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#### Question 1:

From the following pairs of numbers, find the reduced form of ratio of first number to second number.

(i) 72, 60 (ii) 38, 57 (iii) 52, 78

(i)
72 : 60 = $\frac{72}{60}=\frac{72÷12}{60÷12}=\frac{6}{5}$ = 6 : 5             (HCF of 72 and 60 = 12)

Thus, the reduced form of 72 : 60 is 6 : 5.

(ii)
38 : 57 = $\frac{38}{57}=\frac{38÷19}{57÷19}=\frac{2}{3}$ = 2 : 3             (HCF of 38 and 57 = 19)

Thus, the reduced form of 38 : 57 is 2 : 3.

(iii)
52 : 78 = $\frac{52}{78}=\frac{52÷26}{78÷26}=\frac{2}{3}$ = 2 : 3             (HCF of 52 and 78 = 26)

Thus, the reduced form of 52 : 78 is 2 : 3.

#### Question 2:

Find the reduced form of the ratio of the first quantity to second quantity.

(i) 700 rs, 308 rs (ii) 14 rs, 12 rs. 40 paise.   (iii) 5 litre, 2500 ml (iv) 3 years 4 months,   5years 8 months (v) 3.8 kg, 1900 gm

(vi) 7 minutes 20 seconds,  5 minutes 6 seconds.

(i)
Rs 700 : Rs 308 = $\frac{700}{308}=\frac{700÷28}{308÷28}=\frac{25}{11}$ = 25 : 11             (HCF of 700 and 308 = 28)

(ii)
14 rupees = 1400 p

12 rupees 40 paise = 1200 + 40 = 1240 p

∴ 14 rupees : 12 rupees 40 paise = 1400 p : 1240 p = $\frac{1400}{1240}=\frac{1400÷40}{1240÷40}=\frac{35}{31}$ = 35 : 31                (HCF of 1400 and 1240 = 40)

(iii)
5 litre = 5000 mL            (1 L = 1000 mL)

∴ 5 litre : 2500 mL = 5000 mL : 2500 mL = $\frac{5000}{2500}=\frac{5000÷2500}{2500÷2500}=\frac{2}{1}$ = 2 : 1            (HCF of 5000 and 2500 = 2500)

(iv)
3 years 4 months  = 3 years + 4 months = 3 × 12 + 4 = 36 + 4 = 40 months                (1 year = 12 months)

5 years 8 months  = 5 years + 8 months = 5 × 12 + 8 = 60 + 8 = 68 months

∴ 3 years 4 months : 5 years 8 months = 40 months : 68 months = $\frac{40}{68}=\frac{40÷4}{68÷4}=\frac{10}{17}$ = 10 : 17          (HCF of 40 and 68 = 4)

(v)
3.8 kg = 3.8 × 1000 = 3800 g              (1 kg = 1000 g)

∴ 3.8 kg : 1900 g = 3800 g : 1900 g = $\frac{3800}{1900}=\frac{3800÷1900}{1900÷1900}=\frac{2}{1}$ = 2 : 1       (HCF of 3800 and 1900 = 1900)

(iv)
7 minutes 20 seconds = 7 minutes + 20 seconds = 7 × 60 + 20 = 420 + 20 = 440 seconds         (1 minute = 60 seconds)

5 minutes 6 seconds = 5 minutes + 6 seconds = 5 × 60 + 6 = 300 + 6 = 306 seconds

∴ 7 minutes 20 seconds : 5 minutes 6 seconds = 440 seconds : 306 seconds = $\frac{440}{306}=\frac{440÷2}{306÷2}=\frac{220}{153}$ = 220 : 153      (HCF of 220 and 153 = 2)

#### Question 3:

Express the following percentages as ratios in the reduced form.

(i) 75 : 100
(ii) 44 : 100
(iii) 6.25%
(iv) 52 : 100
(v) 0.64%

(i)
75 : 100 = $\frac{75}{100}=\frac{75÷25}{100÷25}=\frac{3}{4}$ = 3 : 4               (HCF of 75 and 100 = 25)

(ii)
44 : 100 = $\frac{44}{100}=\frac{44÷4}{100÷4}=\frac{11}{25}$ = 11 : 25              (HCF of 44 and 100 = 4)

(iii)
6.25% = $\frac{6.25}{100}=\frac{625}{10000}=\frac{625÷625}{10000÷625}=\frac{1}{16}$ = 1 : 16              (HCF of 625 and 10000 = 625)

(iv)
52 : 100 = $\frac{52}{100}=\frac{52÷4}{100÷4}=\frac{13}{25}$ = 13 : 25              (HCF of 52 and 100 = 4)

(v)
0.64% = $\frac{0.64}{100}=\frac{64}{10000}=\frac{64÷16}{10000÷16}=\frac{4}{625}$ = 4 : 625              (HCF of 64 and 10000 = 16)

#### Question 4:

Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?

Let the number of persons required to build the same house in 6 days be x.

The number of persons and the number of days required to build the house are in inverse variation. So, the product of number of persons and the number of days required to build the house is constant.

∴ x × 6 = 3 × 8

⇒ x$\frac{3×8}{6}=\frac{24}{6}$ = 4

Thus, 4 persons are required to build the same house in 6 days.

#### Question 5:

Convert the following ratios into percentage.

(i) 15 : 25 (ii) 47 : 50 ​   (iii) $\frac{7}{10}$    (iv)  $\frac{546}{600}$    (v) $\frac{7}{16}$

(i)
15 : 25 = $\frac{15}{25}=\frac{15}{25}×100%$ = 15 × 4% = 60%

(ii)
47 : 50 = $\frac{47}{50}=\frac{47}{50}×100%$ = 47 × 2% = 94%

(iii)
$\frac{7}{10}=\frac{7}{10}×100%$ = 7 × 10%= 70%

(iv)
$\frac{546}{600}=\frac{546}{600}×100%=\frac{546}{6}%$ = 91%

(v)
$\frac{7}{16}=\frac{7}{16}×100%=\frac{700}{16}%$ = 43.75%

#### Question 6:

The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mothers age was 27 year. Find the present ages of Abha and her mother.

