To do :We have to find the given sums.Solution :We know that, $(+) \times (-) = (-)$$(+) \times (+) = (+)$$(-) \times (-) = (+)$$(-) \times (+) = (-)$(a) $35-(20)=35-20$$=15$Therefore, \( 35-(20)=15 \)(b) $72-(90)=72-90$$=-(90-72)$$=-18$Therefore, \( 72-(90)=-18 \)(c) $(-15)-(-18)=-15+18$$=18-15$$=3$Therefore, \( (-15)-(-18)=3 \)(d) $(-20)-(13)=-20-13$$=-(20+13)$$=-33$Therefore, \( (-20)-(13)=-33 \)(e) $23-(-12)=23+12$$=35$Therefore, \( 23-(-12)=35 \)(f) $(-32)-(-40)=-32+40$$=40-32$$=8$Therefore, ... Read More

To do :We have to fill in the blanks with \( \rangle, < \) or \( = \) sign.Solution :We know that, $(+) \times (-) = (-)$$(+) \times (+) = (+)$$(-) \times (-) = (+)$$(-) \times (+) = (-)$(a) $(-3)+(-6)=-3-6$$=-(3+6)$$=-9$$(-3)-(-6)=-3+6$$=6-3$$=3$$-9-42$Therefore, $(-21)-(-10)>(-31)+(-11)$(c) $45-(-11)=45+11$$=56$$57+(-4)=57-4$$=53$$56>53$Therefore, $45-(-11)>57+(-4)$(d) $(-25)-(-42)=-25+42$$=42-25$$=17$$(-42)-(-25)=-42+25$$=-(42-25)$$=-17$$17>-17$Therefore, $(-25)-(-42)>(-42)-(-25)$ Read More

To do :We have to fill in the blanks.Solution :We know that, $(+) \times (-) = (-)$$(+) \times (+) = (+)$$(-) \times (-) = (+)$$(-) \times (+) = (-)$(a) Let the number in the blank be $x$This implies, $(-8)+x=0$Adding $+8$ on both sides, we get, $(-8)+x+8=0+8$$x+8-8=8$$x=8$Therefore, $(-8)+ 8=0$(b) Let the ... Read More

To do :We have to fill in the blank.Solution :We know that, $(+) \times (-) = (-)$$(+) \times (+) = (+)$$(-) \times (-) = (+)$$(-) \times (+) = (-)$(a) $(-7)-8-(-25)=-7-8+25$$=-(7+8)+25$$=-15+25$$=25-15$$=10$Therefore, \( (-7)-8-(-25)=10 \)(b) $(-13)+32-8-1=-13+32-8-1$$=32-(13+8+1)$$=32-22$$=10$Therefore, \( (-13)+32-8-1=10 \)(c) $(-7)+(-8)+(-90)=-7-8-90$$=-(7+8+90)$$=-105$Therefore, \( (-7)+(-8)+(-90)=-105 \)(d) $50-(-40)-(-2)=50+40+2$$=92$Therefore, \( 50-(-40)-(-2)=92 \) Read More

To do:We have to find(i) $\frac{2}{5}\div\frac{1}{2}$(ii) $\frac{4}{9}\div\frac{2}{3}$(iii) $\frac{3}{7}\div\frac{8}{7}$(iv) $2\frac{1}{3}\div\frac{3}{5}$(v) $3\frac{1}{2}\div\frac{8}{3}$(vi) $\frac{2}{5}\div1\frac{1}{2}$(vii) $3\frac{1}{5}\div1\frac{2}{3}$(viii) $2\frac{1}{5}\div1\frac{1}{5}$Solution:We know that, $\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\times \frac{c}{d}$Therefore, (i) $\frac{2}{5}\div\frac{1}{2}$$=\frac{2}{5}\times\frac{2}{1}$$=\frac{2\times2}{5\times1}$$=\frac{4}{5}$(ii) $\frac{4}{9}\div\frac{2}{3}$$=\frac{4}{9}\times\frac{3}{2}$$=\frac{4\times3}{9\times2}$$=\frac{12}{18}$$=\frac{2}{3}$(iii) $\frac{3}{7}\div\frac{8}{7}$$=\frac{3}{7}\times\frac{7}{8}$$=\frac{3\times7}{7\times8}$$=\frac{3}{8}$(iv) $2\frac{1}{3}\div\frac{3}{5}=\frac{2\times3+1}{3}\times\frac{5}{3}$$=\frac{6+1}{3}\times\frac{5}{3}$$=\frac{7}{3}\times\frac{5}{3}$$=\frac{7\times5}{3\times3}$$=\frac{35}{9}$(v) $3\frac{1}{2}\div\frac{8}{3}=\frac{3\times2+1}{2}\times\frac{3}{8}$$=\frac{6+1}{2}\times\frac{3}{8}$$=\frac{7}{2}\times\frac{3}{8}$$=\frac{7\times3}{2\times8}$$=\frac{21}{16}$(vi) $\frac{2}{5}\div1\frac{1}{2}=\frac{2}{5}\div\frac{1\times2+1}{2}$$=\frac{2}{5}\div\frac{3}{2}$$=\frac{2}{5}\times\frac{2}{3}$  $=\frac{2\times2}{5\times3}$$=\frac{4}{15}$(vii) $3\frac{1}{5}\div1\frac{2}{3}=\frac{3\times5+1}{5}\div\frac{1\times3+2}{3}$$=\frac{15+1}{5}\div\frac{5}{3}$$=\frac{16}{5}\times\frac{3}{5}$$=\frac{16\times3}{5\times5}$$=\frac{48}{25}$(viii) $2\frac{1}{5}\div1\frac{1}{5}=\frac{2\times5+1}{5}\div\frac{1\times5+1}{5}$$=\frac{11}{5}\div\frac{6}{5}$$=\frac{11}{5}\times\frac{5}{6}$$=\frac{11\times5}{5\times6}$$=\frac{11}{6}$Read More

