EXERCISE: 6.1
Question 1
i) Given ratio
$\begin{aligned}=\frac{1}{6}: \frac{1}{9}=\frac{1 / 6}{1 / 9} &=\frac{1}{6} \times 9 \\ &=\frac{3}{2} \\ &=2.2 \end{aligned}$
ii) Given ratio $=4 \frac{1}{2}: 1 \frac{1}{8}$
$\begin{aligned}=\frac{9}{2}: \frac{9}{8} &=\frac{9 / 2}{9 / 8} \\ &=\frac{9}{2} \times \frac{8}{9} \\ &=4: 1 \end{aligned}$
iii) $\frac{1}{5}: \frac{1}{10}: \frac{1}{15}$
Lcm of $5,10,15$ is 30
$=\frac{1}{5} \times 30: \frac{1}{10} \times 30: \frac{1}{15} \times 30$
$=6: 3: 2$
Question 2
i) Rs 5 to 50 paise
$=\frac{5 \times 100 \text { paise }}{50 \text { paise}}$
$=\frac{50}{5}=10: 1$
ii)$3 \mathrm{~km}$ to $300 \mathrm{~m}$
$=\frac{3 \times 1000 \mathrm{~m}}{300}$
$=\frac{3000}{300}$
$=10: 1$
iii)
$\begin{aligned} 9 m & \text { to } 27 \mathrm{~cm} \\ &=\frac{9 \times 100 \mathrm{~cm}}{27} \\ &=\frac{900}{27} \\ &=100: 3 \end{aligned}$
iv)
$\begin{array}{rl}15 & \mathrm{~kg} \text { to } 210 \mathrm{~g} \\ = & \frac{15 \times 1000 \mathrm{~g}}{210} \\ = & \frac{15000}{210} \\ & =500: 7\end{array}$
V) 25 minutes to $1.5$ hour
$=\frac{25 \text { minutes }}{1.5 \times 60 \text { minutes }}$
$25 / 90$ = $5: 18 .$
vi) 30 days to 36 hours
$=\frac{30\times 24}{36 \mathrm{~hours}}$
=20: 1
Question 3
$A: B=3: 4 \quad ; B: C=8: 9$
$\frac{A}{B}=\frac{3}{4} ; \frac{B}{c}=\frac{8}{9}$
Now $\frac{A}{C}=\frac{A}{B} \times \frac{B}{C}=\frac{3}{4} \times \frac{8}{9}$◘
A: C=2: 3
Question 4
$A: B=5: 8$, Value of $B=8$
$B: C=18: 25$, value of $B=18$
LCM of these two value of B i.e 8 and 18 is 72
Thus ,
$A: B=5: 8=\frac{5}{8}=\frac{5 \times 9}{8 \times 9}=\frac{45}{72}=45: 72$ and
$B: C=18: 25=\frac{18}{25}=\frac{18 \times 4}{25 \times 4}=\frac{72}{100}=72: 100$
$A: B: C=45: 72: 100$
Question 5
Let
$3 A=2 B=5 c=k$ (Say), then
$A=\frac{k}{3}, B=\frac{k}{2}, c=\frac{k}{5} .$
$\begin{aligned} \therefore A: B: C=& \frac{K}{3}: \frac{K}{2}: \frac{k}{5} \\ &=\frac{1}{3}: \frac{1}{2}: \frac{1}{5} \end{aligned}$
Lcm of $3,2,5$ is 30
$=\frac{1}{3} \times 30: \frac{1}{2} \times 30: \frac{1}{5} \times 30$
Hence , A:B:C= 10:15:6
Question 6
Given Income =Rs 120
Spendings = Rs 90
Savings = Income -spendings
=120-90
= 30
i) $\frac{\text { Spending }}{\text { Income }}$ = $\frac{90}{120}=\frac{9}{12}=3: 4$
ii) $\frac{\text { Savings }}{\text { Income }}$ = $\frac{30}{120}=\frac{1}{4}=1: 4$
iii) $\frac{\text { Savings }}{\text { Spendings }}=\frac{30}{90}=\frac{1}{3}=1: 3$.
