S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6A

  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.