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Chapter 3

Fractions — Exercise 3(D)

Class - 7 Concise Mathematics Selina



Exercise 3(D)

Question 1

Simplify:

6+{43+(34−13)}6 + \Bigg\lbrace\dfrac{4}{3} + \left(\dfrac{3}{4} - \dfrac{1}{3}\right)\Bigg\rbrace

Answer

Simplifying,

6+{43+(34−13)}=6+{43+(9−412)}=6+{43+512}=6+{1612+512}=6+2112=6+74=24+74=314=7346 + \Bigg\lbrace\dfrac{4}{3} + \left(\dfrac{3}{4} - \dfrac{1}{3}\right)\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\dfrac{4}{3} + \left(\dfrac{9 - 4}{12}\right)\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\dfrac{4}{3} + \dfrac{5}{12}\Bigg\rbrace \\[1em] = 6 + \Bigg\lbrace\dfrac{16}{12} + \dfrac{5}{12}\Bigg\rbrace \\[1em] = 6 + \dfrac{21}{12} \\[1em] = 6 + \dfrac{7}{4} \\[1em] = \dfrac{24 + 7}{4} = \dfrac{31}{4} = 7\dfrac{3}{4}

Hence, the required value is 7347\dfrac{3}{4}.

Question 2

Simplify:

8−{32+(35−12)}8 - \Bigg\lbrace\dfrac{3}{2} + \left(\dfrac{3}{5} - \dfrac{1}{2}\right)\Bigg\rbrace

Answer

Simplifying,

8−{32+(35−12)}=8−{32+(6−510)}=8−{32+110}=8−{1510+110}=8−1610=8−85=40−85=325=6258 - \Bigg\lbrace\dfrac{3}{2} + \left(\dfrac{3}{5} - \dfrac{1}{2}\right)\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\dfrac{3}{2} + \left(\dfrac{6 - 5}{10}\right)\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\dfrac{3}{2} + \dfrac{1}{10}\Bigg\rbrace \\[1em] = 8 - \Bigg\lbrace\dfrac{15}{10} + \dfrac{1}{10}\Bigg\rbrace \\[1em] = 8 - \dfrac{16}{10} \\[1em] = 8 - \dfrac{8}{5} \\[1em] = \dfrac{40 - 8}{5} \\[1em] = \dfrac{32}{5} = 6\dfrac{2}{5}

Hence, the required value is 6256\dfrac{2}{5}.

Question 3

Simplify:

14(14+13)−25\dfrac{1}{4}\left(\dfrac{1}{4} + \dfrac{1}{3}\right) - \dfrac{2}{5}

Answer

Simplifying,

14(14+13)−25=14(3+412)−25=14×712−25=748−25=7×5240−2×48240=35−96240=−61240\dfrac{1}{4}\left(\dfrac{1}{4} + \dfrac{1}{3}\right) - \dfrac{2}{5} \\[1em] = \dfrac{1}{4}\left(\dfrac{3 + 4}{12}\right) - \dfrac{2}{5} \\[1em] = \dfrac{1}{4} \times \dfrac{7}{12} - \dfrac{2}{5} \\[1em] = \dfrac{7}{48} - \dfrac{2}{5} \\[1em] = \dfrac{7 \times 5}{240} - \dfrac{2 \times 48}{240} \\[1em] = \dfrac{35 - 96}{240} = -\dfrac{61}{240}

Hence, the required value is −61240-\dfrac{61}{240}.

Question 4

Simplify:

234−[318÷{5−(423−1112)}]2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \left(4\dfrac{2}{3} - \dfrac{11}{12}\right)\Bigg\rbrace\right]

Answer

Simplifying,

234−[318÷{5−(423−1112)}]=234−[318÷{5−(143−1112)}]=234−[318÷{5−(56−1112)}]=234−[318÷{5−4512}]=234−[318÷{5−154}]=234−[318÷(20−154)]=234−[318÷54]=234−[258×45]=234−52=114−104=142\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \left(4\dfrac{2}{3} - \dfrac{11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \left(\dfrac{14}{3} - \dfrac{11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \left(\dfrac{56 - 11}{12}\right)\Bigg\rbrace\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \dfrac{45}{12}\Bigg\rbrace\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \Bigg\lbrace 5 - \dfrac{15}{4}\Bigg\rbrace\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \left(\dfrac{20 - 15}{4}\right)\right] \\[1em] = 2\dfrac{3}{4} - \left[3\dfrac{1}{8} \div \dfrac{5}{4}\right] \\[1em] = 2\dfrac{3}{4} - \left[\dfrac{25}{8} \times \dfrac{4}{5}\right] \\[1em] = 2\dfrac{3}{4} - \dfrac{5}{2} \\[1em] = \dfrac{11}{4} - \dfrac{10}{4} \\[1em] = \dfrac{1}{4}

Hence, the required value is 14\dfrac{1}{4}.

