Question 13 Marks
If $\frac{a}{b}=\frac{2}{3}$, then find the values of the following expressions : $\frac{7 b-4 a}{7 b+4 a}$
Answer$ \frac{ a }{ b }=\frac{2}{3}$
$\therefore \quad \frac{ b }{ a }=\frac{3}{2}$
$\therefore \quad \frac{b}{a} \times \frac{7}{4}=\frac{3}{2} \times \frac{7}{4}$
...[Multiplying both sides by $\frac{7}{4}$ ]
$ \therefore \quad \frac{7 b}{4 a}=\frac{21}{8}$
$\therefore \quad \frac{7 b+4 a}{7 b-4 a}=\frac{21+8}{21-8}$
$\therefore \quad \frac{7 b+4 a}{7 b-4 a}=\frac{29}{13} \quad \text { [By componendo - dividendo] }$
$\therefore \quad \frac{7 b-4 a}{7 b+4 a}=\frac{13}{29} \quad \ldots \text { [By invertendo] }$
View full question & answer→Question 23 Marks
If $\frac{a}{b}=\frac{2}{3}$, then find the values of the following expressions : $\frac{ a ^3+ b ^3}{ b ^3}$
Answer$\frac{ a }{ b }=\frac{2}{3}$
$\ldots[\text { Given }]$
$\therefore \quad\left(\frac{a}{b}\right)^3=\left(\frac{2}{3}\right)^3$
$\therefore \quad \frac{ a ^3}{ b ^3}=\frac{8}{27}$
$\therefore \quad \frac{a^3+b^3}{b^3}=\frac{8+27}{27}$
$\therefore \quad \frac{a^3+b^3}{b^3}=\frac{35}{27}$
View full question & answer→Question 33 Marks
If $\frac{a}{b}=\frac{2}{3}$, then find the values of the following expressions : $\frac{5 a ^2+2 b ^2}{5 a ^2-2 b ^2}$
View full question & answer→Question 43 Marks
If $\frac{a}{b}=\frac{2}{3}$, then find the values of the following expressions : $\frac{4 a+3 b}{3 b}$
Answer$ \frac{a}{b}=\frac{2}{3}$
$\therefore \quad \frac{a}{b} \times \frac{4}{3}=\frac{2}{3} \times \frac{4}{3} $
...[Given]
...[Multiplying both sides by $\frac{4}{3}$ ]
$ \therefore \quad \frac{4 a}{3 b}=\frac{8}{9}$
$\therefore \quad \frac{4 a+3 b}{3 b}=\frac{8+9}{9}$
$\therefore \quad \frac{4 a+3 b}{3 b}=\frac{17}{9} $
...[By componendo]
View full question & answer→Question 53 Marks
If $a, b, c$ are in continued proportion then show that $\frac{a}{c}=\frac{a^2+a b+b^2}{b^2+b c+c^2}$
Answer$a, b, c$ are in continued proportion
$\therefore \frac{a}{b}=\frac{b}{c}$
Let, $\frac{a}{b}=\frac{b}{c}=k \quad \therefore b=c k$ and $a=c k^2$
$ \text { LHS }=\frac{a}{c}=\frac{c k^2}{c}=k^2$
$\text { RHS }= \frac{a^2+a b+b^2}{b^2+b c+c^2}$
$=\frac{\left(k^2 c\right)^2+k^2 c(c k)+(c k)^2}{(c k)^2+(c k)(c)+c^2}$
$=\frac{k^4 c^2+k^3 c^2+c^2 k^2}{c^2 k^2+c^2 k+c^2}$
$=\frac{c^2 k^2\left(k^2+k+1\right)}{c^2\left(k^2+k+1\right)}$
$=k^2$
$\therefore \quad \text { LHS }=\text { RHS }$
$\therefore \quad \frac{a}{c}=\frac{a^2+a b+b^2}{b^2+b c+c^2}$
View full question & answer→Question 63 Marks
If $\frac{a}{(x-2 y+3 z)}=\frac{b}{(y-2 z+3 x)}=\frac{c}{(z-2 x+3 y)}$ and $x+\mathrm{y}+\mathrm{z} \neq 0$ then prove that each ratio $=\frac{a+b+c}{2(x+y+z)}$
AnswerLet $\frac{a}{(x-2 y+3 z)}=\frac{b}{(y-2 z+3 x)}=\frac{c}{(z-2 x+3 y)}=\mathrm{k}$.
