Concise Mathematics Class 10 ICSE Solutions for Chapter 7 - Ratio and Proportion (Including Properties and Uses)

Exercise – 7(A)

1. If $\mathbf{a~:b=5~:3}$ , find $\mathbf{\frac{5a~-~3b}{5a~+~3b}}$

Ans: Here, we have $a~:b=5~:3$

Therefore, $\frac{a}{b}=\frac{5}{3}$

Now, $\frac{5a~-~3b}{5a~+~3b}~=~\frac{\left( \frac{5a~-~3b}{b} \right)}{\left( \frac{5a~+~3b}{b} \right)}$

$=~\frac{\left( \frac{5a~}{b}-3 \right)}{\left( \frac{5a~}{b}~+~3 \right)}$

$=~\frac{\left( 5\times ~\frac{5}{3}~-3 \right)}{\left( 5\times \frac{5}{~3}~+~3 \right)}$

$=~\frac{\left( \frac{25}{3}~-3 \right)}{\left( \frac{25}{3}~+~3 \right)}$

$=~\frac{\left( \frac{25-9}{3} \right)~}{\left( \frac{25+9}{3} \right)~}$

$=~\left( \frac{\frac{16}{3}~}{\frac{34}{3}~} \right)$

$=\frac{16}{34}$

$=\frac{8}{17}$

Hence, Value of $\frac{5a~-~3b}{5a~+~3b}$ is $\frac{8}{17}~.$

2. If $\mathbf{x~:y=4~:7}$, Find the value of $\mathbf{\left( 3x+2y \right)~:\left( 5x+y \right).}$

Ans: Here, we have $x~:y=4~:7$

Therefore, $\frac{x}{y}=\frac{4}{7}$

Now, $\frac{3x~+~2y}{5x~+~y}~=~\frac{\left( \frac{3x~+2y}{y} \right)}{\left( \frac{5x~+~y}{y} \right)}$

$=~\frac{\left( \frac{3x~}{y}~+~2 \right)}{\left( \frac{5x~}{y}~+~1 \right)}$

$=~\frac{\left( 3\times ~\frac{4}{7}~+~2 \right)}{\left( 5\times ~\frac{4}{7}+1 \right)}$

$=~\frac{\left( \frac{12}{7}~+~2 \right)}{\left( \frac{20}{7}~+~1 \right)}$

$=~\frac{\left( \frac{12~+~14}{7} \right)~}{\left( \frac{20~+~7}{7} \right)~}$

$=~\left( \frac{\frac{26}{7}~}{\frac{27}{7}~} \right)$

$=\frac{26}{27}$

Hence, Value of $\frac{3x~+~2y}{5x~+~y}$ is $\frac{26}{27}.$

3. If $\mathbf{a~:b=3~:8,~}$find the value of  $\mathbf{\frac{4a~+~3b}{6a-~b}}$. 

Ans: Here, we have $a~:b=3~:8$

Therefore, $\frac{a}{b}=\frac{3}{8}$

Now, $\frac{4a~+~3b}{6a-~b}~=~\frac{\left( \frac{4a~+~3b}{b} \right)}{\left( \frac{6a~-~b}{b} \right)}$

$=~\frac{\left( \frac{4a~}{b}~+~3 \right)}{\left( \frac{6a~}{b}~-1 \right)}$

$=~\frac{\left( 4\times ~\frac{3}{8}~+3 \right)}{\left( 6\times \frac{3}{~8}-1~ \right)}$

$=~\frac{\left( \frac{12}{8}~+~3 \right)}{\left( \frac{18}{8}~-~1 \right)}$

$=~\frac{\left( \frac{12~+~24}{8} \right)~}{\left( \frac{18~-~8}{8} \right)~}$

$=~\left( \frac{\frac{36}{8}~}{\frac{10}{8}~} \right)$

$=\frac{36}{10}$

$=\frac{18}{5}$

Hence, Value of  $\frac{4a~+~3b}{6a-~b}$ is $\frac{18}{5}$.

4. If $\mathbf{\left( a-b \right)~:\left( a+b \right)=1~:11}$ find the ratio: 

$\mathbf{\left( 5a+4b+15 \right)~:\left( 5a-4b+3 \right)}$.

Ans: Here, we have $\left( a-b \right)~:\left( a+b \right)=1~:11$

⇒ $\frac{\left( a-b \right)}{\left( a+b \right)}=\frac{1}{11}$

⇒ $11a-11b=a+b$ 

⇒ $11a-a=b+11b$

⇒ $10a=12b$

⇒ $\frac{a}{b}=\frac{12}{10}$

⇒ $\frac{a}{b}=\frac{6}{5}$

Let $a=6x$ and $b=5x$

Now, $\frac{\left( 5a+4b+15 \right)}{\left( 5a-4b+3 \right)}=~\frac{\left( 5~\times ~6x~+~4\times ~5x~+15 \right)}{\left( 5~\times ~6x~-~4\times ~5x~+3 \right)}$ 

$~=~\frac{\left( 30x~+~20x~+15 \right)}{\left( 30x~-~20x~+3 \right)}$

$=~\frac{\left( 50x~+15 \right)}{\left( 10x~+3 \right)}$

$=~\frac{5\left( 10x~+3 \right)}{\left( 10x~+3 \right)}$

$=~5$

Hence, Value of $\frac{\left( 5a+4b+15 \right)}{\left( 5a-4b+3 \right)}$ is 5.

5. Find the number which bears the same ratio to  $\mathbf{\frac{7}{33}~~}$that $\mathbf{~\frac{8}{21}}$ does to $\mathbf{\frac{4}{9}.}$

Ans: Let the required number be $x$.

Therefore, 

⇒ $x~:\frac{7}{33}=\frac{8}{21}~:\frac{4}{9}$

⇒ $\frac{x}{\frac{7}{33}}=\frac{\frac{8}{21}}{\frac{4}{9}}$

⇒ $\frac{x}{\frac{7}{33}}=\frac{8~\times ~9}{21~\times ~4}$

⇒ $x=\frac{7}{33}\times \frac{8~\times ~9}{21~\times ~4}$

⇒ $x=\frac{7}{33}\times \frac{8~\times ~9}{21~\times ~4}$

⇒ $x=\frac{2}{11}$

Hence, the required number is  $\frac{2}{11}$ .

6.If $\mathbf{\frac{m~+~n}{m~+~3n}=\frac{2}{3}~,}$ find $\mathbf{\frac{2{{n}^{2}}}{3{{m}^{2}}+~mn}~.}$

Ans: Here, we have $\frac{m~+~n}{m~+~3n}=\frac{2}{3}$

⇒ $3\left( m+n \right)=2\left( m+3n \right)$

⇒ $3m+3n=2m+6n$

⇒ $3m-2m=6n-3n$

⇒ $m=3n$

Now, $\frac{2{{n}^{2}}}{3{{m}^{2}}+~mn}=~\frac{2{{n}^{2}}}{3\times {{\left( 3n \right)}^{2}}+~3n.n}$

$=~\frac{2{{n}^{2}}}{3\times 9{{n}^{2}}+~3{{n}^{2}}}$

$=~\frac{2{{n}^{2}}}{27{{n}^{2}}+~3{{n}^{2}}}$

$=~\frac{2{{n}^{2}}}{30{{n}^{2}}}$

$=~\frac{1}{15}$

Hence, Value of $\frac{2{{n}^{2}}}{3{{m}^{2}}+~mn}$   is  $\frac{1}{15}$ .

7. Find  $\mathbf{\frac{x}{y}~;$ when ${{x}^{2}}+6{{y}^{2}}=5xy}$.

Ans: Here, we have ${{x}^{2}}+6{{y}^{2}}=5xy$

Divide by ${{y}^{2}}$ in above equation

⇒ $\frac{{{x}^{2}}}{{{y}^{2}}}+\frac{6{{y}^{2}}}{{{y}^{2}}}=\frac{5xy}{{{y}^{2}}}$

⇒ ${{\left( \frac{x}{y} \right)}^{2}}+6=5\times \frac{x}{y}$

Let $a=\frac{x}{y}$

⇒ ${{a}^{2}}+6=5a$

⇒ ${{a}^{2}}-5a+6=0$

⇒ ${{a}^{2}}-2a-3a+6=0$

⇒ $a\left( a-2 \right)-3\left( a-2 \right)=0$

⇒ $\left( a-2 \right)\left( a-3 \right)=0$

⇒ $\left( a-2 \right)=0~or~\left( a-3 \right)=0$

⇒ $a=2~or~a=3$

Thus, $\frac{x}{y}=2$ or  $\frac{x}{y}=3$.

Hence, Value of  $\frac{x}{y}$  is 2 or 3.

8. If the ratio between 8 and 11 is the same as the ratio of $\mathbf{2x-y}$ to $\mathbf{x+2y,}$ find the value of $\mathbf{\frac{7x}{9y}~.~}$

Ans: here, the ratio between 8 and 11 is the same as the ratio of $2x-y$ to $x+2y$.

Therefore, 

⇒ $\frac{8}{11}=\frac{2x-y}{x~+~2y}$

⇒ $22x-11y=8x+16y$

⇒ $22x-8x=16y+11y$

⇒ $14x=27y$

⇒ $\frac{x}{y}=\frac{27}{14}$

Multiply by $~\frac{7}{9}~$ in the above equation, we get

⇒ $\frac{7x}{9y}=\frac{7}{9}\times \frac{27}{14}$

⇒ $\frac{7x}{9y}=\frac{3}{2}$

Hence, value of  $\frac{7x}{9y}$  is  $\frac{3}{2}$ . 

9. Divide Rs. 1290 into A, B and C such that A is $\mathbf{\frac{2}{5}}$ of B and $\mathbf{B:C~=~4~:3}$.

Ans: here, we have $A=\frac{2}{5}B$ and $\frac{B}{C}=\frac{4}{3}$

⇒ $A=\frac{2}{5}B$ and $4C=3B$

⇒ $A=\frac{2}{5}B$ and $C=\frac{3}{4}B$

Now, $A+B+C=1290$

⇒ $\frac{2}{5}B+B+\frac{3}{4}B=1290$

⇒ $\frac{8B+20B+15B}{20}=1290$

⇒ $\frac{43B}{20}=1290$

⇒ $43B=1290\times 20$

⇒ $B=\frac{1290\times 20}{43}$

⇒ $B=30\times 20$

⇒ $B=600$ 

Now, Amount of A = $\frac{2}{5}B$

= $\frac{2}{5}\times 600$

= $2\times 120$

= $240$

And amount of C = $\frac{3}{4}B$

= $\frac{3}{4}\times 600$

= $3\times 150$

= $450$

Hence, amount of A = Rs. 240

Amount of B = Rs. 600

Amount of C = Rs. 450

10. A school has 630 students. The ratio of the number of boys to the number of girls is 3:2. This ratio changes to 7:5 after the admission of 90 new students. Find the number of newly admitted boys.

Ans: Let the number of boys be $3x$.

And the number of girls be $2x$.

Now, Total student = 630

⇒ $3x+2x=630$

⇒ $5x=630$

⇒ $x=126$

Thus, number of boys before new admission = $3x$

= $3\times 126$

= $378$

And number of girls before new admission = $2x$

= $2\times 126$

= $252.$

Now, Let the number of newly admitted boys be $y.$

And the number of newly girls admitted be $\left( 90-y \right)$.

According to the question, 

⇒ $\frac{378~+~y}{252~+~90-y}=\frac{7}{5}$

⇒ $\frac{378~+~y}{342~-y}=\frac{7}{5}$

⇒ $1890+5y=2394-7y$. 

⇒ $5y+7y=2394-1890$

⇒ $12y=504$

⇒ $y=\frac{504}{12}$

⇒ $y=42$

Hence, the number of newly admitted boys is 42.

11. What quantity must be subtracted from each term of the ratio $\mathbf{9~:17}$ to make it equal to 1:3?

Ans: Let the required quantity be $x$.

w, according to question

⇒ $\frac{\text{9 }\!\!~\!\!\text{ - }\!\!~\!\!\text{ x}}{\text{17-x}}\text{=}\frac{\text{1}}{\text{3}}$

⇒ $\text{17-x=27-3x}$

⇒ $\text{-x+3x=27-17}$

⇒ $2x=10$

⇒ $x=\frac{10}{2}$

⇒ $x=5$

Hence, the required quantity is 5.

12. The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. Their expenditures are in the ratio 3:5. If each saves Rs. 80 every month, find their monthly pocket money.

Ans: Let the pocket money of Ravi be Rs. $5x$.

And the pocket money of Sanjeev is Rs. $7x$.

Expenditure of Ravi is $3y.$

Expenditure of Sanjeev is $5y$.

Now, monthly saving of Ravi = Rs. 80

⇒ $5x-3y=80$ ……….. eq (i)

And monthly saving of Sanjeev = Rs. 80

⇒ $7x-5y=80$ ……….. eq (ii)

Now, multiplying by 5 in eq(i) and 3 in eq (ii), we get

⇒ $25x-15y=400$ ……….. eq (iii)

⇒ $21x-15y=240$ ……….. eq (iv)

Now, subtracting eq (iv) from eq (iii)

⇒ $4x=160$

⇒ $x=\frac{160}{4}$

⇒ $x=40$

Thus, pocket money of Ravi = Rs. $5x$

= Rs. $5\times 40$

= Rs. $200$

And , pocket money of Sanjeev = Rs. $7x$

= Rs. $7\times 40$

= Rs. $280$

13. The work done by $\mathbf{\left( x-2 \right)}$ men in $\mathbf{\left( 4x+1 \right)}$ days and the work done by $\mathbf{\left( 4x+1 \right)}$ men in $\mathbf{\left( 2x-3 \right)}$ days are in the ratio 3:8. Find the value of $\mathbf{x}$. 

Ans: here, we have, work done by $\left( x-2 \right)$ men in $\left( 4x+1 \right)$ days and the work done by $\left( 4x+1 \right)$ men in $\left( 2x-3 \right)$ days are in the ratio $3~:~8$

Thus, the amount of work done by $\left( x-2 \right)$ men in $\left( 4x+1 \right)$ days is

(x-2)(4x+1).

And the amount of work done by $\left( 4x+1 \right)$ men in $\left( 2x-3 \right)$ days is 

$\left( 4x+1 \right)\left( 2x-3 \right)$.

Now, according to question

⇒ $\frac{\left( x-2 \right)\left( 4x+1 \right)}{~\left( 4x+1 \right)\left( 2x-3 \right)}=\frac{3}{8}$

⇒ $\frac{\left( x-2 \right)}{~\left( 2x-3 \right)}=\frac{3}{8}$

⇒ $8x-16=6x-9$

⇒ $8x-6x=-9+16$

⇒ $2x=7$

⇒ $x=\frac{7}{2}$

Hence, Value of $x$ is $~\frac{7}{2}~$.

14. The bus fare between two cities is increased in the ratio $\mathbf{7~:9.}$ Find the increase in the fare, if:

The original fare is Rs. 245.

Ans: here, bus fare between two cities is increased in the ratio $7~:9$ 

And original fare is Rs. 245

Now, according to question

⇒ increased bus fare  $=\frac{9}{7}~\times $ original fare

$=\frac{9}{7}~\times 245$ 

$=9\times 35$ 

$=Rs.~315$ 

Thus, increase in fare = increased fare original fare

= Rs. 315 Rs. 245

= Rs. 70.

The increased fare is Rs. 207.

Ans: here, bus fare between two cities is increased in the ratio $7~:9$ 

And increased fare is Rs. 207

Now, according to question

⇒ increased bus fare of $=\frac{9}{7}~\times $ original fare

$\Rightarrow 207=\frac{9}{7}~\times ~original~fare$ 

$\Rightarrow ~original~fare~=207\times \frac{7}{9}$ 

$\Rightarrow ~original~fare~=23\times 7$ 

$\Rightarrow ~original~fare~=Rs.~161$ 

Thus, increase in fare = increased fare $$ original fare

= Rs. 207 $$ Rs. 161

= Rs. 46.

15. By increasing the cost of entry ticket to a fair in the ratio 10:13, the number of visitors to the fair has decreased in the ratio 6:5. In what ratio has the total collection increased or decreased?

Ans: Let the initial entry ticket cost be $Rs.~10x$.

And initial number of visitors be $6y$.

Thus, initial collection = $Rs.~10x~\times 6y$

= $Rs.~60xy$

Now, let the increased entry ticket cost be Rs. $13x$

And decreased visitors are $5y$.

Thus, final collection = $Rs.~13x~\times 5y$

= $Rs.~65xy$

Here, we have a final collection that is greater than initial collection

Thus, required increased ratio = $\frac{60xy}{65~xy}$

= $\frac{60}{65}$ .

16. In a basket, the ratio between the number of oranges and the number of apples is 7:13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes $1~:2.$ Find the original number of oranges and the original number of apples in the basket.

Ans: Let the original number of oranges is $7x$.

And the original number of apples is $13x.$

Now, according to question

⇒ $\frac{7x-8}{13x-11}=\frac{1}{2}$

⇒ $14x-16=13x-11$

⇒ $14x-13x=-11+16$

⇒ $x=5$

Thus, the original number of oranges $=7x$

$=7\times 5$

$~~=35$

And the original number of apples  $=13x$

$=13\times 5$

$=65$.

17. In a mixture of 126 kg of milk and water, milk and water are in the ratio 5:2. How much water must be added to the mixture to make this ratio 3:2?

Ans: Let the initial quantity of milk be $5x$ kg

And the initial quantity of water be $2x$ kg.

Now, total quantity of mixture $=126~$kg

⇒ $5x+2x=126$

⇒ $7x=126$

⇒ $x=\frac{126}{7}$

⇒ $x=18$

Thus, the initial quantity of milk $=5x$

$=5\times 18$

$=90$ kg

And the initial quantity of water $=2x$

$=2\times 18$

$=36$ kg

Now, let $y$ kg water be added to the mixture to get ratio $3~:~2.$

⇒$~\frac{90}{36+y~~}=\frac{3}{2}$

⇒$~108+3y=180$

⇒$~3y=72$

⇒$~y=\frac{72}{3}$

⇒$~y=24$

Hence, 24 kg water must be added to the mixture to make ratio $3~:~2.$

17. 

1. If $\mathbf{A~:B=3~:4}$ and $\mathbf{B~:C=6~:7,}$ find A : B : C

Ans: here, we have $A~:B=3~:4$ and $B~:C=6~:7$

Therefore, 

⇒ $\frac{A}{B}=\frac{3}{4}$  and $~\frac{B}{C}=\frac{6}{7}$

⇒ $\frac{A}{B}=\frac{3\times 3}{4\times 3}$  and $~\frac{B}{C}=\frac{6\times 2}{7\times 2}$

⇒ $\frac{A}{B}=\frac{9}{12}$  and $~\frac{B}{C}=\frac{12}{14}$

On comparing above both equations, we get

⇒ $A~:B~:C=9~:12~:14$

Hence, value of $A~:B~:C~is~9~:12~:14$.

Here, we have $A~:B=3~:4$ and $B~:C=6~:7$.

Now, we can write  $\frac{A}{C}$  as  $\frac{A}{C}~\times \frac{B}{B}$.

⇒ $\frac{A}{C}=\frac{A}{C}\times \frac{B}{B}$

⇒ $\frac{A}{C}=\frac{A}{B}\times \frac{B}{C}$

⇒ $\frac{A}{C}=\frac{3}{4}\times \frac{6}{7}$

⇒ $\frac{A}{C}=\frac{18}{28}$

⇒ $\frac{A}{C}=\frac{9}{14}$

Thus, value of $A~:C$ is $9~:14$.

