The ratio of 72 to 60 = 72 : 60 = \(\frac{72}{60}\) = \(\frac{72÷12}{60÷12}\) = \(\frac{6}{5}\)   ... (HCF of 72 and 60 = 12)

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

The ratio of 38 to 57 = 38 : 57 = \(\frac{38}{57}\) = \(\frac{38÷19}{57÷19}\) = \(\frac{2}{3}\)   ... (HCF of 38 and 57 = 19)

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

The ratio of 52 to 78 = 52 : 78 = \(\frac{52}{78}\) = \(\frac{52÷26}{78÷26}\) = \(\frac{2}{3}\)    ... (HCF of 52 and 78 = 26)

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

700 ₹ : 308 ₹ = \(\frac{700}{308}\) = \(\frac{28×25}{28×11}\) = \(\frac{25}{11}\)

Thus, the reduced form of 700 ₹ : 308 ₹  is 25 : 11.

₹ 14 = 14 x 100 = 1400 paise

₹ 12 p 40 = 12 x 100 + 40 = 1200 + 40 = 1240 paise

₹ 14 : ₹ 12 p 40

= 1400 p: 1240 p = \(\frac{1400}{1240}\) = \(\frac{4×35}{4×31}\) = \(\frac{35}{31}\)

Thus, the reduced form of 14 ₹ : 12 ₹. 40 paise is 35 : 31.

5 litres = 5 x 1000 = 5000 ml

5 litres : 2500 ml = 5000 ml : 2500 ml

= \(\frac{5000}{2500}\) = \(\frac{2500×2}{2500×1}\) = \(\frac{2}{1}\)

Thus, the reduced form of 5 litres : 2500 ml = 2 : 1

3 years 4 months = 3 x 12 + 4 = 40 months

5 years 8 months = 5 x 12 + 8 = 68 months

3 years 4 months : 5 years 8 months

= 40 months : 68 months = \(\frac{40}{68}\) = \(\frac{4×10}{4×17}\) = \(\frac{10}{17}\)

Thus, the reduced form of 3 years 4 months : 5 years 8 months = 10 : 17

3.8 kg = 3.8 x 1000 = 3800 g.

3.8 kg : 1900 g = 3800 g : 1900 g = \(\frac{3800}{1900}\) = \(\frac{1900×2}{1900×1}\) = \(\frac{2}{1}\)

Thus, the reduced form of 3.8 kg : 1900 g = 2 : 1

7 min 20 sec = 7 x 60 + 20 = 440 sec

5 min 6 sec = 5 x 60 + 6 = 306 sec

7 min 20 sec : 5 min 6 sec = 440 sec : 306 sec

= \(\frac{440}{306}\) = \(\frac{2×220}{2×153}\) = \(\frac{220}{153}\)

Thus, the reduced form of 7 min 20 sec : 5 min 6 sec = 220 : 153

75 : 100 = \(\frac{75}{100}\) = \(\frac{25×3}{25×4}\) = \(\frac{3}{4}\)

Answer is : 3 : 4

44 : 100 = \(\frac{44}{100}\) = \(\frac{4×11}{4×25}\) = \(\frac{11}{25}\)

Answer is : 11 : 25

6.25% = \(\frac{6.25}{100}\) = \(\frac{625}{100×100}\) = \(\frac{25×25}{25×4×25×4}\) = \(\frac{1}{16}\)

Answer is : 1 : 16

52 : 100 = \(\frac{52}{100}\) = \(\frac{4×13}{4×25}\) = \(\frac{13}{25}\)

Answer is : 13 : 25

0.64% = \(\frac{0.64}{100}\) = \(\frac{64}{100×100}\) = \(\frac{4×4×4}{25×4×25×4}\) = \(\frac{4}{625}\)

Answer is : 4 : 625

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.

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

∴ x × 6 = 3 x 8

∴ x = \(\frac{3×8}{6}\) = \(\frac{24}{6}\) = 4

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

15 : 25 = \(\frac{15}{25}\) x 100 %  = 15 x 4 % = 60%

Answer is : 60%

47 : 50 = \(\frac{47}{50}\) x 100 %  = 47 x 2 % = 94%

Answer is : 94%

\(\frac{7}{10}\) = \(\frac{7}{10}\) x 100 % = 7 x 10 % = 70%

Answer is : 70%

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

Answer is : 91%

\(\frac{7}{16}\) = \(\frac{7}{16}\) x 100 % = \(\frac{700}{16}\) %= 43.75 %

Answer is : 43.75 %

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

Let the common multiple of the given ratio be x.

Then the present age of Abha is 2x years and that of her mother is 5x years.

The mother's age at the birth of Abha, i.e. 2x years ago was 27 years.

5x - 2x = 27

∴ 3x = 27

x =  \(\frac{27}{3}\)   ∴ x = 9

2x = 2 × 9 = 18 and 5x = 5 × 9 = 45

Answer is : The present ages of Abha and her mother are 18 years and 45 years respectively.

Let the ratio of the ages of Vatsala and Sara will become 5 : 4 after x years.

