EXERCISE:2.1
Question 1
(i) Total 8 parts , 2 were shaded
$\therefore \frac{2}{8} \mathrm{th}$ part is shaded i.e $\frac{2}{8}=\frac{1}{4}$
(ii) Total 10 parts , 3 were shaded
$\therefore \frac{3}{10}$th part is shaded
(iii) Total 12 Parts , 5 were shaded.
$\therefore \frac{5}{12}$ th part is shaded.
(iv) Total 13 parts , 7 were shaded
$\therefore \frac{7}{13}$ th part is shaded.
Question 2
Let the fraction be x
$x \times 60=35$
$x=\frac{35}{60}$
$x=\frac{7}{12}$ th fraction.
Question 3
i) $2 \frac{7}{9}=2+\frac{7}{9}=\frac{2 \times 9+7}{9}=\frac{25}{9}$
ii) $5 \frac{4}{11}=5+\frac{4}{11}=\frac{5 \times 11+4}{11}=\frac{59}{11}$
Question 4
i) $\frac{73}{8}=\frac{9 \times 8+1}{8}=9+\frac{1}{8}=9 \frac{1}{8}$
ii) $\frac{94}{13}=\frac{13 \times 7+3}{13}=7+\frac{3}{13}=7 \frac{3}{13}$
Question 5
$\begin{aligned} \text { i) } \frac{3}{7}=\frac{x}{35} & \\ 3 \times 35=& 2 \times x \\ x &=\frac{3 \times 35}{{x}} \\ & x=15 \text { is missing namber } \end{aligned}$
(ii)
$\begin{aligned} \frac{5}{x}=\frac{30}{18} & \\ 5 \times 18 &=30 \times x \\ & x=\frac{5 \times 18}{30} \end{aligned}$
(iii) $\frac{x}{9}=\frac{56}{32}$
$x \times 72=9 \times 56$
$x=\frac{9 \times 56}{72}$
$x=7$ is Missing number
Question 6
(i) $\frac{48}{72}$
HCF of 48 and 72 is 24
Divide the numerator and denominator of given fraction by 24
$\frac{48}{72}=\frac{48 \div 24}{72 \div 24}=\frac{2}{3} .$
(ii) $\frac{276}{115}$
HCF of 276 , 115 is 23.
(iii) $\frac{72}{336}$
HCF of 72,336 is 24
Divide numerator and denominator by 24
$\frac{72}{336}=\frac{72 \div 29}{336 \div 24}=\frac{3}{14}$
Question 7
(i) $\mathrm{LCM}$ of $4,6,8$ is 24
$\frac{3}{4}=\frac{3 \times 6}{4 \times 6}=\frac{18}{24}$
$\frac{5}{6}=\frac{5 \times 4}{6 \times 4}=\frac{20}{24}$
$\frac{7}{8}=\frac{7 \times 3}{8 \times 3}=\frac{21}{24}$
Thus the given fractions are equivalent to $\frac{18}{24}, \frac{20}{24}, \frac{21}{24}$
(ii) $\mathrm{LCM}$ of $25,10,40$ is 200 .
$\frac{7}{25}=\frac{7 \times 8}{25 \times 8}=\frac{56}{200}$
$\frac{9}{10}=\frac{9 \times 20}{10 \times 20}=\frac{180}{200}$
$\frac{19}{40}=\frac{19 \times 5}{40 \times 5}=\frac{95}{200}$
Thus the given fractions are equivalent to $\frac{56}{200}, \frac{180}{200}, \frac{95}{200}$
Question 8
(i) LCM of 9,3,21 is 63
$\frac{2}{9}=\frac{2 \times 3}{9 \times 7}=\frac{14}{63}$
$\frac{2}{3}=\frac{2 \times 21}{3 \times 21}=\frac{42}{63}$
$\frac{8}{21}=\frac{8 \times 3}{21 \times 3}=\frac{24}{63}$
Descending order is $\frac{2}{3}, \frac{8}{21}, \frac{2}{9}$
(ii) LCM of 5,7,10=70
$\frac{1}{5}=\frac{1 \times 14}{5 \times 14}=\frac{14}{70}$
$\frac{3}{7}=\frac{3 \times 10^{-}}{7 \times 10}=\frac{30}{70}$
$\frac{7}{10}=\frac{7 \times 7}{10 \times 7}=\frac{49}{70}$
Descending order is $\frac{7}{10}, \frac{3}{7}, \frac{1}{5}$
Question 9
(i) LCM of 7,8,14,21 is 168
$\frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168}$
$\frac{3}{8}=\frac{3 \times 21}{8 \times 21}=\frac{63}{168}$
$\frac{9}{14}=\frac{9 \times 12}{14 \times 12}=\frac{108}{168}$
$\frac{20}{21}, \frac{20 \times 8}{21 \times 8}=\frac{160}{168}$
Ascending order is $\frac{3}{8}, \frac{9}{14}, \frac{5}{7}, \frac{20}{21}$
(ii) LCM of $18,15,24,12$ is 360
$\frac{13}{18}=\frac{13 \times 20}{18 \times 20}=\frac{260}{360}$
$\frac{8}{15}=\frac{8 \times 24}{15 \times 24}=\frac{192}{360}$
$\frac{17}{24}=\frac{17 \times 15}{24 \times 15}=\frac{255}{360}$
$\frac{7}{12}=\frac{7 \times 30}{12 \times 30}=\frac{210}{360}$
Ascendig order is $\frac{8}{15}, \frac{7}{12}, \frac{12}{24}, \frac{13}{18}$
EXERCISE:2.2
Question 1
(i) $\frac{4}{3}+\frac{3}{8}$
$\mathrm{LCM}$ of 3,8 is 24
$=\frac{4 \times 8+7 \times 3}{24}$
$=\frac{32+21}{24}=\frac{53}{24}$
(ii) $8 \frac{1}{2}-3 \frac{5}{8}$
$8 \frac{1}{2}=8+\frac{1}{2}=\frac{8 x 2+1}{2}=\frac{16+1}{2}=\frac{17}{2}$