2. If $\mathbf{A~:B=2~:5}$ and $\mathbf{A~:C=3~:4,}$ find $\mathbf{A~:~B~:~C.}$ 

Ans: here, we have $A~:B=2~:5$ and $A~:C=3~:4$

Therefore, 

⇒ $\frac{A}{B}=\frac{2}{5}$  and $~\frac{A}{C}=\frac{3}{4}$

⇒ $\frac{A}{B}=\frac{2\times 3}{5\times 3}$  and $~\frac{A}{C}=\frac{3\times 2}{4\times 2}$

⇒ $\frac{A}{B}=\frac{6}{15}$  and $~\frac{A}{C}=\frac{6}{8}$

On comparing above both equations, we get

⇒ $A~:B~:C=6~:15~:8$

Hence, value of $A~:B~:C~is~6~:15~:8$.

(i) If $\mathbf{3A=4B=6C~;}$ find $\mathbf{A~:B~:C}$

Ans: here, we have $3A=4B=6C$.

Therefore, 

⇒ $3A=4B$ and $4B=6C$

⇒ $\frac{A}{B}=\frac{4}{3}$  and  $\frac{B}{C}=\frac{6}{4}$

⇒ $\frac{A}{B}=\frac{4}{3}$  and  $\frac{B}{C}=\frac{3}{2}$

On comparing above both equations, we get

⇒ $A~:B~:C=4~:3~:2$

Hence, value of $A~:B~:C~is~4~:3~:2$.

(ii) If $\mathbf{2a=3b}$ and $\mathbf{4b=5c}$ Find $\mathbf{a~:c.}$

Ans: here, we have $2a=3b$ and $4b=5c$.

Therefore,

⇒ $\frac{a}{b}=\frac{3}{2}$ and $\frac{b}{c}=\frac{5}{4}$

Now, we can write  $\frac{a}{c}$  as  $\frac{a}{c}~\times \frac{b}{b}$.

⇒ $\frac{a}{c}=\frac{a}{c}\times \frac{b}{b}$

⇒ $\frac{a}{c}=\frac{a}{b}\times \frac{b}{c}$

⇒ $\frac{a}{c}=\frac{3}{2}\times \frac{5}{4}$

⇒ $\frac{a}{c}=\frac{15}{8}$

Thus, value of $a~:c$ is $15~:8$.

18. Find the compound ratio of: 

1. $\mathbf{2~:3,~9~:14}$ and $\mathbf{14~:27.}$

Ans: We know that the compound ratio of $a~:b,~c~:d$ and $e~:f$ is  $\frac{ace}{bdf}$.

Therefore, the compound ratio of $2~:3,~9~:14$ and $14~:27$ = $\frac{2\times 9\times 14}{3\times 14\times 27}$

= $\frac{2}{3\times 3}$

= $\frac{2}{9}$

Thus, the required compound ratio is $2~:9$.

2. $\mathbf{2a~:3b,~~mn~:~{{x}^{2}}~}$and $\mathbf{x~:n}$.

Ans: We know that the compound ratio of $a~:b,~c~:d$ and $e~:f$ is $\frac{ace}{bdf}$.

Therefore, the compound ratio of $2a~:3b,~~mn~:~{{x}^{2}}~$and $x~:n$ = $\frac{2a\times mn\times x}{3b~\times ~{{x}^{2}}~\times ~n}$

 = $\frac{2am}{3bx}$

Thus, the required compound ratio is $2am~:3bx$.

3. $\mathbf{\sqrt{2}~:1,~~3~:~\sqrt{5}}$ and $\mathbf{\sqrt{20}~:9.}$

Ans: We know that the compound ratio of $a~:b,~c~:d$ and $e~:f$ is  $\frac{ace}{bdf}$.

Therefore, the compound ratio of $\sqrt{2}~:1,~~3~:~\sqrt{5}$ and $\sqrt{20}~:9$ = $\frac{\sqrt{2}\times 3\times \sqrt{20}}{1\times \sqrt{5}\times 9}$

 = $\frac{\sqrt{2}\times 2}{3}$

   = $\frac{2\sqrt{2}}{3}$

Thus, the required compound ratio is $2\sqrt{2}~:3$.

19. Find duplicate ratio of:

1. 3:4

Ans: We know that duplicate ratio of $a~:b$  is ${{a}^{2}}~:{{b}^{2}}$

Therefore, duplicate ratio of  $3~:4$  is ${{3}^{2}}~:{{4}^{2}}$

$={{3}^{2}}~:{{4}^{2}}$ 

$=\frac{{{3}^{2}}}{{{4}^{2}}}$ 

$=\frac{9}{16}$ 

$=9~:16$ 

Hence, duplicate ratio of  $3~:4~$ is $9~:16.$

2. $\mathbf{3\sqrt{3}~:2\sqrt{5}}$

Ans: We know that duplicate ratio of $a~:b$  is ${{a}^{2}}~:{{b}^{2}}$

Therefore, duplicate ratio of  $3\sqrt{3}~:2\sqrt{5}$  is ${{\left( 3\sqrt{3} \right)}^{2}}~:{{\left( 2\sqrt{5} \right)}^{2}}$

$={{\left( 3\sqrt{3} \right)}^{2}}~:{{\left( 2\sqrt{5} \right)}^{2}}$ 

$=\frac{{{\left( 3\sqrt{3} \right)}^{2}}}{{{\left( 2\sqrt{5} \right)}^{2}}}$ 

$=\frac{9~\times ~3}{4~\times ~5}$ 

$=\frac{27}{20}$ 

$=27~:20$ 

Hence, duplicate ratio of  $3\sqrt{3}~:2\sqrt{5}~$ is $27~:20.$

20. Find triplicate ratio of:

1. $\mathbf{1~:3}$

Ans: We know that triplicate ratio of $a~:b$  is ${{a}^{3}}~:{{b}^{3}}$

Therefore, triplicate ratio of  $1~:3$  is ${{1}^{3}}~:{{\left( 3b \right)}^{3}}$

$={{1}^{3}}~:{{3}^{3}}$ 

$=\frac{{{1}^{3}}}{{{3}^{3}}}$ 

$=\frac{1}{27}$ 

$=1~:27$ 

Hence, triplicate ratio of  $1~:3~$ is $1~:27.$

2. $\mathbf{\frac{m}{2}~:\frac{n}{3}}$ 

Ans: We know that triplicate ratio of $a~:b$  is ${{a}^{3}}~:{{b}^{3}}$

Therefore, triplicate ratio of  $\frac{m}{2}~:\frac{n}{3}$  is ${{\left( \frac{m}{2} \right)}^{3}}~:{{\left( \frac{n}{3} \right)}^{3}}$

$={{\left( \frac{m}{2} \right)}^{3}}~:{{\left( \frac{n}{3} \right)}^{3}}$ 

$=\frac{{{m}^{3}}}{8}~:~\frac{{{n}^{3}}}{27}$ 

$=\frac{\frac{{{m}^{3}}}{8}}{\frac{{{n}^{3}}}{27}}$ 

$=27{{m}^{3}}~:8{{n}^{3}}$ 

Hence, triplicate ratio of  $\frac{m}{2}~:\frac{n}{3}~$ is $27{{m}^{3}}~:8{{n}^{3}}.$

21. Find sub – duplicate ratio of:

1. 9:16

Ans: We know that sub - duplicate ratio of $a~:b$  is $\sqrt{a}~:\sqrt{b}$

Therefore, sub - duplicate ratio of  $9~:16$  is $\sqrt{9}~:\sqrt{16}$

$=\sqrt{9}~:\sqrt{16}$ 

$=\frac{\sqrt{9}}{\sqrt{16}}$ 

$=\frac{3}{4}$ 

$=3~:4$ 

Hence, sub- duplicate ratio of  $9~:16~$ is $3~:4.$

2. ${{\mathbf{\left( x-y \right)}^{4}}~:~{{\left( x+y \right)}^{6}}}$

Ans: We know that sub - duplicate ratio of $a~:b$  is $\sqrt{a}~:\sqrt{b}$

Therefore, sub - duplicate ratio of  ${{\left( x-y \right)}^{4}}~:~{{\left( x+y \right)}^{6}}~$  is 

 $\sqrt{{{\left( x-y \right)}^{4}}}~:\sqrt{{{\left( x+y \right)}^{6}}}$

$=\sqrt{{{\left( x-y \right)}^{4}}}~:\sqrt{{{\left( x+y \right)}^{6}}}$ 

$=\frac{\sqrt{{{\left( x-y \right)}^{4}}}}{\sqrt{{{\left( x+y \right)}^{6}}}}$ 

$=\frac{{{\left( x-y \right)}^{2}}}{{{\left( x+y \right)}^{3}}}$ 

$={{\left( x-y \right)}^{2}}~:~{{\left( x+y \right)}^{3}}$ 

Hence, sub- duplicate ratio of  ${{\left( x-y \right)}^{4}}~:~{{\left( x+y \right)}^{6}}~$ is

 ${{\left( x-y \right)}^{2}}~:~{{\left( x+y \right)}^{3}}.$

22. Find the sub – triplicate ratio of: 

1. 64:27

Ans: We know that sub - triplicate ratio of $a~:b$  is $\sqrt[3]{a}~:\sqrt[3]{b}$

Therefore, sub - triplicate ratio of  $64~:27$  is $\sqrt[3]{64}~:\sqrt[3]{27}$

$=\sqrt[3]{64}~:\sqrt[3]{27}$ 

$=\frac{\sqrt[3]{64}}{\sqrt[3]{27}}$ 

$=\frac{\sqrt[3]{4~\times ~4~\times ~4}}{\sqrt[3]{3~\times ~3~\times ~3}}~$ 

$=\frac{\sqrt[3]{{{4}^{3}}}}{\sqrt[3]{{{3}^{3}}}}$ 

$=\frac{4}{3}$ 

$=4~:3$ 

Hence, sub - triplicate ratio of $64~:27~$is $4~:3.$

2. $\mathbf{{{x}^{3}}~:125{{y}^{3}}}$

Ans: We know that sub - triplicate ratio of $a~:b$  is $\sqrt[3]{a}~:\sqrt[3]{b}$

Therefore, sub - triplicate ratio of  ${{x}^{3}}~:125{{y}^{3}}$  is $\sqrt[3]{{{x}^{3}}}~:\sqrt[3]{125{{y}^{3}}}$

$=\sqrt[3]{{{x}^{3}}}~:\sqrt[3]{125{{y}^{3}}}$ 

$=\frac{\sqrt[3]{{{x}^{3}}}}{\sqrt[3]{125{{y}^{3}}}}$ $~$ 

$=\frac{\sqrt[3]{{{x}^{3}}}}{\sqrt[3]{{{\left( 5y \right)}^{3}}}}$ 

$=\frac{x}{5y}$ 

$=x~:5y$ 

Hence, sub - triplicate ratio of ${{x}^{3}}~:125{{y}^{3}}~$is $x~:5y.$

23: Find the reciprocal ratio of: 

1. 5 : 8

Ans: We know that reciprocal ratio of $a~:b$  is $\frac{1}{a}~:\frac{1}{b}$.

Therefore, reciprocal ratio of  $5~:8$  is $\frac{1}{5}~:\frac{1}{8}$.

$=\frac{1}{5}~:\frac{1}{8}$ 

$=\frac{\frac{1}{5}}{\frac{1}{8}}$ 

$=\frac{8}{5}$ 

$=8~:5$ 

Hence, reciprocal ratio of  $5~:8~$is $8~:5$.

2. $\mathbf{\frac{x}{3}~:\frac{y}{7}~}$

Ans: We know that reciprocal ratio of $a~:b$  is $\frac{1}{a}~:\frac{1}{b}$.

Therefore, reciprocal ratio of  $\frac{x}{3}~:\frac{y}{7}$  is $\frac{1}{\frac{x}{3}}~:\frac{1}{\frac{y}{7}}$.

$=\frac{1}{\frac{x}{3}}~:\frac{1}{\frac{y}{7}}$ 

$=\frac{3}{x}~:\frac{7}{y}$ 

$=\frac{\frac{3}{x}}{\frac{7}{y}}$ 

$=\frac{3y}{7x}$ 

$=3y~:7x$ 

Hence, reciprocal ratio of  $\frac{x}{3}~:\frac{y}{7}~$is $3y~:7x$      n

23. If $\mathbf{\left( x+3 \right)~:\left( 4x+1 \right)}$ is the duplicate ratio of $\mathbf{3~:5.}$ Find the value of $\mathbf{x.}$

Ans: here, we have $\left( x+3 \right)~:\left( 4x+1 \right)$ is the duplicate ratio of $3~:5$.        

Therefore, 

⇒ $\frac{\left( x+3 \right)}{\left( 4x+1 \right)}=\frac{{{3}^{2}}}{{{5}^{2}}}$ 

⇒ $\frac{\left( x+3 \right)}{\left( 4x+1 \right)}=\frac{9}{25}$  

⇒ $36x+9=25x+75$   

⇒ $36x-25x=75-9$   

⇒ $11x=66$   

⇒ $x=\frac{66}{11}$ 

⇒ $x=6$  

Hence, value of $x$ is 6.

24. If $\mathbf{m~:n}$ is the duplicate ratio of $\mathbf{\left( m+x \right)~:\left( n+x \right),}$ prove that $\mathbf{{{x}^{2}}=mn.}$

 Ans: here, we have $m~:n$ is the duplicate ratio of $\left( m+x \right)~:\left( n+x \right)$.

 Therefore, 

 ⇒ $\frac{m}{n}=~\frac{{{\left( m~+~x \right)}^{2}}}{{{\left( n~+~x \right)}^{2}}}$

 ⇒ $\frac{m}{n}=~\frac{{{m}^{2}}+~2mx+{{x}^{2}}}{{{n}^{2}}+~2nx+{{x}^{2}}}$

 ⇒ $m\left( {{n}^{2}}+~2nx+{{x}^{2}} \right)=~n\left( {{m}^{2}}+~2mx+{{x}^{2}} \right)$

 ⇒ $m{{n}^{2}}+~2mnx+m{{x}^{2}}=~n{{m}^{2}}+~2mnx+n{{x}^{2}}$

 ⇒ $m{{n}^{2}}+m{{x}^{2}}=~n{{m}^{2}}+n{{x}^{2}}$

 ⇒ $m{{x}^{2}}-n{{x}^{2}}=~n{{m}^{2}}-m{{n}^{2}}$

 ⇒ ${{x}^{2}}\left( m-n \right)=~mn\left( m-n \right)$

 ⇒ ${{x}^{2}}=\frac{mn\left( m-n \right)}{\left( m-n \right)}$

 ⇒ ${{x}^{2}}=mn$

 Hence proved.

25. If $\mathbf{\left( 3x-9 \right)~:\left( 5x+4 \right)}$ is the triplicate ratio of $\mathbf{3~:4,}$ Find the value of $\mathbf{x.}$ 

Ans: Here, we have $\left( 3x-9 \right)~:\left( 5x+4 \right)$ is the triplicate ratio of $3~:4$.

Therefore, 

⇒  $\frac{\left( 3x-9 \right)}{\left( 5x+4 \right)}=\frac{{{3}^{3}}}{{{4}^{3}}}$

⇒  $\frac{\left( 3x-9 \right)}{\left( 5x+4 \right)}=\frac{27}{64}$

⇒ $~\frac{\left( 3x-9 \right)}{\left( 5x+4 \right)}=\frac{27}{64}$

⇒$~192x-576=135x+108$

⇒$~192x-135x=108+576$

⇒$~192x-135x=108+576$

⇒$~57x=684$

⇒$~x=\frac{684}{57}$

⇒$~x=12$

Hence, Value of $x$ is 12.

26. Find the ratio compounded of the reciprocal ratio of $\mathbf{15~:28,}$ the sub – duplicate ratio of $\mathbf{36~:49}$ and the triplicate ratio of $\mathbf{5~:4}$.

Ans: here, the reciprocal ratio of $~\frac{15}{28}~$ is  $\frac{28}{15}~$. 

And the sub-duplicate ratio of  $\frac{36}{49}~$ is $\frac{\sqrt{36}}{\sqrt{49}}=\frac{6}{7}$

The triplicate ratio of $\frac{5}{4}$ is $\frac{{{5}^{3}}}{{{4}^{3}}}=\frac{125}{64}$

Now, the required compound ratio $=\frac{28~\times ~6~\times ~125}{15~\times ~7~\times ~64}$

$=\frac{4\times 6\times 25}{3\times 64}$

$=\frac{25}{8}$

Hence, required compound ratio is $\frac{25}{8}$ .

27. 

1. If $\mathbf{{{r}^{2}}=pq,}$ show that $\mathbf{p~:q}$ is the duplicate ratio of $\mathbf{\left( p+r \right):\left( q+r \right).}$ 

Ans: here, we have ${{r}^{2}}=pq$

Therefore, duplicate ratio of $\left( p+r \right):\left( q+r \right)$ is ${{\left( p+r \right)}^{2}}:{{\left( q+r \right)}^{2}}$

$=~\frac{{{\left( p+r \right)}^{2}}}{{{\left( q+r \right)}^{2}}}$

$=~\frac{{{p}^{2}}+2pr+{{r}^{2}}}{{{q}^{2}}+2qr+{{r}^{2}}}$

$=~\frac{{{p}^{2}}+2pr+pq}{{{q}^{2}}+2qr+pq}$

$=~\frac{p\left( p+2r+q \right)}{q\left( q+2r+p \right)}$

$=~\frac{p}{q}$ 

Thus, $p~:q$ is the duplicate ratio of $\left( p+r \right):\left( q+r \right).$

Hence proved.

2. If $\mathbf{\left( p-x \right)~:\left( q-x \right)}$ be the duplicate ratio of $\mathbf{p~:q}$ then, show that $\mathbf{\frac{1}{p}+\frac{1}{q}=\frac{1}{x}.}$

Ans: Here, we have $\left( p-x \right)~:\left( q-x \right)$ be the duplicate ratio of $p~:q$.

Therefore, 

⇒ $\frac{\left( p-x \right)}{\left( q-x \right)}=\frac{{{p}^{2}}}{{{q}^{2}}}$

⇒ ${{q}^{2}}\left( p-x \right)={{p}^{2}}\left( q-x \right)$

⇒ ${{q}^{2}}p-{{q}^{2}}x={{p}^{2}}q-{{p}^{2}}x$

⇒ ${{p}^{2}}x-{{q}^{2}}x={{p}^{2}}q-{{q}^{2}}p$

⇒ $x\left( {{p}^{2}}-{{q}^{2}} \right)=pq\left( p-q \right)$

⇒ $x\left( p-q \right)\left( p+q \right)=pq\left( p-q \right)$

⇒ $x\left( p+q \right)=pq$

⇒ $\frac{\left( p+q \right)}{pq}=\frac{1}{x}$

⇒ $\frac{1}{q}+\frac{1}{p}=\frac{1}{x}$

⇒ $\frac{1}{p}+\frac{1}{q}=\frac{1}{x}$

Hence proved.

Exercise- 7(B)

1. Find the fourth proportional to:

1.5, 4.5 and 3.5

Ans: Let the fourth proportional be x.

Therefore, $1.5~:4.5=3.5~:x$

⇒ $\frac{1.5}{4.5}=\frac{3.5}{x}$

⇒ $\frac{1}{3}=\frac{3.5}{x}$

⇒ $x=10.5$

Thus, fourth proportional of $1.5,~4.5$ and $3.5$ is $10.5$.

$\mathbf{3a,~6{{a}^{2}}}$ and $\mathbf{2a{{b}^{2}}}$

Ans: Let the fourth proportional be $x$.

Therefore, $3a~:6{{a}^{2}}=2a{{b}^{2}}~:x$

⇒ $\frac{3a}{6{{a}^{2}}}=\frac{2a{{b}^{2}}}{x}$

⇒ $\frac{1}{2a}=\frac{2a{{b}^{2}}}{x}$

⇒ $x=4{{a}^{2}}{{b}^{2}}$

Thus, fourth proportional of $3a,~6{{a}^{2}}$ and $2a{{b}^{2}}$ is $4{{a}^{2}}{{b}^{2}}$.

