Exercise 6A
Question 1
Ans: 1
(i) $8: 14:: x: 28$
$\frac{8}{14}=\frac{x}{28} .$
$\therefore 8 \times 28=14 \times x$
$\frac{8 \times 28}{14}=x$
$8 \times 2=x$
$16=x$
$x=16 $
(ii)
$\begin{aligned} x &=9=5: 3 \\ \frac{x}{9} &=\frac{5}{3} . \\ 3 x &=9 \times 5 \end{aligned}$
$x=\frac{9}{3} \times 5$
$x=15$
(iii) $12: x:=4: 15$.
$\begin{aligned}&\frac{12}{x}=\frac{4}{15} \\&12 \times 15=4 \times x\end{aligned}$
$x=\frac{12}{15} \times 4$
$3 \times 15=x$
$45=x$
$x=45$
Question 2
Ans:
(i) 25,15,40
Let the fourth proportional be x
$\begin{aligned}\therefore \quad 25: 15 &: 40: x \\\frac{25}{15} &=\frac{40}{x} .\end{aligned}$
$25\times x =40 \times 15$
$x=\frac{40 \times 15}{25}$
$x=8 \times 3$
x=24
Hence , the fourth proportional is 24.
(ii) $3 a^{2} b^{2}, a^{3}, b^{3}$.
Let the fourth proportional be x
$\begin{aligned} \therefore & 3 a^{2} b^{2}: a^{3}:: b^{3}: x \\ & \frac{3 a^{2} b^{2}}{9^{3}}=\frac{b^{3}}{x} \end{aligned}$
$3 a^{2} b^{2} \times x=b^{3} \times a^{3}$
$x=\frac{b^{3} \times a^{3}}{3 a^{2} b^{2}}$
$x=\frac{b a}{3}$
$x=\frac{a b}{3}$
Hence the fourth proportional is $\frac{a b}{3}$
(iii) $a^{2}-5 a+6, a^{2}+a-6, a^{2}-9$.
let the fourth proportional be x.
$\therefore a^{2}-5 a+6: a^{2}+a-6: a^{2}-9: x$.
$\frac{a^{2}-5 a+6}{a^{2}+a-6}=\frac{a^{2}-9}{x}$
$\left(a^{2}-5 a+6\right) \times x=\left(a^{2}-9\right) \times\left(a^{2}+a-6\right)$
$x=\frac{\left(a^{2}-9\right) \times\left(a^{2}+a^{2}-6\right)}{\left(a^{2}-5 a+6\right)}$
$x=\frac{\left((a)^{2}-(3)^{2}\right)\left(a^{2}+3 a-2 a-6\right)}{\left(a^{2}-3 a-2 a+6\right)}$
$x=\frac{(a+3)(a-3) \cdot(a(a+3)-2(a+3))}{(a(a-3)-2(a-3))}$
$x=(a+3)(a+3)$
$x=(a+3)^{2}$
Hence the fourth proportional $(a+3)^{2}$
Question 3
Ans:
(i) 16 and 36 .
When, $a: b:: 3: c$
Let the third proportional be x
$\therefore \quad 16: 36: 36: x$
$\frac{16}{36}=\frac{36}{x} $
$16 \times x=36 \times 36$
$x=\frac{36 \times 36}{16}$
$x=9 \times 9$
$x=81 $
Hence the third proportional is 81
(ii) $\frac{x}{y}+\frac{y}{x}$ and $\frac{x}{y}$
When a: b:: b:c
Let the third proportional be x,
$\therefore \frac{x}{y}+\frac{y}{x}: \frac{x}{y}:: \frac{x}{y}: x$
$\frac{\frac{x}{y}+\frac{y}{x}}{\frac{x}{y}}=\frac{x}{y}$
$\frac{\frac{x^{2}+y^{2}}{x y}}{\frac{x}{y}}=\frac{x}{\frac{y}{x}} .$
$\frac{x^{2}+y^{2}}{x y} \times x=\frac{x}{y} \times \frac{x}{y}$
$x=\frac{\frac{x^{2}}{y^{2}}}{\frac{x^{2}+y^{2}}{x y}}$
$x=\frac{x^{2} \times x y}{y^{2} \times x^{2}+y^{2}}$
$x=\frac{x^{3} y}{x^{2} y^{2}+y^{4}}$
$x=\frac{x^{3}}{x^{2} y +y^{3}}$
Hence , the third proportional is $\frac{x^{3}}{x^{2} y + y^{3}}$
(ii) $a^{2}-b^{2} a+b$.
when, $a: b:: b: c$.
