ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.6

 Exercise 1.6

Question 1

Sol :
$\begin{aligned} \text { Total fruits } &=20 \mathrm{~kg} \\ \text { oranges } &=7 \frac{1}{6} \mathrm{~kg} \\ \text { Apples } &=8 \frac{2}{3} \mathrm{~kg} \end{aligned}$

let grapes =x

$7 \frac{1}{6}+8 \frac{2}{3}+x=20$

$\frac{43}{6}+\frac{26}{3}+x=20$

$\frac{(43 \times 1)+(26 \times 2)}{6}+x=20$

$\frac{43+52}{6}+x=20$

$\begin{aligned} \frac{95}{6}+x &=20 \\ x &=\frac{20}{1}-\frac{95}{6} \\ x &=\frac{120-95}{6} \\ x &=\frac{25}{6} \mathrm{~kg} \end{aligned}$

                 $\therefore$ Beg contain $4 \frac{1}{6}$ kg of grapes.

Question 2

Sol :
Total population of city =6,63,432

Adult male $=\frac{1}{2}$ of total population

Adult females $2 \frac{1}{3}$ of total population

 Adult male + Adult female terms + children = Total city

$\frac{1}{2}(6,63,432)+\frac{1}{3}(663432)+$ Children $=6163432$

$\frac{5}{6}(663432)+$ children $=663432$

children $=663432\left(1-\frac{5}{6}\right)$

$=663432 \times \frac{1}{6}$

Children =110572

no. of children in city $=110,572$

Question 3

Sol :
Total votes =30
no. of votes for Mr. $x=\frac{2}{5}$ of $30=\frac{2}{5}(30)$

no. of votes for $M r \cdot z=\frac{1}{3} (30)$

Let Mr.Y votes =x

=$\therefore \quad \frac{2}{5}(30)+x+\frac{1}{3}(30)=30$
=12+x+10=30
=x+22=30
=x=30-22
x=8

No. of votes for Mr. y = 8

Question 4

Sol :
Total earnings =₹ 100

Rupees spent on food $=₹ 14 \frac{2}{7}$

Rupees spent on petrol $=₹ 30 \frac{2}{3}$

Let Savings on that day =x

$14 \frac{2}{7}+30 \frac{2}{3}+x=100$

$\frac{100}{7}+\frac{92}{3}+x=100$

$\begin{aligned}\left(\frac{100 \times 3)+(92 \times 7)}{21}\right.&+x=100 \\ x &=100-\frac{944}{21} \end{aligned}$

$x=\frac{1156}{21}=55 \frac{1}{21}$

$\therefore$ savings $=₹ 55 \frac{1}{21}$

Question 5

Sol :
Total students =400
    
no.of girls =130

no. of boys appeared for exam $=400-130$$=270$

no. of boys passed in exam $=\frac{2}{3}(270)$  $=180$

no. of boys failed  in exam = Total boys - passed boys
=270-180
=90 
$\therefore$ no. of boys failed in exam =90

6.speed of Car$ =40 \frac{2}{3} \mathrm{Km} / \mathrm{h}$
time $=\frac{9}{10} \mathrm{hr}$

distance traveled car = speed ｘtime 

$=40 \frac{2}{3} \times \frac{9}{10}$

$=\frac{122}{3} \times \frac{9}{10}$

$=\frac{183}{5}$

$=36 \frac{3}{5} \mathrm{~km}$
 
7. Side of square = $5 \cdot \frac{7}{9} m$

$S=\frac{52}{9} m$

$\begin{aligned} \text { Area of square } &=S^{2} \\ &=\frac{52}{9} \times \frac{52}{9} \\ &=\frac{2704}{81} \\ \text { Area of square } &=33 \frac{31}{81} \mathrm{~m}^{2} \end{aligned}$

Question 8

Sol :
Perimeter of rectangle = $15 \frac{3}{7} m$

length of rectangle (l) = $4 \frac{2}{7} m$

perimeter = 2 (l+b)

=$15 \frac{3}{7}=2\left(4 \frac{2}{7}+b\right)$

=$\frac{108}{7}=2\left(\frac{30}{7}+b\right)$

=$\frac{3 b}{7}+b=\frac{108}{7 \times 2}$

$\frac{30 }{7}+b=\frac{108}{7 \times 2}$

$\frac{30}{7}+b=\frac{54}{7}$

$b=\frac{54}{7}-\frac{30}{7}$

$b=\frac{24}{7}$

$b=3 \frac{3}{7} m$

Question 9

Sol :
Total length of rope = $325 \frac{4}{5} m$

Rahul cut off $150 \frac{3}{5} \mathrm{~m}$ of rope

Remaining length of rope = $325 \frac{4}{5}-150 \frac{3}{5}$

$=\frac{1629}{5}-\frac{753}{5}$

$=\frac{876}{5} \mathrm{~m}$

$=\frac{876}{5} \mathrm{~m}$

Rahul made remaining rope into 3 equal parts 

$\begin{aligned} \therefore \text { length  of each part } &=\frac{876}{5} \div \frac{3}{1} \\ &=\frac{876}{5} \times \frac{1}{3}=\frac{292}{5}=58 \frac{2}{5} \mathrm{~m} \\ \therefore \text { length of each part } & \text { part }=58 \frac{2}{3} \mathrm{~m} \end{aligned}$


