \frac{16}{54} = \frac{- 21 + 16}{54} = \frac{- 5}{54} = \frac{- 5}{54}$

Question 3:

Simplify:
(i) $\frac{8}{9} + \frac{- 11}{6}$
(ii) $3 + \frac{5}{- 7}$
(iii) $\frac{1}{- 12} + \frac{2}{- 15}$
(iv) $\frac{- 8}{19} + \frac{- 4}{57}$
(v) $\frac{7}{9} + \frac{3}{- 4}$
(vi) $\frac{5}{26} + \frac{11}{- 39}$
(vii) $\frac{- 16}{9} + \frac{- 5}{12}$
(viii) $\frac{- 13}{8} + \frac{5}{36}$
(ix) $0 + \frac{- 3}{5}$
(x) $1 + \frac{- 4}{5}$

Answer:

$(i) \frac{8}{9} + \frac{- 11}{6} L.C.M. of the denominators 9 and 6 is 18. Now, we will express \frac{8}{9} and \frac{- 11}{6} in the form in which they take the denominator 18. \frac{8 \times 2}{9 \times 2} = \frac{16}{18} \frac{- 11 \times 3}{6 \times 3} = \frac{- 33}{18} \frac{8}{9} + \frac{- 11}{6} = \frac{16}{18} + \frac{- 33}{18} = \frac{16 + (- 33)}{18} = \frac{16 - 33}{18} = \frac{- 17}{18} (ii) 3 + \frac{5}{- 7} = \frac{3}{1} + \frac{- 5}{7} L.C.M. of the denominators 1 and 7 is 7. Now, we will express \frac{3}{1} in the form in which it takes the denominator 7. \frac{3 \times 7}{1 \times 7} = \frac{21}{7} \frac{3}{1} + \frac{- 5}{7} = \frac{21}{7} + \frac{- 5}{7} = \frac{21 + (- 5)}{7} = \frac{21 - 5}{7} = \frac{16}{7} (iii) \frac{1}{- 12} + \frac{2}{- 15} = \frac{- 1}{12} + \frac{- 2}{15} L.C.M. of the denominators 12 and 15 is 60. Now, we will express \frac{- 1}{12} and \frac{- 2}{15} in the form in which they take the denominator 60. \frac{- 1 \times 5}{12 \times 5} = \frac{- 5}{60} \frac{- 2 \times 4}{15 \times 4} = \frac{- 8}{60} \frac{- 1}{12} + \frac{- 2}{15} = \frac{- 5}{60} + \frac{- 8}{60} = \frac{(- 5) + (- 8)}{60} = \frac{- 5 - 8}{60} = \frac{- 13}{60} (iv) \frac{- 8}{19} + \frac{- 4}{57} L.C.M. of the denominators 19 and 57 is 57. Now, we will express \frac{- 8}{19} in the form in which i t t a k e s the denominator 57. \frac{- 8 \times 3}{19 \times 3} = \frac{- 24}{57} \frac{- 8}{19} + \frac{- 4}{57} = \frac{- 24}{57} + \frac{- 4}{57} = \frac{- 24 - 4}{57} = \frac{- 28}{57} (v) \frac{7}{9} + \frac{3}{- 4} = \frac{7}{9} + \frac{- 3}{4} L.C.M. of the denominators 9 and 4 is 36. Now, we will express \frac{7}{9} and \frac{- 3}{4} in the form in which they take the denominator 36. \frac{7 \times 4}{9 \times 4} = \frac{28}{36} \frac{- 3 \times 9}{4 \times 9} = \frac{- 27}{36} So \frac{7}{9} + \frac{- 3}{4} = \frac{28}{36} + \frac{- 27}{36} = \frac{28 + (- 27)}{36} = \frac{28 - 27}{36} = \frac{1}{36} (vi) \frac{5}{26} + \frac{11}{- 39} = \frac{5}{26} + \frac{- 11}{39} L.C.M. of the denominators 26 and 39 is 78. Now, we will express \frac{5}{26} and \frac{- 11}{39} in the form in which they take the denominator 78. \frac{5 \times 3}{26 \times 3} = \frac{15}{78} \frac{- 11 \times 2}{39 \times 2} = \frac{- 22}{78} So \frac{5}{26} + \frac{- 11}{39} = \frac{15}{78} + \frac{- 22}{78} = \frac{15 - 22)}{78} = \frac{- 7}{78} (vii) \frac{- 16}{9} + \frac{- 5}{12} L.C.M. of the denominators 9 and 12 is 36. Now, we will express \frac{- 16}{9} and \frac{- 5}{12} in the form in which they take the denominator 36. \frac{- 16 \times 4}{9 \times 4} = \frac{- 64}{36} \frac{- 5 \times 3}{12 \times 3} = \frac{- 15}{36} \frac{- 16}{9} + \frac{- 5}{12} = \frac{- 64}{36} + \frac{- 15}{36} = \frac{(- 64) + (- 15)}{36} = \frac{- 64 - 15}{36} = \frac{- 79}{36} (viii) \frac{- 13}{8} + \frac{5}{36} L.C.M. of the denominators 8 and 36 is 72. Now, we will express \frac{- 13}{8} and \frac{5}{36} in the form in which they take the denominator 72. \frac{- 13 \times 9}{8 \times 9} = \frac{- 117}{72} \frac{5 \times 2}{36 \times 2} = \frac{10}{72} \frac{- 13}{8} + \frac{5}{36} = \frac{- 117}{72} + \frac{10}{72} = \frac{- 117 + 10}{72} = \frac{- 107}{72} (ix) 0 + \frac{- 3}{5} Taking L . C . M . of the denominator : = \frac{0 \times 5 - 3}{5} = \frac{- 3}{5} (x) 1 + \frac{- 4}{5} = \frac{1}{1} + \frac{- 4}{5} L.C.M. of the denominators 1 and 5 is 5. Now, we will express \frac{1}{1} in the form in which i t t a k e s the denominator 5. \frac{1 \times 5}{1 \times 5} = \frac{5}{5} \frac{1}{1} + \frac{- 4}{5} = \frac{5}{5} + \frac{- 4}{5} = \frac{5 - 4}{5} = \frac{1}{5}$