$3 \frac{5}{8}=3+\frac{5}{8 .}=\frac{3 \times 8+5}{8}=\frac{24+5}{8}=\frac{29}{8}$
$=\frac{17}{2}-\frac{29}{8}$
$=\frac{17 \times 4-29 \times 1}{8}$
$=\frac{68-29}{8}$
$=\frac{39}{8}$
(iii) $\frac{5}{12}+\frac{1}{18}-\frac{2}{9}$
Lcm of $12,18,9$ is 36
$=\frac{5 \times 3+1 \times 2-2 \times 4}{36}$
$=\frac{15+2-8}{36}$
$=\frac{9}{36}$
$=\frac{1}{4}$
Question 2
(i) $7 \frac{3}{4}=7+\frac{3}{4}=\frac{7 \times 4+3}{4}=\frac{28+3}{4}=\frac{31}{4}$
$3 \frac{5}{6}=3+\frac{5}{6}=\frac{3 \times 6+5}{6}=\frac{18+5}{6}=\frac{23}{6}$
$=\frac{31}{4}-\frac{23}{6}+\frac{7}{8}$
$\begin{array}{r|l}2&4,6,8\\ \hline 2& 2,3,4 \\ \hline&1,3,2\end{array}$
LCM =24
$=\frac{31 \times 6-23 \times 4+7 \times 3}{24}$
$=\frac{186-92+21}{24}$
$=\begin{array}{l}\frac{207-92}{24} \\ =\frac{115}{24} .\end{array}$
(ii)$6 \frac{1}{8}=6+\frac{1}{8}=\frac{6 \times 8+1}{8}=\frac{49}{8}$
$2 \frac{1}{12}=2+\frac{1}{12} = \frac{2 \times 12+1}{12}=\frac{25}{12}$
$5 \frac{1}{10}=5+\frac{1}{10}=\frac{5 \times 10+1}{10}=\frac{51}{10}$
$3 \frac{7}{25}=3+\frac{7}{25}=\frac{3 \times 25+7}{25}=\frac{75+7}{25}=\frac{82}{25}$
$=\frac{49}{8}-\frac{25}{12}-\frac{51}{10}+\frac{82}{25}$
$\begin{array}{r|l}2&8,10,12,25\\ \hline 5& 4,5,6,25 \\ \hline 2&4,1,6,5\\ \hline&2,1,3,5\end{array}$
LCM = 600
$=\frac{49 \times 75-25 \times 50-51 \times 60+82 \times 24}{600}$
$=\frac{1333}{600}$
Question 3
Jaishree studies daily for $5 \frac{2}{3}$ hours
i.e $5+\frac{2}{3}=\frac{5 \times 3+2}{3}=\frac{17}{3}$ hours
She devoted has time for science and mathematics $2 \frac{4}{5}=2+\frac{4}{5}=\frac{2 \times 5+4}{5}=\frac{14}{5}$ hours
Total time = $\frac{17}{3}$ hrs
Let the time for other subject be x
$\begin{aligned} \frac{14}{5}+x &=\frac{17}{3} \\ x &=\frac{17}{3}-\frac{14}{5} \\ x &=\frac{17 \times 5-14 \times 3}{15} \end{aligned}$
$x=\frac{85-42}{15}$
$x=\frac{43}{15}$ hrs for other subjects
Question 4
Ramesh solved $\frac{2}{7}$ part of an exercise
Reshma solved $\frac{4}{5}$ part
LCM of 7, 5 is 35
$\frac{2}{7}=\frac{2 \times 5}{7 \times 5}=\frac{10}{35} ; \frac{4}{5}=\frac{4 \times 7}{5 \times 3}=\frac{28}{35}$
$\frac{4}{5}>\frac{2}{7}$
i.e Ramesh Solved lesser part and by $\frac{28}{35}-\frac{10}{35}$ i.e $\frac{18}{35}$ part
Question 5
Sohali had Rs $35 \frac{3}{5}$ i.e $\frac{35 \times 5+3}{5}$ = Rs $\frac{178}{5}$
She got Rs $16 \frac{1}{15}$ From her mother
i.e $\frac{16 \times 15+1}{15}=₹ \frac{241}{15}$
She spent on food Rs $28 \frac{2}{3}=\frac{28 \times 3+2}{3}$
=Rs $\frac{86}{3}$
Money left = ?
Total money she had = Rs $\frac{178}{5}+\frac{241}{15}$
$=\frac{178 \times 3+241 \times 1}{15}$
$=\frac{534+241}{15}$
$=\frac{275}{15}$
Out of which she spent Rs on food i.e $\frac{86}{3}$
ஃ Money left = Rs $\frac{775}{15}-\frac{86}{3}$
=$\frac{775 \times 1-8 \times 5}{15}$
=$\frac{77-430}{15}$
=$\frac{345}{15}$
=$₹ 23$
ஃ She had left with Rs 23 with her
EXERCISE:2.3
Question 1
(i) $7 \times \frac{3}{5}=\frac{7 \times 3}{5}=\frac{21}{5}$
ii) $21 \times \frac{3}{14}=\frac{3 \times 3}{2}=\frac{9}{2}$
iii) $3 \frac{2}{5} \times 8=\frac{17}{5} \times 8=\frac{136}{5}$
iv) $5 \times 6 \frac{3}{4}= 5 \times \frac{27}{4}=\frac{135}{4}$
Question 2
i)$\frac{2}{3} \times 18=2 \times 6=12$
ii) $\frac{1}{2} \times 4 \frac{2}{9}=\frac{1}{7} \times \frac{38}{9}=\frac{19}{9}$
iii)$\frac{5}{8} \times 9 \frac{2}{3}=\frac{5}{8} \times \frac{29}{3}=\frac{145}{24}$
Question 3
i) $\frac{3}{7} \times \frac{5}{9}=\frac{5}{7 \times 3}=\frac{5}{21}$
ii) $\frac{2}{5} \times 5 \frac{1}{4}=\frac{2}{5} \times \frac{21}{4}=\frac{21}{10}$.
iii) $2 \frac{1}{3} \times 3 \frac{4}{21}=\frac{7}{3} \times \frac{109}{21}=\frac{109}{9}$
iv) $3 \frac{1}{6} \times 7 \frac{4}{23}=\frac{19}6 \times \frac{16555}{23}=\frac{19 \times 55}{2 \times 23}=\frac{1045}{46}$.