Let the third proportional be x,
$\therefore a^{2}-b^{2}: a+b: a+b: x .$
$\frac{a^{2}-b^{2}}{a+b}=\frac{a+b}{x}$
$\left(a^{2}-b^{2}\right) \times x=(a+b) \times(a+b)$
$x=\frac{(a+b) \times (a+b)}{\left(a^{2}-b^{2}\right)}$
$x=\frac{(a+b) \times (a+b)}{(a+b)(a-b)}$
x = $\frac{a+b}{a-b}$
Hence , the third proportional is $\frac{a+b}{a-b}$
Question 4
Ans:
(i) 5 and 80
According to question
a: b:: b:c
5:b::b:80
$\frac{5}{b}=\frac{b}{80}$
$5 \times 80=b \times b $
$\begin{aligned}&400=b^{2} . \\&b^{2}=400 \\&b=\sqrt{400} \\&b=\sqrt{20 \times 20}\end{aligned}$
$b=20$
Hence, the mean proportional is 20
(ii) $360a_{4}$ and $250 a^{2} b^{2}$
According to question,
$\begin{aligned} & a: b:: b: c \\ & 360 a^{4}: b:: b: 250 a^{2} b^{2} \end{aligned}$
$360 a^{4} \times 250 a^{2} b^{2}=b^{2}$
$b^{2}=360 a^{4} \times 250 a^{2} b^{2}$
$b=\sqrt{360 a^{4} \times 250 a^{2} b^{2}}$
$b=\sqrt{360 \times 250 \times a^{4}+^{2} \times b^{2}}$
$b=\sqrt{360 \times 250 \times a^{6} \times b^{2}}$
$b=\sqrt{90000 \times 9^{6} \times b^{2}}$
$b=300 \times a^{3} \times b .$
$b=300 . a^{3} b $
Hence the mean proportional is $300 a^{3} b$
(iii) $(x-y) and \left(x^{3}-x^{2} y\right)$.
According to question,
$\begin{aligned} \therefore a: b: b &: c \\(x-y) &: b: b:\left(x^{3}-x^{2} y\right) . \\ \frac{(x-y)}{b} &=\frac{b}{x^{3}-x^{2} y .} \\(x-y) & x\left(x^{3}-x^{2} y\right)=b^{2} \\ b^{2} &=(x-y)\left(x^{3}-x^{2} y\right) . \\ b &=\sqrt{(x-y)\left(x^{3}-x^{2} y\right)} \\ b &=\sqrt{(x-y) x^{2}(x-y)} \\ b &=\sqrt{(x-y)^{2} x^{2}} \\ b &=(x-y) x \\ b &=x(x-y) \end{aligned}$
Hence, the mean proportional is x (x-y)
Question 5
Ans:
(i) Given,
x, 16,48,y are in continued proportion
$\therefore \quad \frac{x}{16}=\frac{16}{48}=\frac{48}{y} $
$\begin{aligned} \frac{x}{16} &=\frac{16}{48} \\ 48 x &=16 \times 16 \end{aligned}$
$x=\frac{16 \times 16}{48}$
$x=\frac{16}{3}$
and.
$\begin{aligned}&\frac{16}{48}=\frac{48}{y} \\&16 y=48 \times 48\end{aligned}$
$y=48 \times 3 .$
$y=144 .$
Hence, $x=\frac{16}{3}$ and $y=144$.
(ii) Given,
$x, 9, y, 16$ are in continue proportion,
$\begin{aligned} \frac{x}{9} &=\frac{9}{y}=\frac{y}{16} \\ \frac{9}{y} &=\frac{y}{16} . \\ 9 \times 16 &=y^{2} . \\ y^{2} &=1.44 . \\ y &=\sqrt{144} \\ y &=\sqrt{12 \times 12} \\ y &=12 .\end{aligned}$
and. $\frac{x}{9}=\frac{9}{y} $
$\frac{x}{9}=\frac{9}{12} \quad(y=12)$
$12 x=9 \times 9 .$
$x=\frac{27}{4}$
Hence, $x=\frac{27}{4}$ and $y=12$.
Question 6
Ans:
Let x be added to each number
3+x, 5+x,7+x and 10+x are in proportion
$\frac{3+x}{5+x}=\frac{7+x}{10+x}$
$(3+x)(10+x)=(7+1) x)(5+x)$
$30+3 x+16 x+x^{2}=35+7 x+5 x+x^{2}$
$30+13 x+x^{2}=35+12 x+x^{2}$.
$30+13 x+ x^{2}-35-12 x-x^{2}=0 .$
$13 x-12 x+30-35=0$
$x-5=0$
$x=0+5$
$x=5$
Hence 5 is added to each number
Question 7
Ans: Let x be subtracted from each of the number
$\therefore \quad 28 x: 53-x:: 19-x: 35-x$.