2. Find the third proportional to:

$\mathbf{2\frac{2}{3}}$ and 4

Ans: Let the required third proportional be $x$.

Thus, $2\frac{2}{3}~,~4$ and $x$ are in continued proportion.

Therefore, 

⇒ $2\frac{2}{3}~:4=4~:x$ 

⇒ $\frac{8}{3}~:4=4~:x$ 

⇒ $\frac{\frac{8}{3}}{4}=\frac{4}{x}$

⇒ $\frac{8}{12}=\frac{4}{x}$

⇒ $8x=48$

⇒ $x=\frac{48}{8}$

⇒ $x=6$

Hence, required third proportional to $2\frac{2}{3}~$ and $4$ is 6.

3. $\mathbf{\left( a-b \right)}$ and $\mathbf{\left( {{a}^{2}}-{{b}^{2}} \right)}$

Ans: Let the required third proportional be $x$.

Thus, $\left( a-b \right)$ , $\left( {{a}^{2}}-{{b}^{2}} \right)$ and $x~$are in continued proportion.

Therefore, 

⇒ $\left( a-b \right)$ : $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( {{a}^{2}}-{{b}^{2}} \right)~:x$ 

⇒ $\frac{\left( a-b \right)}{\left( {{a}^{2}}-{{b}^{2}} \right)}=\frac{\left( {{a}^{2}}-{{b}^{2}} \right)}{x}$ 

⇒ $x~\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}-{{b}^{2}} \right)$

⇒ $x\left( a-b \right)=\left( a-b \right)\left( a+b \right)\left( {{a}^{2}}-{{b}^{2}} \right)$

⇒ $x=\left( a+b \right)\left( {{a}^{2}}-{{b}^{2}} \right)$

Hence, required third proportional is $\left( a+b \right)\left( {{a}^{2}}-{{b}^{2}} \right)$.

4. Find the mean proportional between: 

1. $\mathbf{6+3\sqrt{3}~}$ and $\mathbf{8-4\sqrt{3}}$ 

Ans: Let the required mean proportional be $x$.

Thus, $6+3\sqrt{3},~~x$ and $8-4\sqrt{3}$ are in continued proportion.

Therefore, 

⇒ $\left( 6+3\sqrt{3} \right)~:x=x~:\left( 8-4\sqrt{3}~ \right)$

⇒ $\frac{\left( 6+3\sqrt{3} \right)}{x}=\frac{x}{\left( 8-4\sqrt{3}~ \right)}$

⇒ ${{x}^{2}}=\left( 6+3\sqrt{3} \right)\left( 8-4\sqrt{3}~ \right)$

⇒ ${{x}^{2}}=48-24\sqrt{3}+24\sqrt{3}-36$

⇒ ${{x}^{2}}=12$

⇒ $x=\sqrt{12}$

⇒ $x=\pm 2\sqrt{3}$

Since, the mean proportional to positive number is positive.

Thus, $x=2\sqrt{3}$

Hence, required mean proportional is $2\sqrt{3}.$ 

2. $\mathbf{\left( a-b \right)}$ and $\mathbf{\left( {{a}^{3}}-{{a}^{2}}b \right).}$

Ans: Let the required mean proportional be $x$.

Thus, $\left( a-b \right),~~x$ and $\left( {{a}^{3}}-{{a}^{2}}b \right)$ are in continued proportion.

Therefore, 

⇒ $\left( a-b \right)~:x=x~:~\left( {{a}^{3}}-{{a}^{2}}b \right)$

⇒ $\frac{\left( a-b \right)}{x}=\frac{x}{\left( {{a}^{3}}-{{a}^{2}}b \right)}$

⇒ ${{x}^{2}}=\left( a-b \right)\left( {{a}^{3}}-{{a}^{2}}b \right)$

⇒ ${{x}^{2}}=\left( a-b \right).~{{a}^{2}}\left( a-b \right)$

⇒ ${{x}^{2}}={{a}^{2}}{{\left( a-b \right)}^{2}}$

⇒ $x=\sqrt{{{a}^{2}}{{\left( a-b \right)}^{2}}}$

⇒ $x=a\left( a-b \right)$

Hence, required mean proportional is $a\left( a-b \right).$

5. If $\mathbf{x+5}$ is the mean proportion between $\mathbf{x+2}$ and $\mathbf{x+9}$; find the value of $\mathbf{x}$. 

Ans: Here, $x+5$ is the mean proportion between $x+2$ and $x+9$

Thus, $\left( x+2 \right),~~(x+5$) and $\left( x+9 \right)$ in continued proportion.

Therefore, 

⇒ $\left( x+2 \right)~:\left( x+5 \right)=\left( x+5 \right)~:\left( x+9 \right)$

⇒ $\frac{\left( x+2 \right)}{\left( x+5 \right)}=\frac{\left( x+5 \right)}{~\left( x+9 \right)}$

⇒ ${{\left( x+5 \right)}^{2}}=\left( x+2 \right)\left( x+9 \right)$

⇒ ${{x}^{2}}+10x+25={{x}^{2}}+9x+2x+18$

⇒ $10x+25=11x+18$

⇒ $10x-11x=18-25$

⇒ $-x=-7$

⇒ $x=7$

Thus, Value of $x$ is $~7~.$

6. If $\mathbf{{{x}^{2}},~4}$ and $\mathbf{9}$ are in continued proportion, find $\mathbf{x}$. 

Ans: Here, ${{x}^{2}},~~4~and~9~ in~continued~proportion$

Therefore, 

⇒ ${{x}^{2}}~~:4=4~:9~$

⇒ $\frac{{{x}^{2}}}{4}=\frac{4}{9}$

⇒ ${{x}^{2}}=\frac{16}{9}$

⇒ $x=\sqrt{\frac{16}{9}}$

⇒ $x=\pm \frac{4}{3}$ 

Hence, Value of $x$ is $\pm \frac{4}{3}$ .

7. What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?

Ans: Let the least number is $x$ to be added to the numbers 6, 15, 20 and 43 to make them proportional

Therefore, 

⇒ $\frac{6+x}{15+x}=\frac{20+x}{43+x}$

⇒ $\left( 6+x \right)\left( 43+x \right)=\left( 20+x \right)\left( 15+x \right)$

⇒ $258+6x+43x+{{x}^{2}}=300+20x+15x+{{x}^{2}}$

⇒ $258+49x=300+35x$

⇒ $49x-35x=300-258$

⇒ $14x=42$

⇒ $x=\frac{42}{14}$

⇒ $x=3$

Thus, the required least number is $3$.

(i) If $\mathbf{a,~b,~c}$ are in continued proportion, 

Show that: $\mathbf{\frac{{{a}^{2}}+{{b}^{2}}}{b\left( a+c \right)}=\frac{b\left( a+c \right)}{{{b}^{2}}+{{c}^{2}}}}$ 

Ans: Here, $a,~b,~c$ are in continued proportion

Therefore, 

⇒$\frac{a}{b}=\frac{b}{c}$

⇒${{b}^{2}}=ac$ ……… (i)

Now, 

⇒$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ = $\left( {{a}^{2}}+ac \right)\left( ac+{{c}^{2}} \right)$            (from eq (i))

⇒$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ = $a\left( a+c \right).c\left( a+c \right)$    

⇒$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ = $ac{{\left( a+c \right)}^{2}}$        

⇒$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ = ${{b}^{2}}{{\left( a+c \right)}^{2}}$                            (${{b}^{2}}=ac$)

⇒$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}} \right)$ = $b\left( a+c \right).~b\left( a+c \right)$

⇒$\frac{{{a}^{2}}+{{b}^{2}}}{b\left( a+c \right)}$ = $\frac{b\left( a+c \right)}{{{b}^{2}}+{{c}^{2}}}$

Hence proved.

(ii) If $\mathbf{a,~b,~c}$ are in continued proportion and $\mathbf{a\left( b-c \right)=2b,~}$

Prove that: $a-c=~\frac{2\left( a+b \right)}{a}$.

Ans: Here, $a,~b,~c$ are in continued proportion and $a\left( b-c \right)=2b$

Therefore, $\frac{a}{b}=\frac{b}{c}$

⇒${{b}^{2}}=ac$  and $a\left( b-c \right)=2b$

⇒${{b}^{2}}=ac$  and $ab-ac=2b$

⇒${{b}^{2}}=ac$  and $ab-{{b}^{2}}=2b$

⇒${{b}^{2}}=ac$  and $b\left( a-b \right)=2b$

⇒${{b}^{2}}=ac$  and $\left( a-b \right)=2$

Now, L.H.S  $=a-c$

$~~=\frac{a\left( a-c \right)}{a}$ 

$=\frac{{{a}^{2}}-ac}{a}$

$=\frac{{{a}^{2}}-{{b}^{2}}}{a}$                                           ($~{{b}^{2}}=ac$  )

$=\frac{\left( a+b \right)\left( a-b \right)}{a}$                                       (${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$)

$=\frac{2\left( a+b \right)}{a}$                                              ($a-b=2$)                        

$=R.H.S$

Therefore, L.H.S = R.H.S         

Hence proved.

(iii). If $\mathbf{~\frac{a}{b}=\frac{c}{d},}$ show that:  $\mathbf{\frac{{{a}^{3}}c+a{{c}^{3}}}{{{b}^{3}}d+b{{d}^{3}}}=\frac{{{\left( a+c \right)}^{4}}}{{{\left( b+d \right)}^{4}}}}$

Ans: Here, we have $\frac{a}{b}=\frac{c}{d}$

Let $\frac{a}{b}=\frac{c}{d}=k$, then we get

⇒$a=kb~and~c=kd$

Now, L.H.S $=\frac{{{a}^{3}}c+a{{c}^{3}}}{{{b}^{3}}d+b{{d}^{3}}}$

 $=~\frac{{{\left( kb \right)}^{3}}.kd+kb.{{\left( kd \right)}^{3}}}{{{b}^{3}}d+b{{d}^{3}}}$                      $=~\frac{{{k}^{3}}{{b}^{3}}.kd+kb.{{k}^{3}}{{d}^{3}}}{{{b}^{3}}d+b{{d}^{3}}}$

$=~\frac{{{k}^{4}}{{b}^{3}}d+{{k}^{4}}b{{d}^{3}}}{{{b}^{3}}d+b{{d}^{3}}}$

$=~\frac{{{k}^{4}}\left( {{b}^{3}}d+b{{d}^{3}} \right)}{\left( {{b}^{3}}d+b{{d}^{3}} \right)}$ 

=$~{{k}^{4}}$

⇒     R.H.S  $=\frac{{{\left( a+c \right)}^{4}}}{{{\left( b+d \right)}^{4}}}~$

$=\frac{{{\left( kb+kd \right)}^{4}}}{{{\left( b+d \right)}^{4}}}$         

$=\frac{{{\left[ k\left( b+d \right) \right]}^{4}}}{{{\left( b+d \right)}^{4}}}$    

$=\frac{{{k}^{4}}{{\left( b+d \right)}^{4}}}{{{\left( b+d \right)}^{4}}}$     

$={{k}^{4}}$

Therefore, L.H.S = R.H.S         

Hence proved.

8. What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?

Ans: Let the least number is $x$ to be subtracted from each of the numbers 7, 17 and 47 to make them in continued proportional

Therefore, 

⇒ $\frac{7-x}{17-x}=\frac{17-x}{47-x}$

⇒ $\left( 7-x \right)\left( 47-x \right)=\left( 17-x \right)\left( 17-x \right)$

⇒ $329-7x-47x+{{x}^{2}}=289-17x-17x+{{x}^{2}}$

⇒ $329-54x=289-34x$

⇒ $-54x+34x=289-329$

⇒ $-20x=-40$

⇒ $x=\frac{-40}{-20}$

⇒ $x=2$

Thus, the required least number is $2$.

9. If $\mathbf{y}$ is the mean proportional between $\mathbf{x}$ and $\mathbf{z;}$ show that $xy+yz$ is the mean proportional between $\mathbf{{{x}^{2}}+{{y}^{2}}}$ and $\mathbf{{{y}^{2}}+{{z}^{2}}}$.

Ans: Here, $y$ is the mean proportional between $x$ and $z$

Therefore, 

⇒${{y}^{2}}=xz$   ……… (i)

Now, the mean proportional between ${{x}^{2}}+{{y}^{2}}$ and ${{y}^{2}}+{{z}^{2}}$ will be, 

$=~\sqrt{({{x}^{2}}+{{y}^{2}})({{y}^{2}}+{{z}^{2}})}$ 

$=~\sqrt{({{x}^{2}}+xz)\left( xz+{{z}^{2}} \right)}$                                ($~{{y}^{2}}=xz$   )

$=~\sqrt{x\left( x+z \right).z\left( x+z \right)}$ 

$=~\sqrt{xz{{\left( x+z \right)}^{2}}}~~$                                             

$=~\sqrt{{{y}^{2}}{{\left( x+z \right)}^{2}}}$                                               ($~~~{{y}^{2}}=xz$   )

$=y\left( x+z \right)$ 

$=xy+yz$ 

Hence, $xy+yz$ is the mean proportional between ${{x}^{2}}+{{y}^{2}}$ and ${{y}^{2}}+{{z}^{2}}$...

10. If q is the mean proportional between $\mathbf{p}$ and $\mathbf{r,}$ show that:

$\mathbf{pqr{{\left( p+q+r \right)}^{3}}={{\left( pq+qr+pr \right)}^{3}}.}$ 

Ans: Here, $q$ is the mean proportional between $x$ and $z$

Therefore, 

⇒${{q}^{2}}=pr$   ……… (i)

Now, show that $pqr{{\left( p+q+r \right)}^{3}}={{\left( p+q+r \right)}^{3}}$

L.H.S $=pqr{{\left( p+q+r \right)}^{3}}$

$=q.pr{{\left( p+q+r \right)}^{3}}$

$=q.{{q}^{2}}{{\left( p+q+r \right)}^{3}}$

$={{q}^{3}}{{\left( p+q+r \right)}^{3}}$

$={{\left[ q\left( p+q+r \right) \right]}^{3}}$

$={{\left[ pq+{{q}^{2}}+qr \right]}^{3}}$

$={{\left[ pq+pr+qr \right]}^{3}}$                           ($~~{{q}^{2}}=pr$   )

$=R.H.S$ 

Therefore, $pqr{{\left( p+q+r \right)}^{3}}={{\left( pq+qr+pr \right)}^{3}}.$ 

Hence proved.

11. If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.

Ans: Let $x,~y$ and $z$ are in continued proportion.

Therefore, 

⇒${{y}^{2}}=xz$ 

Now, multiply by $x$ in the above equation, we get

⇒$x{{y}^{2}}={{x}^{2}}z$ 

⇒$\frac{x}{z}=\frac{{{x}^{2}}}{{{y}^{2}}}$ 

Hence, the ratio of the first to the third is the duplicate ratio of the first to the second.

12. If $\mathbf{y}$ is the mean proportional between $\mathbf{x}$ and $\mathbf{z}$, Prove that: $\mathbf{\frac{{{x}^{2}}-{{y}^{2}}+{{z}^{2}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}={{y}^{4}}.}$

Ans: Here, $y$ is the mean proportional between $x$ and $z$

Therefore, 

⇒${{y}^{2}}=xz$   ……… (i)

Now, L.H.S $=\frac{{{x}^{2}}-{{y}^{2}}+{{z}^{2}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}$

$=\frac{{{\left( \frac{{{y}^{2}}}{z} \right)}^{2}}-{{\left( \frac{{{y}^{2}}}{y} \right)}^{2}}+{{\left( \frac{{{y}^{2}}}{x} \right)}^{2}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}$                    

$=\frac{\frac{{{y}^{4}}}{{{z}^{2}}}-\frac{{{y}^{4}}}{{{y}^{2}}}+{{\frac{y}{{{x}^{2}}}}^{4}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}$

$=\frac{{{y}^{4}}{{z}^{-2}}-{{y}^{4}}{{y}^{-2}}+{{y}^{4}}{{x}^{-2}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}$

$=\frac{{{y}^{4}}\left( {{z}^{-2}}-{{y}^{-2}}+{{x}^{-2}} \right)}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}$

$=\frac{{{y}^{4}}\left( {{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}} \right)}{\left( {{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}} \right)}$

$={{y}^{4}}$

L.H.S  $=R.H.S$

Therefore,

$\frac{{{x}^{2}}-{{y}^{2}}+{{z}^{2}}}{{{x}^{-2}}-{{y}^{-2}}+{{z}^{-2}}}={{y}^{4}}$

Hence proved.

13. Given four quantities $\mathbf{a,~b,~c and d}$ are in the proportion. Show that:

$\mathbf{\left( a-c \right){{b}^{2}}~:\left( b-d \right)~cd=\left( {{a}^{2}}-{{b}^{2}}-ab \right)~:\left( {{c}^{2}}-{{d}^{2}}-cd \right)}$ 

Ans: Here $a,~b,~c$ and $d$ are in the proportion

Therefore, $\frac{a}{b}=\frac{c}{d}$

Let $\frac{a}{b}=\frac{c}{d}=k$, then we get

⇒$a=kb~and~c=kd$

Now, L.H.S $=$  $\frac{\left( a-c \right){{b}^{2}}}{\left( b-d \right)~cd}$  

$=$  $\frac{\left( kb-kd \right){{b}^{2}}}{\left( b-d \right)~kd.d}$

$=$  $\frac{k\left( b-d \right){{b}^{2}}}{k\left( b-d \right){{d}^{2}}~}$

=$~\frac{{{b}^{2}}}{{{d}^{2}}}$

⇒ R.H.S  $=\frac{\left( {{a}^{2}}-{{b}^{2}}-ab \right)}{\left( {{c}^{2}}-{{d}^{2}}-cd \right)}~~$

$=\frac{\left( {{k}^{2}}{{b}^{2}}-{{b}^{2}}-kb.b \right)}{\left( {{k}^{2}}{{d}^{2}}-{{d}^{2}}-kd.d \right)}$         

$=\frac{\left( {{k}^{2}}{{b}^{2}}-{{b}^{2}}-k{{b}^{2}} \right)}{\left( {{k}^{2}}{{d}^{2}}-{{d}^{2}}-k{{d}^{2}} \right)}$   

$=\frac{{{b}^{2}}\left( {{k}^{2}}-1-k \right)}{{{d}^{2}}\left( {{k}^{2}}-1-k \right)}$ 

$=\frac{{{b}^{2}}}{{{d}^{2}}}$

Therefore, L.H.S = R.H.S         

Hence proved.

14. Find two numbers such that the mean proportionality between them is 12 and the third proportional to them is 96.

Ans: Let the two numbers are $x$ and $y.$

According to 1st condition,

The mean proportionality between $x$ and $y$ is 12.

⇒ $\sqrt{xy}=12$ 

On squaring both sides, we get

⇒ $xy=144$ 

⇒ $x=\frac{144}{y}$  ……………. Eq(i)

According to the 2nd condition,

⇒ $\frac{x}{y}=\frac{y}{96}$

⇒ ${{y}^{2}}=96x$

⇒ ${{y}^{2}}=96\times \frac{144}{y}$

⇒ ${{y}^{3}}=96\times 144$

⇒ ${{y}^{3}}=12\times 8\times 12\times 12$

⇒ ${{y}^{3}}={{12}^{3}}\times {{2}^{3}}$

⇒ ${{y}^{3}}={{\left( 24 \right)}^{3}}$

⇒ $y=24$

Now, put the value of $y$ in eq(i), we get

⇒ $x=\frac{144}{24}$

⇒ $x=6$

Hence, two required numbers are 6 and 24.