$\therefore$ by theorem of equal ratios
$
\begin{aligned}
k & =\frac{a+b+c}{(x-2 y+3 z)+(y-2 z+3 x)+(z-2 x+3 y)} \\
& =\frac{a+b+c}{2 x+2 y+2 z} \\
& =\frac{a+b+c}{2(x+y+z)} \\
& \therefore \frac{a}{x-2 y+3 z}=\frac{b}{y-2 z+3 x}=\frac{c}{z-2 x+3 y}=\frac{a+b+c}{2(x+y+z)}
\end{aligned}
$
View full question & answer→Question 73 Marks
If $\frac{a}{3}=\frac{b}{2}$ then find the value of the ratio $\frac{5 a+3 b}{7 a-2 b}$.
AnswerMethod I
$\frac{a}{3} =\frac{b}{2}$
$\therefore \frac{a}{b} =\frac{3}{2}\dots$ (using Alternando)
Now dividing each term of $\frac{5 a+3 b}{7 a-2 b}$ by $b$.
$ \frac{\frac{5 a}{b}+\frac{3 b}{b}}{\frac{7 a}{b}-\frac{2 b}{b}}=\frac{5\left(\frac{a}{b}\right)+3}{7\left(\frac{a}{b}\right)-2}$
$=\frac{5\left(\frac{3}{2}\right)+3}{7\left(\frac{3}{2}\right)-2}$
$=\frac{\frac{15}{2}+3}{\frac{21}{2}-2}$
$=\frac{15+6}{21-4}$
$=\frac{21}{17}$
Method II
$\frac{a}{3}=\frac{b}{2}$
Let $\frac{a}{3}=\frac{b}{2}=t$.
$\therefore$ by substituting $a=3 \mathrm{t}$ and $b=2 t$,
$ \frac{5 a+3 b}{7 a-2 b} =\frac{5(3 t)+3(2 t)}{7(3 t)-2(2 t)} \quad(t \neq 0)$
$ =\frac{15 t+6 t}{21 t-4 \mathrm{t}}$
$ =\frac{21 t}{17 t}$
$ =\frac{21}{17} $
View full question & answer→Question 83 Marks
If $a: b=2: 1$ and $b: c=4: 1$ then find the value of $\left(\frac{a^4}{32 b^2 c^2}\right)^3$.
Answer$\frac{a}{b}=\frac{2}{1} \quad \therefore a=2 b \quad \frac{b}{c}=\frac{4}{1} \quad \therefore b=4 c$
$
a=2 b=2 \times 4 c=8 c \quad \therefore a=8 c
$
Now substituting the values $a=8 c, b=4 c$
$
\begin{aligned}
\left(\frac{a^4}{32 b^2 c^2}\right)^3 & =\left(\frac{(8 c)^4}{32 \times 4^2 \times c^2 \times c^2}\right)^3 \\
& =\left[\frac{8 \times 8 \times 8 \times 8 \times c^4}{32 \times 16 \times c^2 \times c^2}\right]^3 \\
& =(8)^3 \\
\therefore \quad\left(\frac{a^4}{32 b^2 c^2}\right)^3 & =512
\end{aligned}
$
View full question & answer→Question 93 Marks
The ratio of number of mango trees to chikoo trees in an orchard is 2 : 3. If 5 more trees
of each type are planted the ratio of trees would be 5 : 7. Then find the number of mango
and chickoo trees in the orchard.
AnswerThe ratio of trees is $2: 3$.
Let the number of mango trees $=2 x$ and chikoo trees $=3 x$
From given condition, $\frac{2 x+5}{3 x+5}=\frac{5}{7}$
$
\begin{aligned}
14 x+35 & =15 x+25 \\
\therefore x & =10
\end{aligned}
$
$\therefore$ number of mango trees in the orchard $=2 x=2 \times 10=20$
and number of chikoo trees $=3 x=3 \times 10=30$
View full question & answer→Question 103 Marks
The ratio of ages of Seema and Rajashree is 3 : 1. The ratio of ages of Rajashree and
Atul is 2 : 3. Then find the ratio of ages of Seema, Rajashree and Atul.
