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.
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