10.  $3 \frac{1}{2}$ liters of petrol cost= ₹ $270 . \frac{3}{8}$

cost for 1 liter of petrol $=\frac{270 \frac{3}{8}}{3 \frac{1}{2}}$

$=\frac{2163}{8} \times \frac{2}{7}$

Cost for 1 liter of petrol $=\frac{309}{4}$

cost for 4 liters of petrol $=\frac{309}{4} \times 4$

$\therefore$ cost for 4 liters of petrol = ₹ 309

Question 11

Sol :
Ramesh Spends $\frac{3}{8}$ of income on food $=\frac{3}{8} \times 40000$

=15000

Remaining money =40000-15000
=25000


remaining spend $\frac{1}{5}$ of remaining on LIC 

$=\frac{1}{5}(25000)$

$=5000$

Remaining money $=25000-195000$
$=20000$

other expenses are $\frac{1}{2} of$ remaining money
= $=\frac{1}{2} \times 20000$
= 10,000

$\begin{aligned} \text { Remaining money } &=20,000-10,000 \\ &=10,0001\end{aligned}$

12. Let total bill Amount as $x$
Amount paid A will be $=\frac{x}{2}$
Amount paid $B_{1} C_{1} D$ will be $=x-\frac{x}{2}$

$=\frac{x}{2}$

given bill is shared equally among three 

 let bill paid by each one = y 

$y+y+y=\frac{x}{2}$

$3 y=\frac{x}{2}$

$y=\frac{x}{6}$

$\therefore$ Each paid $\frac{1}{6}$ th of total bill.


Question 13

Sol :
Let total no. of student s = х
no. of students of school come by car $=\frac{2}{5} x$

no. of Students of school come by bus e $\frac{1}{4} x$

no of students of school come by walk $=x-\left(\frac{2}{5} x+\frac{1}{4} x\right)$
 = x- $\left(\frac{(2 \times  4)+(1 \times 5)}{20} \cdot x\right)$
$=x-\frac{13}{20} \cdot x$

no. of students of school come by walk $=\frac{7}{20} \cdot x$
no. of students of school come by walk on their own

$\begin{aligned} &=\frac{1}{3} \ of \left(\frac{7}{20} x\right) \\ &=\frac{7}{60} \cdot x \end{aligned}
  $\begin{aligned} \therefore \quad \frac{7}{60} \cdot x &=224 \\ x &=\frac{224}{1} \times \frac{60}{7} \\ x &=1920 \\ \text { Total Students in school }=1920 \end{aligned}$ 

Question 14

Sol :
Total cost of Room = ₹ 60,000

let Mother's contribution = ₹ x

Elder son contribution $= ₹ \frac{3}{8} x$

younger son contribution $= ₹ \frac{1}{2} x$

$\begin{aligned} \therefore \quad x+\frac{3}{8} x+\frac{1}{2} x &=60,000 \\ \frac{( 8)+(3)+(4)}{8} \cdot x &=60,000 \\ \frac{15}{8} \cdot x &=60000 \\ x &=60000 \times \frac{8}{15} \\ x &=32000 \end{aligned}$

; Mother's contribution 2232,000 

; Elder son's contribution e $\frac{3}{8} \times 32000=₹ 12,000$

; Younger son's Contribution = $\frac{1}{2} \times 32000=₹ 16,000$ 

Question 15

Sol :
Total students =56

let no.of girls =x

no. of boys $=\frac{2}{5} x$

$\therefore \quad x+\frac{2}{5} x=56$

$\frac{7}{5} x=56$

$x=\frac{56}{1} \times \frac{5}{7}$

x=40

$\therefore$ no. of girls =40
$\therefore$ no. of boys =56-40=16

Question 16

Sol :
Let  money posses by man 2x

$\frac{1}{10}$th money donated to school $=\frac{x}{10}$

Remaining money $=x-\frac{x}{10}=\frac{9 x}{10}$

$\frac{1}{6}$th remaining money  $=\left(\frac{9 x}{10}\right) \times \frac{1}{6}$

Remaining money $=\frac{9 x}{10}-\frac{9 x}{10 \times 6}$

$=\frac{45 x}{60}$

Now, man distributed This money equally to hin three sons and each one gets $=₹50,000$

$\frac{45 x}{60} \div 3=50,000$

$\frac{45 x}{60} \times \frac{1}{3}=50000$

$\frac{3}{4} x \times \frac{1}{8}=50000$

$x=50000 \times 4$

x=2,00,000

Question 17

Sol :
Let a number be 'x'

$\frac{1}{4}$ of a number is added to $\frac{1}{3}$ of number

$\frac{x}{4}+\frac{x}{3}$ is 15 greater thaw half of  number

$\begin{aligned} \frac{x}{4}+\frac{x}{3}=15 &+\frac{x}{2} \\ \frac{7 x}{12}=15 &+\frac{x}{2} \\ \frac{7 x}{12}-\frac{x}{2} &=15 \\ \frac{x}{12} &=15 \\ x &=15 \times 12 \\ x &=180 \end{aligned}$

Question 18

Sol :
Let the number be 'x'

$\frac{x}{1} \div \frac{4}{5}=36+\left(x+\frac{4}{5}\right)$

$\frac{5 x}{4}=36+\frac{4 x}{5} .$

$\frac{5 x}{4}-\frac{4 x}{5}=36$

$\frac{(5 \times 5)-(4 \times 4)}{20} \cdot x=36$

$\frac{25-16}{20} \cdot x=36$

$\frac{9}{20} \cdot x=36$

$x=\frac{36}{9} \times 20$

x=80

$\therefore$ The given number is 80

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