Question 4
i) $\frac{1}{3} \times 42=Rs 14$
ii) $\frac{3}{7} \times 5 \frac{1}{4}=\frac{3}{7} \times \frac{21}{4}=\frac{3 \times 3}{4}=\frac{9}{4} \mathrm{~kg}$
iii) $4 \frac{1}{2} \times 5 \frac{1}{2}=\frac{9}{2} \times \frac{11}{2}=\frac{99}{4} \cdot$ metres
Question 5
(i)$\frac{2}{2} \times \frac{3}{4} \quad ; \quad \frac{3}{8} \times \frac{3}{8}$
$=\frac{3}{7 \times 2} \quad ; \quad \frac{3 \times 1}{1 \times 8}$
$=\frac{3}{14} \quad ; \quad \frac{3}{8}$
$\frac{3}{14}=\frac{3 \times 4}{14 \times 4}=-\frac{12}{56} \quad ; \frac{3}{8}=\frac{3 \times 7}{8 \times 7}=\frac{21}{56} .$
$\therefore 21>12 .$
$\therefore \frac{3}{8}$ is greater than $\frac{3}{14}$
(ii) $\frac{1}{7} \times \frac{6}{7} ; \quad \frac{2}{3} \times \frac{3}{7}$
$=\frac{1 \times 3}{7} \quad ; \quad \frac{2 \times 1}{7}$
$=\frac{3}{7} \quad ; \frac{2}{7}$
As $\frac{3}{7},>\frac{2}{7} \quad(\because 3>2)$
Question 6
Given, 1 metre cloth costs Rs $31 \frac{3}{4}$ = Rs $\frac{127}{4}$
$5 \frac{1}{2}$ metres cloth costs = ?
Let it be equal = x
$5 \frac{1}{2}=\frac{5 \times 2+1}{2}=\frac{11}{2}$
$\frac{11 / 2}{1}=\frac{x}{(127 / 4)}$
$\frac{11}{2}=\frac{x \times 4}{127}$
$x=\frac{11 \times 127}{2 \times 4}$
X = $=\frac{1397}{8}$ = Rs $174 \frac{5}{8}$
$\therefore 5 \frac{1}{2}$ metre cloth costs is $174 \frac{5}{8}$
Question 7
Speed of a car = $\operatorname{105} \frac{1}{5} \mathrm{kmph}$
$=\frac{526}{5} \mathrm{kmph}$
Time taken = $3 \frac{3}{5}$ hours$=\frac{18}{5}$ hours
$\begin{aligned} \text { Distance } &=\text { speed } \times \text { Time } \\ &=\frac{526}{5} \times \frac{18}{5} \mathrm{~km} \end{aligned}$
Distance $=\frac{9468}{25} \mathrm{~km}$
Question 8
For 1 litre of petrol , the car runs 16km
For $2 \frac{3}{4}$ litres of petrol , the car runs ? km
Let the car runs be x km
$2 \frac{3}{4}=\frac{11}{4}$
i.e $\frac{x}{16}=\frac{11 / 4}{1}$
$\frac{x}{16}=\frac{11}{4}$
$x=\frac{11 \times 16}{4}$
$x=44 \mathrm{~km}$
Question 9
In one hour, Sushant reads $\frac{1}{3}$ part of book
Let In $2 \frac{1}{5}=\frac{11}{5}$ Hours, he reads x part of book
i.e
$\begin{aligned} \frac{x}{(1 / 3)} &=\frac{(11 / 5)}{(1)} \\ 3 x &=\frac{11}{5} \\ x &=\frac{11}{3 \times 5} \end{aligned}$
$x=\frac{11}{15}$ part of book
Question 10
Gold and copper ornament weigh = 52grams
Out of which, part of copper is $\frac{2}{13} \mathrm{th}$
Then the copper weighs -$\frac{2}{13}, \times 52$
= 8gms
Remaining Gold weighs = 52- 8
= 44grams
Question 11
Total no.of students in a class = 40
(i) Out of them, students like to study = $\frac{1}{5}th$ of total
$=\frac{1}{5} \times 40$
$=8$
(ii) Students like to study mathematics = $\frac{2}{5} th$ of Total
$=\frac{2}{5} \times 40$
=16
(iii) Fraction of total no . of students like to study
Science = $1-\left\{\frac{1}{5}+\frac{2}{5}\right\}$
$1-\left\{\frac{1+2}{5}\right\}$
=$1-\frac{3}{5}$
$=\frac{5-3}{5}$
$=\frac{2}{5}$
Fraction of students who like to study science = $\frac{2}{5} \th$ of total
Question 12
A rectangular sheet is having length = $12 \frac{1}{2}(m)=\frac{25}{2} c m=L$
Width B= $10 \frac{2}{3} \mathrm{~cm}=\frac{32}{3}$Cm
(i) We know perimeter of rectangle = $2 \times($ Length + width $)$
$=2 \times\left(\frac{25}{2}+\frac{32}{3}\right)$
$=2 \times\left(\frac{25 \times 3+32 x^{2}}{6}\right)$
$=2 \times\left(\frac{75+64}{6}\right)$
$=2 \times\left(\frac{139}{6}\right)$
∴ Perimeter of rectangle = $\frac{139}{3}$ cm
(ii) Area of rectangle = length $\times$ width
$=\frac{25}{2} \times \frac{32}{3}$
$=\frac{25 \times 16}{7 \times 1}$
$=\frac{400}{3}$
$=133 \frac{1}{3} \mathrm{~cm}^{2}$
∴ Area of rectangle = $133 \frac{1}{3} \mathrm{~cm}^{2}$.