AnswerSeema's age : rajashree's age $=3: 1$ rajashree's age : Atul's age $=2: 3$
The second term of the first ratio should be the same as the first term of the second ratio.
Hence to get the continuous ratio, multiply each term of the first ratio by 2. We get $3: 1=6: 2$.
$
\begin{aligned}
& \frac{\text { Seema's age }}{\text { rajashree's age }}=\frac{6}{2}, \quad \frac{\text { rajashree's age }}{\text { Atul's age }}=\frac{2}{3} \\
& \, therefore, \text { Seema's age: Rajashree's age: Atul's age } 6: 2: 3.
\end{aligned}
$
View full question & answer→Question 113 Marks
The proportion of compounds of nitrogen, phosphorous and potassium in certain
fertilizer is 18 : 18 : 10. Here compound of nitrogen is 18%, compound of phosphorous
is 18% and that of potassium is 10%. Remaining part is of other substances. Find the
weight of each of the above compounds in 20 kg of fertilizer.
AnswerLet the weight of nitrogen compound in $20 \mathrm{~kg}$ of fertilizer be $x \mathrm{~kg}$.
$
\therefore \quad \frac{18}{100}=\frac{x}{20}
$
$
\therefore \quad x=\frac{18 \times 20}{100}=3.6
$
$\therefore$ weight of nitrogen compound is $3.6 \mathrm{~kg}$
The percentage of the phosphorous compound is also $18 \%$.
$\therefore$ Weight of compound of phosphorous is $3.6 \mathrm{~kg}$
If we assume the weight of potassium compound $y \mathrm{~kg}$ then
$
\frac{10}{100}=\frac{y}{20} \quad \therefore y=2 \quad \therefore \text { weight of potassium compound is } 2 \mathrm{~kg} \text {. }
$
View full question & answer→Question 123 Marks
If $\frac{3 x-5 y}{5 z+3 y}=\frac{x+5 z}{y-5 x}=\frac{y-z}{x-z}$, then show that every ratio $=\frac{x}{y}$.
AnswerLet $\frac{3 x-5 y}{5 z +3 y}=\frac{x+5 z }{y-5 x}=\frac{y- z }{x- z }= k$
$\therefore \quad k =\frac{3 x-5 y}{5 z +3 y}=\frac{x+5 z }{y-5 x}=\frac{5 y-5 z }{5 x-5 z}$
...[Multiplying numerator and denominator of third ratio by $5$ ]
$\therefore \quad k =\frac{(3 x-5 y)+(x+5 z)+(5 y-5 z)}{(5 z+3 y)+(y-5 x)+(5 x-5 z)}$
... [Theorem on equal ratios]
$=\frac{4 x}{4 y}$
$=\frac{x}{y}$
$\therefore \quad \text { Each ratio }=\frac{x}{y} $
View full question & answer→Question 133 Marks
$\frac{x}{3 x-y-z}=\frac{y}{3 y-z-x}=\frac{z}{3 z-x-y}$ and $x+y+z \neq 0$, then show that $\frac{a+b}{2}$.
AnswerLet $\frac{ a + b y}{x+y}=\frac{ b x+ az }{x+ z }=\frac{ a y+ bz }{y+ z }= k$
$
\begin{aligned}
\therefore \quad k & =\frac{( ax + b y)+( b x+ az )+( a y+ bz )}{(x+y)+(x+ z )+(y+ z )} \\
& =\frac{ a x+ b x+ a y+ b y+ az + bz }{2 x+2 y+2 z } \\
& =\frac{x( a + b )+y( a + b )+ z ( a + b )}{2(x+y+ z )} \\
& =\frac{( a + b )(x+y+ z )}{2(x+y+ z )} \\
\therefore \quad k & =\frac{ a + b }{2} \quad \ldots[\because x+y+ z \neq 0] \\
\therefore \quad & \text { Each ratio }=\frac{ a + b }{2}
\end{aligned}
$
View full question & answer→Question 143 Marks
If $\frac{x}{3 x-y-z}=\frac{y}{3 y-z-x}=\frac{z}{3 z-x-y}$ and $x + y + z \neq 0$, then show that the value of each ratio is equal to 1.