Question 7
An alloy contains = 5 grams
and of which copper was = $3 \frac{3}{4} \mathrm{grams}$
=$\frac{15}{4}$ grams
Now nickel contains = $5-\frac{15}{4}$
=$\frac{20-15}{4}$
=$\frac{5}{4}$
Question 8
Let the two pieces will be $x, \frac{x}{2}$
Total height of pole. $=3$ metres
i.e $x+\frac{x}{2}=3$
$\frac{2 x+1}{2}=3$
$\frac{3 x}{2}=3$
$x=\frac{3 \times 2}{3}$
$x=2, \frac{x}{2}=\frac{2}{2}=1$
∴ Length of two pieces are 2, 1 metres
Question 9
Given height of Anshul and Dhruv are 1.04 m and 78cm
Ratio of their height = $\frac{\text { Height of Anshul }}{\text { Height of Dhruv }}$
$=\frac{1.04 \mathrm{~m}}{78 \mathrm{~cm}}$
=$\begin{aligned} & \frac{1.04 \times 100}{78} \\=& \frac{104}{78} \\=& 4: 3 \end{aligned}$ $(\because 1 m=100 \mathrm{~cm})$
Question 10
Total money to be shared = Rs 180
Ratio of these children = $\frac{1}{3}: \frac{1}{4}: \frac{1}{6}$
LCM of 3, 4 and 6 is 12
$=\frac{1}{3} \times 12: \frac{1}{4} \times 12: \frac{1}{6} \times 12$
$=4: 3: 2$
Total sum of ratio = 4+3+2= 9
1st children share = $\frac{4}{9} \times 180=4 \times 20=\varepsilon 80 .$
2nd children share = $=\frac{3}{9} \times 180=3 \times 20=₹ 60$
3rd children share = $\frac{2}{9} \times 180=2 \times 20$=Rs 40
Question 11
Let the two part be 7x, 11x
Given difference of two parts = 20
i.e $11 x-7 x=20$
$4 x=20$
$x=\frac{20}{4}$
$x=5$
$\therefore 7x=7 \times 5=35 \quad ; 11 x=11 \times 5.55$
Sum of two numbers = 35+55= 90
Question 12
Let the total amount be Rs x
The amount has been divided in two parts in the ratio 9:13
Sum of ratios = 9+ 13= 22
According to given condition $\frac{13}{22}$ of Rs x=260
$\Rightarrow \frac{13}{22} \times x=260 \Rightarrow x=\frac{260 \times 22}{13}$
x = Rs 440
Hence, the total amount = Rs 440
Question 13
As the present ages of anjali and ashu are in the ratio 2:3
Let their present ages be 2x, years and 3x years resp
After 5 years ,
The age of anjali will be (2x+ 5) years and the age of ashu will be (3x+ 5) years
According to given information , $\frac{2 x+5}{3 x+5}=\frac{3}{4}$
$3(3 x+5)=4(2 x+5)$
$9 x+15=8 x+20$
$9 x-8 x=20-15$
$x=5$
Hence, the presentage of anjali = $2\times 5$ = 10 years
and the present age of ashu = $3\times 5$0 = 15 years
Question 14
Let their present ages of A and B be 5x years and 6x years resp
Three years ago ,
The age of A will be (5x- 3) and the age of B will be (6x- 3) years
According to given information, $\frac{5 x-3}{6 x-3}=\frac{4}{5}$
$5(5 x-3)=4(6 x-3)$
$25 x-15=24 x-12$
$25 x-24 x=15-12$
$x=3$
Hence, the present age of A = $5\times 3$= 15 years and the present age of B = $6\times 3$ = 18 years
Question 15
Let the numbers be 5 x, 6 x
2 is added to first =5 x+2
3 is added to second =6 x+3
Their ratio, $\frac{5 x+2}{6 x+3}=\frac{4}{5}$
$5(5 x+2)=4(6 x+3)$
$25 x+10=24 x+12$
$25 x-24 x=12-10$
$\begin{aligned} x=2 \Rightarrow 5 x &=5 \times 2=10 \\ 6 x &=6 x^{2}=12 \end{aligned}$
∴ The numbers are 10, 12
Question 16
Let the number of boys and girls be 7x, 6x
Total no. of students = 1430
$7 x+6 x=1430$
$13 x=1430$
$x=\frac{1430}{13}$
$x=110$
∴ Number of boys = 7x = 7$\times 110$=770
Number of girls = 6x = $6 \times 110$ = 660
As , given 26 girls are admitted
i.e 660+ 26 = 686
Let the new boys are admitted be x
Ratio of number of boys to girls = 8:7
i.e $\frac{x+770}{686}=\frac{8}{7} .$
$x+770=\frac{8}{7} \times 686$
$x+770=784$
$x=784-7.70$
$x=14$ new boys are admitted.