The ratio of the present ages of Abha and her mother is 2 : 5.

Let the present age of Abha and her mother be 2x years and 5years, respectively.

∴ Abha's mother age at the time of Abha's birth = 5x − 2x = 3x years

It is given that at the time of Abha's birth her mothers age was 27 year.

∴ 3x = 27

⇒ x = 9

∴ Present age of Abha = 2x = 2 × 9 = 18 years

Present age of Abha's mother = 5x = 5 × 9 = 45 years

Thus, the present age of Abha and her mother is 18 years and 45 years, respectively.

#### Question 7:

Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?

Let after x years, the ratio of their ages will become 5 : 4.

Age of Vatsala after x years = (14 + x) years

Age of Sara after x years = (10 + x) years

$\therefore \frac{14+x}{10+x}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒56+4x=50+5x\phantom{\rule{0ex}{0ex}}⇒5x-4x=56-50\phantom{\rule{0ex}{0ex}}⇒x=6$
Thus, the ratio of their ages will become 5 : 4 after 6 years.

#### Question 8:

The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age ?

Let the present ages of Rehana and her mother be 2x years and 7x years, respectively.

After 2 years,

Age of Rehana = (2x + 2) years

Age of Rehana's mother = (7x + 2) years

It is given that after 2 years, the ratio of their ages will be 1 : 3.

$\therefore \frac{2x+2}{7x+2}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒6x+6=7x+2\phantom{\rule{0ex}{0ex}}⇒7x-6x=6-2\phantom{\rule{0ex}{0ex}}⇒x=4$
∴ Rehana's present age = 2x = 2 × 4 = 8 years

Thus, the present age of Rehana is 8 years.

#### Question 1:

Using the property  , fill in the blanks substituting proper numbers in the following.

(i)      (ii)

(i)
28 = 7 × 4, 35 = 5 × 4, 3.5 = 7 × 0.5

Now,

(ii)
4.5 = 9 × 0.5, 42 = 14 × 3, 3.5 = 14 × 0.25

Now,

#### Question 2:

Find the following ratios.

(i) The ratio of radius to circumference of the circle.
(ii) The ratio of circumference of circle with radius r to its area.
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.

(i)
Let the radius of the circle be r units.

∴ Circumference of the circle = 2$\mathrm{\pi }$r units

Radius of the circle : Circumference of the circle = r : 2$\mathrm{\pi }$r = $\frac{r}{2\mathrm{\pi }r}=\frac{1}{2\mathrm{\pi }}$ = 1 : 2$\mathrm{\pi }$

Thus, the ratio of radius to cirumference of the circle is 1 : 2$\mathrm{\pi }$.

(ii)
Radius of the circle = r units

∴ Circumference of the circle = 2$\mathrm{\pi }$r units

Area of the circle = $\mathrm{\pi }$r2 square units

Circumference of the circle : Area of the circle = 2$\mathrm{\pi }$$\mathrm{\pi }$r$\frac{2\mathrm{\pi }r}{\mathrm{\pi }{r}^{2}}=\frac{2}{r}$ = 2 : r

Thus, the ratio of circumference of circle with radius r to its area is 2 : r.

(iii)
Side of the square = 7 cm

∴ Length of diagonal of the square = $\sqrt{2}$ × Side of the square = $7\sqrt{2}$ cm

Length of diagonal of the square : Side of the square = $7\sqrt{2}$ cm : 7 cm = $\frac{7\sqrt{2}}{7}=\frac{\sqrt{2}}{1}$ = $\sqrt{2}$ : 1

Thus, the ratio of diagonal of the square to its side is $\sqrt{2}$ : 1.

(iv)
Length of the rectangle, l = 5 cm

Breadth of the rectangle, b = 3.5 cm

∴ Perimeter of the rectangle = 2(lb) = 2 × (5 + 3.5) = 2 × 8.5 = 17 cm

Area of the rectangle = l × b = 5 × 3.5 = 17.5 cm2

Perimeter of the rectangle : Area of the rectangle = 17 : 17.5 = $\frac{17}{17.5}=\frac{170}{175}=\frac{170÷5}{175÷5}=\frac{34}{35}$ = 34 : 35

Thus, the ratio of perimeter to area of the rectangle is 34 : 35.

#### Question 3:

Compare the following pairs of ratios.

(i)       (ii)       (iii)       (iv)       (v)

(i)
$\sqrt{5}×\sqrt{7}=\sqrt{5×7}=\sqrt{35}\phantom{\rule{0ex}{0ex}}3×3=9=\sqrt{81}$
Now, $\sqrt{35}<\sqrt{81}$

$\therefore \frac{\sqrt{5}}{3}<\frac{3}{\sqrt{7}}$

(ii)
$3\sqrt{5}×\sqrt{125}=3\sqrt{5}×\sqrt{25×5}=3\sqrt{5}×5\sqrt{5}=75\phantom{\rule{0ex}{0ex}}5\sqrt{7}×\sqrt{63}=5\sqrt{7}×\sqrt{9×7}=5\sqrt{7}×3\sqrt{7}=105$
Now, 75 < 105