Question 13
Part of students are girls = $\frac{25}{54}$
Then part of students are boys = $1-\frac{25}{54}$
$=\frac{54-25}{54}$
$=\frac{29}{54}$
Given number of boys is 2030
Let Total no. of students = x
Now, $\frac{29}{54} \times x=2030$
$x=\frac{2030 \times 54}{29}$
$x=3780$
∴ Total no. of students = 3780
No. of girls = Total -No of boys
= 3780 - 2030 = 1750
Question 14
Let number of trees in an archard = 'x'
Given banana trees are 148 in number
Also orange trees are $\frac{1}{5}$th total
Mango trees are $\frac{3}{13}$th total
Now, the part of banana trees= $1=\left\{\frac{1}{5}+\frac{3}{13}\right\}$
=$1-\frac{13+3 \times 5}{65}$
=$1-\frac{13+15}{65}$
$=1-\frac{28}{65}$
$=\frac{65-28}{65}$
$=\frac{37}{65}$
Now , as banana Trees = 148
ஃ $\frac{37}{65}$ th of total = 148
i.e
$\begin{aligned} \frac{37}{65} \times x &=148 \\ x &=\frac{148 \times 65}{3.7}=260 . \end{aligned}$
ஃ Total no. of trees in an archard = 260
EXERCISE:2.4
Question 1
(i) Reciprocal of $\frac{3}{7}=\frac{1}{(3 / 7)}=\frac{7}{3} .$
ii) Reciprocal of $\frac{13}{9}=\frac{1}{\left(\frac{13}{9}\right)}=\frac{9}{13}$
iii) Reciprocal of $8=\frac{1}{8}$
Question 2
i)
$\begin{aligned} 14 \div \frac{5}{6} &=\frac{14}{(5 / 6)} \\ &=\frac{14 \times 6}{5} \\ &=\frac{84}{5}=16 \frac{4}{5} . \end{aligned}$
ii)
$\begin{aligned} 5 \div 3 \frac{4}{7} &=5 \div \frac{25}{7} \\ &=\frac{5}{(25 / 7)} \\ &=\frac{15 \times 7}{255} \\ &=\frac{7}{5} . \end{aligned}$
iii)$3 \frac{1}{5} \div 1 \frac{2}{3}=\frac{16}{5} \div \frac{5}{3} .$
$=\frac{16 / 5}{5 / 3}$
$=\frac{16}{5} \times \frac{3}{5}$
$=\frac{48}{25}=1 \frac{23}{25}$
iv)
$\begin{aligned} 2 \frac{5}{8} \div 1 \frac{1}{6} &=\frac{21}{8} \div \frac{7}{6} \\ &=\frac{21 / 8}{7 / 6} \\ &=\frac{27}{8} \times \frac{6}{7} \\ &=\frac{9}{4}=2 \frac{1}{4} . \end{aligned}$
Question 3
Total length of cloth = 7 7 $\frac{1}{2}$ metres
$=\frac{155}{2}$
Each piece of cloth length requried = $5 \frac{1}{6}$
$\frac{31}{6}$
∴ Now total no. of pieces of cloth = $\frac{155 / 2}{31 / 6}$
$\frac{155}{2} \times \frac{6}{31}$
= 15pieces
Question 4
Let the number multiplied to $4 \frac{7}{8}$ be $x$
$4 \frac{7}{8}=4+\frac{7}{8}=\frac{4 \times 8+7}{8}=\frac{32+7}{8}=\frac{39}{8}$
Required answer = $87 \frac{3}{4}$
i.e $\frac{39}{8} \times x=87+\frac{3}{4}$
$\frac{39 x}{8}=\frac{87 \times 4+3}{4}$
$\frac{39x}{8}=\frac{351}{4}$
$x=\frac{351+8}{4 \times 39}$
$x=9 \times 2$
x=18
∴ Number multiplied = 18
Question 5
Part of milk each students gets = $\frac{1}{3}$ th of Total
Total consumption of milk = $57 \frac{2}{3}$ litres
$=\frac{57 \times 3+2}{3}$
$=\frac{123}{3}$
Let the number of students in hosted b 'x '
$\frac{1}{3} \mathrm{th}$ of Total $=57 \frac{2}{3} .$
$\frac{1}{3} \times x=\frac{173}{3}$
$x=\frac{173 \times 3}{3 \times 1}$
x=173
∴ Total number of students in hostel = 173
Question 6
Given cost of $5 \frac{1}{4}$ kg apple =Rs 336
Let Cost of 1kg apples = x
$5 \frac{1}{4}=5+\frac{1}{4}=\frac{5 \times 4+1}{4}=\frac{21}{4} .$
$\begin{aligned} \frac{x}{336} &=\frac{1}{(21 / 4)} \\ \frac{x}{336} &=\frac{4}{21} \\ x &=\frac{4 \times 336}{21} \\ x &=64 \end{aligned}$
∴ Cost of apples = Rs 64 per kg
Question 7
Given area of rectangular plot = $68 \frac{3}{4} \operatorname{sq} \cdot m$
$=\frac{68 \times 4+3}{4}$
$=\frac{275}{4} \mathrm{sqm}$
Also length of rectangular plot = $12 \frac{1}{2} \mathrm{~m}$
$=\frac{12 \times 2+1}{2}$
$=\frac{25}{2} m$
We know Area of rectangle $A=$ Lensth $\times$ width
width $\times \frac{27}{2}=\frac{275}{4}$
$\quad$ width $=\frac{275 \times 2}{25 \times 4}$
$=\frac{11}{2}$
$\therefore$ Width $=5 \frac{1}{2} m$
Question 8
Given Cost of $5 \frac{1}{2}=\frac{11}{2} \mathrm{~kg}$ of sugar $=Rs 206 \frac{1}{4}$
$=\frac{206 \times 4+1}{4}$
=Rs $\frac{825}{4}$
Let Cost of $8 \frac{1}{4}=\frac{33}{4}$ Kg of sugar be 'x '
$\frac{x}{\left(\frac{825}{4}\right)}=\frac{33 / 4}{11 / 2}$
$\frac{4 x}{825}=\frac{33}{4} \times \frac{2}{11}$
$x=\frac{33}{4} \times \frac{2}{11} \times \frac{825}{4}$
$x=\frac{3 \times 825}{2 \times 4}$
$x= \frac{2425}{8}= 309 \frac{3}{8}$
∴ Cost of $8 \frac{1}{4}$ kg sugar = Rs $309 \frac{3}{8}$
Question 9
In 2 hours, Renu Cimpleted $\frac{2}{2}$ part of work.
Let
In $1 \frac{1}{4}$ hours, RenU Completes $x^{\prime}$ past of work
1 $\frac{1}{4}=1+\frac{1}{4}=\frac{\mid x y+1}{4}=\frac{5}{y}$
i.e $\frac{x}{(2 / 3)}=\frac{5 / 4}{2}$
$x=\frac{5}{4 \times 2} \times \frac{2}{3} .$
$x=\frac{5}{12}$
EXERCISE:2.5
Question 1
i) $20.03=2 \times 10+0 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{100}$
ii) $200.03=2 \times 100+0 \times 10+0 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{100}$
iii) $2.034=2 \times 1+0 \times \frac{1}{10}+3 \times \frac{1}{100}+4 \times \frac{1}{1000}$.