Question 17
i) $5: 6$ or $6: 7$
$\frac{5}{6} ; \frac{6}{7}$
LCM of 6,7 is 42
i.e $\frac{5}{6} \times \frac{7}{7}=\frac{35}{42} ; \frac{6}{7} \times \frac{6}{6}=\frac{36}{42}$
As $36>35$, so $6: 7$ is greater than $5: 6$.
ii) $13: 24$ (or) $17: 32$
$\frac{13}{24}$ (or) $\frac{17}{32}$
$\mathrm{L} \mathrm{CM}$ of 24,32 is 96
i.e $\frac{13}{24} \times \frac{4}{4}=\frac{52}{96} ; \frac{17}{32} \times \frac{3}{3}=\frac{51}{96}$
As $52>51$, So $13: 24$ is greater than 17:32.
EXERCISE: 6.2
Question 1
i)
$\begin{aligned} 2.5: 1.5 &=\frac{2.5}{1.5} \times \frac{10}{10} \\ &=\frac{25}{15}=\frac{5}{3} . \end{aligned}$
$\begin{aligned} 7.0: 4 \cdot 2 &=\frac{7.0}{4.2} \times \frac{10}{10} \\ &=\frac{70}{42}=\frac{5}{3} . \end{aligned}$
$\frac{2.5}{1.5}:: \frac{7.0}{4 \cdot 2}=\frac{5}{4}:: \frac{5}{3}$ , True
ii) $\frac{1}{2}: \frac{1}{3}$
LCM of 2,3 is 6
$\frac{1}{2} \times 6: \frac{1}{3} \times 6=3: 2$
$\frac{1}{3}: \frac{1}{4}$
LCm of 3,4 is 12
$\frac{1}{3} \times 12: \frac{1}{4} \times 12=4: 3$
$3: 2 \neq 4: 3$, False.
iii) 24 men: $16 \mathrm{men}$ ; 33 horses: 22 horses
$=\frac{24}{16}$ ; $=\frac{33}{22}$
$=\frac{3}{2}$ ; $=\frac{3}{2}$
$\frac{3}{2}=\frac{3}{2}$, True.
Question 2
i) Product of extremes = $18 \times 5=90$
Product of means = $9 \times 10=90 .$
By cross product rule, the number 18, 10,9, 5 are in proportion
ii) $3 \frac{1}{2}=3+\frac{1}{2}=\frac{6+1}{2}=\frac{3}{2}$.
$4 \frac{1}{2}=4+\frac{1}{2}=\frac{8+1}{2}-\frac{9}{2}$
product of extremes $=3 \times \frac{9}{2}=\frac{27}{2}$.
product of means $=4 \times \frac{7}{2}=\frac{28}{2}$.