$\therefore \frac{3\sqrt{5}}{5\sqrt{7}}<\frac{\sqrt{63}}{\sqrt{125}}$

(iii)
$5×121=605\phantom{\rule{0ex}{0ex}}18×17=306$
Now, 605 > 306

$\therefore \frac{5}{18}>\frac{17}{121}$

(iv)
$\sqrt{80}×\sqrt{27}=\sqrt{16×5}×\sqrt{9×3}=4\sqrt{5}×3\sqrt{3}=12\sqrt{15}\phantom{\rule{0ex}{0ex}}\sqrt{45}×\sqrt{48}=\sqrt{9×5}×\sqrt{16×3}=3\sqrt{5}×4\sqrt{3}=12\sqrt{15}$
Now, $12\sqrt{15}=12\sqrt{15}$

$\therefore \frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}$

(v)
$9.2×7.1=65.32\phantom{\rule{0ex}{0ex}}5.1×3.4=17.34$
Now, 65.32 > 17.34

$\therefore \frac{9.2}{5.1}>\frac{3.4}{7.1}$

#### Question 4:

(i)    $\square$ABCD is a parallelogram. The ratio of $\angle$A and $\angle$B of this parallelogram is 5 : 4. Find the measure of  $\angle$B.

(ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
(iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
(iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
(v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.

(i)

Let the measure of $\angle$A and $\angle$B be 5x and 4x, respectively.

Now,

$\angle$A + $\angle$B = 180º          (Adjacent angles of a parallelogram are supplementary)

∴ 5x + 4x = 180º

⇒ 9x = 180º

⇒ x = 20º

∴ Measure of $\angle$B = 4x = 4 × 20º = 80º

Thus, the measure of $\angle$B is 80º.

(ii)
Let the present ages of Albert and Salim be 5x years and 9x years, respectively.

5 years hence,

Age of Albert = (5x + 5) years

Age of Salim = (9x + 5) years

It is given that five year hence, the ratio of their ages will be 3 : 5.

$\therefore \frac{5x+5}{9x+5}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}⇒25x+25=27x+15\phantom{\rule{0ex}{0ex}}⇒27x-25x=25-15\phantom{\rule{0ex}{0ex}}⇒2x=10\phantom{\rule{0ex}{0ex}}⇒x=5$
∴ Present age of Albert = 5x = 5 × 5 = 25 years

Present age of Salim = 9x = 9 × 5 = 45 years

Thus, the present age of Albert is 25 years and the present age of Salim is 45 years.

(iii)
Let the length and breadth of the rectangle be 3x cm and cm, respectively.

Perimeter of the rectangle = 36 cm

∴ 2(Length + Breadth) = 36 cm

⇒ 2(3xx) = 36

⇒ 2 × 4x = 36

⇒ 8x = 36

⇒ x = 4.5

∴ Length of the rectangle = 3x = 3 × 4.5 = 13.5 cm

Breadth of the rectangle = x = 4.5 cm

Thus, the length and breadth of the rectangle is 13.5 cm and 4.5 cm, respectively.

(iv)
Let the two numbers be 31x and 23x.

Sum of the two numbers = 216

∴ 31x + 23x = 216

⇒ 54x = 216

⇒ x = $\frac{216}{54}$ = 4

∴ One number = 31x = 31 × 4 = 124

Other number = 23x = 23 × 4 = 92

Thus, the two numbers are 92 and 124.

(v)
Let the two numbers be 10x and 9x.

Product of the two numbers = 360

∴ 10x × 9x = 360

⇒ 90x2 = 360

⇒ x2 = 4

⇒ x = 2

∴ One number = 10x = 10 × 2 = 20

Other number = 9x = 9 × 2 = 18

Thus, the two numbers are 18 and 20.

#### Question 5:

If a : b = 3 : 1 and b : c = 5 : 1 then find the value of

(i) ${\left(\frac{{a}^{3}}{15{b}^{2}c}\right)}^{3}$    (ii) $\frac{{a}^{2}}{7bc}$

ab = 3 : 1

.....(1)

b : c = 5 : 1

.....(2)

From (1) and (2), we have

a = 3 × 5c = 15c

(i)
${\left(\frac{{a}^{3}}{15{b}^{2}c}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left[\frac{{\left(15c\right)}^{3}}{15×{\left(5c\right)}^{2}×c}\right]}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{15×15×15{c}^{3}}{15×25{c}^{2}×c}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(9\right)}^{3}\phantom{\rule{0ex}{0ex}}=729$
(ii)
$\frac{{a}^{2}}{7bc}\phantom{\rule{0ex}{0ex}}=\frac{{\left(15c\right)}^{2}}{7×5c×c}\phantom{\rule{0ex}{0ex}}=\frac{225}{35}\phantom{\rule{0ex}{0ex}}=\frac{45}{7}$

#### Question 6:

If   then find the ratio $\frac{a}{b}$.

$⇒\frac{a}{b}=0.4×0.04\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{4}{10}×\frac{4}{100}=\frac{16}{1000}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{2}{125}$
∴ ab = 2 : 125

#### Question 7:

( x + 3) : ( x + 11) = ( x $-$ 2) : ( x + 1) then find the value of x.

$\left(x+3\right):\left(x+11\right)=\left(x-2\right):\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x+3}{x+11}=\frac{x-2}{x+1}\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)×\left(x+1\right)=\left(x-2\right)×\left(x+11\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x+3x+3={x}^{2}+11x-2x-22$
$⇒4x+3=9x-22\phantom{\rule{0ex}{0ex}}⇒9x-4x=22+3\phantom{\rule{0ex}{0ex}}⇒5x=25\phantom{\rule{0ex}{0ex}}⇒x=5$
Thus, the value of x is 5.

#### Question 1:

If  then find the values of the following  ratios.

