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Chapter 2

Rational Numbers — Exercise 2(D)

Class - 7 Concise Mathematics Selina

Exercise 2(D)

Question 1(i)

Evaluate:

54×37\dfrac{5}{4} \times \dfrac{3}{7}

Answer

Solving,

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

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

Question 1(ii)

Evaluate:

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

Answer

Solving,

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

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

Question 1(iii)

Evaluate:

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

Answer

Solving,

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

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

Question 1(iv)

Evaluate:

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

Answer

Solving,

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

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

Question 1(v)

Evaluate:

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

Answer

Solving,

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

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

Question 1(vi)

Evaluate:

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

Answer

Solving,

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

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

Question 1(vii)

Evaluate:

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

Answer

Solving,

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

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

Question 1(viii)

Evaluate:

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

Answer

Solving,

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

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

Question 2(i)

Multiply:

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

Answer

Solving,

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

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

Question 2(ii)

Multiply:

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

Answer

Solving,

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

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

Question 2(iii)

Multiply:

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

Answer

Solving,

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

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

Question 2(iv)

Multiply:

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

Answer

Solving,

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

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

Question 2(v)

Multiply:

1161\dfrac{1}{6} and 18

Answer

Solving,

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

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

Question 2(vi)

Multiply:

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

Answer

Solving,

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

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

Question 2(vii)

Multiply:

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

Answer

Solving,

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

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

Question 2(viii)

Multiply:

35 and −1825\dfrac{-18}{25}

Answer

Solving,

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

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

Question 2(ix)

Multiply:

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

Answer

Solving,

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

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

Question 2(x)

Multiply:

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

Answer

Solving,

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

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

Question 2(xi)

Multiply:

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

Answer

Solving,

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

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

Question 2(xii)

Multiply:

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

Answer

Solving,

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

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

Question 3(i)

Evaluate:

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

Answer

Solving,

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

LCM of 3 and 9 = 9

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

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

Question 3(ii)

Evaluate:

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

Answer

Solving,

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

LCM of 1 and 15 = 15

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

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

Question 3(iii)

Evaluate:

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

Answer

Solving,

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

LCM of 12 and 20 = 60

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

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

Question 3(iv)

Evaluate:

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

Answer

Solving,

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

LCM of 15 and 1 = 15

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

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

Question 3(v)

Evaluate:

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

Answer

Solving,

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

LCM of 3 and 10 = 30

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

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

Question 3(vi)

Evaluate:

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

Answer

Solving,

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

LCM of 5, 1 and 8 = 40

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

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

Question 4

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

Answer

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

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

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

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

Question 5

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

Answer

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

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

Distance covered = Speed × Time

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

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

Question 6(i)

Divide:

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

Answer

Solving,

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

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

Question 6(ii)

Divide:

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

Answer

Solving,

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

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

Question 6(iii)

Divide:

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

Answer

Solving,

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

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

Question 6(iv)

Divide:

-7 by −145\dfrac{-14}{5}

Answer

Solving,

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

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

Question 6(v)

Divide:

-14 by 7−2\dfrac{7}{-2}

Answer

Solving,

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

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

Question 6(vi)

Divide:

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

Answer

Solving,

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

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

Question 6(vii)

Divide:

35 by −79\dfrac{-7}{9}

Answer

Solving,

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

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

Question 6(viii)

Divide:

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

Answer

Solving,

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

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

Question 7(i)

Evaluate:

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

Answer

Solving,

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

LCM of 12 and 3 is 2 × 2 × 3 = 12

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

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

Question 7(ii)

Evaluate:

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

Answer

Solving,

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

LCM of 12 and 3 = 12

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

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

Question 7(iii)

Evaluate:

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

Answer

Solving,

From parts (i) and (ii):

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

Solving,

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

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

Question 8

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

Answer

Product of two numbers = 14

Let the other number = x

One number = −87\dfrac{-8}{7}

According to question,

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

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

Question 9

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

Answer

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

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

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

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

Question 10

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

Answer

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

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

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

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

Question 11

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

Answer

Let the required number be x.

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

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

Question 12

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

Answer

Let the required number be x.

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

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

Question 13

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

Answer

By Division Method,

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

LCM of 15 and 5 = 3 × 5 = 15

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

Solving,

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

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

Question 14

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

Answer

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

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

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

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

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