$\frac{28-x}{53-x}=\frac{19-x}{35-x} \text {}$
$(28-x)(35-x)=(19-x)(53-x) .$
$980-28 x-35 x+x^{2}=1007-19 x-53 x+x^{2}$
$980-28 x-35 x+ x^{2}-1007+19 x+53 x-x^{2}=0$
$980-63 x-1007+72 x=0 .$
$980-1007-63 x+72 x=0 .$
$-27+9 x=0$
$9 x=0+27$
$9 x=27$
x=3
Hence, 3 is subtracted from each number
Question 8
Ans:
(i) Given
Mean proportional = 14 and third proportional= 122
Let a and b be the two numbers
$\begin{aligned} a &: 14: 14: b \\ \frac{a}{14} &=\frac{14}{b} \\ a b &=14 \times 14 \\ a b &=196 . \end{aligned}$
$a=\frac{19 6}{b}$.............(i)
and a: b::b:112
$\begin{aligned} \frac{a}{b} &=\frac{b}{112} . \\ 112 a &=b^{2} . \\ 112 \times \frac{196}{b} &=b^{2} \end{aligned}$
$\begin{aligned} 21952 &=b^{3} \\ b^{3} &=21952 \\ b &=\sqrt[3]{21952} . \\ b &=\sqrt{28 \times 28 \times 28} . \\ b &=28 . \end{aligned}$
put the value of bin equation (i)
$a=\frac{196}{28}$
a=7
Hence the two number a and b is 7 and 28 respectively
(ii) Given,
Mean proportional =18 and
Third proportional =144
Let a and b be the two number
$\begin{aligned} \because a &: 18:: 18: b \\ \frac{a}{18} &=\frac{18}{b} \\ a b &=18 \times 18 \\ a b &=324 \end{aligned}$
a = $\frac{324}{b}$ ...........(i)
and a: b :: b:144
$\frac{a}{b}=\frac{b}{144}$
$144 a=b^{2}$
$144 \times \frac{324}{b}=b^{2}$
$b^{2}=\frac{144 \times 324}{b^{\circ}}$
$b^{2}=\frac{46656}{b}$
$b^{3}=46656$
$b=\sqrt[3]{46656}$
$b=\sqrt{36 \times 36 \times 36}$
$b=36$
put the value of bin equation (i)
$a=\frac{324}{36}$
$a=9$
Hence, the two number a and b is 9 and 36 respectively.
Question 9
Ans: Given
$p+r=2 q$..........(i)
$\frac{1}{q}+\frac{1}{s}=\frac{2}{r}$..........(ii)
From eq (ii)
$\frac{s+q}{qs}=\frac{2}{r }$
$r(s+q)=2 q s$
$r(s+2)=(p+r) s$ (from (i))
rs +rq =ps +rs
rq = ps +rs -rs
rq = ps
$\frac{r}{s}=\frac{p}{q} .$
$r: s=p: q .$
$p: q=r: s$
Hence proved
Question 10
Ans:
Given
B is the mean proportional between a and c,
$\begin{aligned} \therefore \quad a &=b:: b: c \\ \frac{a}{b} &=\frac{b}{c} . \\ a c &=b^{2} . \end{aligned}$
According to the question
$a, c, a^{2}+b^{2}$ and $b^{2}+c^{2}$
$\frac{a}{c}=\frac{a^{2}+b^{2}}{b^{2}+c^{2}}$
$\frac{a}{c}=\frac{a^{2}+a c}{a c+c^{2}} \quad\left(b^{2}=a c\right)$
$\frac{a}{c}=\frac{a(a+c)}{c(a+c)}$
$\frac{a}{c}=\frac{a}{c}$
Hence proved
Question 11
Ans:
Given
x+ 7 is the mean proportional between (x+3) and (x+12)
∴ $x+3: x+7: \therefore x+7: x+12$
$\frac{x+3}{x+7}=\frac{x+7}{x+12}$
$\left(x^{2}+12 x+2 x+12\right)=(x+7)(x+7)$
$\left(x^{2}+12 x+3 x+36\right)=(x+7)^{2}$
$x^{2}+15 x+36=x^{2}+(7)^{2}+2 \times x \times 7$
$x^{2}+15x+36= $x^{2}+49+14x
$x^{2}+15 x+36$$-x^{2}-49-4 x=0$
$15 x-14 x+36-49=0$
$x=13=0$
$x=0+13$
$x=13$
Hence, the value of x is 13
Question 12
Ans:
Given
$\frac{a^{2}+c^{2}}{a b+c d} = \frac{a b+c d}{b^{2}+d^{2}}$
$\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)=(a b+c d)(a b+c d)$
$\left(a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}\right)=\left(a b+(d)^{2}\right.$
$\left(a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}\right)=(a b)^{2}+(c d)^{2}+2 x a b \times c d$
$a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}=a^{2} b^{2}+c^{2} d^{2}+2 a b c d $
$a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}-y^{2} b^{2}-c^{2} d^{2}-2 a b c d=0$
$a^{2} d^{2}+b^{2} c^{2}-2 a b c d=0 $
$(a d-b c)^{2}=0$
$a d-b c=0$
a d=b c
$\frac{a}{b}=\frac{c}{d}$, Hence Proved.
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