(i)   (ii)    (iii)     (iv)

(i)
$\frac{5a+3b}{5a-3b}\phantom{\rule{0ex}{0ex}}=\frac{5×7k+3×3k}{5×7k-3×3k}\phantom{\rule{0ex}{0ex}}=\frac{35k+9k}{35k-9k}\phantom{\rule{0ex}{0ex}}=\frac{44k}{26k}\phantom{\rule{0ex}{0ex}}=\frac{22}{13}$
(ii)
$\frac{2{a}^{2}+3{b}^{2}}{2{a}^{2}-3{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2×{\left(7k\right)}^{2}+3×{\left(3k\right)}^{2}}{2×{\left(7k\right)}^{2}-3×{\left(3k\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{98{k}^{2}+27{k}^{2}}{98{k}^{2}-27{k}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{125{k}^{2}}{71{k}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{125}{71}$
(iii)
$\frac{{a}^{3}-{b}^{3}}{{b}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(7k\right)}^{3}-{\left(3k\right)}^{3}}{{\left(3k\right)}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{343{k}^{3}-27{k}^{3}}{27{k}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{316{k}^{3}}{27{k}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{316}{27}$
(iv)
$\frac{7a+9b}{7a-9b}\phantom{\rule{0ex}{0ex}}=\frac{7×7k+9×3k}{7×7k-9×3k}\phantom{\rule{0ex}{0ex}}=\frac{49k+27k}{49k-27k}\phantom{\rule{0ex}{0ex}}=\frac{76k}{22k}\phantom{\rule{0ex}{0ex}}=\frac{38}{11}$

#### Question 2:

If  then find the values of following ratios.

(i) $\frac{a}{b}$   (ii)  (iii)    (iv)

(i)
$\frac{15{a}^{2}+4{b}^{2}}{15{a}^{2}-4{b}^{2}}=\frac{47}{7}$

Applying componendo and dividendo, we get

$\frac{\left(15{a}^{2}+4{b}^{2}\right)+\left(15{a}^{2}-4{b}^{2}\right)}{\left(15{a}^{2}+4{b}^{2}\right)-\left(15{a}^{2}-4{b}^{2}\right)}=\frac{47+7}{47-7}\phantom{\rule{0ex}{0ex}}⇒\frac{15{a}^{2}+4{b}^{2}+15{a}^{2}-4{b}^{2}}{15{a}^{2}+4{b}^{2}-15{a}^{2}+4{b}^{2}}=\frac{54}{40}\phantom{\rule{0ex}{0ex}}⇒\frac{30{a}^{2}}{8{b}^{2}}=\frac{27}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{{b}^{2}}=\frac{27×8}{20×30}=\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
(ii)

$=\frac{\frac{21-15}{5}}{\frac{21+15}{5}}\phantom{\rule{0ex}{0ex}}=\frac{6}{36}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}$
(iii)

$=\frac{7}{43}$

(iv)

$=\frac{71}{179}$

#### Question 3:

If  then find the value of the  ratio

$\frac{3a+7b}{3a-7b}=\frac{4}{3}$

Applying componendo and dividendo, we get

Now,

#### Question 4:

Solve the following equations.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i)
$\frac{{x}^{2}+12x-20}{3x-5}=\frac{{x}^{2}+8x+12}{2x+3}$

Multiplying both sides by $\frac{1}{4}$, we get

$\frac{{x}^{2}+12x-20}{12x-20}=\frac{{x}^{2}+8x+12}{8x+12}$

Using dividendo, we get

$\frac{\left({x}^{2}+12x-20\right)-\left(12x-20\right)}{12x-20}=\frac{\left({x}^{2}+8x+12\right)-\left(8x+12\right)}{8x+12}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+12x-20-12x+20}{12x-20}=\frac{{x}^{2}+8x+12-8x-12}{8x+12}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{12x-20}=\frac{{x}^{2}}{8x+12}$
This equation is true for x = 0. Therefore, x = 0 is a solution of the given equation.

If x ≠ 0, then x2 ≠ 0.

Dividing both sides by x2, we get

$\frac{1}{12x-20}=\frac{1}{8x+12}\phantom{\rule{0ex}{0ex}}⇒12x-20=8x+12\phantom{\rule{0ex}{0ex}}⇒12x-8x=20+12\phantom{\rule{0ex}{0ex}}⇒4x=32\phantom{\rule{0ex}{0ex}}⇒x=8$
Thus, the solutions of the given equation are x = 0 and x = 8.

(ii)

Multiplying both sides by $\frac{1}{5x}$, we get

$\frac{10{x}^{2}+15x+63}{10{x}^{2}+15x}=\frac{5{x}^{2}-25x+12}{5{x}^{2}-25x}$

Using dividendo, we get

$\frac{\left(10{x}^{2}+15x+63\right)-\left(10{x}^{2}+15x\right)}{10{x}^{2}+15x}=\frac{\left(5{x}^{2}-25x+12\right)-\left(5{x}^{2}-25x\right)}{5{x}^{2}-25x}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{10{x}^{2}+15x}=\frac{12}{5{x}^{2}-25x}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{5x\left(2x+3\right)}=\frac{12}{5x\left(x-5\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{2x+3}=\frac{12}{x-5}$
$⇒63x-315=24x+36\phantom{\rule{0ex}{0ex}}⇒63x-24x=315+36\phantom{\rule{0ex}{0ex}}⇒39x=351\phantom{\rule{0ex}{0ex}}⇒x=\frac{351}{39}=9$
Thus, the solution of the give  equation is x = 9.

(iii)

Applying componendo and dividendo, we get

$⇒6x+3=10x-5\phantom{\rule{0ex}{0ex}}⇒10x-6x=5+3\phantom{\rule{0ex}{0ex}}⇒4x=8\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, the solution of the given equation is x = 2.