Question 4:

Add and express the sum as a mixed fraction:
(i) $\frac{- 12}{5} and \frac{43}{10}$

(ii) $\frac{24}{7} and \frac{- 11}{4}$

(iii) $\frac{- 31}{6} and \frac{- 27}{8}$

(iv) $\frac{101}{6} and \frac{7}{8}$

Answer:

$(i) We have \frac{- 12}{5} + \frac{43}{10} . L.C.M. of the denominators 5 and 10 is 10. Now, we will express \frac{- 12}{5} in the form in which it takes the denominator 10. \frac{- 12 \times 2}{5 \times 2} = \frac{- 24}{10} ∴ \frac{- 12}{5} + \frac{43}{10} = \frac{- 24}{10} + \frac{43}{10} = \frac{- 24 + 43}{10} = \frac{19}{10} = 1 \frac{9}{10} (ii) We have \frac{24}{7} + \frac{- 11}{4} . L.C.M. of the denominators 7 and 4 is 28. Now, we will express \frac{24}{7} and \frac{- 11}{4} in the form in which they take the denominator 28. \frac{24 \times 4}{7 \times 4} = \frac{96}{28} \frac{- 11 \times 7}{4 \times 7} = \frac{- 77}{28} ∴ \frac{24}{7} + \frac{- 11}{4} = \frac{96}{28} + \frac{- 77}{28} = \frac{96 - 77}{28} = \frac{19}{28} (iii) We have \frac{- 31}{6} + \frac{- 27}{8} . L.C.M. of the denominators 6 and 8 is 24. Now, we will express \frac{- 31}{6} and \frac{- 27}{8} in the form in which they take the denominator 24. \frac{- 31 \times 4}{6 \times 4} = \frac{- 124}{24} \frac{- 27 \times 3}{8 \times 3} = \frac{- 81}{24} ∴ \frac{- 31}{6} + \frac{- 27}{8} = \frac{- 124}{24} + \frac{- 81}{24} = \frac{- 124 - 81}{24} = \frac{- 205}{24} = - 8 \frac{13}{24} (iv) We have \frac{101}{6} + \frac{7}{8} . L.C.M. of the denominators 6 and 8 is 24. Now, we will express \frac{101}{6} and \frac{7}{8} in the form in which they take the denominator 24. \frac{101 \times 4}{6 \times 4} = \frac{404}{24} \frac{7 \times 3}{8 \times 3} = \frac{21}{24} ∴ \frac{101}{6} + \frac{7}{8} = \frac{404}{24} + \frac{21}{24} = \frac{404 + 21}{24} = \frac{425}{24} = 17 \frac{17}{24}$

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