∴ The number are not in proportion
Question 3
i)
$\begin{aligned} \frac{x}{4}=\frac{9}{12} \\ \text { Cross product } \\ 12 \times x &=9 \times 4 \\ x=\frac{9 \times 4}{12}=\frac{36}{12} \\ x=3 \end{aligned}$
ii) $\frac{1}{13}: x:: \frac{1}{2}: \frac{1}{5}$
$\Rightarrow \frac{1}{13}: x=\frac{1}{2}: \frac{1}{5}$'
$\frac{1 / 13}{x}=\frac{1 / 2}{1 / 5}$
$\frac{1}{13 \times x}=\frac{1}{2} \times 5$
$\frac{1}{13 x} \cdot \frac{5}{2} .$
Cross multiplication.
$2 \times 1=13 \times 55$
$65 x=2$
$x=2 / 65$
iii) $3.6: 0.4=x: 0.5$
$\frac{3.6}{0.4}=\frac{x}{0.5}$
$\frac{3.6 \times 10}{0.4 \times 10}=\frac{x \times 10}{0.5 \times 10}$
$\frac{36}{4}=\frac{10 x}{5}$
$9=\frac{10 x}{5}$
$10 x=9 \times 5$
$x=\frac{9 \times 5}{10}=\frac{45}{10}$
$x=4.5$
Question 4
If a,b,c and are in proportion then ad= bc
d is the first proportion
i)
$\begin{aligned} a=42 ; b=12: c=7 ; d=? \\ 42 \times d=& 12 \times 7 \\ 42 d=84 \\ d=\frac{8 y}{42} \end{aligned}$
$d=2$ is fourth propotion.
ii) $a=\frac{1}{3}, b=\frac{1}{4}, \quad c-\frac{1}{5}, d=7$
$\frac{1}{3} \times d=\frac{1}{4} \times \frac{1}{5}$
$\frac{d}{3}=\frac{1}{20}$
$d=\frac{3}{20}$. is fourth propotion
iii) $\begin{aligned} a=3, b=12, & c=15, d=? \\ 3 \times d &=12 \times 15 \\ d &=\frac{12 \times +5^{5}}{8} \\ d &=12 \times 5 \\ d &=60 \text { is fourth proportion. } \end{aligned}$
Question 5
Continued proportion
If a,b,c are said to be in continued proportion
If a:b= b: c i.e $b^{2}=a c$
$a=3, b=49, c=343$
$b^{2}=49^{2}=2401$
$a c=7 \times 343=2401$
$\therefore b^{2}=a c$
∴ They are said to be in continued proportion
Question 6
As we know $b^{2}=a c$ if $a, b, c$ are in proportim
Third proportion $C=\frac{b^{2}}{a}$.
i) $a=36: b=18, c=?$
$C=\frac{18^{2}}{36}$
$C \cdot \frac{18 \times 18}{36}$
$C=9$ is third propdtion
ii) $a=5 \frac{1}{4}=\frac{21}{4} ; b=7 ; c=?$
$c=\frac{7^{2}}{(21 / 4)}$
$c=\frac{49 \times 4}{21}$
$C=\frac{28}{3}$ is third proportion
iii) $a=3 \cdot 2=\frac{32}{10}=\frac{16}{5}, \quad b=0.8=\frac{8}{10}=\frac{4}{5} .$
$\begin{aligned} c &=\frac{(4 / 5)^{2}}{16 / 5)} \\ &=\frac{16}{25} \times \frac{5}{16} \end{aligned}$
$C=\frac{1}{5}$ is third proportion
Question 7
Given Ratio of length to width is $7: 5$
Width of sheet $=20.5 \mathrm{~cm}$
Length of sheef $=?
