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Chapter 6

Fractions — Exercise 6(C)

Class - 6 Concise Mathematics Selina

Exercise 6(C)

Question 1(i)

Add the following fractions:

134 and 381\dfrac{3}{4} \text{ and } \dfrac{3}{8}

Answer

134 and 381\dfrac{3}{4} \text{ and } \dfrac{3}{8}

⇒134+38=74+38 L.C.M. of 4 and 8 = 8 =7×24×2+38=148+38=178=218.\Rightarrow 1\dfrac{3}{4} + \dfrac{3}{8} = \dfrac{7}{4} + \dfrac{3}{8} \\[1em] \text{ L.C.M. of 4 and 8 = 8 }\\[1em] = \dfrac{7 \times 2}{4 \times 2} + \dfrac{3}{8} \\[1em] =\dfrac{14}{8} + \dfrac{3}{8} \\[1em] = \dfrac{17}{8} \\[1em] = 2\dfrac{1}{8}.

Hence, the required sum = 2182\dfrac{1}{8}.

Question 1(ii)

Add the following fractions:

25,2315 and 710\dfrac{2}{5}, 2\dfrac{3}{15} \text{ and } \dfrac{7}{10}

Answer

25,2315 and 710\dfrac{2}{5}, 2\dfrac{3}{15} \text{ and } \dfrac{7}{10}

LCM of 5, 15 and 10 = 30.

⇒25+2315+710=25+3315+710=1230+6630+2130=9930=3310=3310.\Rightarrow \dfrac{2}{5} + 2\dfrac{3}{15} + \dfrac{7}{10} = \dfrac{2}{5} + \dfrac{33}{15} + \dfrac{7}{10} \\[1em] = \dfrac{12}{30} + \dfrac{66}{30} + \dfrac{21}{30} \\[1em] = \dfrac{99}{30}\\[1em] = \dfrac{33}{10}\\[1em] = 3\dfrac{3}{10}.

Hence, the required sum = 33103\dfrac{3}{10}.

Question 1(iii)

Add the following fractions:

178,112 and 1341\dfrac{7}{8}, 1\dfrac{1}{2} \text{ and } 1\dfrac{3}{4}

Answer

178,112 and 1341\dfrac{7}{8}, 1\dfrac{1}{2} \text{ and } 1\dfrac{3}{4}

LCM of 8, 2 and 4 = 8.

⇒178+112+134=158+32+74=158+128+148=418=518.\Rightarrow 1\dfrac{7}{8} + 1\dfrac{1}{2} + 1\dfrac{3}{4} = \dfrac{15}{8} + \dfrac{3}{2} + \dfrac{7}{4} \\[1em] = \dfrac{15}{8} + \dfrac{12}{8} + \dfrac{14}{8} \\[1em] = \dfrac{41}{8} \\[1em] = 5\dfrac{1}{8}.

Hence, the required sum = 5185\dfrac{1}{8}.

Question 1(iv)

Add the following fractions:

334,216 and 1583\dfrac{3}{4}, 2\dfrac{1}{6} \text{ and } 1\dfrac{5}{8}

Answer

334,216 and 1583\dfrac{3}{4}, 2\dfrac{1}{6} \text{ and } 1\dfrac{5}{8}

LCM of 4, 6 and 8 = 24.

⇒334+216+158=154+136+138=9024+5224+3924=18124=71324.\Rightarrow 3\dfrac{3}{4} + 2\dfrac{1}{6} + 1\dfrac{5}{8} = \dfrac{15}{4} + \dfrac{13}{6} + \dfrac{13}{8} \\[1em] = \dfrac{90}{24} + \dfrac{52}{24} + \dfrac{39}{24} \\[1em] = \dfrac{181}{24} \\[1em] = 7\dfrac{13}{24}.

Hence, the required sum = 713247\dfrac{13}{24}.

Question 1(v)

Add the following fractions:

289,1118 and 3562\dfrac{8}{9}, \dfrac{11}{18} \text{ and } 3\dfrac{5}{6}

Answer

289,1118 and 3562\dfrac{8}{9}, \dfrac{11}{18} \text{ and } 3\dfrac{5}{6}

LCM of 9, 18 and 6 = 18.