Question 2
i) Place value of digit 2 in $2.56$ is $2 \times 1=2$
ii) place value of digit 2 in $21.37$ is $2 \times 10=20$
iii) place value of digit 2 in $10.25$ is $2 \times \frac{1}{10}=\frac{2}{10}$.
iv) place Value of digit 2 in $63 \cdot 352$ is $2 \times \frac{1}{1000}=\frac{2}{1000}$.
Question 3
i) $0.8=\frac{8}{10}=\frac{4}{5}$
ii) $0.225=\frac{225}{1000}=\frac{9}{40}$
iii) $0.0092=\frac{92}{10000}=\frac{23}{2500}$
iv) $3.025=\frac{3025}{1000}=\frac{121}{40}$
Question 4
i) $5.05=\frac{505}{100}=5+\frac{5}{100}=5+\frac{1}{20}=5 \frac{1}{20}$
ii) $63.125=63+\frac{125}{1000}=63+\frac{1}{8}=63 \frac{1}{8}$
iii) $17.075=17+\frac{75}{1000}=17+\frac{3}{40}=17 \frac{3}{40}$.
iv) $317.0006=312+\frac{6}{10000}=317+\frac{3}{5000} = 317 \frac{3}{5000}$
Question 5
i) $\frac{3}{5}=\frac{3 \times 2}{5 \times 2}=\frac{6}{10}=0.6$
ii) $\frac{7}{8}=\frac{7 \times 125}{8 \times 125}=\frac{875}{1000}=0.875$
iii) $3 \frac{5}{16}=\frac{3 \times 16 \p 5}{16}=\frac{53}{16}=\frac{53 \times 625}{16 \times 10000}=\frac{33125}{10000}=3.3125$
iv) $137 \frac{13}{625}=137+\frac{13}{625}=137+\frac{13 \times 16}{625 \times 16}=137+\frac{208}{10000}$
$=137+0.0208$
$=137.0208$
Question 6
(i) 0.5 or 0.05 are like decimal mumbers
We know that whole number parts of given number are equal
Let us their digits at tenth place
In both number , digits at tenth place in 0.5 is 5 and digits at tenth place in 0.05 is 0
Since $5>0$, So 0.5 $>0.05 .$
(ii)Given numbers are 7,0.7
Whole numbers parts of given numbers are not equal
In number 7 , the whole number is 7
In number $0.7$, the whole number is 0 .
since $7>0$, so $7>0.7$
(iii) Given numbers are 2.03 or 2.30
We note that whole number parts of given number are equal
Let us compare their tenth digits
In 2.03 digits at tenth place is 0
In 2.30 digits at tenth place is 3
Since $3>0$, so $2.30>2.03$
(iv) Given numbers are 0.80 or 0.88
We know that whole number parts of given number are equal
Let us compare their tenth digits
In both number , digits at tenth place = 8
So, we compare their hundreth place
In 0.80 digits at hundreth place = 0
In 0.88 , digits at hundredth place = 8
Since $8>0$, so $0.88>0.80$.
Question 7
i) Campare whole numbers
$83>38>3 \quad$ (or) $\quad 3<38<83$
Compare tenth digits numbers in $38,02,38 \cdot 021,38.002$
In $38.02$, the tenth digit is 0
In $38.021,38.002$, the tenth digit is 0
Compare hundredth digits
In $38.02,38.021$, he hundred th digit is 2
But In $38.002$, the hundred th digit is 0
$\quad since \quad 0<2, \quad 38 \cdot 002<(38.02,38\cdot 021)$
Compare thousandth's place in 38.020, 38.021
In 38 .020 the thousandth place is 0
In 38.021 the thousandth place is 1
Since $0<1$, So $38.020<38.021$
∴ So the ascending order is $3.802<38.002<38.420<38.21$
$<83: 02$
(ii) In all the given numbers except $64.542$, all are having
46 as their whele number.
So $64.542$ is the highest $(m)$ largest number
Comparing tenth digits
In $46.542$, the tenth digit is 5
In $46.452$, the tenth digit is 4
In $46.254$, the tenth digh is 2
Since $2<4<5$, so $46.254<46.452<46542$
Compare hundredth and thousand place in 46.05, 46.0542
In both numbers hundredth and tenth digits is same 5,0
In 46.050,the thousandth digit is 0
In 46.0542, the thousandth digits is 4
Since $0<4$, so $46.050<46.0542$
∴ Ascending order is $46.050<46.0542<46.254<46.452<46.542<64.542$
Question 8
(i) We know that $5>1>0$. (whole numbers)
Now $1.87,1.9,1.78$
The tenth digit are 8 , 9 , 7
Since $9>8>7$, so $1.9>1.87>1.78$
Now In 0.93, 0.39
The tenth digits are 9, 3
Since $9>3$, so $0.93>0.39$
∴ Desending order is $5.6>1.9>1.87>1.78>0.93>0.39$
(ii) We know that $71 \geqslant 20>3>2>0$ in whole number
Compare Tenth digits in $2.01,2.14$
In $2.01$, The tenth digit is 0
In $2.14$, the tenth disit is 1
since $1>0$, so $2.14>2.01$
∴ Descending order is $71.201>20.1>3.1>2.14>2.01>0.652$
Question 9
(i)
$\begin{aligned} 1 \text { Rupee } &=100 \text { paise } \\ \text { So } 1 \text { paise } &=\frac{1}{100} \text { rupee } \end{aligned}$
$\begin{aligned} 7 \text { paise }=7 \times \frac{1}{100} \text { rupee } &=\frac{7}{160} \text { rypee } \\ &=0.07 \text { rupee } \end{aligned}$
(ii) 77 rupees 77 paise