Answer$\begin{array}{ll} & \text { Let } \frac{x}{3 x-y- z }=\frac{y}{3 y- z -x}=\frac{ z }{3 z -x-y}= k \\ \therefore \quad k = & \frac{x+y+ z }{(3 x-y- z )+(3 y- z -x)+(3 z -x-y)} \\ \therefore \quad k = & \frac{x+\text { Theorem on equal ratios] }}{3 x-x-x+3 y-y-y+3 z - z - z } \\ & =\frac{x+y+ z }{x+y+ z } \\ \therefore \quad & k =1 \\ \therefore \quad & \text { Each ratio }=1\end{array}$
View full question & answer→Question 153 Marks
Fill in the blanks of the following : $\quad \frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2 b+3 c}{\ldots \ldots . .}=\frac{\ldots \ldots .}{6-8+14}$
Answer$ \frac{ a }{3}=\frac{ b }{4}=\frac{ c }{7}=\frac{ a -2 b +3 c }{\ldots \ldots}=\frac{\ldots \ldots .}{6-8+14}$
$\frac{ a }{3}=\frac{ b }{4}=\frac{ c }{7}=\frac{ a -2 b +3 c }{3-2 \times 4+3 \times 7} $
... [Theorem on equal ratios]
$\begin{aligned}
& =\frac{a-2 b+3 c}{3-8+21}=\frac{a-2 b+3 c}{16} \\
\frac{a}{3}=\frac{b}{4}=\frac{c}{7} & =\frac{2 a-2 b+2 c}{2 \times 3-2 \times 4+2 \times 7} \\
& =\frac{2 a-2 b+2 c}{6-8+14} \\
\therefore \quad \frac{a}{3}=\frac{b}{4}=\frac{c}{7} & =\frac{a-2 b+3 c}{16}=\frac{2 a-2 b+2 c}{6-8+14}
\end{aligned}$
View full question & answer→Question 163 Marks
Fill in the blanks of the following : $\quad \frac{x}{7}=\frac{y}{3}=\frac{3 x+5 y}{\ldots \ldots . .}=\frac{7 x-9 y}{\ldots \ldots .}$
Answer$\begin{aligned}
\frac{x}{7} & =\frac{y}{3}=\frac{3 x+5 y}{\ldots \ldots}=\frac{7 x-9 y}{\ldots \ldots .} \\
\frac{x}{7} & =\frac{y}{3}=\frac{3 x+5 y}{3 \times 7+5 \times 3}
\end{aligned}$
... [Theorem on equal ratios]
$\begin{aligned}
& =\frac{3 x+5 y}{21+15}=\frac{3 x+5 y}{36} \\
\frac{x}{7}=\frac{y}{3} & =\frac{7 x-9 y}{7 \times 7-9 \times 3} \\
& =\frac{7 x-9 y}{49-27}=\frac{7 x-9 y}{22} \\
\therefore \quad \frac{x}{7}=\frac{y}{3} & =\frac{3 x+5 y}{36}=\frac{7 x-9 y}{22}
\end{aligned}$
View full question & answer→Question 173 Marks
If $\frac{15 a^2+4 b^2}{15 a^2-4 b^2}=\frac{47}{7}$, then find the value of the following ratios : $\frac{b^3-2 a^3}{b^3+2 a^3}$
Answer$\frac{a}{b}=\frac{3}{5}$
$\therefore \quad\left(\frac{a}{b}\right)^3=\left(\frac{3}{5}\right)^3$
$\therefore \quad \frac{a^3}{b^3}=\frac{27}{125}$
$\therefore \quad \frac{b^3}{a^3}=\frac{125}{27}$
$\therefore \quad \frac{b^3}{a^3} \times \frac{1}{2}=\frac{125}{27} \times \frac{1}{2}$
$\text {... }\left[\text { Multiplying both sides by } \frac{1}{2}\right]$
$\therefore \quad \frac{b^3}{2 a^3}=\frac{125}{54}$
$\therefore \quad \frac{b^3+2 a^3}{b^3-2 a^3}=\frac{125+54}{125-54}$
$\text {... [By componendo - dividendo] }$
$\therefore \quad \frac{b^3+2 a^3}{b^3-2 a^3}=\frac{179}{71}$
$\therefore \quad \frac{b^3-2 a^3}{b^3+2 a^3}=\frac{71}{179}$
$\therefore \quad \frac{b^3-2 a^3}{b^3+2 a^3}=71: 179$
View full question & answer→Question 183 Marks
If $\frac{15 a^2+4 b^2}{15 a^2-4 b^2}=\frac{47}{7}$, then find the value of the following ratios : $\frac{b^2-2 a^2}{b^2+2 a^2}$