⇒289+1118+356=269+1118+236=5218+1118+6918=13218=223=713.\Rightarrow 2\dfrac{8}{9} + \dfrac{11}{18} + 3\dfrac{5}{6} = \dfrac{26}{9} + \dfrac{11}{18} + \dfrac{23}{6} \\[1em] = \dfrac{52}{18} + \dfrac{11}{18} + \dfrac{69}{18} \\[1em] = \dfrac{132}{18} \\[1em] = \dfrac{22}{3} \\[1em] = 7\dfrac{1}{3}.

Hence, the required sum = 7137\dfrac{1}{3}.

Question 2(i)

Simplify:

11112−13161\dfrac{11}{12} - \dfrac{13}{16}

Answer

11112−13161\dfrac{11}{12} - \dfrac{13}{16}

LCM of 12 and 16 = 48.

⇒11112−1316=2312−1316=9248−3948=5348=1548.\Rightarrow 1\dfrac{11}{12} - \dfrac{13}{16} = \dfrac{23}{12} - \dfrac{13}{16} \\[1em] = \dfrac{92}{48} - \dfrac{39}{48} \\[1em] = \dfrac{53}{48} \\[1em] = 1\dfrac{5}{48}.

Hence, 11112−1316=15481\dfrac{11}{12} - \dfrac{13}{16} = 1\dfrac{5}{48}.

Question 2(ii)

Simplify:

234−1562\dfrac{3}{4} - 1\dfrac{5}{6}

Answer

234−1562\dfrac{3}{4} - 1\dfrac{5}{6}

LCM of 4 and 6 = 12.

⇒234−156=114−116=3312−2212=1112.\Rightarrow 2\dfrac{3}{4} - 1\dfrac{5}{6} = \dfrac{11}{4} - \dfrac{11}{6} \\[1em] = \dfrac{33}{12} - \dfrac{22}{12} \\[1em] = \dfrac{11}{12}.

Hence, 234−156=11122\dfrac{3}{4} - 1\dfrac{5}{6} = \dfrac{11}{12}.

Question 2(iii)

Simplify:

257+314−13212\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21}

Answer

257+314−13212\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21}

LCM of 7, 14 and 21 = 42.

⇒257+314−1321=197+314−1321=11442+942−2642=9742=21342.\Rightarrow 2\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21} = \dfrac{19}{7} + \dfrac{3}{14} - \dfrac{13}{21} \\[1em] = \dfrac{114}{42} + \dfrac{9}{42} - \dfrac{26}{42} \\[1em] = \dfrac{97}{42} \\[1em] = 2\dfrac{13}{42}.

Hence, 257+314−1321=213422\dfrac{5}{7} + \dfrac{3}{14} - \dfrac{13}{21} = 2\dfrac{13}{42}.

Question 2(iv)

Simplify:

356−16−11123\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12}

Answer

356−16−11123\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12}

LCM of 6 and 12 = 12.

⇒356−16−1112=236−16−1312=4612−212−1312=3112=2712.\Rightarrow 3\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12} = \dfrac{23}{6} - \dfrac{1}{6} - \dfrac{13}{12} \\[1em] = \dfrac{46}{12} - \dfrac{2}{12} - \dfrac{13}{12} \\[1em] = \dfrac{31}{12} \\[1em] = 2\dfrac{7}{12}.

Hence, 356−16−1112=27123\dfrac{5}{6} - \dfrac{1}{6} - 1\dfrac{1}{12} = 2\dfrac{7}{12}.

Question 2(v)

Simplify:

6+310−18156 + \dfrac{3}{10} - 1\dfrac{8}{15}

Answer

6+310−18156 + \dfrac{3}{10} - 1\dfrac{8}{15}

LCM of 1, 10 and 15 = 30.

⇒6+310−1815=6+310−2315=18030+930−4630=14330=42330.\Rightarrow 6 + \dfrac{3}{10} - 1\dfrac{8}{15} = 6 + \dfrac{3}{10} - \dfrac{23}{15} \\[1em] = \dfrac{180}{30} + \dfrac{9}{30} - \dfrac{46}{30} \\[1em] = \dfrac{143}{30} \\[1em] = 4\dfrac{23}{30}.

Hence, 6+310−1815=423306 + \dfrac{3}{10} - 1\dfrac{8}{15} = 4\dfrac{23}{30}.

Question 2(vi)

Simplify:

134+257−13141\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14}

Answer

134+257−13141\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14}

LCM of 4, 7 and 14 = 28.