$=\left[77+77 \times \frac{1}{100}\right] \mathrm{rupee}$
$=\left[77+\frac{77}{100}\right]$ rupee
$=[27+0.77]$ ruper
$=77.77$ rupee
(iii) 235 paise $=235 \times \frac{1}{100}$ ruper $=$ 2.35. rupee
Question 10
We know 1m =100cm $\Rightarrow k m=\frac{1}{100} m$
$1 \mathrm{~km}=1000 \mathrm{~m} \Rightarrow \mathrm{1m}=\frac{1}{1000} \mathrm{~km}$
5cm =$5 \times \frac{1}{100} m=0.05 \mathrm{~m}$
$0.05 m=\frac{5}{100} m=\frac{5}{100} \times \frac{1}{1000} km=0.00005 \mathrm{~km}$\
Question 11
We know 1kg =100g
So $\quad 1 g=\frac{1}{1000} k_{g}$
i) $200 g=200 \times \frac{1}{1000} k_{g}=\frac{2}{10} k_{9}=0.2 \mathrm{~kg}$
ii) $3470 g=347 \times \times \frac{1}{1000} k_{9}=\frac{347}{100} \mathrm{~kg}=3.47 \mathrm{~kg}$
iii)$4 \mathrm{~kg} 8 \mathrm{~g}= \left(4+\frac{8}{1000}\right) \mathrm{kg}=(4+0.008) \mathrm{kg}=4.008 \mathrm{~kg}$
Question 12
i) $S u m=5.765+9.2+3.08$
$\begin{array}{r}5.765 \\+9.200 \\+3.080 \\\hline 18.045\end{array}$
(ii) $\begin{array}{r}15.4900 \\+8.3572 \\+0.9030 \\+\quad 7.8000 \\\hline 32.5502\end{array}$
Question 13
(i)
$\begin{array}{r}72.530 \\-\quad 46.782 \\\hline 25.748\end{array}$
(ii)Given the positive and negative numbers seperately
Positive numbers Negative numbers
18.376 $\begin{array}{r}5.4300 \\+8.8976 \\\hline 14.3276\end{array}$
$\begin{array}{r}18.3760 \\-14.3276 \\\hline4.0484\end{array}$
(iii) positive number
28. 5
$\begin{array}{l}\text { Negative number } \\9.708 \\+6.234 \\\hline 15.942\end{array}$
$\begin{array}{r}28.500 \\-15.942 \\\hline=12.558\end{array}$
(iv) Positive numbers
$\begin{array}{l}8.20 \\2.67 \\\hline 10.87\end{array}$
Negative numbers
$\begin{array}{r}4.5600 \\+0.7912 \\\hline 5.3512\end{array}$
$\begin{array}{r}10.8700 \\-5.3512 \\\hline5.5188\end{array}$
Question 14
i) Let the number added = x
$\begin{aligned} x+3.56 &=13.016 \\ x &=13.016-3.56 \end{aligned}$
$\begin{array}{r}13.016 \\-3.560 \\\hline 9.456\end{array}$
x = 9.456
(ii) Let the number Substracted be x
$\begin{aligned} 30-x &=23.709 \\ x &=30-23.709 \end{aligned}$
$\begin{array}{r}30.000 \\-23.709 \\\hline 6.291\end{array}$
x=6.291
(iii) Excess of 20.4 over 9.7403 is
= 20.4-9.7403
$\begin{array}{r}20.4000 \\-9.4403 \\\hline 10.6597\end{array}$
EXERCISE:2.6
Question 1
\text { i) } 2.7 \times 4=\frac{27}{10} \times 4
$=\frac{108}{10}=10.8$
ii) $2.71 \times 5=\frac{271}{100} \times 5$
$=\frac{1355}{100}=13.55 .$
iii) $2.5 \times 0.3=\frac{25}{10} \times \frac{3}{10}$
$=\frac{75}{100}=0.75$
iv)$2.3 \times 4.35=\frac{23}{10} \times \frac{435}{100}$
$=\frac{10005}{1000}=10: 005$
v) $238.06 \times 7.5=\frac{23806}{100} \times \frac{75}{10}=\frac{1785450}{1000}=1785.45 .$
vi)$0.79 \times 32.4=\frac{79}{100} \times \frac{324}{10}=\frac{25,596}{1000}=25.596$
vii) $1.07 \times 0.02=\frac{107}{100} \times \frac{2}{100}=\frac{214}{10000}=0.0214 .$
viii) $10.05 \times 1.05=\frac{1005}{100} \times \frac{105}{100}=\frac{105.525}{10000}=10.5525$
Question 2
i) (diagram to be added)
= 2.7
ii) $126.35 \div 7$
(diagram to be added)
Ans = 18.05
iii) $22.5 \div 1.5=\frac{22.5}{1.5} \times \frac{10}{10}=\frac{225}{15}$
(diagram to be added)
Hence, $22.5 \div 1.5=15$.
iv) $4 \cdot 28 \div 0.02=\frac{4.28}{0.02} \times \frac{100}{100}=\frac{428}{2}$
(diagram to be added)
Hence, $4.28 \div 0.02=214$.
v)$3.645 \div 1.35=\frac{3.645}{1.350} \times \frac{1000}{1000}=\frac{3645}{1350}$
(diagram to be added)
Hence, $3.645 \div 1.35=2.7$
vi)$0.728 \div 0.04=\frac{0.728}{0.040} \times \frac{1000}{1000}=\frac{782}{40}$
$=\frac{182}{10}$
$=18.2$
vii)$13.06 \div 0.08=\frac{13.01}{0.08} \times \frac{100}{100}=\frac{1306}{8}$
(diagram to be added)
Hence, $13.06 \div 0.08=163.25$.
viii)$58.635 \div 4.5=\frac{58.635}{4.500} \times \frac{1000}{1000}=\frac{58635}{4500}$
(diagram to be added)
Hence, $58.635 \div 4.5=13.03$
Question 3
i) $5.9 \times 10=59$
$5.9 \times 100=590$
$5.9 \times 1000=5900$
To Multiply by 10, shift decimal point to the right by one place
ii) $3.76 \times 10=37.6$
$3.76 \times 100=376$
$3.76 \times 1000=3760$
iii) $0.549 \times 10=5.49$
$0.549 \times 100=54.9$
$0.549 \times 1000=549$
Question 4
To divide a decimals number by 10,100,1000 Shift decimal point to the left by one, two three places.