$\frac{\text { Lenght }}{\text { width }}=\frac{7}{5}$
$\frac{\text { height }}{20.5}=\frac{7}{5}$
Length $=\frac{7}{5} \times 20.1$
Length $=28.7 \mathrm{~cm}$
Question 8
Amit age is 4 years months
i.e ($4+\frac{8}{12}$) years
$=4 \frac{2}{3}$ years $=\frac{14}{3}$ years
Ages of amit and archana are in ratio 4: 5
i.e $\frac{\text { Amit }}{\text { Archans }}=\frac{4}{5}$
$\frac{14 / 3}{\text { Archana }}=\frac{4}{5}$
Archana age = $\frac{14}{3} \times \frac{5}{4}=\frac{35}{6}$ years
∴Archana's age is 5 years 10 months
Since $\frac{35}{6} \times \frac{2}{2}=\frac{70}{12}$
(daigram to be added)
$\frac{7_{0}}{12}=5 \frac{10}{12}$ years
i.e 5 years 10 months.
∴ Archana's age is 5 years 10 months
EXERCISE: 6.3
Question 1
Cost of 6 bowls = Rs 90
Then Cost of 10 Such bowls = ?
Let cost of 10 bowls = Rs x
1.e $\begin{aligned} \frac{x}{90} &=\frac{10}{6} \\ x &=\frac{10 \times 90}{ 6} \\ x &=₹ 150 \end{aligned}$
∴ Cost of 10 Such bowls = Rs 150
Question 2
With the cost of Rs 15 ,we bought = 10 pencils
Then with the cost of Rs 72 ,we bought =? Pencils
Let it be = x pencils
i.e $\frac{x}{10}=\frac{72}{15}$
$x=\frac{32 \times 10}{15}$
$x=\frac{720}{15}$
$x=48$ pencils
Question 3
Given 400 grams Cake costs Rs 80
Then 1.5 kg Cake Cost = ?
Let the cost be Rs x
i.e
$\begin{aligned} \frac{x}{80} &=\frac{1.5 \mathrm{k}{g}}{400} \\ \frac{x}{80} &=\frac{1.5 \times 1000}{400} \\ x &=\frac{1500 \times 80}{400} \\ x &=15 \times 20 \\ x &=₹ 300 \end{aligned}$
∴ 1.5 kg cake cost = Rs 300
Question 4
i) In 3 months, a man earns rs 18000
In $x$ months, a man earns Rs 30,000
i.e $\frac{x}{3}=\frac{30,000}{18000}$
$x=\frac{30 \times 3}{18}=\frac{90}{18}$
$x=5$
∴ In 5 months , a man earns Rs 30,000
ii)In 3 months , he earns Rs 18000
Let In 7 months , he earns Rs x
$\therefore \frac{x}{18,000}=\frac{7}{3}$
$x=\frac{7 \times 18000}{3}$
$x=7 \times 6,000$
$x=42000$
∴ In 7 months , he earns Rs 42,000
Question 5
Given 12 mangoes weight $24 \mathrm{~kg}$
Then 8 mangoes weight $x \mathrm{~kg}$
$\frac{x}{2 \cdot 4}=\frac{8}{12}$
$x=\frac{8 \times 2.4}{12}$
$x=\frac{8 \times 24}{12 \times 10}=\frac{16}{10}$☻
x = 1.6kg
∴ 8 mangoes weight 1.6kg
Question 6
Given 12 sheets of thick paper = 40 grams
Let x sheets of thick paper $=2 \frac{1}{2} k g$
=$\frac{5}{2} k g$
$\therefore \frac{x}{12}=\frac{5 / 2 k g}{40 g}$
$\frac{x}{12}=\frac{\frac{5}{2} \times 1000}{40} \mathrm{~g}$
$\frac{x}{12}=\frac{5 \times 500}{40}$
$\frac{x}{12}=\frac{5 \times 125}{10}$
$x=\frac{5 \times 125 \times 12}{10}$
x = 750
Question 7
For a distance of 90 km , a bus consumer = 25 litres
Let for a distance of 288km, a bus consumer = x litres
$\begin{aligned} \therefore \frac{x}{25} &=\frac{288}{90} \\ x &=\frac{288 \times 25}{90} \end{aligned}$
$x=\frac{32 \times 25}{10}$
$x=16 \times 5$
$x=80$ litres.