View full question & answer→Question 193 Marks
If $\frac{15 a^2+4 b^2}{15 a^2-4 b^2}=\frac{47}{7}$, then find the value of the following ratios : $\quad \frac{7 a-3 b}{7 a+3 b}$
Answer$\quad \frac{a}{b}=\frac{3}{5}$
$\therefore \quad \frac{a}{b} \times \frac{7}{3}=\frac{3}{5} \times \frac{7}{3}$
$\therefore \quad \ldots\left[\text { Multiplying both sides by } \frac{7}{3}\right]$
$\therefore \quad \frac{7 a}{3 b}=\frac{7}{5}$
$\therefore \quad \frac{7 a+3 b}{7 a-3 b}=\frac{7+5}{7-5}$
$\therefore \quad \frac{7 a+3 b}{7 a-3 b}=\frac{12}{2}$
$\therefore \quad \frac{7 a+3 b}{7 a-3 b}=\frac{6}{1}$
$\therefore \quad \frac{7 a-3 b}{7 a+3 b}=\frac{1}{6}$
$\therefore \quad \frac{7 a-3 b}{7 a+3 b}=1: 6$
View full question & answer→Question 203 Marks
If $\frac{15 a^2+4 b^2}{15 a^2-4 b^2}=\frac{47}{7}$, then find the value of the following ratios : $\quad \frac{a}{b}$
View full question & answer→Question 213 Marks
(x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Answer$\begin{aligned} & (x+3):(x+11)=(x-2):(x+1) \\ & \frac{x+3}{x+11}=\frac{x-2}{x+1} \\ \therefore & (x+3)(x+1)=(x-2)(x+11) \\ \therefore & x(x+1)+3(x+1)=x(x+11)-2(x+11) \\ \therefore & x^2+x+3 x+3=x^2+1 \mid x-2 x-22 \\ \therefore & x^2+4 x+3=x^2+9 x-22 \\ \therefore & 4 x+3=9 x-22 \\ \therefore & 3+22=9 x-4 x \\ \therefore & 25=5 x \\ \therefore & x=5\end{aligned}$
View full question & answer→Question 223 Marks
if$\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}$, then find the ratio $\frac{a}{b}$.
Answer$\sqrt{0.04} \times 0.4 \times a=0.4 \times 0.04 \times \sqrt{b} \ldots \text { [Given] }$
$ \therefore 0.04 \times 0.4 \times a=(0.4)^2 \times(0.04)^2 \times b \ldots \text { [Squaring both sides] }$
$\therefore \quad \frac{a}{b}=\frac{(0.04)^2 \times(0.4)^2}{0.04 \times 0.4}$
$=0.04 \times 0.4=0.016$
$=\frac{16}{1000}$
$=\frac{8 \times 2}{8 \times 125}$
$\therefore \quad \frac{a}{b}=\frac{2}{125}$
View full question & answer→Question 233 Marks
If $a : b = 3 : 1$ and $b : c = 5 : 1,$ then find the value of
i. $\left(\frac{a^3}{15 b^2 c}\right)^3$
ii. $\frac{a^2}{7 b c}$
AnswerGiven, $a: b=3: 1$
$\therefore \frac{a}{b}=\frac{3}{1}$
$\therefore a=3 b \text {...(i) }$
$\text { and } b: c=5: 1$
$\therefore \frac{b}{c}=\frac{5}{1}$
$b=5 c \ldots \text {..ii) }$
Substituting (ii) in (i),
we get $a=3(5 c)$
$\therefore a=15 c \ldots \text { (iii) }$
i. $\left(\frac{ a ^3}{15 b^2 c }\right)^3 =\left[\frac{(15 c )^3}{15(5 c )^2 c }\right]^3 \ldots[\text { From (ii) and (iii)] }$
$ =\left[\frac{15 c \times 15 c \times 15 c }{15 \times 5 c \times 5 c \times c }\right]^3$
$ =(3 \times 3)^3$
$ =(9)^3$
$\therefore \quad\left(\frac{a^3}{15 b^2 c }\right)^3 =729$
ii.$ \frac{ a ^2}{7 bc } =\frac{(15 c )^2}{7 \times 5 c \times c } \quad \ldots[\text { From (ii) and (iii)] }$
$ =\frac{15 c \times 15 c }{7 \times 5 c \times c }$
$\therefore \quad \frac{ a ^2}{7 bc } =\frac{45}{7}$