i) $4.8 \div 10=0.48 \quad 4.8 \div 1000=0.0048$
$4.8 \div 100=0.048$
ii)
$\begin{aligned} 38.53 \div 10 &=3.853 \\ 38.53 \div 100 &=0.3853 \\ 38.53 \div 1000 &=0.03853 . \end{aligned}$
iii) $128.9 \div 10=12.89$
$128.9 \div 100=1.289$
$128.9 \div 1000=0.1289$
Question 5
Given length = 5.7cm ; Breadth = 3.5cm
Area of rectangle A =Length $\times$ Breadth
$=5.3 \times 3.5$
$=\frac{57}{10} \times \frac{35}{10}$
$=\frac{1995}{100}$
$=19.95 \mathrm{Cm}^{2} .$
Question 6
Given, cost of one metre cloth is Rs 38.50
Cost of 3.6 metre cloth is $3.6 \times 38.5$
$=\frac{36}{10} \times \frac{355}{10}$
$=\frac{13860}{100}$
$=₹ 138.60$
Question 7
One liter of petrol covers distance $=45.3 \mathrm{~km}$
$5.9$ litres of petrol covers distance $=5.9 \times 45.3 \mathrm{~km}$
$=\frac{59}{10} \times \frac{453}{10} \mathrm{~km}$
$=\frac{26727}{100} \mathrm{~km}$
$=267.27 \mathrm{~km}$
Question 8
$1 \mathrm{~kg}$ of pure milk contains $=0.245 \mathrm{kg}$ fat
$12.8 \mathrm{~kg}$ of milk contains $=12.8 \times 0.245$'
=$\frac{128}{10} \times \frac{245}{1000}$
=$\frac{31360}{10000}$
=$3.136 \mathrm{~kg}$
Question 9
6 Children Shared equally
Total rupees wik them ar $\overline{2} 242 \cdot 46$.
Then each Childred shared $=5 \frac{242.46}{6}$
$=\frac{\left(\frac{24246}{100}\right)}{6}$
$= \frac{24246}{100 \times 6}$
$= \frac{4041}{100}$
Rs $40.41$
∴ Each Children shared Rs 40.41
Question 10
$2.4$ litres of petrol coress a distance $=43.2 \mathrm{~km}$
one litre of petrol Covers a distance $=\frac{43.2}{2 \cdot 4} \mathrm{~km}$
$=\frac{432 / 10}{24 / 10}$
$=\frac{432}{10} \times \frac{10}{24}$
$=\frac{432}{24}$
=18km
Question 11
Total 8.4 litres of icecream
One cone can be filled with 35 millilitres of icecream
Number of icecream cones can be filled
$=\frac{8.4 \times 1000 \mathrm{ml}}{35 \mathrm0{ml}}$ (As 1 litre = 1000ml)
$=\frac{8400}{35}$
Number of icecream cones = 240
Question 12
Product of two decimal numbers = 38.745
One of the number is 2.7
Let the other number be x
i.e $x \times 2.7=38.745$
$-x=\frac{38.745}{2.7}$
$x=\frac{38745 / 1000}{27 / 10}$
$x=\frac{38745}{1000} \times \frac{10}{27}$
$x=\frac{1435}{100}$
$x=14.75$
The other number is 14.35
Question 13
Let the number be 'x'
Given $\frac{2}{3}$ of a number is 10
$\begin{aligned} \frac{2}{3} \times x &=10 \\ x &=\frac{50 \times 3}{2} \\ x &=15 \end{aligned}$
The number x = 15
∴ Now 1.75 times of number = $=1.75 \times 15$
$=\frac{135}{100} \times 15$
$=\frac{2625}{100}$
=26.25
EXERCISE:2.7
Question 1
i) $1 \frac{1}{9}=1+\frac{1}{9}=\frac{9+1}{9}=\frac{10}{9}$
$3 \frac{1}{2}=3+\frac{1}{2}=\frac{6+1}{2}=\frac{7}{2}$
$\frac{3}{5}$ if $1 \frac{1}{9}=\frac{3}{5} \times \frac{10}{9}=\frac{2}{3}$
So $\frac{2}{3}+\frac{7}{2}=\frac{2 x^{2}+7 \times 3}{6}=\frac{4+21}{6}=\frac{25}{6}=4 \frac{1}{6}$
ii)$2 \frac{3}{8}=2+\frac{3}{8}=\frac{16+3}{8}=\frac{19}{8}$
$\frac{4}{5} \times 2 \frac{3}{8}=\frac{4}{5} \times \frac{19}{82}=\frac{19}{10}$
$2 \times \frac{3}{5}=\frac{6}{5} \times \frac{2}{2}=\frac{12}{10}$
$50, \frac{19}{10}-\frac{12}{10}=\frac{19-12}{10}=\frac{7}{10}$
iii) $\frac{4}{5}+2=\frac{4+2 \times 5}{5}=\frac{4+10}{5}=\frac{14}{5}$
$3-\frac{2}{3}=\frac{3 \times 3-2}{3}=\frac{9-2}{3}=\frac{7}{3}$
So, $\frac{14}{5} \times \frac{7}{3}=\frac{98}{15}=6 \frac{8}{15}$
Question 2
i) $2 \frac{2}{3}=2+\frac{2}{7}=\frac{2 x-2}{7}=\frac{16}{7} .$
$\frac{1}{4}$ of $2 \frac{2}{7}=\frac{1}{4} \times \frac{16}{7}=\frac{4}{7} .$
So, $\frac{4}{7} \div \frac{3}{5}=\frac{4 / 7}{3 / 5}$
$=\frac{4}{7} \times \frac{5}{3}=\frac{20}{21} .$
ii) $\frac{3}{7} \div \frac{1}{2}=\frac{3 / 7}{1 / 2}=\frac{3}{7} \times 2=\frac{6}{7}$
So $\frac{6}{7} \div \frac{7}{8}=\frac{6 / 7}{7 / 8}=\frac{6}{7} \times \frac{8}{7}=\frac{48}{49} .$
iii) $\frac{5}{8} \div \frac{3}{4}=\frac{5 / 8}{3 / 4}=\frac{5}{8} \times \frac{4}{3}=\frac{5}{6}$
so, $\frac{5}{6}+\frac{2}{5}=\frac{5 \times 5+2 \times 6}{30}=\frac{25+12}{30}=\frac{37}{30}=1$