15. Find the third proportional to $\mathbf{~\frac{x}{y}+\frac{y~}{x}}$  and $\mathbf{\sqrt{{{x}^{2}}+{{y}^{2}}}~.}$

Ans: Let the required third proportional be $z.$ 

Therefore, $\frac{x}{y}+\frac{y~}{x}$  , $\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $z$ are in continued proportion.

⇒$\frac{\frac{x}{y}+\frac{y~}{x}}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{z}$

⇒ $z\left( \frac{x}{y}+\frac{y~}{x} \right)={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}$

⇒ $z\left( \frac{{{x}^{2}}+{{y}^{2}}}{xy} \right)=({{x}^{2}}+{{y}^{2}})$

⇒ $z=\frac{xy({{x}^{2}}+{{y}^{2}})}{({{x}^{2}}+{{y}^{2}})}$

⇒ $z=xy$

Hence, the required third proportional is $xy$.

16. If $\mathbf{p~:q=r~:s;}$ then show that: $\mathbf{\left( mp+nq \right)~:q=\left( mr+ns \right)~:s}$.

Ans: Here, we have  $\frac{p}{q}=\frac{r}{s}~$ 

⇒ $\frac{p}{q}=\frac{r}{s}$

Multiply by $m$ in the above equation, we get

⇒ $\frac{mp}{q}=\frac{mr}{s}$

Now, adding $n$ to both sides

⇒ $\frac{mp}{q}+n=\frac{mr}{s}+n$

⇒ $\frac{mp+nq}{q}=\frac{mr+ns}{s}$

⇒ $\left( mp+nq \right)~:q=\left( mr+ns \right)~:s$

Hence proved.

17. If $\mathbf{p+r=mq}$ and $\mathbf{\frac{1}{q}+\frac{1}{s}=\frac{m}{r};}$ then prove that $\mathbf{p~:q=r~:s}$.

Ans: Here, we have $p+r=mq$ and $\frac{1}{q}+\frac{1}{s}=\frac{m}{r}$

⇒ $mq=p+r$ 

⇒ $m=\frac{p+r}{q}$ 

Now, put the value of $m$ in $\frac{1}{q}+\frac{1}{s}=\frac{m}{r}$, we get

⇒ $\frac{1}{q}+\frac{1}{s}=\frac{\frac{p+r}{q}}{r}$

⇒ $\frac{s+q}{qs}=\frac{p+r}{qr}$

⇒ $\frac{s+q}{s}=\frac{p+r}{r}$

⇒ $r\left( s+q \right)=s\left( p+r \right)$

⇒ $rs+qr=sp+rs$

⇒ $qr=sp$

⇒ $\frac{r}{s}=\frac{p}{q}$

⇒ $\frac{p}{q}=\frac{r}{s}$

Hence proved.

Exercise- 7(C)

1. If $\mathbf{a~:b=c~:d}$, prove that: 

$\mathbf{\left( 5a+7b \right)~:\left( 5a-7b \right)=\left( 5c+7d \right)~:\left( 5c-7d \right)}$ 

Ans: Here, we have $~a~:b=c~:d$

Therefore, $\frac{a}{b}=\frac{c}{d}$

Now, multiplying by $\frac{5}{7}$ in the above equation, we get

⇒ $\frac{5}{7}\left( \frac{a}{b} \right)=\frac{5}{7}\left( \frac{c}{d} \right)$

⇒ $\frac{5a}{7b}=\frac{5c}{7d}$

Applying componendo and dividendo rule 

⇒ $\frac{5a+7b}{5a-7b}=\frac{5c+7d}{5c-7d}~$

$\Rightarrow ~\left( 5a+7b \right)~:\left( 5a-7b \right)=\left( 5c+7d \right)~:\left( 5c-7d \right)$ 

Hence proved.

2. $\mathbf{\left( 9a+13b \right)\left( 9c-13d \right)=\left( 9c+13d \right)\left( 9a-13b \right)}$.

Ans: Here, we have $~a~:b=c~:d$

Therefore, $\frac{a}{b}=\frac{c}{d}$

Now, multiplying by $\frac{9}{13}$ in the above equation, we get

 $\frac{9}{13}\left( \frac{a}{b} \right)=\frac{9}{13}\left( \frac{c}{d} \right)$

⇒ $\frac{9a}{13b}=\frac{9c}{13d}$

Applying componendo and dividendo rule 

⇒ $\frac{9a+13b}{9a-13b}=\frac{9c+13d}{9c-13d}~$

$~~~~\Rightarrow ~\left( 9a+13b \right)\left( 9c-13d \right)=\left( 9c+13d \right)\left( 9a-13b \right)$ 

 Hence proved.

3. $\mathbf{\left( xa+yb \right)~:\left( xc+yd \right)=b~:d}$

Ans: Here, we have $~a~:b=c~:d$

Therefore, $\frac{a}{b}=\frac{c}{d}$

Now, multiplying by $\frac{x}{y}~$ in the above equation, we get

⇒ $\frac{x}{y}\left( \frac{a}{b} \right)=\frac{x}{y}\left( \frac{c}{d} \right)$

⇒ $\frac{xa}{yb}=\frac{xc}{yd}$

Adding $1$ to both sides, we get

⇒ $\frac{xa}{yb}+1=\frac{xc}{yd}+1$

⇒ $\frac{xa+yb}{yb}=\frac{xc+yd}{yd}~$

⇒ $\frac{xa+yb}{xc+yd}=\frac{yb}{yd}$

⇒ $\frac{xa+yb}{xc+yd}=\frac{b}{d}$

⇒ $\left( xa+yb \right)~:\left( xc+yd \right)=b~:d$

Hence proved.

4. If $\mathbf{a~:b=c~:d}$, prove that:

$\mathbf{\left( 6a+7b \right)\left( 3c-4d \right)=\left( 6c+7d \right)\left( 3a-4b \right)}$. 

Ans: Here, we have $~a~:b=c~:d$

Therefore, $\frac{a}{b}=\frac{c}{d}$                  ……… eq (i)

Now, multiplying by $\frac{6}{7}~$ in the above equation, we get

⇒ $\frac{6}{7}\left( \frac{a}{b} \right)=\frac{6}{7}\left( \frac{c}{d} \right)$

⇒ $\frac{6a}{7b}=\frac{6c}{7d}$

Adding $1$ to both sides, we get

⇒ $\frac{6a}{7b}+1=\frac{6c}{7d}+1$

⇒ $\frac{6a+7b}{7b}=\frac{6c+7d}{7d}$

⇒ $\frac{6a+7b}{6c+7d}=\frac{7b}{7d}$

⇒ $\frac{6a+7b}{6c+7d}=\frac{b}{d}$  ………… eq (ii)

Now, multiplying by $\frac{3}{4}~$in eq (i), we get

⇒ $\frac{3}{4}\left( \frac{a}{b} \right)=\frac{3}{4}\left( \frac{c}{d} \right)$

⇒ $\frac{3a}{4b}=\frac{3c}{4d}$

Subtracting $1$ from both sides, we get

⇒ $\frac{3a}{4b}-1=\frac{3c}{4d}-1$

⇒ $\frac{3a-4b}{4b}=\frac{3c-4d}{4d}$

⇒ $\frac{3a-4b}{3c-4d}=\frac{4b}{4d}$

⇒ $\frac{3a-4b}{3c-4d}=\frac{b}{d}$  ………….. eq (iii)

Now, from eq (ii) and (iii), we get

⇒ $\frac{6a+7b}{6c+7d}=\frac{3a-4b}{3c-4d}$

⇒  $\left( 6a+7b \right)\left( 3c-4d \right)=\left( 6c+7d \right)\left( 3a-4b \right)$

Hence proved.

5. Given $\mathbf{\frac{a}{b}=\frac{c}{d}~,~}$prove that: 

$\mathbf{\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}}$ .

Ans: Here, we have $\frac{a}{b}=\frac{c}{d}$.

Now, multiplying by $\frac{3}{5}$ in the above equation, we get

⇒ $\frac{3}{5}\left( \frac{a}{b} \right)=\frac{3}{5}\left( \frac{c}{d} \right)$

⇒ $\frac{3a}{5b}=\frac{3c}{5d}$

Applying componendo and dividendo rule

 ⇒ $\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}~$

⇒ $\frac{3a-5b}{3a+5b}=\frac{3c-5d}{3c+5d}$

Hence proved.

6. If  $\mathbf{\frac{5x+6y}{5u+6v}=\frac{5x-6y}{5u-6v}}$ ; then prove that $\mathbf{x~:y=u~:v.}$

Ans: Here, we have $\frac{5x+6y}{5u+6v}=\frac{5x-6y}{5u-6v}$.

⇒ $\frac{5x+6y}{5x-6y}=\frac{5u+6v}{5u-6v}$

Applying componendo and dividendo rule

⇒ $\frac{5x+6y+5x-6y}{5x+6y-\left( 5x-6y \right)}=\frac{5u+6v+5u-6v}{5u+6v-\left( 5u-6v \right)}$

⇒ $\frac{10x}{5x+6y-5x+6y}=\frac{10u}{5u+6v-5u+6v}$

⇒ $\frac{10x}{12y}=\frac{10u}{12v}$

⇒ $\frac{x}{y}=\frac{u}{v}$

⇒ $x~:y=u~:v$

Hence proved.

7. If $\mathbf{\left( 7a+8b \right)\left( 7c-8d \right)=\left( 7a-8b \right)\left( 7c+8d \right)}$ ; prove that $\mathbf{a~:b=c~:d.}$

Ans: Here, we have $\left( 7a+8b \right)\left( 7c-8d \right)=\left( 7a-8b \right)\left( 7c+8d \right)$.

⇒ $\frac{\left( 7a+8b \right)}{\left( 7a-8b \right)}=\frac{\left( 7c+8d \right)}{\left( 7c-8d \right)}$

Applying componendo and dividendo rule

⇒ $\frac{\left( 7a+8b \right)+\left( 7a-8b \right)}{\left( 7a+8b \right)-\left( 7a-8b \right)}=\frac{\left( 7c+8d \right)+\left( 7c-8d \right)}{\left( 7c+8d \right)-\left( 7c-8d \right)}$

⇒ $\frac{7a+8b+7a-8b}{7a+8b-7a+8b}=\frac{7c+8d+7c-8d}{7c+8d-7c+8d}$

⇒ $\frac{14a}{16b}=\frac{14c}{16d}$

⇒ $\frac{a}{b}=\frac{c}{d}$

⇒ $a~:b=c~:d$

Hence proved.

8. (i) If $\mathbf{x=\frac{6ab}{a+b},~}$find the value of  $\mathbf{\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}}$ .

Ans: Here, we have $x=\frac{6ab}{a+b}$ ……… eq (i)

Now, dividing by $3a$ in the above equation, we get

⇒ $\frac{x}{3a}=\frac{6ab}{3a\left( a+b \right)}$

⇒ $\frac{x}{3a}=\frac{2b}{\left( a+b \right)}$

Applying componendo and dividendo rule

⇒ $\frac{x+3a}{x-3a}=\frac{2b+a+b}{2b-\left( a+b \right)}$

⇒ $\frac{x+3a}{x-3a}=\frac{3b+a}{2b-a-b}$

⇒ $\frac{x+3a}{x-3a}=\frac{3b+a}{b-a}$  ……….. eq (ii) 

Now, dividing by $3b$ in eq (i), we get 

⇒ $\frac{x}{3b}=\frac{6ab}{3b\left( a+b \right)}$

⇒ $\frac{x}{3b}=\frac{2a}{\left( a+b \right)}$

Applying componendo and dividendo rule

⇒ $\frac{x+3b}{x-3b}=\frac{2a+a+b}{2a-\left( a+b \right)}$

⇒ $\frac{x+3b}{x-3b}=\frac{3a+b}{2a-a-b}$

⇒ $\frac{x+3b}{x-3b}=\frac{3a+b}{a-b}$ ……… eq (iii)

Adding eq (ii) and (iii)

⇒ $\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}=\frac{3b+a}{b-a}+\frac{3a+b}{a-b}$

$=\frac{3b+a}{b-a}-\frac{3a+b}{b-a}$ 

$=\frac{3b+a-3a-b}{b-a}$ 

$=\frac{2b-2a}{b-a}$

$=\frac{2\left( b-a \right)}{\left( b-a \right)}$

$=2$

Hence value of  $\frac{x+3a}{x-3a}+\frac{x+3b}{x-3b}$  is 2. 

(ii) If $\mathbf{=~\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}}$ , find the value of  $\mathbf{\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}.~}$

Ans: Here, we have $a=~\frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ ……… eq (i)

Now, dividing by $2\sqrt{2}$ in the above equation, we get

⇒ $\frac{a}{2\sqrt{2}}=\frac{4\sqrt{6}}{2\sqrt{2}\left( \sqrt{2}+\sqrt{3} \right)}$

⇒ $\frac{a}{2\sqrt{2}}=\frac{2\sqrt{3}}{\left( \sqrt{2}+\sqrt{3~} \right)}$

Applying componendo and dividendo rule

⇒ $\frac{a+2\sqrt{2}}{a-2\sqrt{2}}=\frac{2\sqrt{3}+\left( \sqrt{2}+\sqrt{3} \right)}{2\sqrt{3}-(\sqrt{2}+\sqrt{3~)}}$

⇒$\frac{a+2\sqrt{2}}{a-2\sqrt{2}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}}{2\sqrt{3}-\sqrt{2}-\sqrt{3}}$

⇒ $\frac{a+2\sqrt{2}}{a-2\sqrt{2}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$  ……….. eq (ii) 

Now, dividing by $2\sqrt{3}$ in eq (i), we get 

⇒ $\frac{a}{2\sqrt{3}}=\frac{4\sqrt{6}}{2\sqrt{3}\left( \sqrt{2}+\sqrt{3} \right)}$

⇒ $\frac{a}{2\sqrt{3}}=\frac{2\sqrt{2}}{\left( \sqrt{2}+\sqrt{3~} \right)}$

Applying componendo and dividendo rule

⇒ $\frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{2\sqrt{2}+\left( \sqrt{2}+\sqrt{3} \right)}{2\sqrt{2}-(\sqrt{2}+\sqrt{3~)}}$

⇒ $\frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{2\sqrt{2}+\sqrt{2}+\sqrt{3}}{2\sqrt{2}-\sqrt{2}-\sqrt{3}}$

⇒ $\frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$  ……….. eq (iii) 

Adding eq (ii) and (iii)

⇒$\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$                             $=\frac{3\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{3\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=\frac{3\sqrt{3}+\sqrt{2}-3\sqrt{2}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}$

$=~\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=~\frac{2\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}-\sqrt{2} \right)}$

$=2$

Hence value of  $\frac{a+2\sqrt{2}}{a-2\sqrt{2}}+\frac{a+2\sqrt{3}}{a-2\sqrt{3}}$  is 2. 

9. If $\mathbf{\left( a+b+c+d \right)~\left( a-b-c+d \right)=\left( a+b-c-d \right)\left( a-b+c-d \right);}$ Prove that $\mathbf{a~:b=c~:d}$. 

Ans: Here, we have 

$\left( a+b+c+d \right)~\left( a-b-c+d \right)=\left( a+b-c-d \right)\left( a-b+c-d \right)$ 

Therefore, 

⇒ $\frac{\left( a+b+c+d \right)}{\left( a+b-c-d \right)}=\frac{\left( a-b+c-d \right)}{\left( a-b-c+d \right)}$

Applying componendo and dividendo rule

⇒ $\frac{\left( a+b+c+d \right)+\left( a+b-c-d \right)}{\left( a+b+c+d \right)-\left( a+b-c-d \right)}=\frac{\left( a-b+c-d \right)+\left( a-b-c+d \right)}{\left( a-b+c-d \right)-\left( a-b-c+d \right)}$

⇒$\frac{a+b+c+d+a+b-c-d}{a+b+c+d-a-b+c+d}=\frac{a-b+c-d+a-b-c+d}{a-b+c-d-a+b+c-d}$

⇒ $\frac{2a+2b}{2c+2d}=\frac{2a-2b}{2c-2d}$

⇒ $\frac{2\left( a+b \right)}{2\left( c+d \right)}=\frac{2\left( a-b \right)}{2\left( c-d \right)}$

⇒ $\frac{\left( a+b \right)}{\left( c+d \right)}=\frac{\left( a-b \right)}{\left( c-d \right)}$

⇒ $\frac{\left( a+b \right)}{\left( a-b \right)}=\frac{\left( c+d \right)}{\left( c-d \right)}$

Applying componendo and dividendo rule

⇒ $\frac{\left( a+b \right)+\left( a-b \right)}{\left( a+b \right)-\left( a-b \right)}=\frac{\left( c+d \right)+\left( c-d \right)}{\left( c+d \right)-\left( c-d \right)}$

⇒ $\frac{a+b+a-b}{a+b-a+b}=\frac{c+d+c-d}{c+d-c+d}$

⇒ $\frac{2a}{2b}=\frac{2c}{2d}$

⇒ $\frac{a}{b}=\frac{c}{d}$

⇒ $a~:b=c~:d$

Hence proved.

10. If  $\mathbf{\frac{a-2b-3c+4d}{a+2b-3c-4d}=\frac{a-2b+3c-4d}{a+2b+3c+4d}}$ , show that: $\mathbf{2ad=3bc}$

Ans: Here, we have $\frac{a-2b-3c+4d}{a+2b-3c-4d}=\frac{a-2b+3c-4d}{a+2b+3c+4d}$

⇒ $\frac{a-2b-3c+4d}{a+2b-3c-4d}=\frac{a-2b+3c-4d}{a+2b+3c+4d}$

Applying componendo and dividendo rule

⇒ $\frac{\left( a-2b-3c+4d \right)+\left( a+2b-3c-4d \right)}{\left( a-2b-3c+4d \right)-\left( a+2b-3c-4d \right)}=\frac{\left( a-2b+3c-4d \right)+\left( a+2b+3c+4d \right)}{\left( a-2b+3c-4d \right)-\left( a+2b+3c+4d \right)}$

⇒ $\frac{a-2b-3c+4d+a+2b-3c-4d}{a-2b-3c+4d-a-2b+3c+4d}=\frac{a-2b+3c-4d+a+2b+3c+4d}{a-2b+3c-4d-a-2b-3c-4d}$

⇒ $\frac{2a-6c}{-4b+8d}=\frac{2a~+~6c}{-4b-8d}$

⇒ $\frac{2\left( a-3c \right)}{-4\left( b-2d \right)}=\frac{2\left( a~+~3c \right)}{-4\left( b+2d \right)}$

⇒ $\frac{\left( a-3c \right)}{\left( b-2d \right)}=\frac{\left( a~+~3c \right)}{\left( b+2d \right)}$

⇒ $\frac{\left( a-3c \right)}{\left( a+3c \right)}=\frac{\left( b-2d \right)}{\left( b+2d \right)}$

Applying componendo and dividendo rule

⇒ $\frac{\left( a-3c \right)+\left( a+3c \right)}{\left( a-3c \right)-\left( a+3c \right)}=\frac{\left( b-2d \right)+\left( b+2d \right)}{\left( b-2d \right)-\left( b+2d \right)}$

⇒ $\frac{a-3c~+~a+3c}{a-3c-~a-~3c}=\frac{b-2d+b+2d}{b-2d-b-2d}$

⇒ $\frac{2a}{-6c}=\frac{2b}{-4d}$

⇒ $\frac{a}{3c}=\frac{b}{2d}$

⇒ $2ad=3bc$

Hence proved.

11. If $\mathbf{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( ax+by \right)}^{2}};}$ prove that $\mathbf{\frac{a}{x}=\frac{b}{y}}$.