(iv)
$\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$

Applying componendo and dividendo, we get

$\frac{\left(\sqrt{4x+1}+\sqrt{x+3}\right)+\left(\sqrt{4x+1}-\sqrt{x+3}\right)}{\left(\sqrt{4x+1}+\sqrt{x+3}\right)-\left(\sqrt{4x+1}-\sqrt{x+3}\right)}=\frac{4+1}{4-1}\phantom{\rule{0ex}{0ex}}⇒\frac{2\sqrt{4x+1}}{2\sqrt{x+3}}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{4x+1}}{\sqrt{x+3}}=\frac{5}{3}$
Squaring on both sides, we get

$\frac{4x+1}{x+3}={\left(\frac{5}{3}\right)}^{2}=\frac{25}{9}\phantom{\rule{0ex}{0ex}}⇒36x+9=25x+75\phantom{\rule{0ex}{0ex}}⇒36x-25x=75-9\phantom{\rule{0ex}{0ex}}⇒11x=66\phantom{\rule{0ex}{0ex}}⇒x=6$
Thus, the solution of the given equation is x = 6.

(v)

Applying dividendo, we get

$\frac{{\left(4x+1\right)}^{2}+{\left(2x+3\right)}^{2}-{\left(2x+3\right)}^{2}}{{\left(2x+3\right)}^{2}}=\frac{61-36}{36}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(4x+1\right)}^{2}}{{\left(2x+3\right)}^{2}}=\frac{25}{36}$
Taking square root on both sides, we get

$\frac{4x+1}{2x+3}=\sqrt{\frac{25}{36}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}⇒24x+6=10x+15\phantom{\rule{0ex}{0ex}}⇒24x-10x=15-6\phantom{\rule{0ex}{0ex}}⇒14x=9\phantom{\rule{0ex}{0ex}}⇒x=\frac{9}{14}$
Thus, the solution of the given equation is $x=\frac{9}{14}$.

(vi)
$\frac{{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}}{{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}}=\frac{61}{189}$

Applying componendo and dividendo, we get

$\frac{\left[{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}\right]+\left[{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}\right]}{\left[{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}\right]-\left[{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}\right]}=\frac{61+189}{61-189}\phantom{\rule{0ex}{0ex}}⇒\frac{2{\left(3x-4\right)}^{3}}{-2{\left(x+1\right)}^{3}}=\frac{250}{-128}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(3x-4\right)}^{3}}{{\left(x+1\right)}^{3}}=\frac{125}{64}$
Taking cube root on both sides, we get

$\frac{3x-4}{x+1}=\sqrt[3]{\frac{125}{64}}\phantom{\rule{0ex}{0ex}}⇒\frac{3x-4}{x+1}=\sqrt[3]{{\left(\frac{5}{4}\right)}^{3}}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒12x-16=5x+5\phantom{\rule{0ex}{0ex}}⇒12x-5x=16+5\phantom{\rule{0ex}{0ex}}⇒7x=21\phantom{\rule{0ex}{0ex}}⇒x=3$
Thus, the solution of the given equation is x = 3.

#### Question 1:

Fill in the blanks of the following

(i)

(ii)

(i)

Also,

$\therefore \frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{\overline{)\mathbf{36}}}=\frac{7x-9y}{\overline{)\mathbf{22}}}$

(ii)

Also,

$\therefore \frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2b+3c}{\overline{)\mathbf{16}}}=\frac{\overline{)\mathbf{2}\mathbf{a}\mathbf{-}\mathbf{2}\mathbf{b}\mathbf{+}\mathbf{2}\mathbf{c}}}{6-8+14}$

#### Question 2:

5   then find the values of the following expressions.

(i)

(ii)

$5m-n=3m+4n\phantom{\rule{0ex}{0ex}}⇒5m-3m=4n+n\phantom{\rule{0ex}{0ex}}⇒2m=5n\phantom{\rule{0ex}{0ex}}⇒\frac{m}{n}=\frac{5}{2}$
(i)

(ii)
$\frac{3m+4n}{3m-4n}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3m}{n}+\frac{4n}{n}}{\frac{3m}{n}-\frac{4n}{n}}\phantom{\rule{0ex}{0ex}}=\frac{3×\frac{5}{2}+4}{3×\frac{5}{2}-4}\phantom{\rule{0ex}{0ex}}=\frac{15+8}{15-8}\phantom{\rule{0ex}{0ex}}=\frac{23}{7}$

#### Question 3:

(i)  If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal  then show that,

(ii)  If    and then show that the value of each ratio is equal to 1.

(iii)   If    and    then show that  .

(iv)  If    then show that   .

(v)  If   then show that every ratio = $\frac{x}{y}$.

(i)
Let $a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)=k$

Similarly,

From (1), (2) and (3), we have

(ii)
$\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{\left(3x-y-z\right)+\left(3y-z-x\right)+\left(3z-x-y\right)}$          (Theorem of equal ratios)

$⇒\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{x+y+z}$

$⇒\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1$

(iii)
$\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(ax+by\right)+\left(bx+az\right)+\left(ay+bz\right)}{\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}$          (Theorem of equal ratios)

$⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)x+\left(a+b\right)y+\left(a+b\right)z}{2x+2y+2z}\phantom{\rule{0ex}{0ex}}⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)\left(x+y+z\right)}{2\left(x+y+z\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{a+b}{2}$
(iv)

(By invertendo)

Now,

Also,

From (1), (2) and (3), we have

(v)

#### Question 4:

Solve

(i)

(ii)

(i)
$\frac{16{x}^{2}-20x+9}{8{x}^{2}+12x+21}=\frac{4x-5}{2x+3}$

If x = 0, then $\frac{16×0-20×0+9}{8×0+12×0+21}=\frac{4×0-5}{2×0+3}$ ⇒ $\frac{9}{21}=\frac{-5}{3}$, which is not true.

So, x = 0 is not a solution of the given equation.