∴ For a distance of 288 km, a bus consumer = 80 litres
Question 8
Given $\frac{4}{5}$ metre cloth costs $₹ 36 .$
Then $2 \frac{1}{5}$ metre cloth costs x
$2 \frac{1}{5}=2+\frac{1}{5}=\frac{10+1}{5}=\frac{11}{5}$
$\begin{aligned} \therefore \frac{x}{36} &=\frac{11 / 5}{4 / 5} \\ \frac{x}{36} &=\frac{11}{5} \times \frac{8}{4} \end{aligned}$
$\frac{x}{36}=\frac{11}{4}$
$x=\frac{11}{4} \times 36$
∴ $2 \frac{1}{5}$ metre cloth costs Rs 99
Question 9
Given 15 men can pack 540 parcels per day
Let x men can pack 396 parcels per day
i.e $\frac{x}{15}=\frac{396}{540}$
$x=\frac{396 \times 15}{540}$
x= 11
∴ 11men can pack 396 parcels per day
Question 10
Given $12 \mathrm{~kg}$ potatoes costs Rs 132
Then $1 \mathrm{~kg}$ potates costs Rs x
i.e $\frac{x}{132}=\frac{1}{12}$
$x=\frac{132 \times 1}{12}$
x=Rs 11
ஃ 1 kg potatoes cost Rs 11
Given $16 \mathrm{~kg}$ potatoes cost Rs 168
then $1 \mathrm{~kg}$ potatoes costs Rs ?
Let the cost $₹ x$.
i.e $\frac{x}{168}=\frac{1}{16}$
$x=\frac{168}{16}$
$x=₹\frac{21}{2}$
x = ₹ 10.5
∴ 1kg potatoes = Rs 10.5
Compared to 1st case , 2nd case is better buy
EXERCISE: 6.4
Question 1
$1 \mathrm{~km} / \mathrm{h}=\frac{5}{18} \mathrm{~m} / \mathrm{sec}$
i) $72 \mathrm{~km} / \mathrm{h}=72 \times \frac{5}{18}=4 \times 5=20 \mathrm{~m} / \mathrm{sec}$
ii) $9 \mathrm{~km} / \mathrm{h}=9 \times \frac{5}{18}=\frac{5}{2} \mathrm{~m} / \mathrm{sec}=2.5 \mathrm{~m} / \mathrm{sec}$
iii) $1.2 \mathrm{~km}= / \mathrm{min}, 1.2 \times \frac{1000 \mathrm{~m}}{60 \mathrm{sec}}=\frac{1200 \mathrm{~m} / \mathrm{sec}}{60}=20 \mathrm{~m} / \mathrm{sec}$
iv) $600 \mathrm{~m} / \mathrm{hour}$ = $600 \times \frac{1 \mathrm{~m}}{60\times60 \mathrm{sec}}$ = $\frac{1}{6} \mathrm{~m} / \mathrm{sec}$
Question 2
$1 \mathrm{~m} / \mathrm{sec}=\frac{18}{5} \mathrm{~km} / \mathrm{h}$
i) $15 \mathrm{~m} / \mathrm{sec}=1 \mathrm{5} \times \frac{18}{5} \mathrm{km} / \mathrm{h}=54 \mathrm{~km} / \mathrm{h}$
ii) $5 \mathrm{~m} / \mathrm{sec}$ =$\frac{1{5}}{10} \times \frac{18}{5}$= $\frac{54}{10} \mathrm{~km} / \mathrm{h}=5.4 \mathrm{~km} / \mathrm{h} .$
Question 3
$\begin{aligned} 30 \mathrm{~m} / \mathrm{sec}=30 \times \frac{18}{8} \mathrm{~km} / \mathrm{h} &=108 \mathrm{~km} / \mathrm{h} . \\ \text { since, } 108 \mathrm{~km} / \mathrm{h} &>30 \mathrm{~km} / \mathrm{h} \\ \text { So, } 30 \mathrm{~m} / \mathrm{sec} &>30 \mathrm{~km} / \mathrm{L} \end{aligned}$