Ans: Here, we have $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( ax+by \right)}^{2}}$

On simplifying, we get

⇒ ${{a}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}+{{b}^{2}}{{y}^{2}}={{a}^{2}}{{x}^{2}}+{{b}^{2}}{{y}^{2}}+2abxy$

⇒ ${{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}=2abxy$

⇒ ${{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}-2abxy=0$

⇒ ${{\left( ay-bx \right)}^{2}}=0$

⇒ $ay-bx=0$

⇒ $ay=bx$

⇒ $\frac{a}{x}=\frac{b}{y}$

Hence proved.

12. If $\mathbf{a,~b and c}$ are in continued proportions, prove that:

1. $\mathbf{\frac{{{a}^{2}}+~ab~+~{{b}^{2}}}{{{b}^{2}}~+~bc~+~{{c}^{2}}}=\frac{a}{c}~}$

Ans: Here, $a,~b$ and $c$ are in continued proportions

Therefore, $\frac{a}{b}=\frac{b}{c}~$

⇒ ${{b}^{2}}=ac$ ……… eq (i)

Now, L.H.S  $=\frac{{{a}^{2}}+~ab~+~{{b}^{2}}}{{{b}^{2}}~+~bc~+~{{c}^{2}}}$

$=\frac{{{a}^{2}}+~ab~+~ac}{ac~+~bc~+~{{c}^{2}}}$                              [From eq (i)]

$=\frac{a\left( a~+~b~+~c \right)}{c\left( a~+~b~+~c \right)}$

$=~\frac{a}{c}$

$=R.H.S$ 

Thus, L.H.S = R.H.S

Hence proved.

2. $\mathbf{\frac{{{a}^{2}}~+~{{b}^{2}}~+~{{c}^{2}}}{{{\left( a~+b~+~c~ \right)}^{2}}}=\frac{a~-b~+~c}{a~+~b~+~c}}$

Ans: Here, $a,~b$ and $c$ are in continued proportions

Therefore, $\frac{a}{b}=\frac{b}{c}~$

⇒ ${{b}^{2}}=ac$ ……… eq (i)

Now, L.H.S  $=~\frac{{{a}^{2}}~+~{{b}^{2}}~+~{{c}^{2}}}{{{\left( a~+b~+~c~ \right)}^{2}}}$

$=~\frac{{{a}^{2}}~+~{{b}^{2}}~+~{{c}^{2}}+~2ab~+~~2bc+~2ac-2\left( ab+bc+ac \right)}{{{\left( a~+b~+~c~ \right)}^{2}}}$

$=\frac{{{\left( ~a~+~b~+~c~ \right)}^{2}}-2\left( ab+bc+{{b}^{2}} \right)}{{{\left( a~+b~+~c~ \right)}^{2}}}$                              

$=\frac{{{\left( ~a~+~b~+~c~ \right)}^{2}}-2b\left( a+c+b \right)}{{{\left( a~+b~+~c~ \right)}^{2}}}$

$=\frac{\left( ~a~+~b~+~c~ \right)~\left[ a~+~b~+~c~-2b \right]}{{{\left( a~+b~+~c~ \right)}^{2}}}$

$=\frac{\left( ~a-~b~+~c~ \right)~}{\left( a~+~b~+~c~ \right)}$ 

 $=R.H.S$ 

Thus, L.H.S = R.H.S

Hence proved.

13. Using properties of proportion, solve for x: 

1. $\mathbf{\frac{\sqrt{x~+~5}~+~\sqrt{x-16}}{\sqrt{x~+~5}-~\sqrt{x-16}}=\frac{7}{3}}$

Ans: Here, we have  $\frac{\sqrt{x~+~5}~+~\sqrt{x-16}}{\sqrt{x~+~5}-~\sqrt{x-16}}=\frac{7}{3}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{\sqrt{x~+~5}~+~\sqrt{x-16~}+\left( \sqrt{x~+~5}-~\sqrt{x-16}~ \right)}{\sqrt{x~+~5}~+~\sqrt{x-16~}-\left( \sqrt{x~+~5}-~\sqrt{x-16}~ \right)}=\frac{7~+~3}{7-~3}$

⇒$\frac{\sqrt{x~+~5}~+~\sqrt{x-16~}+~\sqrt{x~+~5}-~\sqrt{x-16}~}{\sqrt{x~+~5}~+~\sqrt{x-16~}-\sqrt{x~+~5}~+~\sqrt{x-16}}=\frac{10}{4}$

⇒ $\frac{2~\sqrt{x~+~5}~}{2\sqrt{x-16}}=\frac{5}{2}$

⇒ $\frac{\sqrt{x~+~5}~}{\sqrt{x-16}}=\frac{5}{2}$

Now, squaring both sides, we get

⇒ ${{\left( \frac{\sqrt{x~+~5}~}{\sqrt{x-16}} \right)}^{2}}={{\left( \frac{5}{2} \right)}^{2}}$

⇒ $\frac{x~+~5}{x-16}=\frac{25}{4}$

⇒ $25\left( x-16 \right)=4\left( x+5 \right)$

⇒ $25x-400=4x+20$

⇒ $25x-4x=20+400$

⇒ $21x=420$

⇒ $x=\frac{420}{21}$

⇒ $x=20$

Hence value of $x$ is 20. 

2. $\mathbf{\frac{\sqrt{x~+~1}~+~\sqrt{x-1}}{\sqrt{x~+~1}-~\sqrt{x-1}}=\frac{4x-1}{2}}$

Ans: Here, we have  $\frac{\sqrt{x~+~1}~+~\sqrt{x-1}}{\sqrt{x~+~1}-~\sqrt{x-1}}=\frac{4x-1}{2}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{\sqrt{x~+~1}~+~\sqrt{x-1~}+\left( \sqrt{x~+~1}-~\sqrt{x-1}~ \right)}{\sqrt{x~+~1}~+~\sqrt{x-1~}-\left( \sqrt{x~+~1}-~\sqrt{x-1}~ \right)}=\frac{4x-1~+~2}{4x~-~1-2}$

⇒$\frac{\sqrt{x~+~1}~+~\sqrt{x-1~}+~\sqrt{x~+~1}-~\sqrt{x-1}~}{\sqrt{x~+~1}~+~\sqrt{x-1~}-\sqrt{x~+~1}~+~\sqrt{x-1}}=\frac{4x~+~1}{4x~-~3}$

⇒ $\frac{2~\sqrt{x~+~1}~}{2\sqrt{x-1}}=\frac{4x~+~1}{4x~-~3}$

⇒ $\frac{\sqrt{x~+~1}~}{\sqrt{x~-1}}=\frac{4x~+~1}{4x~-~3}$

Now, squaring both sides, we get

⇒ ${{\left( \frac{\sqrt{x~+~1}~}{\sqrt{x-1}} \right)}^{2}}={{\left( \frac{4x~+~1}{4x~-~3} \right)}^{2}}$

⇒ $\frac{x~+~1}{x~-1}=\frac{16{{x}^{2}}~+~1~+~8x}{16{{x}^{2}}~+~9-24x}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{x~+~1+~x~-~1}{x+1-\left( x~-1 \right)}=\frac{16{{x}^{2}}~+~1~+~8x~+16{{x}^{2}}~+~9-24x}{16{{x}^{2}}~+~1~+~8x-\left( 16{{x}^{2}}~+~9-24x \right)}$

⇒$\frac{2x}{x+1-x~+1~}=\frac{32{{x}^{2}}~+~10-16x~}{16{{x}^{2}}~+~1~+~8x-16{{x}^{2}}-~9+24x}$

⇒ $\frac{2x}{2~}=\frac{32{{x}^{2}}~+~10-16x~}{-8~+~32x}$

⇒ $x=\frac{32{{x}^{2}}~+~10-16x~}{-8~+~32x}$

⇒ $-8x+32{{x}^{2}}=32{{x}^{2}}~+~10-16x$

⇒ $-8x=~10-16x$

⇒ $-8x+16x=~10$

⇒ $8x=~10$

⇒$~x=\frac{10}{8}$

⇒$~x=\frac{5}{4}$ 

Hence value of $x$ is $\frac{5}{4}~$. 

3. $\mathbf{\frac{3x~+~\sqrt{9{{x}^{2}}-5}}{3x-~\sqrt{9{{x}^{2}}-5}}=5.}$

Ans: Here, we have  $\frac{3x~+~\sqrt{9{{x}^{2}}-5}}{3x-~\sqrt{9{{x}^{2}}-5}}=\frac{5}{1}$

Applying componendo and dividendo rule in above equation

⇒$\frac{3x~+~\sqrt{9{{x}^{2}}-5}~+~3x-~\sqrt{9{{x}^{2}}-5}}{3x~+~\sqrt{9{{x}^{2}}-5}-\left( 3x-~\sqrt{9{{x}^{2}}-5} \right)}=\frac{5~+~1}{5-~1}$

⇒$~\frac{6x}{~3x~+~\sqrt{9{{x}^{2}}-5}-3x+~\sqrt{9{{x}^{2}}-5}}=\frac{6}{4}$

⇒$~\frac{6x}{~2~\sqrt{9{{x}^{2}}-5}}=\frac{6}{4}$

⇒$~\frac{x}{~\sqrt{9{{x}^{2}}-5}}=\frac{1}{2}$

Now, squaring both sides, we get

⇒${{\left( \frac{x}{\sqrt{9{{x}^{2}}-5}} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}}$

⇒$~\frac{{{x}^{2}}}{9{{x}^{2}}-5}=\frac{1}{4}$

⇒$~9{{x}^{2}}-5=4{{x}^{2}}$

⇒$~9{{x}^{2}}-4{{x}^{2}}=5$

⇒$~5{{x}^{2}}=5$

⇒$~{{x}^{2}}=1$

⇒$~x=\sqrt{1}$

⇒$~x=\pm 1$

Hence value of $x$ is $\pm 1$.

14. If $\mathbf{x=~\frac{\sqrt{a~+~3b}~+\sqrt{a-~3b}~}{\sqrt{a~+~3b}~-\sqrt{a-~3b}}}$, prove that $\mathbf{3b{{x}^{2}}-2ax+3b=0}$.

Ans: Here, we have  $x=\frac{\sqrt{a~+~3b}~+\sqrt{a-~3b}~}{\sqrt{a~+~3b}~-\sqrt{a-~3b}}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{x~+~1}{x~-~1}=\frac{\sqrt{a~+~3b}~+\sqrt{a-~3b}~+~\sqrt{a~+~3b}~-\sqrt{a-~3b}~}{\sqrt{a~+~3b}~+\sqrt{a-~3b}-\left( \sqrt{a~+~3b}~-\sqrt{a-~3b~} \right)}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{2\sqrt{a~+~3b}~~}{\sqrt{a~+~3b}~+\sqrt{a-~3b}-\sqrt{a~+~3b}+\sqrt{a-~3b~}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{2\sqrt{a~+~3b}~~}{2\sqrt{a-~3b~}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{\sqrt{a~+~3b}~~}{\sqrt{a-~3b~}}$

Now, squaring both sides, we get

⇒$~{{\left( \frac{x~+~1}{x~-~1} \right)}^{2}}={{\left( \frac{\sqrt{a~+~3b}~~}{\sqrt{a-~3b~}} \right)}^{2}}$

⇒$~\frac{{{x}^{2}}+~2x~+~1}{{{x}^{2}}-~2x~+~1}=\frac{a~+~3b}{a~-~3b}$

⇒$\left( a-3b \right)~({{x}^{2}}+~2x~+~1)=\left( a+3b \right)~\left( {{x}^{2}}-~2x~+~1~ \right)$

⇒$~a{{x}^{2}}+2ax+a-3b{{x}^{2}}-6bx-3b=~a{{x}^{2}}-2ax+a+3b{{x}^{2}}-6bx+3b$

⇒$~2ax-3b{{x}^{2}}-3b=~-2ax+3b{{x}^{2}}+3b$

⇒$~6b{{x}^{2}}-4ax+6b~=~0$

⇒$~2\left( 3b{{x}^{2}}-2ax+3b \right)~=~0$

⇒$~3b{{x}^{2}}-2ax+3b~=~0$

Hence proved.

15. Using the properties of proportion, solve for x, given $\mathbf{\frac{{{x}^{4}}~+~1}{2{{x}^{2}}}=\frac{17}{8}}$.

Ans: Here, we have  $\frac{{{x}^{4}}~+~1}{2{{x}^{2}}}=\frac{17}{8}$

Applying componendo and dividendo rule in above equation

⇒$\frac{{{x}^{4}}~+~1~+~2{{x}^{2}}}{{{x}^{4}}~+~1-~2{{x}^{2}}}=\frac{17~+~8}{17-~8}$

⇒ $\frac{{{\left( {{x}^{2}}~+~1 \right)}^{2}}}{{{\left( {{x}^{2}}-~1 \right)}^{2}}}=\frac{25}{9}$

⇒ ${{\left( \frac{{{x}^{2}}~+~1}{{{x}^{2}}~-~~1} \right)}^{2}}={{\left( \frac{5}{3} \right)}^{2}}$

⇒ $\frac{{{x}^{2}}~+~1}{{{x}^{2}}~-~~1}=\frac{5}{3}~$

⇒ $5{{x}^{2}}-5=3{{x}^{2}}+3~$

⇒ $5{{x}^{2}}-3{{x}^{2}}=3+5$

⇒ $2{{x}^{2}}=8$

⇒ ${{x}^{2}}=\frac{8}{2}$

⇒ ${{x}^{2}}=4$

⇒ $x=\sqrt{4}$

⇒ $x=\pm ~2$

Hence value of $x$ is $\pm ~2$.

16. If $\mathbf{=~\frac{\sqrt{m~+~n}~+~\sqrt{m~-~n~}}{\sqrt{m~+~n}~-~\sqrt{m~-~n~}}}$ , express $\mathbf{n}$ in terms of $\mathbf{x}$ and $\mathbf{m}$.

Ans: Here, we have  $x=~\frac{\sqrt{m~+~n}~+~\sqrt{m~-~n~}}{\sqrt{m~+~n}~-~\sqrt{m~-~n~}}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{x~+~1}{x~-~1}=\frac{\sqrt{m~+~n}~+~\sqrt{m~-~n~}+\sqrt{m~+~n}~-~\sqrt{m~-~n~}}{\sqrt{m~+~n}~+~\sqrt{m~-~n~}-(\sqrt{m~+~n}~-~\sqrt{m~-~n~~)}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{2\sqrt{m~+~n}~}{\sqrt{m~+~n}~+~\sqrt{m~-~n~}-\sqrt{m~+~n}~+~\sqrt{m~-~n~~}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{2\sqrt{m~+~n}~}{2~\sqrt{m~-~n~~}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{\sqrt{m~+~n}~}{~\sqrt{m~-~n~~}}$

Now, squaring both sides, we get

⇒$~{{\left( \frac{x~+~1}{x~-~1} \right)}^{2}}={{\left( \frac{\sqrt{m~+~n}~}{~\sqrt{m~-~n~~}} \right)}^{2}}$

⇒$~\frac{{{x}^{2}}+~2x~+~1}{{{x}^{2}}-~2x~+~1}=\frac{m~+~n}{m~-~n}$

Applying componendo and dividendo rule in above equation

⇒$~\frac{{{x}^{2}}+~2x~+~1+~{{x}^{2}}-~2x~+~1}{{{x}^{2}}+~2x~+~1-~~\left( {{x}^{2}}-~2x~+~1 \right)}=\frac{m~+~n+~m~-~n}{m~+~n-\left( m~-~n \right)}$

⇒$~\frac{2{{x}^{2}}+~2}{{{x}^{2}}+~2x~+~1-~~{{x}^{2}}+~2x-~1}=\frac{2m}{m~+~n-m~+~n}$

⇒$~\frac{2{{x}^{2}}+~2x~}{4x}=\frac{2m}{2n}$

⇒$~\frac{2({{x}^{2}}+1)~}{4x}=\frac{m}{n}$

⇒$~\frac{({{x}^{2}}+1)~}{2x}=\frac{m}{n}$

⇒$~n=\frac{2mx}{({{x}^{2}}+1)}$

Hence value of $n$ is $\frac{2mx}{({{x}^{2}}+1)}$ .

17. If  $\mathbf{\frac{{{x}^{3}}~+~3x{{y}^{2}}}{3{{x}^{2}}y~+~{{y}^{3}}}=\frac{{{m}^{3}}~+~3m{{n}^{2}}}{3{{m}^{2}}n~+~{{n}^{3}}}}$ , show that $\mathbf{nx=my}$.

Ans: Here, we have  $\frac{{{x}^{3}}~+~3x{{y}^{2}}}{3{{x}^{2}}y~+~{{y}^{3}}}=\frac{{{m}^{3}}~+~3m{{n}^{2}}}{3{{m}^{2}}n~+~{{n}^{3}}}$

Applying componendo and dividendo rule in above equation

⇒$\frac{{{x}^{3}}~+~3x{{y}^{2}}+~3{{x}^{2}}y~+~{{y}^{3}}}{{{x}^{3}}~+~3x{{y}^{2}}-\left( 3{{x}^{2}}y~+~{{y}^{3}} \right)}=\frac{{{m}^{3}}~+~3m{{n}^{2}}+~3{{m}^{2}}n~+~{{n}^{3}}}{{{m}^{3}}~+~3m{{n}^{2}}-\left( 3{{m}^{2}}n~+~{{n}^{3}} \right)}$

⇒$\frac{{{x}^{3}}~+~{{y}^{3}}+~3{{x}^{2}}y~+~3x{{y}^{2}}~}{{{x}^{3}}~+~3x{{y}^{2}}-3{{x}^{2}}y-~{{y}^{3}}}=\frac{{{m}^{3}}~~+~{{n}^{3}}~+~3{{m}^{2}}n~+~3m{{n}^{2}}}{{{m}^{3}}~+~3m{{n}^{2}}-3{{m}^{2}}n-~{{n}^{3}}}$

⇒$\frac{{{x}^{3}}~+~{{y}^{3}}+~3{{x}^{2}}y~+~3x{{y}^{2}}~}{{{x}^{3}}~-~{{y}^{3}}~-3{{x}^{2}}y+~3x{{y}^{2}}}=\frac{{{m}^{3}}~~+~{{n}^{3}}~+~3{{m}^{2}}n~+~3m{{n}^{2}}}{{{m}^{3}}-~{{n}^{3}}-3{{m}^{2}}n~+~3m{{n}^{2}}}$

⇒ $\frac{{{x}^{3}}~+~{{y}^{3}}+~3xy\left( x~+~y~ \right)}{{{x}^{3}}~-~{{y}^{3}}~-3xy~\left( ~x-~y \right)}=\frac{{{m}^{3}}~~+~{{n}^{3}}~+~3mn~\left( ~m~+~n \right)}{{{m}^{3}}-~{{n}^{3}}-3mn~\left( ~m~-~n \right)}$

⇒ $\frac{{{\left( x~+~y \right)}^{3}}}{{{\left( x~-~y \right)}^{3}}}=\frac{{{\left( m~\mp ~n \right)}^{3}}}{{{\left( m~-~n \right)}^{3}}}$

⇒ ${{\left( \frac{x~+~y}{x~-~y} \right)}^{3}}={{\left( \frac{m~+~n}{m~-~n} \right)}^{3}}$

⇒ $\frac{x~+~y}{x~-~y}=\frac{m~+~n}{m~-~n}$

Applying componendo and dividendo rule in above equation

⇒$\frac{x~+~y~+~x~-y}{x~+~y-\left( x~-~y \right)}=\frac{m~+~n~+~m-n}{m~+~n-\left( m~-~n \right)}$

⇒  $\frac{2x~}{x~+~y-~x~+~~y}=\frac{2m}{m~+~n-m~+~n}$

⇒  $\frac{2x~}{2y}=\frac{2m}{2n}$

⇒  $\frac{x~}{y}=\frac{m}{n}$

⇒  $nx=my$

Hence proved.