Now,

$\therefore \frac{4x-5}{2x+3}=\frac{9}{21}\phantom{\rule{0ex}{0ex}}⇒\frac{4x-5}{2x+3}=\frac{3}{7}\phantom{\rule{0ex}{0ex}}⇒28x-35=6x+9\phantom{\rule{0ex}{0ex}}⇒28x-6x=35+9\phantom{\rule{0ex}{0ex}}⇒22x=44\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, the solution of the given equation is x = 2.

(ii)
$\frac{5{y}^{2}+40y-12}{5y+10{y}^{2}-4}=\frac{y+8}{1+2y}$

If y = 0, then $\frac{5×0+40×0-12}{5×0+10×0-4}=\frac{0+8}{1+2×0}$ ⇒ $\frac{-12}{-4}=\frac{8}{1}$, which is not true.

So, y = 0 is not a solution of the given equation.

Now,

$\therefore \frac{y+8}{1+2y}=\frac{-12}{-4}=3\phantom{\rule{0ex}{0ex}}⇒y+8=3+6y\phantom{\rule{0ex}{0ex}}⇒6y-y=8-3\phantom{\rule{0ex}{0ex}}⇒5y=5\phantom{\rule{0ex}{0ex}}⇒y=1$
Thus, the solution of the given equation is y = 1.

#### Question 1:

Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?

Le the number x be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.

∴ (12 − x), (16 − x), (21 − x) are in continued proportion.

$⇒\frac{12-x}{16-x}=\frac{16-x}{21-x}\phantom{\rule{0ex}{0ex}}⇒{\left(16-x\right)}^{2}=\left(12-x\right)×\left(21-x\right)\phantom{\rule{0ex}{0ex}}⇒256+{x}^{2}-32x=252-12x-21x+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒256-32x=252-33x\phantom{\rule{0ex}{0ex}}⇒33x-32x=252-256\phantom{\rule{0ex}{0ex}}⇒x=-4$
Thus, the number −4 must be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.

#### Question 2:

If (28$-x$) is the mean proportional of (23$-x$) and (19$-x$) then find the vaue of x.

It is given that (28 − x) is the mean proportional of (23 − x) and (19 − x).

$\therefore {\left(28-x\right)}^{2}=\left(23-x\right)×\left(19-x\right)\phantom{\rule{0ex}{0ex}}⇒784+{x}^{2}-56x=437-23x-19x+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒784-56x=437-42x\phantom{\rule{0ex}{0ex}}⇒56x-42x=784-437\phantom{\rule{0ex}{0ex}}⇒14x=347\phantom{\rule{0ex}{0ex}}⇒x=\frac{347}{14}$
Thus, the value of x is $\frac{347}{14}$.

#### Question 3:

Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.

Let the remaining two numbers be and y.

So, the numbers x, 12, y are in continued proportion.

​∴ xy = (12)2 = 144      .....(1)

Also,

xy = 26                    .....(2)

Solving (1) and (2), we get

$x+\frac{144}{x}=26\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+144=26x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-26x+144=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-18x-8x+144=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-18\right)-8\left(x-18\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-8\right)\left(x-18\right)=0$ ​​
⇒ x − 8 = 0 or x − 18 = 0

⇒ x = 8 or x = 18

When x = 8, y = 26 − 8 = 18            [Using (2)]

When  x = 18, y = 26 − 18 = 8

Thus, the numbers are 8, 12, 18 or 18, 12, 8.

#### Question 4:

If (a + b + c) (a $-$ b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion.

$\left(a+b+c\right)\left(a-b+c\right)={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(a+c\right)}^{2}-{b}^{2}={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{c}^{2}+2ac-{b}^{2}={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒2{b}^{2}=2ac\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$
Therefore, abc are in continued proportion.

#### Question 5:

If   and  $a,b,c$ > 0 then show that,

(i) (a + b + c) ($-$ c) = ab $-$ c2

(ii) (a2 + b2) (b2 + c2 ) = (ab + bc)2

(iii)

(i)

(ii)

(iii)

#### Question 6:

Find mean proportional of

If b is the mean proportional of a and c, then ${b}^{2}=ac$ or $b=\sqrt{ac}$.

Mean proportional of $\frac{x+y}{x-y}$ and $\frac{{x}^{2}-{y}^{2}}{{x}^{2}{y}^{2}}$

$=\sqrt{\frac{x+y}{x-y}×\frac{{x}^{2}-{y}^{2}}{{x}^{2}{y}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{x+y}{x-y}×\frac{\left(x-y\right)\left(x+y\right)}{{x}^{2}{y}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{\left(x+y\right)}^{2}}{{\left(xy\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{x+y}{xy}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{x+y}{xy}$

#### Question 1:

Select the appropriate alternative answer for the following questions.

(i) If 6 : 5 = y : 20 then what will be the value of y ?
(A) 15 (B) 24 (C) 18 (D) 22.5

(ii) What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100 (B) 10 : 1 (C) 1 : 10 (D) 100 : 1

(iii * ) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(A) 3 : 2 (B) 2 : 3 (C) 4 : 3 (D) 3 : 4

(iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get ?
(A) 8 (B) 15 (C) 12 (D) 9

(v) What is the mean proportional of 4 and 25 ?
(A) 6 (B) 8 (C) 10 (D) 12

(i)
6 : 5 = y : 20

$\therefore \frac{6}{5}=\frac{y}{20}\phantom{\rule{0ex}{0ex}}⇒y=\frac{6×20}{5}=24$
Thus, the value of y is 24.

Hence, the correct answer is option (B).

(ii)
Ratio of 1 mm to 1 cm

= 1 mm : 1 cm

= 1 mm : 10 mm            (1 cm = 10 mm)

= 1 : 10

Hence, the correct answer is option (C).