Question 4
i) Given speed of aeroplane = 72km\h
Also distance between two cities = 1800km
We know time= $\frac{\text { Distance }}{\text { Speed }}$
$=\frac{1800}{720}$ hour
$=\frac{5}{2}$ honss
Time $=2 \frac{1}{2}$ hour
ii) Here time = 40 min, speed = 720km\h
$\frac{720 \times 1000}km{60 \mathrm{~min}} \mathrm{~m}$
Now Distance $=$ Speed $\times$ Time
$=12,000 \times 40 \mathrm{~m}$
$=480000 \mathrm{~m}$
$=\frac{480000}{1000} \mathrm{~km}$
Distance $=480 \mathrm{~km}$
iii) Given Time $=15 \mathrm{sec}$, Speed $=720 \mathrm{~km} / \mathrm{h}$
$=720 \times \frac{5}{18} \mathrm{~m} / \mathrm{sec}$
$=200 \mathrm{~m} / \mathrm{sec}$
Distance =speed $\times$ Time
$=15 \times 200 \mathrm{~m}$
$\begin{aligned} &=3000 \mathrm{~m} \\ &=\frac{3000}{1000} \mathrm{~km} \\ \therefore \quad \text { Distance } &=3 \mathrm{~km} \end{aligned}$
Question 5
Given speed = 6km\h
$=\frac{6 \times 1000 \mathrm{~m}}{60 \mathrm{~min}}$
$=100 \mathrm{~m} / \mathrm{min}$
i) Distance
$\begin{aligned} &=\text { Speed } \times \text { Time } \\ &=100 \times 5 \\ &=500 \text { metres } \\ &=0.5 \mathrm{~km} \end{aligned}$
ii)
$\begin{aligned} \text { Time }=\frac{\text { Distance }}{\text { speed }} &=\frac{200}{100} \\ \text { Time } &=2 \mathrm{~min} \end{aligned}$
Question 6
Given Distance
$\begin{aligned} &=50 \text { metres } \\ &=\frac{50}{1000} \mathrm{~km} \\ &=0.05 \mathrm{~km} \\ &=\frac{1}{20} \mathrm{~km} \end{aligned}$
Total Distance $=2 \times \frac{1}{20} \mathrm{~km}=\frac{1}{10} \mathrm{~km}$
Time taken
$\begin{aligned} &=5 \mathrm{~min} \\ &=\frac{5}{60} \text { hours } \end{aligned}$
Now, speed $=\frac{\text { Distance }}{\text { Time }}$
$=\frac{(1 / 10) \mathrm{km}}{(5 / 60) \mathrm{hour}}$
$=\frac{1}{10} \times \frac{60}{5}$
$=\frac{6}{5}$
Speed $=1.2 \mathrm{~km} / \mathrm{k}$
Question 7
Given Time $=48$ minutes $=\frac{48}{60}$ hours
$=\frac{4}{5}$ hours.
Speed $=50 \mathrm{~km} / \mathrm{h}$
$\begin{aligned} \text { Distance } &=\text { Speed } \times \text { Jime } \\ &=50 \times \frac{4}{5} \\ &=40 \mathrm{~km} . \end{aligned}$
$\begin{aligned} \text { Time }=? & \text { : Speed }=30 \mathrm{~km} / \mathrm{k} \\ \text { Time }=& \frac{\text { Distance }}{\text { speed }} \end{aligned}$
$=\frac{4}{3}$ hours
$=\frac{4}{3} \times 60$ min
∴ Time $=80 \mathrm{~min}$
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