View full question & answer→Question 243 Marks
Compare the following : $\quad \frac{9.2}{5.1}, \frac{3.4}{7.1}$
Answer$\begin{array}{ll}\quad & \frac{9.2}{5.1}, \frac{3.4}{7.1} \\ & 9.2 \times 7.1=65.32 \\ & 5.1 \times 3.4=17.34 \\ & \text { Here, } 65.32>17.34 \\ \therefore \quad & \frac{9.2}{5.1}>\frac{3.4}{7.1}\end{array}$
View full question & answer→Question 253 Marks
Compare the following : $\frac{\sqrt{80}}{\sqrt{48}}, \frac{\sqrt{45}}{\sqrt{27}}$
Answer$\begin{array}{ll}
& \frac{\sqrt{80}}{\sqrt{48}}, \frac{\sqrt{45}}{\sqrt{27}} \\
& \sqrt{80} \times \sqrt{27}=\sqrt{2160} \\
& \sqrt{48} \times \sqrt{45}=\sqrt{2160} \\
& \text { Here, } 2160=2160 \\
\therefore \quad & \sqrt{2160}=\sqrt{2160} \\
\therefore \quad & \frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}
\end{array}$
Alternate method:
$\frac{\sqrt{80}}{\sqrt{48}}, \frac{\sqrt{45}}{\sqrt{27}}$
Consider, $\frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{16 \times 5}}{\sqrt{16 \times 3}}=\frac{4 \sqrt{5}}{4 \sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}$
$\frac{\sqrt{45}}{\sqrt{27}}=\frac{\sqrt{9 \times 5}}{\sqrt{9 \times 3}}=\frac{3 \sqrt{5}}{3 \sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}$
Here, $\frac{\sqrt{5}}{\sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}$
$\therefore \quad \frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}$
View full question & answer→Question 263 Marks
Compare the following : $\frac{5}{18}, \frac{17}{121}$
Answer$\begin{array}{ll}\quad & \frac{5}{18}, \frac{17}{121} \\ & 5 \times 121=605 \\ & 18 \times 17=306 \\ & \text { Here, } 605>306 \\ \therefore \quad & \frac{5}{18}>\frac{17}{121}\end{array}$
View full question & answer→Question 273 Marks
Compare the following : $\frac{3 \sqrt{5}}{5 \sqrt{7}}, \frac{\sqrt{63}}{\sqrt{125}}$
Answer$\begin{aligned}
\frac{3 \sqrt{5}}{5 \sqrt{7}}, \frac{\sqrt{63}}{\sqrt{125}} & \\
3 \sqrt{5} \times \sqrt{125} & =3 \times \sqrt{5 \times 125} \\
& =3 \times \sqrt{5 \times 25 \times 5} \\
& =3 \times 5 \times 5 \\
& =3 \times 25=75 \\
5 \sqrt{7} \times \sqrt{63} & =5 \times \sqrt{7 \times 63} \\
& =5 \times \sqrt{7 \times 9 \times 7} \\
& =5 \times 3 \times 7=105
\end{aligned}$
Here, $75<105$
$\therefore \quad \frac{3 \sqrt{5}}{5 \sqrt{7}}<\frac{\sqrt{63}}{\sqrt{125}}$
View full question & answer→Question 283 Marks
Compare the following : $\quad \frac{\sqrt{5}}{3}, \frac{3}{\sqrt{7}}$
Answer$\begin{array}{ll}\quad & \frac{\sqrt{5}}{3}, \frac{3}{\sqrt{7}} \\ & \sqrt{5} \times \sqrt{7}=\sqrt{35} \\ & 3 \times 3=9=\sqrt{9^2}=\sqrt{81} \\ & \text { Here, } 35<81 \\ \therefore \quad & \sqrt{35}<\sqrt{81} \\ \therefore \quad & \frac{\sqrt{5}}{3}<\frac{3}{\sqrt{7}}\end{array}$
View full question & answer→Question 293 Marks
Find the following ratios : The lengths of sides of a rectangle are $5 cm$ and $3.5 cm$. Find the ratio of numbers denoting its perimeter to area.