(iii)
Age of Nitin = 24 years

Age of Mohasin = 36 years

∴ Ratio of Nitin's age to Mohansin's age

= Age of Nitin : Age of Mohasin

= 24 years : 36 years

= 2 : 3

Thus, the ratio of Nitin's age to Mohasin's age is 2 : 3.

Hence, the correct answer is option (B).

(iv)
The number of bananas received by Shubham and Anil is the ratio 3 : 5.

Let the number of bananas received by Shubham and Anil be 3x and 5x, respectively.

∴ 3x + 5x = 24

⇒ 8x = 24

⇒ x = 3

∴ Number of bananas received by Shubham = 3x = 3 × 3 = 9

Thus, Shubham gets 9 bananas.

Hence, the correct answer is option (D).

(v)
Mean proportional of 4 and 25

$=\sqrt{4×25}\phantom{\rule{0ex}{0ex}}=\sqrt{100}\phantom{\rule{0ex}{0ex}}=10$
Thus, the mean proportional of 4 and 25 is 10.

Hence, the correct answer is option (C).

#### Question 2:

For the following numbers write the ratio of first number to second number in the reduced form.

(i) 21, 48 (ii) 36, 90 (iii) 65, 117 (iv) 138, 161 (v) 114, 133

(i)
21 : 48 = $\frac{21}{48}=\frac{21÷3}{48÷3}=\frac{7}{16}$ = 7 : 16                    (HCF of 21 and 48 = 3)

Thus, the reduced form of 21 : 48 is 7 : 16.

(ii)
36 : 90 = $\frac{36}{90}=\frac{36÷18}{90÷18}=\frac{2}{5}$ = 2 : 5                    (HCF of 36 and 90 = 18)

Thus, the reduced form of 36 : 90 is 2 : 5.

(iii)
65 : 117 = $\frac{65}{117}=\frac{65÷13}{113÷13}=\frac{5}{9}$ = 5 : 9                    (HCF of 65 and 113 = 13)

Thus, the reduced form of 65 : 117 is 5 : 9.

(iv)
138 : 161 = $\frac{138}{161}=\frac{138÷23}{161÷23}=\frac{6}{7}$ = 6 : 7                    (HCF of 138 and 161 = 23)

Thus, the reduced form of 138 : 161 is 6 : 7.

(v)
114 : 133 = $\frac{114}{133}=\frac{114÷19}{133÷19}=\frac{6}{7}$ = 6 : 7                    (HCF of 114 and 133 = 19)

Thus, the reduced form of 114 : 133 is 6 : 7.

#### Question 3:

Write the following ratios in the reduced form.

(i) Radius to the diameter of a circle.
(ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
(iii) The ratio of perimeter to area of a square, having side 4 cm.

(i)
Let the radius of the circle be r units.

Diameter of the circle = 2 × Radius of the circle = 2r units

∴ Ratio of radius to diameter of a circle = Radius of the circle : Diameter of the circle = r : 2r = 1 : 2

(ii)
Length of the rectangle, l = 4 cm

Breadth of the rectangle, b = 3 cm

Now,

Diagonal of the rectangle = $\sqrt{{l}^{2}+{b}^{2}}=\sqrt{{3}^{2}+{4}^{2}}=\sqrt{9+16}=\sqrt{25}$ = 5 cm

∴ Ratio of diagonal to the length of the rectangle = Diagonal of the rectangle : Length of the rectangle = 5 cm : 4 cm = 5 : 4

(iii)
Side of the squre = 4 cm

Perimeter of the square = 4 ×  Side of the squre = 4 × 4 cm = 16 cm

Area of the squre = (Side of the square)2 = (4 cm)2 = 16 cm2

∴ Ratio of perimeter to area of the square = Perimeter of the square : Area of the squre = 16 : 16 = 1 : 1

#### Question 4:

Check whether the following numbers are in continued proportion.

(i) 2, 4, 8   (ii) 1, 2, 3   (iii) 9, 12, 16  (iv) 3, 5, 8

The numbers abc are in continued proportion if $\frac{a}{b}=\frac{b}{c}$.

(i)
$\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{4}{8}=\frac{1}{2}$
Since $\frac{2}{4}=\frac{4}{8}$, so the numbers 2, 4, 8 are in continued proportion.

(ii)
Since $\frac{1}{2}\ne \frac{2}{3}$, so the numbers 1, 2, 3 are not in continued proportion.

(iii)
$\frac{9}{12}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{12}{16}=\frac{3}{4}$
Since $\frac{9}{12}=\frac{12}{16}$, so the numbers 9, 12, 16 are in continued proportion.

(iv)
Since $\frac{3}{5}\ne \frac{5}{8}$, so the numbers 3, 5, 8 are not in continued proportion.

#### Question 5:

a, b, c are in continued proportion. If a = 3 and c = 27 then find b.

It is given that 3, b, 27 are in continued proportion.

$\therefore \frac{3}{b}=\frac{b}{27}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=3×27\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=81\phantom{\rule{0ex}{0ex}}⇒b=\sqrt{81}=9$
Thus, the value of b is 9.

#### Question 6:

Convert the following ratios into percentages..

(i) 37 : 500    (ii)  $\frac{5}{8}$ (iii) $\frac{22}{30}$ (iv)  $\frac{5}{16}$ ( v) $\frac{144}{1200}$

(i)
37 : 500 = $\frac{37}{500}×100%=\frac{37}{5}%=7.4%$

(ii)
$\frac{5}{8}=\frac{5}{8}×100%=\frac{500}{8}%=62.5%$

(iii)
$\frac{22}{30}=\frac{22}{30}×100%=\frac{220}{3}%=73.33%$

(iv)
$\frac{5}{16}=\frac{5}{16}×100%=\frac{500}{16}%=31.25%$

(v)
$\frac{144}{1200}=\frac{144}{1200}×100%=\frac{144}{12}%=12%$

#### Question 7:

Write the ratio of first quantity to second quantity in the reduced form.

