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
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