AnswerLength of rectangle = (l) =$ 5 cm,$
Breadth of rectangle = (b) =$ 3.5 cm$
Perimeter of the rectangle = $2(l + b)$
$= 2(5 + 3.5)$
$= 2 x 8.5$
$= 17 cm$
Area of the rectangle$ = l x b$
$= 5 x 3.5$
$= 17.5 cm^2$
Ratio of numbers denoting perimeter to the area of rectangle
$\begin{aligned}
& =\frac{\text { perimeter }}{\text { area }} \\
& =\frac{17}{17.5} \\
& =\frac{17 \times 10}{17.5 \times 10} \\
& =\frac{170}{175} \\
& =\frac{5 \times 34}{5 \times 35} \\
& =\frac{34}{35} \\
& =34: 35
\end{aligned}$
$\therefore$ Ratio of numbers denoting perimeter to the area of rectangle is $34: 35$.
View full question & answer→Question 303 Marks
Find the following ratios : The ratio of diagonal of a square to its side, if the length of side is 7 cm.
AnswerLength of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side
$=\frac{\text { diagonal }}{\text { side }}$
$=\frac{7 \sqrt{2}}{7}$
$=\frac{\sqrt{2}}{1}$
$=\sqrt{2}: 1$
$\therefore$ The ratio of diagonal of a square to its side is $\sqrt{ } 2: 1$.
View full question & answer→Question 313 Marks
Find the following ratios $:$ The ratio of circumference of circle with radius $r$ to its area.
AnswerLet the radius of the circle is $r.$
$\therefore$ circumference $= 2\pi r$ and area $= \pi r^2$
Ratio of circumference to the area of circle
$ =\frac{\text { circumference }}{\text { area }}$
$=\frac{2 \pi r }{\pi r ^2}$
$=\frac{2}{ r }$
$=2: r$
$\therefore$ The ratio of circumference of circle with radius $r$ to its area is $2: r$.
View full question & answer→Question 323 Marks
Find the following ratios : The ratio of radius to circumference of the circle.
AnswerLet the radius of circle be $r.$
then, its circumference $= 2\pi r$
Ratio of radius to circumference of the circle
$=\frac{\text { radius }}{\text { circumference }}$
$=\frac{ r }{2 \pi r }$
$=\frac{1}{2 \pi}$
$=1: 2 \pi $
The ratio of radius to circumference of the circle is $1: 2 \pi$.
View full question & answer→Question 333 Marks
If a : b = 3 : 1 and b : c = 5 : 1, then find the value of
i. $\left(\frac{a^3}{15 b^2 c}\right)^3$
ii. $\frac{a^2}{7 b c}$
AnswerGiven, $a: b=3: 1$
$\therefore \frac{a}{b}=\frac{3}{1}$
$\therefore a=3 b \text {...(i) }$
$\text { and } b: c=5: 1$
$\therefore \frac{b}{c}=\frac{5}{1}$
$b=5 c \ldots \text {..ii) }$
Substituting (ii) in (i),
we get $a=3(5 c)$
$\therefore a=15 c \ldots \text { (iii) }$
i. $\left(\frac{ a ^3}{15 b^2 c }\right)^3=\left[\frac{(15 c )^3}{15(5 c )^2 c }\right]^3 \ldots[\text { From (ii) and (iii)] }$
$=\left[\frac{15 c \times 15 c \times 15 c }{15 \times 5 c \times 5 c \times c }\right]^3$
$=(3 \times 3)^3$
$=(9)^3$
$\therefore \quad\left(\frac{a^3}{15 b^2 c }\right)^3 =729$
ii. $\frac{ a ^2}{7 bc } =\frac{(15 c )^2}{7 \times 5 c \times c } \quad \ldots[\text { From (ii) and (iii)] }$
$ =\frac{15 c \times 15 c }{7 \times 5 c \times c }$
$\therefore \quad \frac{ a ^2}{7 bc } =\frac{45}{7}$
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