(i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)]        (ii) 17 Rupees, 25 Rupees 60 paise (iii) 5 dozen, 120 units
(iv) 4 sq.m, 800 sq.cm (v) 1.5 kg, 2500 gm

(i)
Ratio of 1024 MB to 1.2 GB

= 1024 MB : 1.2 × 1024 MB

$=\frac{1024}{1.2×1024}\phantom{\rule{0ex}{0ex}}=\frac{1}{1.2}\phantom{\rule{0ex}{0ex}}=\frac{10}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{6}$
= 5 : 6

(ii)
17 Rupees = 1700 paise            (Re 1 = 100 paise)

25 Rupees 60 paise = 25 Rupees + 60 paise = 2500 paise + 60 paise = 2560 paise

∴ Ratio of 17 Rupees to 25 Rupees 60 paise

= 1700 paise : 2560 paise

= 85 : 128

(iii)
Ratio of 5 dozen to 120 units

= 5 × 12 units : 120 units

= 60 units : 120 units

$=\frac{60}{120}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$
= 1 : 2

(iv)
1 m2 = 10000 cm2

4 m2 = 4 × 10000 cm2 = 40000 cm2

∴ Ratio of 4 m2 to 800 cm2

= 40000 cm: 800 cm2

$=\frac{40000}{800}\phantom{\rule{0ex}{0ex}}=50$
= 50 : 1

(v)
1.5 kg = 1.5 × 1000 g = 1500 g

∴ Ratio of 1.5 kg to 2500 g

= 1500 g : 2500 g

$=\frac{1500}{2500}\phantom{\rule{0ex}{0ex}}=\frac{3}{5}$
= 3 : 5

#### Question 8:

If   then find the values of the following expressions.

(i)    (ii)      (iii)      (iv)

(i)
$\frac{a}{b}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4a}{3b}=\frac{4×2}{3×3}=\frac{8}{9}$
Applying componendo, we get

$\frac{4a+3b}{3b}=\frac{8+9}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{4a+3b}{3b}=\frac{17}{9}$
(ii)

Applying componendo and dividendo, we get

$\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=\frac{20+18}{20-18}\phantom{\rule{0ex}{0ex}}⇒\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=\frac{38}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=19$
(iii)

Applying componendo, we get

$\frac{{a}^{3}+{b}^{3}}{{b}^{3}}=\frac{8+27}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{3}+{b}^{3}}{{b}^{3}}=\frac{35}{27}$
(iv)
$\frac{a}{b}=\frac{2}{3}$

Applying invertendo, we get

Applying componendo and dividendo, we get

$\frac{7b+4a}{7b-4a}=\frac{21+8}{21-8}\phantom{\rule{0ex}{0ex}}⇒\frac{7b+4a}{7b-4a}=\frac{29}{13}$
Applying invertendo, we get

$\frac{7b-4a}{7b+4a}=\frac{13}{29}$

#### Question 9:

If a, b, c, d   are in proportion , then prove that

(i)

(ii)

(iii)

It is given that abcd are in proportion.

(i)

From (1) and (2), we get

$\frac{11{a}^{2}+9ac}{11{b}^{2}+9bd}=\frac{{a}^{2}+3ac}{{b}^{2}+3bd}$

(ii)

From (1) and (2), we get

$\sqrt{\frac{{a}^{2}+5{c}^{2}}{{b}^{2}+5{d}^{2}}}=\frac{a}{b}$

(iii)

From (1) and (2), we get

$\frac{{a}^{2}+ab+{b}^{2}}{{a}^{2}-ab+{b}^{2}}=\frac{{c}^{2}+cd+{d}^{2}}{{c}^{2}-cd+{d}^{2}}$

#### Question 10:

If   a, b, c  are in continued proportion , then prove that

(i)

(ii)

abc are in continued proportion.

(i)

From (1) and (2), we get

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

(ii)

From (1) and (2), we get

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

#### Question 11:

Solve :

If x = 0, then $\frac{12×0+18×0+42}{18×0+12×0+58}=\frac{2×0+3}{3×0+2}⇒\frac{42}{58}=\frac{3}{2}$, which is not true.

So, x = 0 is not a solution of the given equation.

Now,

$\therefore \frac{2x+3}{3x+2}=\frac{42}{58}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+3}{3x+2}=\frac{21}{29}\phantom{\rule{0ex}{0ex}}⇒58x+87=63x+42\phantom{\rule{0ex}{0ex}}⇒63x-58x=87-42\phantom{\rule{0ex}{0ex}}⇒5x=45\phantom{\rule{0ex}{0ex}}⇒x=9$
Thus, the solution of the given equation is x = 9.

#### Question 12:

If   then prove that every ratio = $\frac{x}{y}$.

$⇒\frac{2x-3y}{3z+y}=\frac{3\left(z-y\right)}{3\left(z-x\right)}=\frac{x+3z}{2y-3x}=\frac{2x-3y-3z+3y+x+3z}{3z+y-3z+3x+2y-3x}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-3y}{3z+y}=\frac{3\left(z-y\right)}{3\left(z-x\right)}=\frac{x+3z}{2y-3x}=\frac{3x}{3y}\phantom{\rule{0ex}{0ex}}\therefore \frac{2x-3y}{3z+y}=\frac{z-y}{z-x}=\frac{x+3z}{2y-3x}=\frac{x}{y}$

#### Question 13:

(13 * ) If     then prove that  .

$⇒\frac{-ax}{-{a}^{2}}=\frac{-by}{-{b}^{2}}=\frac{-cz}{-{c}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$