To do:We have to find(i) $\frac{7}{3}\div2$(ii) $\frac{4}{9}\div5$(iii) $\frac{6}{13}\div7$(iv) $4\frac{1}{3}\div3$(v) $3\frac{1}{2}\div4$(vi) $4\frac{3}{7}\div7$Solution:We know that, $\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\times \frac{c}{d}$Therefore, (i) $\frac{7}{3}\div2=\frac{7}{3}\times\frac{1}{2}$$=\frac{7\times1}{3\times2}$$=\frac{7}{6}$(ii) $\frac{4}{9}\div5=\frac{4}{9}\times\frac{1}{5}$$=\frac{4\times1}{9\times5}$$=\frac{4}{45}$(iii) $\frac{6}{13}\div7=\frac{6}{13}\times\frac{1}{7}$$=\frac{6\times1}{13\times7}$$=\frac{6}{91}$(iv) $4\frac{1}{3}\div3=\frac{4\times3+1}{3}\times\frac{1}{3}$$=\frac{12+1}{3}\times\frac{1}{3}$$=\frac{13}{3}\times\frac{1}{3}$$=\frac{13\times1}{3\times3}$$=\frac{13}{9}$(v) $3\frac{1}{2}\div4$$=\frac{3\times2+1}{2}\times\frac{1}{4}$$=\frac{6+1}{2}\times\frac{1}{4}$$=\frac{7}{2}\times\frac{1}{4}$$=\frac{7\times1}{2\times4}$$=\frac{7}{8}$(vi) $4\frac{3}{7}\div7$$=\frac{4\times7+3}{7}\times\frac{1}{7}$$=\frac{28+3}{7}\times\frac{1}{7}$$=\frac{31}{7}\times\frac{1}{7}$$=\frac{31\times1}{7\times7}$$=\frac{31}{49}$Read More

To do:We have to find the reciprocal of each of the given fractions and classify the reciprocals as proper fractions, improper fractions and whole numbers.Solution:Proper fraction: A proper fraction is a fraction in which the denominator is greater than the numerator of the fraction.Improper fraction:An improper fraction is a fraction in ... Read More

To do:We have to find(i) $12\div\frac{3}{4}$(ii) $14\div\frac{5}{6}$(iii) $8\div\frac{7}{3}$(iv) $4\div\frac{8}{3}$(v) $3\div2\frac{1}{3}$(vi) $5\div3\frac{4}{7}$Solution:We know that, $\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}$Therefore, (i) $12\div\frac{3}{4}=12\times\frac{1}{\frac{3}{4}}$$=12\times\frac{4}{3}$$=\frac{12\times4}{3}$$=\frac{48}{3}$$=16$(ii) $14\div\frac{5}{6}=14\times\frac{1}{\frac{5}{6}}$$=14\times\frac{6}{5}$$=\frac{14\times6}{5}$$=\frac{84}{5}$(iii) $8\div\frac{7}{3}=8\times\frac{1}{\frac{7}{3}}$$=8\times\frac{3}{7}$$=\frac{8\times3}{7}$$=\frac{24}{7}$(iv) $4\div\frac{8}{3}=4\times\frac{1}{\frac{8}{3}}$$=4\times\frac{3}{8}$$=\frac{4\times3}{8}$$=\frac{3}{2}$(v) $3\div2\frac{1}{3}=3\div\frac{2\times3+1}{3}$$=3\div\frac{6+1}{3}$$=3\div\frac{7}{3}$$=3\times\frac{1}{\frac{7}{3}}$$=3\times\frac{3}{7}$$=\frac{3\times3}{7}$$=\frac{9}{7}$(vi) $5\div3\frac{4}{7}=5\div\frac{3\times7+4}{7}$$=5\div\frac{21+4}{7}$$=5\div\frac{25}{7}$$=5\times\frac{1}{\frac{25}{7}}$$=5\times\frac{7}{25}$$=\frac{5\times7}{25}$$=\frac{7}{5}$Read More

To do:We have to find the missing numbers.Solution:(a) (i) $\frac{2}{3}\times$________$=\frac{10}{30}$Let the missing number be $\frac{x}{y}$This implies, $\frac{2}{3}\times \frac{x}{y}=\frac{10}{30}$$\frac{2\times x}{3\times y}=\frac{10}{30}$$2x=10$ and $3y=30$$x=\frac{10}{2}$ and $y=\frac{30}{3}$$x=5$ and $y=10$Therefore, $\frac{5}{10}$ is the number in the box, such that $\frac{2}{3}\times\frac{5}{10}=\frac{10}{30}$.(ii) The simplest form of $\frac{5}{10}$ is $\frac{5}{10}=\frac{5\times1}{5\times2}$$=\frac{1}{2}$(b) (i) $\frac{3}{5}\times$______$=\frac{24}{75}$Let the missing number be $\frac{x}{y}$This implies, $\frac{3}{5}\times ... Read More

Given: A car runs 16 km using 1 litre of petrol. To do: We have to find the distance covered using $2\frac{3}{4}$ litres of petrol.Solution: Distance covered by the car using $1$ litre of petrol$=16\ km$Total quantity of petrol $=2\frac{3}{4}$$=\frac{2\times4+3}{4}$$=\frac{11}{4}$ litresDistance covered by using $2\frac{3}{4}$ litres of petrol$=16\ km\times \frac{11}{4}$ $=4\times11$$=44\ km$Thus, ... Read More