Question 5

Simplify:

1212−[812+{9−(5−3−2‾)}]12\dfrac{1}{2} - \left[8\dfrac{1}{2} + \Bigg\lbrace 9 - \left(5 - \overline{3 - 2}\right)\Bigg\rbrace\right]

Answer

Simplifying,

1212−[812+{9−(5−3−2‾)}]=1212−[812+{9−(5−1)}]=1212−[812+{9−4}]=1212−[812+5]=1212−1312=252−272=−22=−112\dfrac{1}{2} - \left[8\dfrac{1}{2} + \Bigg\lbrace 9 - \left(5 - \overline{3 - 2}\right)\Bigg\rbrace\right] \\[1em] = 12\dfrac{1}{2} - \left[8\dfrac{1}{2} + \Bigg\lbrace 9 - \left(5 - 1\right)\Bigg\rbrace\right] \\[1em] = 12\dfrac{1}{2} - \left[8\dfrac{1}{2} + \Bigg\lbrace 9 - 4\Bigg\rbrace\right] \\[1em] = 12\dfrac{1}{2} - \left[8\dfrac{1}{2} + 5\right] \\[1em] = 12\dfrac{1}{2} - 13\dfrac{1}{2} \\[1em] = \dfrac{25}{2} - \dfrac{27}{2} \\[1em] = -\dfrac{2}{2} = -1

Hence, the required value is -1.

Question 6

Simplify:

115÷{213−(5+2−3‾)}−3121\dfrac{1}{5} \div \Bigg\lbrace 2\dfrac{1}{3} - \left(5 + \overline{2 - 3}\right)\Bigg\rbrace - 3\dfrac{1}{2}

Answer

Simplifying,

115÷{213−(5+2−3‾)}−312=115÷{213−(5+(−1))}−312=115÷{213−4}−312=115÷(73−123)−312=115÷(−53)−312=65×(−35)−312=−1825−72=−3650−17550=−21150=−411501\dfrac{1}{5} \div \Bigg\lbrace 2\dfrac{1}{3} - \left(5 + \overline{2 - 3}\right)\Bigg\rbrace - 3\dfrac{1}{2} \\[1em] = 1\dfrac{1}{5} \div \Bigg\lbrace 2\dfrac{1}{3} - \left(5 + (-1)\right)\Bigg\rbrace - 3\dfrac{1}{2} \\[1em] = 1\dfrac{1}{5} \div \Bigg\lbrace 2\dfrac{1}{3} - 4\Bigg\rbrace - 3\dfrac{1}{2} \\[1em] = 1\dfrac{1}{5} \div \left(\dfrac{7}{3} - \dfrac{12}{3}\right) - 3\dfrac{1}{2} \\[1em] = 1\dfrac{1}{5} \div \left(-\dfrac{5}{3}\right) - 3\dfrac{1}{2} \\[1em] = \dfrac{6}{5} \times \left(-\dfrac{3}{5}\right) - 3\dfrac{1}{2} \\[1em] = -\dfrac{18}{25} - \dfrac{7}{2} \\[1em] = -\dfrac{36}{50} - \dfrac{175}{50} \\[1em] = -\dfrac{211}{50} = -4\dfrac{11}{50}

Hence, the required value is −41150-4\dfrac{11}{50}.

Question 7

Simplify:

(12+23)÷(34−29)\left(\dfrac{1}{2} + \dfrac{2}{3}\right) \div \left(\dfrac{3}{4} - \dfrac{2}{9}\right)

Answer

Simplifying,

(12+23)÷(34−29)=(3+46)÷(27−836)=76÷1936=76×3619=7×366×19=252114=4219=2419\left(\dfrac{1}{2} + \dfrac{2}{3}\right) \div \left(\dfrac{3}{4} - \dfrac{2}{9}\right) \\[1em] = \left(\dfrac{3 + 4}{6}\right) \div \left(\dfrac{27 - 8}{36}\right) \\[1em] = \dfrac{7}{6} \div \dfrac{19}{36} \\[1em] = \dfrac{7}{6} \times \dfrac{36}{19} \\[1em] = \dfrac{7 \times 36}{6 \times 19} = \dfrac{252}{114} \\[1em] = \dfrac{42}{19} = 2\dfrac{4}{19}

Hence, the required value is 24192\dfrac{4}{19}.