Exercise- 7(D)

1. If $\mathbf{~:b=3~:5}$ , find $\mathbf{\left( 10a+3b \right)~:\left( 5a+2b \right).}$

Ans: Here, we have $~a~:b=3~:5$

Therefore, $\frac{a}{b}=\frac{3}{5}$           ……… (i)

Now, multiplying by $\frac{10}{3}$ in the above equation, we get

⇒ $\frac{10}{3}\left( \frac{a}{b} \right)=\frac{10}{3}\left( \frac{3}{5} \right)$

⇒ $\frac{10a}{3b}=2$

Adding 1 to both sides, we get

⇒ $\frac{10a}{3b}+1=2+1$

⇒ $\frac{10a+3b}{3b}=3$

⇒ $10a+3b=9b$ ……………… eq (ii)

Now, multiplying by  $\frac{5}{2}$  in the above equation, we get

⇒ $\frac{5}{2}\left( \frac{a}{b} \right)=\frac{5}{2}\left( \frac{3}{5} \right)$

⇒ $\frac{5a}{2b}=\frac{3}{2}$

Adding 1 to both sides, we get

⇒ $\frac{5a}{2b}+1=\frac{3}{2}+1$

⇒ $\frac{5a+2b}{2b}=\frac{3~+~2}{2}$

⇒ $\frac{5a+2b}{b}=5~$

⇒ $5a+2b=5b$ ……………. eq (iii)

Now, divide eq (i) by (ii), we get

⇒ $\frac{10a~+~3b}{5a~+~2b}=\frac{9b}{5b~}$

⇒ $\frac{10a~+~3b}{5a~+~2b}=\frac{9}{5}$

Hence value of $\left( 10a+3b \right)~:\left( 5a+2b \right)$ is $9~:~5$.

2. If $\mathbf{\left( 5x+6y \right)~:\left( 8x+5y \right)=8~:9}$, find $\mathbf{x~:y}$.

Ans: Here, we have  $\left( 5x+6y \right)~:\left( 8x+5y \right)=8~:9$

Therefore, $\frac{\left( 5x+6y \right)}{\left( 8x+5y \right)}=\frac{8}{9}$

⇒ $8~\left( 8x+5y \right)=9\left( 5x+6y \right)~$

⇒ $64~x+40~y=45x+54y~$

⇒ $64~x-45x=54y-40y$

⇒ $19x=14y$

⇒ $\frac{x}{y}=\frac{14}{19}$

⇒ $x~:y=14~:19$

Hence, value of $x~:y~is~14~:19$.

3. If $\mathbf{\left( 3x-4y \right)~:\left( 2x-3y \right)=\left( 5x-6y \right)~:\left( 4x-5y \right).}$ find $\mathbf{x~:y}$

Ans: Here, we have  $\left( 3x-4y \right)~:\left( 2x-3y \right)=\left( 5x-6y \right)~:\left( 4x-5y \right)$

Therefore, $\frac{\left( 3x-4y \right)}{\left( 2x-3y \right)}=\frac{\left( 5x-6y \right)}{\left( 4x-5y \right)}$

⇒ $\left( 3x-4y \right)\left( 4x-5y \right)=\left( 5x-6y \right)~\left( 2x-3y \right)~$

⇒ $12{{x}^{2}}-15xy-16xy+20{{y}^{2}}=10{{x}^{2}}-15xy-12xy+18{{y}^{2}}~$

⇒ $12{{x}^{2}}-31xy+20{{y}^{2}}=10{{x}^{2}}-27xy+18{{y}^{2}}$

⇒  $2{{x}^{2}}-4xy+2{{y}^{2}}=0~$

⇒  $2\left( {{x}^{2}}-2xy+{{y}^{2}} \right)=0$

⇒ $\left( {{x}^{2}}-2xy+{{y}^{2}} \right)=0~$

⇒ ${{\left( x-y \right)}^{2}}=0$

⇒ $x-y=0$

⇒ $x=y$

⇒ $\frac{x}{y}=1$

⇒ $x~:y=1~:1$

Hence, value of $x~:y~is~1~:1$.

4. Find the: 

1. Duplicate ratio of $\mathbf{2\sqrt{2}~:3\sqrt{5}}$

Ans: We know that duplicate ratio of $a~:b$  is ${{a}^{2}}~:{{b}^{2}}$

Therefore, duplicate ratio of  $2\sqrt{2}~:3\sqrt{5}$  is ${{\left( 2\sqrt{2} \right)}^{2}}~:{{\left( 3\sqrt{5} \right)}^{2}}$

$={{\left( 2\sqrt{2} \right)}^{2}}~:{{\left( 3\sqrt{5} \right)}^{2}}$ 

$=\frac{{{\left( 2\sqrt{2} \right)}^{2}}}{{{\left( 3\sqrt{5} \right)}^{2}}}$ 

$=\frac{4~\times ~2}{9~\times ~5}$ 

$=\frac{8}{45}$ 

$=8~:45$ 

Hence, duplicate ratio of  $2\sqrt{2}~:3\sqrt{5}~$ is $8~:45.$

2. Triplicate ratio of $\mathbf{2a~:3b.~}$

Ans:  We know that triplicate ratio of $a~:b$  is ${{a}^{3}}~:{{b}^{3}}$

Therefore, triplicate ratio of  $2a~:3b$  is ${{\left( 2a \right)}^{3}}~:{{\left( 3b \right)}^{3}}$

$={{\left( 2a \right)}^{3}}~:{{\left( 3b \right)}^{3}}$ 

$=\frac{{{\left( 2a \right)}^{3}}}{{{\left( 3b \right)}^{3}}}$ 

$=\frac{8~{{a}^{3}}}{27{{b}^{3}}}$ 

$=8{{a}^{3}}~:27{{b}^{3}}$ 

Hence, triplicate ratio of  $2a~:3b~$ is $8{{a}^{3}}~:27{{b}^{3}}.$

3. Sub – duplicate ratio of $\mathbf{~9{{x}^{2}}{{a}^{4}}~:25{{y}^{6}}{{b}^{2}}.}$

Ans: We know that sub - duplicate ratio of $a~:b$  is $\sqrt{a}~:\sqrt{b}$

Therefore, sub - duplicate ratio of  $9{{x}^{2}}{{a}^{4}}~:25{{y}^{6}}{{b}^{2}}$  is $\sqrt{9{{x}^{2}}{{a}^{4}}}~:\sqrt{25{{y}^{6}}{{b}^{2}}}$

$=\sqrt{9{{x}^{2}}{{a}^{4}}}~:\sqrt{25{{y}^{6}}{{b}^{2}}}$ 

$=\frac{\sqrt{9{{x}^{2}}{{a}^{4}}}}{\sqrt{25{{y}^{6}}{{b}^{2}}}}$ 

$=\frac{3x{{a}^{2}}}{5{{y}^{3}}b}$ 

$=3x{{a}^{2}}~:5{{y}^{3}}b$ 

Hence, sub- duplicate ratio of  $9{{x}^{2}}{{a}^{4}}~:25{{y}^{6}}{{b}^{2}}~$ is $3x{{a}^{2}}~:5{{y}^{3}}b.$

4. Sub – triplicate ratio of $\mathbf{216~:343.}$

Ans: We know that sub - triplicate ratio of $a~:b$  is $\sqrt[3]{a}~:\sqrt[3]{b}$

Therefore, sub - triplicate ratio of  $216~:343$  is $\sqrt[3]{216}~:\sqrt[3]{343}$

$=\sqrt[3]{216}~:\sqrt[3]{343}$ 

$=\frac{\sqrt[3]{216}}{\sqrt[3]{343}}$ 

$=\frac{\sqrt[3]{6~\times ~6~\times ~6}}{\sqrt[3]{7~\times ~7~\times ~7}}~$ 

$=\frac{\sqrt[3]{{{6}^{3}}}}{\sqrt[3]{{{7}^{3}}}}$ 

$=\frac{6}{7}$ 

$=6~:7$ 

Hence, sub - triplicate ratio of  $216~:343~$is $6~:7.$

5. Reciprocal ratio of $\mathbf{3~:5.}$

Ans: We know that reciprocal ratio of $a~:b$  is $\frac{1}{a}~:\frac{1}{b}$.

Therefore, reciprocal ratio of  $3~:5$  is $\frac{1}{3}~:\frac{1}{5}$.

$=\frac{1}{3}~:\frac{1}{5}$ 

$=\frac{\frac{1}{3}}{\frac{1}{5}}$ 

$=\frac{5}{3}$ 

$=5~:3$ 

Hence, reciprocal ratio of  $3~:5~$is $5~:3.$

5. Ratio compounded of the duplicate ratio of 5:6, the reciprocal ratio of $\mathbf{25~:42}$ and the sub – duplicate ratio of $\mathbf{36~:49.}$

Ans: The duplicate ratio of $5~:6$ is ${{5}^{2}}~:~{{6}^{2}}$

$=\frac{{{5}^{2}}}{{{6}^{2}}}$ 

$=\frac{25}{36}$ 

And the reciprocal ratio of $25~:42$ is $42~:25$

$=\frac{42}{25}$ 

Now, sub – duplicate ratio of $36~:49$ is $\sqrt{36}~:~\sqrt{49}$

$=\frac{\sqrt{36}}{\sqrt{49}}$ 

$=\frac{\sqrt{{{6}^{2}}}}{\sqrt{{{7}^{2}}}}$ 

$=\frac{6}{7}$ 

Thus, the compound ratio of $\frac{25}{36}~~,\frac{42}{25}~~$and $\frac{6}{7}$ will be, 

$=\frac{25\times 42\times 6}{36\times 25\times 7}$ 

$=\frac{6\times 6}{36}$ 

$=\frac{36}{36}$ 

$=1$ 

Hence, the required compound ratio is $1~:1$.

6. Find the value of $x$, if: 

1. $\mathbf{\left( 2x+3 \right)~:\left( 5x-38 \right)~}$ is the duplicate ratio of $\mathbf{\sqrt{5~}~:\sqrt{6}}$. 

Ans: We have $\left( 2x+3 \right)~:\left( 5x-38 \right)~$ is the duplicate ratio of $\sqrt{5~}~:\sqrt{6}$

Therefore, 

⇒ $\frac{~\left( 2x+3 \right)}{\left( 5x-38 \right)}=\frac{{{\left( \sqrt{5~} \right)}^{2}}}{{{\left( \sqrt{6} \right)}^{2}}}$

⇒ $\frac{~\left( 2x+3 \right)}{\left( 5x-38 \right)}=\frac{5}{6}$

⇒ $5\left( 5x-38 \right)=6\left( 2x+3 \right)$

⇒ $25x-190=12x+18$

⇒ $25x-12x=18+190$

⇒ $13x=208$

⇒ $x=\frac{208}{13}$

⇒ $x=16$

Hence, value of $x$ is 16.

2. $\mathbf{\left( 2x+1 \right)~:\left( 3x+13 \right)~}$ is the sub - duplicate ratio of $\mathbf{9~:25.}$

Ans: We have $\left( 2x+1 \right)~:\left( 3x+13 \right)~$ is the sub - duplicate ratio of $9~:25$

Therefore, 

⇒ $\frac{~\left( 2x+1 \right)}{\left( 3x+13 \right)}=\frac{\sqrt{9}}{\sqrt{25}}$

⇒ $\frac{~\left( 2x+1 \right)}{\left( 3x+13 \right)}=\frac{\sqrt{{{3}^{2}}}}{\sqrt{{{5}^{2}}}}$

⇒ $\frac{~\left( 2x+1 \right)}{\left( 3x+13 \right)}=\frac{3}{5}$

⇒ $5\left( 2x+1 \right)=3\left( 3x+13 \right)$

⇒ $10x+5=9x+39$

⇒ $x=34$

Hence, value of $x$ is 34.

3. $\mathbf{\left( 3x-7 \right)~:\left( 4x+3 \right)~}$ is the sub - triplicate ratio of $\mathbf{8~:27.}$

Ans: We have $\left( 3x-7 \right)~:\left( 4x+3 \right)~$ is the sub - triplicate ratio of $8~:27$

Therefore, 

⇒ $\frac{~\left( 3x-7 \right)}{\left( 4x+3 \right)}=\frac{\sqrt[3]{8}}{\sqrt[3]{27}}$

⇒ $\frac{~\left( 3x-7 \right)}{\left( 4x+3 \right)}=\frac{\sqrt[3]{{{2}^{3}}}}{\sqrt[3]{{{3}^{3}}}}$

⇒ $\frac{~\left( 3x-7 \right)}{\left( 4x+3 \right)}=\frac{2}{3}$

⇒ $3\left( 3x-7 \right)=2\left( 4x+3 \right)$

⇒ $9x-21=8x+6$

⇒ $9x-8x=6+21$

⇒ $x=27$

Hence, value of $x$ is 27.

7. What quantity must be added to each term of the ratio $\mathbf{x~:y}$ so that it may become equal to $\mathbf{c~:d}$.

Ans: Let the quantity must be added is $p$.

Therefore, $\frac{x~+~p}{y~+~p}=\frac{c}{d}$

⇒ $dx+dp=cy+cp$

⇒ $dp-cp=cy-dx$

⇒ $p\left( d-c \right)=cy-dx$

⇒ $p=\frac{cy-dx}{d-c}$

Hence, required quantity is $\frac{cy-dx}{d-c}$.

8. A woman reduces her weight in the ratio 7:5. What does her weight become if originally it was 84 kg?

Ans:  Let the reduced weight of the woman be $x$ kg

Therefore, $\frac{84}{x}=\frac{7}{5}$

⇒ $7x=84\times 5$

⇒ $x=\frac{84~\times ~5}{7}$

⇒ $x=12\times 5$

⇒ $x=60$ kg

Hence, the reduced weight of the woman is 60 kg.

9. If $\mathbf{15\left( 2{{x}^{2}}-{{y}^{2}} \right)=7xy,}$ find $\mathbf{x~:y~;}$ if $\mathbf{x~}$and $\mathbf{y}$ both are positive.

Ans: Here, we have $15\left( 2{{x}^{2}}-{{y}^{2}} \right)=7xy$

⇒ $30{{x}^{2}}-15{{y}^{2}}=7xy$

Divide by ${{y}^{2}}$ in the above equation, 

⇒ $\frac{30{{x}^{2}}}{~{{y}^{2}}}-\frac{15{{y}^{2}}}{~{{y}^{2}}}=\frac{7xy}{~{{y}^{2}}}$

⇒ $30~{{\left( \frac{x}{y} \right)}^{2}}-15=7\left( \frac{x}{y} \right)$

Now, let $a=\frac{x}{y}$ 

⇒ $30~{{a}^{2}}-15=7a$

⇒ $30~{{a}^{2}}-7a-15=0$

⇒ $30~{{a}^{2}}-25a+18a-15=0$

⇒ $5a\left( 6a-5 \right)+3\left( 6a-5 \right)=0$

⇒ $\left( 6a-5 \right)\left( 5a+3 \right)=0$

⇒ $\left( 6a-5 \right)=0~or~\left( 5a+3 \right)=0$

⇒ $a=\frac{5}{6}~or~a=-\frac{3}{5}~$

Now, put $a=\frac{x}{y}~$,  we get

⇒ $\frac{x}{y}=\frac{5}{6}~or\frac{x}{y}=-\frac{3}{5}~$

⇒ $x~:y=5~:6~or~x~:y=-3~:5$

Since, $x~$and $y$ Both are positive. Thus, value of  $x~:y~$will be positive.

Hence, Value of $x~:y~$is $5~:6.$

10. Find the:

1. Fourth proportional to $\mathbf{2xy,~~{{x}^{2}}}$ and $\mathbf{{{y}^{2}}.}$

Ans: Let the fourth proportional be $z.$

Therefore, $2xy,~{{x}^{2}},~{{y}^{2}}$ and $z$ are in continued proportion.

⇒ $2xy~:~~{{x}^{2}}=~{{y}^{2}}~:z$

⇒ $\frac{2xy}{{{x}^{2}}}=\frac{{{y}^{2}}}{z}$

⇒ $\frac{2}{x}=\frac{y}{z}$

⇒ $z=\frac{xy}{2}$

Hence, the required fourth proportion is $\frac{xy}{2}$.

2. Third proportional to $\mathbf{{{a}^{2}}-{{b}^{2}}}$ and $\mathbf{a+b}$.

Ans: Let the third proportional be $x.$

Therefore, ${{a}^{2}}-{{b}^{2}},~a+b$ and $x~$ are in continued proportion.

⇒ ${{a}^{2}}-{{b}^{2}}~:~~a+b=~a+b~:x$

⇒ $\frac{{{a}^{2}}-{{b}^{2}}}{a~+~b}=~\frac{a~+~b}{x}$

⇒ $x=~\frac{{{\left( a~+~b \right)}^{2}}}{{{a}^{2}}-{{b}^{2}}}$

⇒ $x=~\frac{\left( a+b \right)~\left( a+b \right)}{\left( a+b \right)\left( a-b \right)}$                                       $({{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right))$

⇒ $x=~\frac{\left( a+b \right)~}{\left( a-b \right)}$

Hence, the required third proportion is  $\left( \frac{a+b~}{a-b~} \right)~.$

3. Mean proportional to $\mathbf{\left( x-y \right)}$ and ($\mathbf{{{x}^{3}}-{{x}^{2}}y.}$)

Ans: Let the third proportional be $x.$

Therefore, ${{a}^{2}}-{{b}^{2}},~a+b$ and $x~$ are in continued proportion.

⇒ ${{a}^{2}}-{{b}^{2}}~:~~a+b=~a+b~:x$

⇒ $\frac{{{a}^{2}}-{{b}^{2}}}{a~+~b}=~\frac{a~+~b}{x}$

⇒ $x=~\frac{{{\left( a~+~b \right)}^{2}}}{{{a}^{2}}-{{b}^{2}}}$

⇒ $x=~\frac{\left( a+b \right)~\left( a+b \right)}{\left( a+b \right)\left( a-b \right)}$                                       (${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$)

⇒ $x=~\frac{\left( a+b \right)~}{\left( a-b \right)}$

Hence, the required third proportion is  $\left( \frac{a+b~}{a-b~} \right)~.$

11. Find two numbers such that the mean proportional between them is 14 and third proportional to them 112. 

Ans: Let two required numbers be $x$ and $y$.

Now, according to the 1st condition,

The mean proportional between $x$ and $y$ is 14

⇒ $\sqrt{xy~}$ = 14

Squaring both sides, 

⇒ $xy$ = 196

⇒ $x=\frac{196}{y}$ ………….. eq (i)

Now, according to 2nd condition,

Third proportional of $x$ and $y$ is 112.

Therefore, 

⇒ $\frac{x}{y}=\frac{y}{112}$

⇒ ${{y}^{2}}=112x$

⇒ ${{y}^{2}}=112\times \frac{196}{y}$

⇒ ${{y}^{3}}=112\times 196$

⇒ ${{y}^{3}}=7\times 4\times 4\times 7\times 7\times 4$

⇒ ${{y}^{3}}={{7}^{3}}{{.4}^{3}}$

⇒ ${{y}^{3}}={{\left( 28 \right)}^{3}}$

⇒ $y=28$ 

Now, put the value of $y$ in equation (i),

⇒ $x=\frac{196}{28}$

⇒ $x=7$

Hence, two required numbers are 7 and 28.

12. If $\mathbf{x}$ and $\mathbf{y}$.  be unequal and $\mathbf{x~:y}$ is the duplicate ratio of $\mathbf{x+z}$ and y+z, prove that $\mathbf{z}$ is mean proportional between x and y.

Ans: We have $x~:y$ is the duplicate ratio of $\left( x+z \right)$ and ($y+z)$.

