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Chapter 2

Rational Numbers — Exercise 2(C)

Class - 7 Concise Mathematics Selina

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}.

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