⇒134+257−1314=74+197−1714=4928+7628−3428=9128=134=314.\Rightarrow 1\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14} = \dfrac{7}{4} + \dfrac{19}{7} - \dfrac{17}{14} \\[1em] = \dfrac{49}{28} + \dfrac{76}{28} - \dfrac{34}{28} \\[1em] = \dfrac{91}{28} \\[1em] = \dfrac{13}{4} \\[1em] = 3\dfrac{1}{4}.

Hence, 134+257−1314=3141\dfrac{3}{4} + 2\dfrac{5}{7} - 1\dfrac{3}{14} = 3\dfrac{1}{4}.

Question 2(vii)

Simplify:

4+318−3164 + 3\dfrac{1}{8} - 3\dfrac{1}{6}

Answer

4+318−3164 + 3\dfrac{1}{8} - 3\dfrac{1}{6}

LCM of 1, 8 and 6 = 24.

⇒4+318−316=4+258−196=9624+7524−7624=9524=32324.\Rightarrow 4 + 3\dfrac{1}{8} - 3\dfrac{1}{6} = 4 + \dfrac{25}{8} - \dfrac{19}{6} \\[1em] = \dfrac{96}{24} + \dfrac{75}{24} - \dfrac{76}{24} \\[1em] = \dfrac{95}{24} \\[1em] = 3\dfrac{23}{24}.

Hence, 4+318−316=323244 + 3\dfrac{1}{8} - 3\dfrac{1}{6} = 3\dfrac{23}{24}.

Question 2(viii)

Simplify:

6−312−2156 - 3\dfrac{1}{2} - 2\dfrac{1}{5}

Answer

6−312−2156 - 3\dfrac{1}{2} - 2\dfrac{1}{5}

LCM of 1, 2 and 5 = 10.

⇒6−312−215=6−72−115=6010−3510−2210=310.\Rightarrow 6 - 3\dfrac{1}{2} - 2\dfrac{1}{5} = 6 - \dfrac{7}{2} - \dfrac{11}{5} \\[1em] = \dfrac{60}{10} - \dfrac{35}{10} - \dfrac{22}{10} \\[1em] = \dfrac{3}{10}.

Hence, 6−312−215=3106 - 3\dfrac{1}{2} - 2\dfrac{1}{5} = \dfrac{3}{10}.

Question 2(ix)

Simplify:

158−216+3341\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4}

Answer

158−216+3341\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4}

LCM of 8, 6 and 4 = 24.

⇒158−216+334=138−136+154=3924−5224+9024=7724=3524.\Rightarrow 1\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4} = \dfrac{13}{8} - \dfrac{13}{6} + \dfrac{15}{4} \\[1em] = \dfrac{39}{24} - \dfrac{52}{24} + \dfrac{90}{24} \\[1em] = \dfrac{77}{24} \\[1em] = 3\dfrac{5}{24}.

Hence, 158−216+334=35241\dfrac{5}{8} - 2\dfrac{1}{6} + 3\dfrac{3}{4} = 3\dfrac{5}{24}.

Question 2(x)

Simplify:

312+123−2143\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4}

Answer

312+123−2143\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4}

LCM of 2, 3 and 4 = 12.

⇒312+123−214=72+53−94=4212+2012−2712=3512=21112.\Rightarrow 3\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4} = \dfrac{7}{2} + \dfrac{5}{3} - \dfrac{9}{4} \\[1em] = \dfrac{42}{12} + \dfrac{20}{12} - \dfrac{27}{12} \\[1em] = \dfrac{35}{12} \\[1em] = 2\dfrac{11}{12}.

Hence, 312+123−214=211123\dfrac{1}{2} + 1\dfrac{2}{3} - 2\dfrac{1}{4} = 2\dfrac{11}{12}.

Question 2(xi)

Simplify:

435−279−1215−2454\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45}

Answer

435−279−1215−2454\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45}

LCM of 5, 9, 15 and 45 = 45.

⇒435−279−1215−245=235−259−1715−245=20745−12545−5145−245=2945.\Rightarrow 4\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45} = \dfrac{23}{5} - \dfrac{25}{9} - \dfrac{17}{15} - \dfrac{2}{45} \\[1em] = \dfrac{207}{45} - \dfrac{125}{45} - \dfrac{51}{45} - \dfrac{2}{45} \\[1em] = \dfrac{29}{45}.

Hence, 435−279−1215−245=29454\dfrac{3}{5} - 2\dfrac{7}{9} - 1\dfrac{2}{15} - \dfrac{2}{45} = \dfrac{29}{45}.

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