$ \frac{7}{30} .$
Question 3
i) $4 \frac{1}{2}=4+\frac{1}{2}=\frac{4 \times 2+1}{2}=\frac{9}{2}$
$2 \frac{2}{3}=2+\frac{2}{3}=\frac{2 \times 3+2}{3}=\frac{8}{3}$
$5 \frac{1}{2}=5+\frac{1}{2}=\frac{5 \times 2+1}{2}=\frac{11}{2}$
$3 \frac{5}{6}=3+\frac{5}{6}=\frac{3 \times 6+5}{6}=\frac{18+5}{6}=\frac{23}{6}$
$\left(4 \frac{1}{2}-2 \frac{2}{3}\right)=\frac{9}{2}-\frac{8}{3}=\frac{9 \times 3-8 \times {2}}{6}=\frac{27-16}{6}=\frac{11}{6}$
$5 \frac{1}{2}$ of $3 \frac{5}{6}=\frac{11}{2} \times \frac{23}{6}=\frac{253}{12} .$
So, therefore $\frac{11}{6} \div \frac{7}{12}+\frac{253}{12}$
$=\frac{11}{6} \times \frac{12}{3}+\frac{253}{12}$
$=\frac{22}{7}+\frac{253}{12}$
$=\frac{22 \times 12+253 \times 7}{84}$
$=\frac{2075}{84}$
$=24 \frac{19}{84}$
ii)$5 \frac{1}{3}=5+\frac{1}{3}=\frac{5 x]+1}{3}=\frac{16}{3} .$
=$2 \frac{1}{2}=2+\frac{1}{2} \cdot \frac{2 x^{2}+1}{2}=\frac{5}{2} .$
=$\left(\frac{1}{2}+\frac{1}{3}\right) \div\left(\frac{1}{4}-\frac{1}{6}\right)-\left[8-\left\{5 \frac{1}{3}-\left(3-2 \frac{1}{2}\right)\right\}\right]$
$=\left(\frac{3+2}{6}\right) \div\left(\frac{3-2}{12}\right)-\left[8-\left\{\frac{16}{3}-\left(3-\frac{5}{2}\right)\right\}\right]$
$=\left(\frac{5}{6}\right) \div\left(\frac{1}{12}\right)-\left[8-\left\{\frac{16}{3}-\left(\frac{6-5}{2}\right)\right\}\right]$
$\left(4 \frac{1}{2}-2 \frac{2}{3}\right)=\frac{9}{2}-\frac{8}{3}=\frac{9 \times 3-8 x^{2}}{6}=\frac{27-16}{6}=\frac{11}{6} .$
$5 \frac{1}{2}$ of $3 \frac{5}{6}=\frac{11}{2} \times \frac{23}{6}=\frac{253}{12} .$
So, therefore $\frac{11}{6} \div \frac{7}{12}+\frac{253}{12}$
$=\frac{11}{8} \times \frac{12}{7}+\frac{253}{12}$
$\frac{22}{7}+\frac{253}{12}$
$=\frac{22 \times 12+253 \times 7}{84}$
$=\frac{2075}{84}$
$=24 \frac{19}{84} .$
iii) $5 \frac{1}{3}=5+\frac{1}{3}=\frac{5 \times 3+1}{3}=\frac{16}{3}$
$2 \frac{1}{2}=2+\frac{1}{2} \cdot \frac{2 \times 2+1}{2}=\frac{5}{2}$
=$\left(\frac{1}{2}+\frac{1}{3}\right) \div\left(\frac{1}{4}-\frac{1}{6}\right)-\left[8-\left\{5 \frac{1}{3}-\left(3-2 \frac{1}{2}\right)\right\}\right] .$
$=\left(\frac{3+2}{6}\right) \div\left(\frac{3-2}{12}\right)-\left[8-\left\{\frac{16}{3}-\left(3-\frac{5}{2}\right)\right\}\right]$
$=\left(\frac{5}{6}\right) \div\left(\frac{1}{12}\right)-\left[8-\left\{\frac{16}{3}-\left(\frac{6-5}{2}\right)\right]\right]$
$=\frac{516}{1 / 12}-\left[8-\left(\frac{16}{3}-\frac{1}{2}\right)\right]$
$=\frac{5}{6} \times 1^{2}-\left[8-\left(\frac{16 \times 2-3 \times 1}{6}\right)\right]$
$=10-\left[8-\left(\frac{32-3}{6}\right)\right]$
$=10-\left[8-\frac{29}{6}\right]$
$10-\left[\frac{8 \times 6-29}{6}\right]$
$=10-\frac{48-29}{6}$
$=10-\frac{19}{6}$
$=\frac{60-19}{6}$
$=\frac{41}{6}=6 \frac{5}{6} .$
Question 4
(i)
$\begin{aligned} & 2.3-[1.89-\{3.6-(2.7-0.8-0.03)\}] \\ & 2.3-[1.89-\{3.6-(2.7-0.77)\}] \\ &=2.3-[1.89-\{3.6-1.93\}] \\=& 2.3-[1.89-1.67] \\=& 2.3-0.22 \\=& 2.08 \end{aligned}$
ii) $4.5-\frac{1}{2}$ of $(7.6-3.5)+2.3 \times 4.05$
$=4.5-\frac{1}{2} \times(4.1)+2.3 \times 4.05$
$=4.5-\frac{4.1}{2}+9.315$
$=4.5+9.315-2.05$
$=13.815-2.05$
$=11.765$
Question 5
i) $2 \frac{1}{2}=\frac{5}{2}$
$2 \frac{1}{2}+\frac{1}{5}=\frac{5}{2}+\frac{1}{5}$
$=\frac{-5 \times 5+1 \times 2}{10}=\frac{27}{10} .$
$2 \frac{1}{2} \div \frac{1}{5}=\frac{5}{2} \div \frac{1}{5}$
$\frac{5}{2} \times 5=\frac{25}{2}$
So, $\frac{2 \frac{1}{2}+\frac{1}{5}}{2 \frac{1}{2} \div \frac{1}{5}}=\frac{27 / 10}{25 / 2}$
$=\frac{27}{10} \times \frac{2}{25}$
$=\frac{27}{125} .$
(ii) $\frac{3.5 \times 0.24}{0.21}-0.037$
$3 \cdot 5=\frac{35}{10}=\frac{7}{2} .$
$0.24= \frac{24}{100}=\frac{6}{25}$
$0 .21=\frac{21}{100}$
$=\frac{\frac{7}{2} \times \frac{6}{25}}{\frac{21}{100}}-\frac{37}{1000}$
$=\frac{7}{2} \times \frac{6}{25} \times \frac{100}{21}-\frac{37}{1000}$
$=4-\frac{37}{1000}$
$=4-0.037$
$=3.963 .$
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