Therefore, 

⇒ $\frac{x}{y}=\frac{{{\left( x+z \right)}^{2}}}{{{\left( y+z \right)}^{2}}}$

⇒ $x{{\left( y+z \right)}^{2}}=y{{\left( x+z \right)}^{2}}$

⇒ $x~\left( {{y}^{2}}+2yz+{{z}^{2}} \right)=y\left( {{x}^{2}}+2xz+{{z}^{2}} \right)$

⇒ $x{{y}^{2}}+2xyz+x{{z}^{2}}=y{{x}^{2}}+2xyz+y{{z}^{2}}$

⇒ $x{{y}^{2}}+x{{z}^{2}}=y{{x}^{2}}+y{{z}^{2}}$

⇒ $x{{z}^{2}}-y{{z}^{2}}=y{{x}^{2}}-x{{y}^{2}}$

⇒ ${{z}^{2}}\left( x-y \right)=xy\left( x-y \right)$

⇒ ${{z}^{2}}=xy$

⇒ $z=\sqrt{xy}$

Thus, $z$ is mean proportional between $x$ and $y.$

Hence proved.

13. If $\mathbf{x=\frac{2ab}{a~+~b}}$, find the value of  $\mathbf{\frac{x~+~a}{x~-~a}+\frac{x~+~b}{x~-~b}}$.

Ans: Here, we have $x=\frac{2ab}{a+b}$ ……… eq (i)

Now, dividing by $a$ in the above equation, we get

⇒ $\frac{x}{a}=\frac{2ab}{a\left( a+b \right)}$

⇒ $\frac{x}{a}=\frac{2b}{\left( a+b \right)}$

Applying componendo and dividendo rule

⇒ $\frac{x~+~a}{x~-~a}=\frac{2b+a+b}{2b-\left( a+b \right)}$

⇒ $\frac{x~+~a}{x~-~a}=\frac{3b+a}{2b-a-b}$

⇒ $\frac{x~+~a}{x~-~a}=\frac{3b+a}{b-a}$  ……….. eq (ii) 

Now, dividing by $b$ in eq (i), we get 

⇒ $\frac{x}{b}=\frac{2ab}{b\left( a+b \right)}$

⇒ $\frac{x}{b}=\frac{2a}{\left( a+b \right)}$

Applying componendo and dividendo rule

⇒ $\frac{x+~b}{x-~b}=\frac{2a+a+b}{2a-\left( a+b \right)}$

⇒ $\frac{x+~b}{x-~b}=\frac{3a+b}{2a-a-b}$

⇒ $\frac{x+~b}{x-~b}=\frac{3a+b}{a-b}$ ……… eq (iii)

Adding eq (ii) and (iii)

⇒ $\frac{x+~a}{x-~a}+\frac{x+~b}{x-~b}=\frac{3b+a}{b-a}+\frac{3a+b}{a-b}$

$=\frac{3b+a}{b-a}-\frac{3a+b}{b-a}$ 

$=\frac{3b+a-3a-b}{b-a}$ 

$=\frac{2b-2a}{b-a}$

$=\frac{2\left( b-a \right)}{\left( b-a \right)}$

$=2$

Hence value of $\frac{x+~a}{x-~a}+\frac{x+~b}{x-~b}$  is 2. 

14. If $\mathbf{\left( 4a+9b \right)\left( 4c-9d \right)=\left( 4a-9b \right)\left( 4c+9d \right)}$, prove that $\mathbf{a~:b=~c~:d}$.

Ans: Here, we have $\left( 4a+9b \right)\left( 4c-9d \right)=\left( 4a-9b \right)\left( 4c+9d \right)$

Therefore, 

⇒ $\frac{\left( 4a+9b \right)}{\left( 4a-9b \right)}=\frac{\left( 4c+9d \right)}{\left( 4c-9d \right)}$

Applying componendo and dividendo rule

⇒ $\frac{\left( 4a+9b \right)+\left( 4a-9b \right)}{\left( 4a+~9b \right)-\left( 4a-9b \right)}=\frac{\left( 4c+9d \right)+\left( 4c-9d \right)}{\left( 4c~+~9d \right)-\left( 4c-9d \right)}$

⇒$\frac{4a+~9b~~+~4a-9b}{4a+~9b~-~4a~+~9b}=\frac{4c~+~9d~+~4c-9d}{4c~+~9d-~4c~+~9d}$

⇒ $\frac{8a}{18b}=\frac{8c}{18d}$

⇒ $\frac{a}{b}=\frac{c}{d}$

⇒ $a~:b=c~:d$

Hence proved.

15. If  $\mathbf{\frac{a}{b}=\frac{c}{d}~}$, show that $\mathbf{\left( a+b \right)~:\left( c+d \right)=~\sqrt{{{a}^{2}}+{{b}^{2}}}~:~\sqrt{{{c}^{2}}+{{d}^{2}}}}$.

Ans: Here, we have $\frac{a}{b}=\frac{c}{d}$  ……… eq (i)

Adding 1 to both sides in the above equation,

⇒ $\frac{a}{b}+1=\frac{c}{d}+1$

⇒ $\frac{a~+~b}{b}=\frac{c~+~d}{d}$

⇒ $\frac{a~+~b}{c~+~d}=\frac{b}{d}$ ……… eq (ii)

Now, squaring both sides in eq (i), we get

⇒ ${{\left( \frac{a}{b} \right)}^{2}}={{\left( \frac{c}{d} \right)}^{2}}$

⇒ $\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{{{c}^{2}}}{{{d}^{2}}}$

Adding 1 to both sides in the above equation,

⇒ $\frac{{{a}^{2}}}{{{b}^{2}}}+1=\frac{{{c}^{2}}}{{{d}^{2}}}+1$

⇒ $\frac{{{a}^{2}}+~{{b}^{2}}}{{{b}^{2}}}=\frac{{{c}^{2}}+~{{d}^{2}}}{{{d}^{2}}}$

⇒ $\frac{{{a}^{2}}+~{{b}^{2}}}{{{c}^{2}}+~{{d}^{2}}}=\frac{{{b}^{2}}}{{{d}^{2}}}$ 

Taking square root to both sides, we get,

⇒ $\sqrt{\frac{{{a}^{2}}+~{{b}^{2}}}{{{c}^{2}}+~{{d}^{2}}}}=\frac{b}{d}$

⇒ $\frac{\sqrt{{{a}^{2}}+~{{b}^{2}}}}{\sqrt{{{c}^{2}}+~~{{d}^{2}}}}=\frac{b}{d}$ ……….. eq (iii)

Now, from eq (ii) and (iii), 

⇒$\frac{a~+~b}{c~+~d}=\frac{\sqrt{{{a}^{2}}+~{{b}^{2}}}}{\sqrt{{{c}^{2}}+~~{{d}^{2}}}}$

⇒$~\left( a+b \right)~:\left( c+d \right)=~\sqrt{{{a}^{2}}+{{b}^{2}}}~:~\sqrt{{{c}^{2}}+{{d}^{2}}}$

Hence proved.

16. There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3:1. How many more girls should be added to the council so that the ratio of number of boys to the number of girls may be 9:5?

Ans: here, we have the ratio   of the number of boys to the number of girls is $3~:1$.

Let the number of boys be $3x.$

 And number of girls be x.

Now, according to question

⇒ Total no. of members in council $=~36$

⇒ $3x+x=36$

⇒ $4x=36$

⇒ $x=\frac{36}{4}$

⇒ $x=9$

Thus, the number of boys $=3\times 9=27$

And the number of girls $=9$

Let $y$ more girls should be added to get required ratio $9~:5.$. 

⇒ $\frac{27}{9~+~y}=\frac{9}{5}$

⇒ $81+9y=135$

⇒ $9y=135-81$

⇒ $9y=54$

⇒ $y=\frac{54}{9}$

⇒ $y=6$

Thus, 6 girls should be added.

7. If $\mathbf{7x-15y=4x+y}$, find the value of $\mathbf{x~:y}$. Hence, use componendo and dividendo to find the values of:

1. $\mathbf{\frac{9x~+~5y}{9x~-5y}}$

Ans: here, we have $7x-15y=4x+y$.

⇒ $7x-4x=y+15y$

⇒ $3x=16y$

⇒ $\frac{x}{y}=\frac{16}{3}$ …………. Eq (i)

Thus, value of $x~:y~is~16~:3$.

Now, multiplying by $~\frac{9}{5}~$ in eq (i), we get

⇒ $\frac{9}{5}\left( \frac{x}{y} \right)=\frac{9}{5}\left( \frac{16}{3} \right)$

⇒ $\frac{9x}{5y}=\frac{48}{5}$

Applying componendo and dividendo rule

⇒ $\frac{9x~+~5y}{9x~-~5y}=\frac{48~+~5}{48~-5}$

⇒ $\frac{9x~+~5y}{9x~-~5y}=\frac{53}{43}$

Hence, value of $~~\frac{9x~+~5y}{9x~-~5y}~is~\frac{53}{43}$ . 

2. $\mathbf{\frac{3{{x}^{2}}+~2{{y}^{2}}}{3{{x}^{2~}}-2{{y}^{2}}}}$

Ans: here, we have  $\frac{x}{y}=\frac{16}{3}$ …………. Eq (i)

Now, squaring both sides in eq (i), 

⇒ ${{\left( \frac{x}{y} \right)}^{2}}={{\left( \frac{16}{3} \right)}^{2}}~$

⇒ $\frac{{{x}^{2}}}{{{y}^{2}}}=\frac{256}{9}$

Multiplying by $\frac{3}{2}$ in the above equation, we get

⇒ $\frac{3}{2}\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)=\frac{3}{2}\left( \frac{256}{9} \right)$

⇒ $\frac{3{{x}^{2}}}{2{{y}^{2}}}=\frac{128}{3}$

Applying componendo and dividendo rule

⇒$\frac{3{{x}^{2}}~+~2{{y}^{2}}}{3{{x}^{2}}~-~2{{y}^{2}}}=\frac{128~+~3~}{128~-~3}$

⇒ $\frac{3{{x}^{2}}~+~2{{y}^{2}}}{3{{x}^{2}}~-~2{{y}^{2}}}=\frac{~131~}{125}$

Hence, Value of  $\frac{3{{x}^{2}}~+~2{{y}^{2}}}{3{{x}^{2}}~-~2{{y}^{2}}}~is~\frac{~131~}{125}$.

18. If  $\frac{4m~+~3n}{4m~-~3n}=\frac{7}{4}$ , use properties of proportion to find: 

1. $\mathbf{m~:n}$

Ans: Here, we have $\frac{4m~+~3n}{4m~-~3n}=\frac{7}{4}$

⇒ $7\left( 4m-~3n \right)=4~\left( 4m+~3n \right)$

⇒ $28m-21n=16m+12n$

⇒ $28m-16m=12n+21n$

⇒ $12m=33n$

⇒ $\frac{m}{n}=\frac{33}{12}$

⇒ $\frac{m}{n}=\frac{11}{4}$

Hence, value of $m~:n$ is $11~:4$.

2. $\mathbf{\frac{2{{m}^{2}}-11{{n}^{2}}}{2{{m}^{2}}+11{{n}^{2}}}}$

Ans: Here, we have $\frac{m}{n}=\frac{11}{4}$

Squaring both sides, we get

⇒ ${{\left( \frac{m}{n} \right)}^{2}}={{\left( \frac{11}{4} \right)}^{2}}$

⇒ $\frac{{{m}^{2}}}{{{n}^{2}}}=\frac{121}{16}$

Multiplying by $~\frac{2}{11}~$ in the above equation, we get

⇒ $\frac{2}{11}\left( \frac{{{m}^{2}}}{{{n}^{2}}} \right)=\frac{2}{11}\left( \frac{121}{16} \right)$

⇒ $\frac{2{{m}^{2}}}{11{{n}^{2}}}=\frac{11}{8}$

Applying componendo and dividendo rule

⇒$\frac{2{{m}^{2}}~+~11{{n}^{2}}}{2{{m}^{2~}}-~11{{n}^{2}}}=\frac{11~+~8}{11~-~8}$

⇒ $\frac{2{{m}^{2}}~+~11{{n}^{2}}}{2{{m}^{2~}}-~11{{n}^{2}}}=\frac{19}{3}$

⇒ $\frac{2{{m}^{2}}-~11{{n}^{2}}}{2{{m}^{2~}}+~11{{n}^{2}}}=\frac{3}{19}$

Hence, Value of  $\frac{2{{m}^{2}}-~11{{n}^{2}}}{2{{m}^{2~}}+~11{{n}^{2}}}~is~\frac{3}{19}$.

19. If $\mathbf{x,~y,~z}$ are in continued proportion, prove that $\mathbf{\frac{{{\left( x+y \right)}^{2}}}{{{\left( y~+z \right)}^{2}}}=\frac{x}{z}~.}$

Ans: Here,  $x,~y,~z$ are in continued proportion

Therefore, 

⇒ $\frac{x}{y}=\frac{y}{z}~$

⇒ ${{y}^{2}}=xz~$………….. eq (i)

Now, L.H.S $=~\frac{{{\left( x+y \right)}^{2}}}{{{\left( y~+z \right)}^{2}}}$

$=~\frac{{{x}^{2}}~+~2xy~+~{{y}^{2}}}{{{y}^{2}}~+~2yz~+~{{z}^{2}}}$

$=~\frac{{{x}^{2}}~+~2xy~+~xz}{xz~+~2yz~+~{{z}^{2}}}~$                             [from eq (i)]

$=~\frac{x\left( x~+~2y~+~z \right)}{z~\left( x~+~2y~+~z \right)}$

$=\frac{x}{z}~$

= R.H.S

Thus, L.H.S = R.H.S

Hence proved.

20.Given,$\mathbf{x=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+~\sqrt{{{a}^{2}}-{{b}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}-~\sqrt{{{a}^{2}}-{{b}^{2}}}}}$. Use componendo and dividendo to prove that $\mathbf{{{b}^{2}}=\frac{2{{a}^{2}}x}{{{x}^{2}}~+~1}}$. 

Ans: Here, we have $x=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+~\sqrt{{{a}^{2}}-{{b}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}-~\sqrt{{{a}^{2}}-{{b}^{2}}}}$

Applying componendo and dividendo rule

⇒$\frac{x~+~1}{x~-~1}=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}+~\sqrt{{{a}^{2}}-{{b}^{2~}}}+\sqrt{{{a}^{2}}+{{b}^{2}}}-~\sqrt{{{a}^{2}}-{{b}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}+~\sqrt{{{a}^{2}}-{{b}^{2}}~~}-\left( \sqrt{{{a}^{2}}+{{b}^{2}}}-~\sqrt{{{a}^{2}}-{{b}^{2}}} \right)}$

⇒$\frac{x~+~1}{x~-~1}=\frac{2\sqrt{{{a}^{2}}+{{b}^{2}}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}+~\sqrt{{{a}^{2}}-{{b}^{2}}~~}-\sqrt{{{a}^{2}}+{{b}^{2}}}~+~\sqrt{{{a}^{2}}-{{b}^{2}}}}$

⇒$\frac{x~+~1}{x~-~1}=\frac{2\sqrt{{{a}^{2}}+{{b}^{2}}}}{2~\sqrt{{{a}^{2}}-{{b}^{2}}~}}$

⇒ $\frac{x~+~1}{x~-~1}=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{\sqrt{{{a}^{2}}-{{b}^{2}}~}}$

Squaring both sides, we get

⇒ ${{\left( \frac{x~+~1}{x~-~1} \right)}^{2}}={{\left( \frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{\sqrt{{{a}^{2}}-{{b}^{2}}~}} \right)}^{2}}$

⇒$\frac{{{x}^{2}}~+~2x~+~1}{{{x}^{2}}-~2x~+~1}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$

Applying componendo and dividendo rule

⇒ $\frac{{{x}^{2}}~+~2x~+~1+~{{x}^{2}}-~2x~+~1}{{{x}^{2}}\mp ~2x~+~1-\left( ~{{x}^{2}}-~2x~+~1 \right)}=\frac{{{a}^{2}}+{{b}^{2}}~+~{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+~{{b}^{2}}-\left( {{a}^{2}}-{{b}^{2}} \right)}$

⇒$\frac{~2{{x}^{2}}+~2}{{{x}^{2}}+~~2x~+~1-~{{x}^{2}}+~2x-~1}=\frac{2{{a}^{2}}}{{{a}^{2}}+~{{b}^{2}}-{{a}^{2}}~+~{{b}^{2}}}$

⇒ $\frac{~2{{x}^{2}}+~2}{4x}=\frac{2{{a}^{2}}}{2~{{b}^{2}}}$

⇒ $\frac{~2\left( {{x}^{2}}+~1 \right)}{4x}=\frac{{{a}^{2}}}{~{{b}^{2}}}$

⇒ $\frac{~\left( {{x}^{2}}+~1 \right)}{2x}=\frac{{{a}^{2}}}{~{{b}^{2}}}$

⇒ ${{b}^{2}}=\frac{2{{a}^{2}}x}{~\left( {{x}^{2}}+~1 \right)}$

Hence proved. 

21. If $~\frac{{{x}^{2}}+~{{y}^{2}}}{{{x}^{2}}-~{{y}^{2}}}=2\frac{1}{8}$,  find:

1. $\mathbf{\frac{x}{y}~}$

Ans: Here, we have $\frac{{{x}^{2}}+~{{y}^{2}}}{{{x}^{2}}-~{{y}^{2}}}=2\frac{1}{8}$

⇒ $\frac{{{x}^{2}}+~{{y}^{2}}}{{{x}^{2}}-~{{y}^{2}}}=\frac{17}{8}$

⇒ $17~\left( {{x}^{2}}-~{{y}^{2}} \right)=8\left( {{x}^{2}}+~{{y}^{2}} \right)~$

⇒ $17{{x}^{2}}-17{{y}^{2}}=8{{x}^{2}}+~8{{y}^{2}}$

⇒ $17{{x}^{2}}-8{{x}^{2}}=~8{{y}^{2}}+17{{y}^{2}}$

⇒ $9{{x}^{2}}=~25~{{y}^{2}}$

⇒ $\frac{{{x}^{2}}}{{{y}^{2}}}=\frac{25}{9}$

⇒ ${{\left( \frac{x}{y} \right)}^{2}}={{\left( \frac{5}{3} \right)}^{2}}$

⇒ $\frac{x}{y}=\pm \frac{5}{3}$

Hence, Value of $\frac{x}{y}~is~\pm \frac{5}{3}$ .

2. $\mathbf{\frac{{{x}^{3}}~+~{{y}^{3}}}{{{x}^{3}}~-~{{y}^{3}}}}$

Ans: Here, we have $~\frac{x}{y}=\frac{5}{3}$ from part (i)

Now, cubing both sides in above equation

⇒ ${{\left( \frac{x}{y} \right)}^{3}}={{\left( \frac{5}{3} \right)}^{3}}$

⇒  $\frac{{{x}^{3}}}{{{y}^{3}}}=\frac{125}{27}$

Applying componendo and dividendo rule

⇒$\frac{{{x}^{3}}~+~{{y}^{3}}~}{{{x}^{3}}~-~{{y}^{3}}}=\frac{125~+~27}{125-~27}$

⇒  $\frac{{{x}^{3}}~+~{{y}^{3}}~}{{{x}^{3}}~-~{{y}^{3}}}=\frac{152}{98}$

⇒  $\frac{{{x}^{3}}~+~{{y}^{3}}~}{{{x}^{3}}~-~{{y}^{3}}}=\frac{76}{49}$

Hence, Value of $\frac{{{x}^{3}}~+~{{y}^{3}}~}{{{x}^{3}}~-~{{y}^{3}}}~is~\frac{76}{49}$ .

22. Using componendo and dividendo, find the value of $x$: 

$\mathbf{\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=9}$ .  