Question 8

Simplify:

65 of (313−212)÷(2521−2)\dfrac{6}{5} \text{ of } \left(3\dfrac{1}{3} - 2\dfrac{1}{2}\right) \div \left(2\dfrac{5}{21} - 2\right)

Answer

Simplifying,

65 of (313−212)÷(2521−2)=65 of (103−52)÷(4721−2)=65 of (20−156)÷(47−4221)=65 of 56÷521=65×56÷521=1÷521=1×215=215=415\dfrac{6}{5} \text{ of } \left(3\dfrac{1}{3} - 2\dfrac{1}{2}\right) \div \left(2\dfrac{5}{21} - 2\right) \\[1em] = \dfrac{6}{5} \text{ of } \left(\dfrac{10}{3} - \dfrac{5}{2}\right) \div \left(\dfrac{47}{21} - 2\right) \\[1em] = \dfrac{6}{5} \text{ of } \left(\dfrac{20 - 15}{6}\right) \div \left(\dfrac{47 - 42}{21}\right) \\[1em] = \dfrac{6}{5} \text{ of } \dfrac{5}{6} \div \dfrac{5}{21} \\[1em] = \dfrac{6}{5} \times \dfrac{5}{6} \div \dfrac{5}{21} \\[1em] = 1 \div \dfrac{5}{21} \\[1em] = 1 \times \dfrac{21}{5} = \dfrac{21}{5} = 4\dfrac{1}{5}

Hence, the required value is 4154\dfrac{1}{5}.

Question 9

Simplify:

1018 of 45÷3536 of 204910\dfrac{1}{8} \text{ of } \dfrac{4}{5} \div \dfrac{35}{36} \text{ of } \dfrac{20}{49}

Answer

Simplifying,

1018 of 45÷3536 of 2049=(818×45)÷(3536×2049)=8110÷2563=8110×6325=81×6310×25=5103250=2010325010\dfrac{1}{8} \text{ of } \dfrac{4}{5} \div \dfrac{35}{36} \text{ of } \dfrac{20}{49} \\[1em] = \left(\dfrac{81}{8} \times \dfrac{4}{5}\right) \div \left(\dfrac{35}{36} \times \dfrac{20}{49}\right) \\[1em] = \dfrac{81}{10} \div \dfrac{25}{63} \\[1em] = \dfrac{81}{10} \times \dfrac{63}{25} \\[1em] = \dfrac{81 \times 63}{10 \times 25} = \dfrac{5103}{250} \\[1em] = 20\dfrac{103}{250}

Hence, the required value is 2010325020\dfrac{103}{250}.

Question 10

Simplify:

534−37×1534+2235÷111255\dfrac{3}{4} - \dfrac{3}{7} \times 15\dfrac{3}{4} + 2\dfrac{2}{35} \div 1\dfrac{11}{25}

Answer

Simplifying,

534−37×1534+2235÷11125=534−37×634+7235÷3625=534−37×634+7235×2536=534−274+107=234−274+107=−44+107=−1+107=−7+107=375\dfrac{3}{4} - \dfrac{3}{7} \times 15\dfrac{3}{4} + 2\dfrac{2}{35} \div 1\dfrac{11}{25} \\[1em] = 5\dfrac{3}{4} - \dfrac{3}{7} \times \dfrac{63}{4} + \dfrac{72}{35} \div \dfrac{36}{25} \\[1em] = 5\dfrac{3}{4} - \dfrac{3}{7} \times \dfrac{63}{4} + \dfrac{72}{35} \times \dfrac{25}{36} \\[1em] = 5\dfrac{3}{4} - \dfrac{27}{4} + \dfrac{10}{7} \\[1em] = \dfrac{23}{4} - \dfrac{27}{4} + \dfrac{10}{7} \\[1em] = -\dfrac{4}{4} + \dfrac{10}{7} \\[1em] = -1 + \dfrac{10}{7} \\[1em] = \dfrac{-7 + 10}{7} = \dfrac{3}{7}

Hence, the required value is 37\dfrac{3}{7}.

Question 11

Simplify:

34 of 737−535÷3415\dfrac{3}{4} \text{ of } 7\dfrac{3}{7} - 5\dfrac{3}{5} \div 3\dfrac{4}{15}

Answer

Simplifying,

34 of 737−535÷3415=34×527−285÷4915=34×527−285×1549=397−127=277=367\dfrac{3}{4} \text{ of } 7\dfrac{3}{7} - 5\dfrac{3}{5} \div 3\dfrac{4}{15} \\[1em] = \dfrac{3}{4} \times \dfrac{52}{7} - \dfrac{28}{5} \div \dfrac{49}{15} \\[1em] = \dfrac{3}{4} \times \dfrac{52}{7} - \dfrac{28}{5} \times \dfrac{15}{49} \\[1em] = \dfrac{39}{7} - \dfrac{12}{7} \\[1em] = \dfrac{27}{7} = 3\dfrac{6}{7}

Hence, the required value is 3673\dfrac{6}{7}.

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