Ans: Here, we have $\frac{\sqrt{3x+4}+\sqrt{3x-5}}{\sqrt{3x+4}-\sqrt{3x-5}}=\frac{9}{1}$

Applying componendo and dividendo rule in above equation

⇒$\frac{\sqrt{3x+4}+\sqrt{3x-5}~+~\sqrt{3x+4}-\sqrt{3x-5}}{\sqrt{3x+4}+\sqrt{3x-5}-\left( \sqrt{3x+4}-\sqrt{3x-5}~ \right)}=\frac{9~+~1}{9~-~1}$

⇒$\frac{2\sqrt{3x+4}}{\sqrt{3x+4}+\sqrt{3x-5}-\sqrt{3x+4}~+~\sqrt{3x-5}~}=\frac{10}{8}$

⇒ $\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}~}=\frac{5}{4}$

⇒ $\frac{\sqrt{3x+4}}{\sqrt{3x-5}~}=\frac{5}{4}$

Squaring both sides, we get

⇒ ${{\left( \frac{\sqrt{3x+4}}{\sqrt{3x-5}~} \right)}^{2}}={{\left( \frac{5}{4} \right)}^{2}}$

⇒ $\frac{3x+4}{3x-5}=\frac{25}{16}$

⇒ $25~\left( 3x-5~ \right)=16\left( 3x+4~ \right)~$

⇒ $75x-125=48x+64~$

⇒ $75x-48x=64+125$

⇒ $27x=189$

⇒ $x=\frac{189}{27}$

⇒ $x=7$

Hence, Value of $x$ is 7.

23. If $\mathbf{x=~\frac{\sqrt{a~+~1}~+~\sqrt{a~-~1}~}{\sqrt{a~+~1}~-~\sqrt{a~-~1}}}$, using properties of proportion show that: ${{x}^{2}}-2ax+1=0$. 

Ans: Here, we have $x=~\frac{\sqrt{a~+~1}~+~\sqrt{a~-~1}~}{\sqrt{a~+~1}~-~\sqrt{a~-~1}}$

Applying componendo and dividendo rule in above equation

$\Rightarrow \frac{x~+~1}{x~-~1}=~\frac{\sqrt{a~+~1}~+~\sqrt{a~-~1}~+~\sqrt{a~+~1}~-~\sqrt{a~-~1}~}{\sqrt{a~+~1}~+~\sqrt{a~-~1}-\left( \sqrt{a~+~1}~-~\sqrt{a~-~1} \right)}$

$\Rightarrow\frac{x~+~1}{x~-~1}=~\frac{2\sqrt{a~+~1}~~}{\sqrt{a~+~1}~+~\sqrt{a~-~1}-\sqrt{a~+~1}+~\sqrt{a~-~1}}$

$\Rightarrow \frac{x~+~1}{x~-~1}=~\frac{2\sqrt{a~+~1}~~}{2~\sqrt{a~~1}}$

$\Rightarrow \frac{x~+~1}{x~-~1}=~\frac{\sqrt{a~+~1}~~}{\sqrt{a~-~1}}$

Squaring both sides, we get

⇒ ${{\left( \frac{x~+~1}{x~-~1} \right)}^{2}}={{\left( \frac{\sqrt{a~+~1}~~}{\sqrt{a~-~1}} \right)}^{2}}$

⇒ $\frac{{{x}^{2}}~+~2x~+~1}{{{x}^{2}}-~2x~+~1}=\frac{a~+~1}{a~-~1}$

Applying componendo and dividendo rule

⇒ $\frac{{{x}^{2}}~+~2x~+~1+~{{x}^{2}}-~2x~+~1}{{{x}^{2}}\mp ~2x~+~1-\left( ~{{x}^{2}}-~2x~+~1 \right)}=\frac{a~+~1+~a~-~1}{a~+1-\left( a-~1 \right)}$

⇒$\frac{~2{{x}^{2}}+~2}{{{x}^{2}}+~~2x~+~1-~{{x}^{2}}+~2x-~1}=\frac{2a}{a~+1-a+~1}$

⇒ $\frac{~2{{x}^{2}}+~2}{4x}=\frac{2a}{2~}$

⇒ $\frac{~2\left( {{x}^{2}}+~1 \right)}{4x}=a$

⇒ $\frac{~\left( {{x}^{2}}+~1 \right)}{2x}=a$

⇒ ${{x}^{2}}+1=2ax$

⇒ ${{x}^{2}}-2ax+1=0$

Hence proved.

24. Given $\mathbf{\frac{{{x}^{3}}+12x}{6{{x}^{2}}~+~8}=\frac{{{y}^{3}}~+~27y}{9{{y}^{2}}~+~27}.}$ Using componendo and dividendo, find $\mathbf{x~:y.}$

Ans:  Here, we have $\frac{{{x}^{3}}+12x}{6{{x}^{2}}~+~8}=\frac{{{y}^{3}}~+~27y}{9{{y}^{2}}~+~27}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{{{x}^{3}}+12x~+~6{{x}^{2}}~+~8}{{{x}^{3}}+12x-\left( 6{{x}^{2}}~+~8 \right)}=\frac{{{y}^{3}}~+~27y~+~9{{y}^{2}}~+~27}{{{y}^{3}}~+~27y~-\left( 9{{y}^{2}}~+~27~ \right)}$

⇒$\frac{{{x}^{3}}+12x~+~6{{x}^{2}}~+~8}{{{x}^{3}}+12x-6{{x}^{2}}~-~8}=\frac{{{y}^{3}}~+~27y~+~9{{y}^{2}}~+~27}{{{y}^{3}}~+~27y~-9{{y}^{2}}-~27~}$

⇒$\frac{{{x}^{3}}+~8+~6{{x}^{2}}~+~12x~}{{{x}^{3}}-~8-~6{{x}^{2}}~+~12x}=\frac{{{y}^{3}}~+~27~+~9{{y}^{2}}+~27y~}{{{y}^{3}}-~27~-~9{{y}^{2}}+~27y~}$

⇒ $\frac{{{x}^{3}}+~~{{2}^{3~}}+~6x~\left( x~+~2 \right)~}{{{x}^{3}}-~~{{2}^{3~}}-~6x~\left( x-~2 \right)}=\frac{{{y}^{3}}~+~{{3}^{3}}~+~~9y~\left( y+~3 \right)~}{{{y}^{3}}-~{{3}^{3}}-~~9y~\left( y~-3 \right)~}$

⇒ $\frac{{{\left( x~+~2 \right)}^{3}}}{{{\left( x-~2 \right)}^{3}}}=\frac{{{\left( y~+~3 \right)}^{3}}}{{{\left( y-~3 \right)}^{3}}}$

⇒ ${{\left( \frac{x~+~2}{x-~2} \right)}^{3}}={{\left( \frac{y~+~3}{y-~3} \right)}^{3}}$

⇒ $\frac{x~+~2}{x-~2}=\frac{y~+~3}{y-~3}$

Applying componendo and dividendo rule in above equation

⇒ $\frac{x~+~2~+~x-~2}{x~+~2-\left( x-~2 \right)}=\frac{y~+~3~+~y-~3}{y-~3-\left( y-~3 \right)}$

⇒ $\frac{2x}{x~+~2-x~+~2}=\frac{2y}{y-~3-y\mp ~3}$

⇒ $\frac{2x}{4}=\frac{2y}{6}$

⇒ $\frac{x}{2}=\frac{y}{3}$

⇒ $\frac{x}{y}=\frac{2}{3}$

Hence, Value of $x~:y$ is $2~:3$.

25. If  $\mathbf{\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}$ , show that  $\mathbf{\frac{{{x}^{3}}}{{{a}^{3}}}+\frac{{{y}^{3}}}{{{b}^{3}}}+\frac{{{z}^{3}}}{{{c}^{3}}}=\frac{3xyz}{abc}.}$ 

Ans: Here, we have $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ ……….  Eq(i) 

Now, L.H.S  $=~\frac{{{x}^{3}}}{{{a}^{3}}}+\frac{{{y}^{3}}}{{{b}^{3}}}+\frac{{{z}^{3}}}{{{c}^{3}}}$

$={{\left( \frac{x}{a} \right)}^{3}}+{{\left( \frac{y}{b} \right)}^{3}}+{{\left( \frac{z}{c} \right)}^{3}}$

$={{\left( \frac{x}{a} \right)}^{3}}+{{\left( \frac{x}{a} \right)}^{3}}+{{\left( \frac{z}{a} \right)}^{3}}$                       [from eq (i) ]

$=~\frac{{{x}^{3}}}{{{a}^{3}}}+\frac{{{x}^{3}}}{{{a}^{3}}}+\frac{{{x}^{3}}}{{{a}^{3}}}$

$=~3\times \frac{{{x}^{3}}}{{{a}^{3}}}$

$=~3\times \frac{x}{a}\times \frac{x}{a}\times \frac{x}{a}$

$=~3\times \frac{x}{a}\times \frac{y}{b}\times \frac{z}{c}$                     [from eq (i)]

$=\frac{3xyz}{abc}$

= R.H.S

Thus, L.H.S = R.H.S

Hence proved.

26. If b is the mean proportion between a and c, show that: 

$\mathbf{\frac{{{a}^{4}}+{{a}^{2}}{{b}^{2}}+{{b}^{4}}}{{{b}^{4}}+{{b}^{2}}{{c}^{2}}+{{c}^{4}}}=\frac{{{a}^{2}}}{{{c}^{2}}}}$ .

Ans: Here, we have $b$ is the mean proportion between $a$ and $c$

Therefore, 

⇒ ${{b}^{2}}=ac$

Now, squaring both sides

⇒ ${{b}^{4}}={{a}^{2}}{{c}^{2}}$

Thus, L.H.S

$~=~\frac{{{a}^{4}}+{{a}^{2}}{{b}^{2}}+{{b}^{4}}}{{{b}^{4}}+{{b}^{2}}{{c}^{2}}+{{c}^{4}}}$

$=~\frac{{{a}^{4}}+{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}}{{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{4}}}$                              (${{b}^{4}}={{a}^{2}}{{c}^{2}}~$)

$=~\frac{{{a}^{2}}\left( {{a}^{2}}~+{{b}^{2}}+{{c}^{2}} \right)}{{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$

$=~\frac{{{a}^{2}}}{{{b}^{2}}}$

= R.H.S

Thus, L.H.S = R.H.S

Hence proved.

27. If $\mathbf{~\frac{7m~+~2n}{7m~-~2n}=\frac{5}{3}~}$, use properties of proportion to find:

1. $\mathbf{m~:n}$

Ans: Here, we have $\frac{7m~+~2n}{7m~-~2n}=\frac{5}{3}$

 Applying componendo and dividendo rule in above equation

⇒$\frac{7m~+~2n~+~7m~-~2n}{7m+~2n-\left(7m~-~2n\right)}=\frac{5~+~3}{5~-3}$

 ⇒ $\frac{14m}{7m+~2n-7m~+~2n}=\frac{8}{2}$

⇒ $\frac{14m}{4n}=4$

⇒ $\frac{7m}{2n}=4$

⇒ $\frac{m}{n}=\frac{8}{7}~$

Hence, Value of $m~:n$ is $8~:7$.

2. $\mathbf{\frac{{{m}^{2}}~+~{{n}^{2}}}{{{m}^{2}}-~{{n}^{2}}}}$

Ans: Here, we have  $\frac{m}{n}=\frac{8}{7}$  from the part (i)

Now, squaring both sides in above equation,

⇒ ${{\left( \frac{m}{n} \right)}^{2}}={{\left( \frac{8}{7} \right)}^{2}}$

⇒ $\frac{{{m}^{2}}}{{{n}^{2}}}=\frac{64}{49}$

Applying componendo and dividendo rule in above equation

⇒$\frac{{{m}^{2}}~+~{{n}^{2}}}{{{m}^{2}}~-~{{n}^{2}}}=\frac{64~+~49}{64~-~49}$

⇒ $\frac{{{m}^{2}}~+~{{n}^{2}}}{{{m}^{2}}~-~{{n}^{2}}}=\frac{113}{15}$

Hence, Value of $~\frac{{{m}^{2}}~+~{{n}^{2}}}{{{m}^{2}}~-~{{n}^{2}}}~is~\frac{113}{15}$.

28. (i) If x and y both are positive and $\mathbf{\left( 2{{x}^{2}}-5{{y}^{2}} \right)~:xy=1~:3,}$ find $\mathbf{x~:y}$. 

Ans: Here, we have $\left( 2{{x}^{2}}-5{{y}^{2}} \right)~:xy=1~:3$

Therefore, 

⇒ $\frac{\left( 2{{x}^{2}}-5{{y}^{2}} \right)}{xy}=\frac{1}{3}$

⇒ $6{{x}^{2}}-15{{y}^{2}}=xy$

Divide by ${{y}^{2}}$ in above equation, 

⇒ $\frac{6{{x}^{2}}}{~{{y}^{2}}}-\frac{15{{y}^{2}}}{~{{y}^{2}}}=\frac{xy}{~{{y}^{2}}}$

⇒ $6{{\left( \frac{x}{y} \right)}^{2}}-15=\left( \frac{x}{y} \right)$

Now, let $a=\frac{x}{y}$ 

⇒ $6{{a}^{2}}-15=a$

⇒ $6{{a}^{2}}-a-15=0$

⇒ $6{{a}^{2}}-10a+9a-15=0$

⇒ $2a\left( 3a-5 \right)+3\left( 3a-5 \right)=0$

⇒ $\left( 3a-5 \right)\left( 2a+3 \right)=0$

⇒ $\left( 3a-5 \right)=0~or~\left( 2a+3 \right)=0$

⇒ $a=\frac{5}{3}~or~a=-\frac{3}{2}~$

Now, put $a=\frac{x}{y}~$,  we get

⇒ $\frac{x}{y}=\frac{5}{3}~or\frac{x}{y}=-\frac{3}{2}~$

⇒ $x~:y=5~:3~or~x~:y=-3~:2$

Since, $x~$and $y$ both are positive. Thus, value of  $x~:y~$will be positive.

Hence, Value of $x~:y~$is $5~:3.$ 

(ii). Find x, if $\mathbf{16{{\left( \frac{a~-~x}{a~+~x} \right)}^{3}}=\frac{a~+~~x}{a~-~x}}$. 

Ans:  Here, we have $16{{\left( \frac{a~-~x}{a~+~x} \right)}^{3}}=\frac{a~+~x}{a~-~x}$

⇒ ${{\left( \frac{a+~x}{a-~x} \right)}^{4}}=16$

⇒ ${{\left( \frac{a+~x}{a-~x} \right)}^{4}}={{2}^{4}}$

⇒ $\frac{a+~x}{a-~x}=\pm 2$  

Case 1: When  $\frac{a+~x}{a-~x}=2$  

⇒ $a+x=2a-2x$

⇒ $x+2x=2a-a$

⇒ $3x=a$

⇒ $x=~\frac{a}{3}~$ 

Now, Case 2:   $\frac{a+~x}{a-~x}=-2$  

⇒ $-2a+2x=a+x$

⇒ $2x-x=a+2a$

⇒ $x=3a$

Hence, Value of $x$ is  $\frac{a}{3}~$ or $3a.$

ICSE Class 10 Mathematics Chapter 7 Selina Concise Solutions - Free PDF Download

Definition of Ratio

We can say that the comparison of two quantities of the same kind by division is referred to as ratio.

The two numbers in a ratio can only be compared when they have the same unit. We make use of ratios to compare two things. 

Representation of Ratio

The Sign Used to Denote a Ratio is ‘:’.

The ratio of any two same quantities p and q can be expressed as either p/q or p : q. Here p is referred to 'antecedent' and q is called 'consequent'.

Types of Ratios

There are various types of ratios in Maths they are as follows:

• Compounded Ratio: The ratio of antecedent as product of antecedents of the ratios and consequent as product of consequents, then the ratio thus formed is called compounded ratio.

For example, the compound ratio of m : n and y : x is my : nx and that of a : b, c : d and e : f is the ratio ace : bdf.

• Duplicate Ratio: The duplicate ratio is the  copy of the given ratio. For example The duplicate ratio of the ratio a : b 

= (a × a) : (b × b)

= a2 : b2

• Reciprocal Ratio: The reciprocal ratio of a:b is the ratio (1/a):(1/b), where a≠0 and b≠0. Which can also be written as 1/a : 1/b = b: a. Hence reciprocal ratio is the inverse ratio of the previous ratio.

For example the reciprocal ratio of 2 : 3 is ½ : ⅓ = 3 : 2

• Ratio of Equalities: If the antecedent and consequent are equal, the ratio is called ratio of equality, for example 5 : 5.

• Ratio of Inequalities: If the antecedent and consequent are not equal, the  ratio is called the ratio of inequality, for example 5:7.

What is Proportion

The proportion is the relation between two ratios such as a : b :: c : d or a/b = c/d, where a,b,c, and d are integers. In simple words, proportion compares two ratios. When two ratios are equal, then they are said to be in proportion. Proportions are denoted by the symbol  ‘::’ or ‘=’.

Suppose we have two numbers and we have to find the ratio of these two, then the formula for ratio is defined as;

p : q = p/q

where p and q  are any two numbers,  “p” is called the antecedent, and “q” is called the consequent.

For example given that the two ratios are p : q & r : s . The two terms ‘q’ and ‘r’ are called ‘means or mean term,’ whereas the terms ‘p’ and ‘s’ are known as ‘extremes or extreme terms.’

Important Properties of Proportion

The following are the important properties of proportion used in solving problems related to ratio and proportions.

• Addendo :

If m : n = o : p, then m + p : n + o

• Subtrahendo :

• If m : n = o : p, then m – p : n – o

• Dividendo :

If m : n = o : p, then m – n : n = o – p : p

• Componendo :

If m : n = o : p, then m + n : n = o + p : p

• Alternendo :

 If m : n = o : p, then m : o = n : p

• Invertendo :

If m : n = o : p, then n : m = p : o

• Componendo and Dividendo:

If m : n = o : p, then m + n : m – n = o + p : o – p

Ratio and Proportion Tricks

Let us learn some rules and tricks to solve problems based on ratio and proportion.

• If u/v = x/y, then uy = vx

• If u/v = x/y, then u/x = v/y

• If u/v = x/y, then v/u = y/x

• If u/v = x/y, then (u+v)/v = (x+y)/y

• If u/v = x/y, then (u-v)/v = (x-y)/y

• If u/v = x/y, then (u+v)/ (u-v) = (x+y)/(x-y), which is known as componendo -Dividendo Rule

• If a/(b+c) = b/(c+a) = c/(a+b) and a+b+ c ≠0, then a =b = c

For a student of ICSE class 10, mathematics is one of the most scoring subjects. That is why it is referred to as a crucial subject to shape a student's total percentage of marks in the ICSE class 10 annual examination. So, to increase the total marks that students can achieve, they should focus on studying maths thoroughly. 

Chapter 7 (Ratio and proportion) in the ICSE class 10 textbook is one of the most crucial chapters from the examination point of view. Vedantu has come up with a solution to help students achieve mastery in this chapter. The solution is ICSE Class 10 Mathematics Chapter 7 Selina Concise Solutions. Using this, a student can learn and master the chapter concepts of ratio and proportion. 

Here is a list of concepts Vedantu's ICSE Class 10 Mathematics Chapter 7 Selina Concise Solutions covers:

1. Composition of ratios 

2. Properties of proportion

3. Direct applications of proportions

4. Continued proportion 

5. Increase and decrease in a ratio

The ICSE Class 10 Mathematics Chapter 7 Selina Concise Solutions provided by Vedantu have been designed in an easy step-by-step way that helps students understand the topic regardless of their understanding capability. 

Students can directly download the ICSE Class 10 Mathematics Selina Concise Solutions for Chapter 7 using the download link given on the same page. These study materials are proven to improve the problem-solving capability and boost the confidence of the ICSE class 10 students. 

These problems are solved by highly experienced teachers and mathematicians. A student is expected to study the ICSE Class 10 Mathematics Chapter 7 Selina Concise Solutions provided by Vedantu after learning and revising the whole Chapter 7 (Ratio and Proportion) from the textbook.

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