Then after x years, Vatsala's age will be (14 + x) years and that of Sara will be (10 + x) years.

From the given condition, (14 + x) : (10 + x) = 5 : 4

∴ \(\frac{14 + x}{10 + x}=\frac{5}{4}\)

∴ 4(14 + x) = 5(10 + x)

∴ 56 + 4x = 50 + 5x

∴ 50 + 5x = 56 + 4x

∴ 5x - 4x = 56 - 50

∴ x = 6

Answer is : After 6 years.

The ratio of the ages is 2 : 7

Let the common multiple of the given ratio be x.

Then Rehana's present age is 2x years and that of her mother is 7x years.

From the given condition, (2x + 2) : (7x + 2) = 1 : 3.

∴ \(\frac{2x + 2}{7x + 2}=\frac{1}{3}\)

∴ 3(2x + 2) = 1(7x + 2)

∴ 6x + 6 = 7x + 2

∴ 7x + 2 = 6x + 6

∴ 7x - 6x = 6 - 2

∴ x = 4 and 2x = 4 x 2 = 8

Answer is : The present age of Rehana is 8 years.

\(\frac{5}{7}=\frac{....}{28}=\frac{5×4}{7×4}=\frac{20}{28}\)

\(\frac{5}{7}=\frac{35}{....}=\frac{5×7}{7×7}=\frac{35}{49}\)

\(\frac{5}{7}=\frac{....}{3.5}=\frac{5×\frac{1}{2}}{7×\frac{1}{2}}=\frac{2.5}{3.5}\)

Answer is : \(\frac{5}{7}=\frac{20}{28}=\frac{35}{49}=\frac{2.5}{3.5}\)

\(\frac{9}{14}=\frac{4.5}{...}=\frac{9×\frac{1}{2}}{14×\frac{1}{2}}=\frac{4.5}{7}\)

\(\frac{9}{14}=\frac{....}{42}=\frac{9×3}{14×3}=\frac{27}{42}\)

\(\frac{9}{14}=\frac{....}{3.5}=\frac{9×\frac{1}{4}}{14×\frac{1}{4}}=\frac{2.25}{3.5}\)

Answer is :\(\frac{9}{14}=\frac{4.5}{7}=\frac{27}{42}=\frac{2.5}{3.5}\)

Let the radius of the circle be r units

∴ Circumference of the circle = 2πr units

Radius of the circle : Circumference of the circle = r : 2πr = \(\frac{r}{2πr}=\frac{1}{2π}\) = 1 : 2π

Answer is : The ratio of radius to circumference of the circle is 1 : 2π

Radius of the circle = r units

∴ Circumference of the circle = 2πr units

Area of the circle = πr² square units

Circumference of the circle : Area of the circle = 2πr : πr²= \(\frac{2πr}{πr^2}=\frac{2}{r}\) = 2 : r

Answer is : The ratio of circumference of circle with radius r to its area is 2 : r.

Side of the square = 7 cm

∴ Length of diagonal of the square = \(\sqrt{2}\) x Side of the square = 7\(\sqrt{2}\)  cm

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

Answer is : The ratio of diagonal of the square to its side is \(\sqrt{2}\)  : 1.

Length of the rectangle, l = 5 cm

Breadth of the rectangle, b = 3.5 cm

∴ Perimeter of the rectangle = 2(l + b) = 2 x (5 + 3.5) = 2 x 8.5 = 17 cm

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

Perimeter of the rectangle : Area of the rectangle = 17 : 17.5

= \(\frac{17}{17.5}\) = \(\frac{170}{175}\) = \(\frac{34×5}{35×5}\) = \(\frac{34}{35}\) = 34 : 35

Answer is : The ratio of perimeter to area of the rectangle is 34 : 35.

\(\sqrt{5}×\sqrt{7}\) = \(\sqrt{35}\)

3 x 3 = 9 = \(\sqrt{81}\)

\(\sqrt{35}\) < \(\sqrt{81}\)

∴ \(\frac{\sqrt{5}}{3}<\frac{3}{\sqrt{7}}\)

\(3\sqrt{5}×\sqrt{125}\) = \(3\sqrt{5}×\sqrt{5×25}\) = \(3\sqrt{5}×5\sqrt{5}\) = 3 x 5 x 5 = 75

\(5\sqrt{7}×\sqrt{63}\) = \(5\sqrt{7}×\sqrt{7×9}\) = \(5\sqrt{7}×3\sqrt{7}\) = 5 x 3 x 7  = 105

75 < 105

∴ \(\frac{3\sqrt{5}}{5\sqrt{7}}<\frac{\sqrt{63}}{\sqrt{125}}\)

5 × 121 = 605

17 x 18 = 306

605 > 306

∴ \(\frac{5}{18}>\frac{17}{121}\)

\(\frac{\sqrt{80}}{\sqrt{48}}\) = \(\frac{\sqrt{16×5}}{\sqrt{16×3}}\)

= \(\frac{4\sqrt{5}}{4\sqrt{3}}\)

= \(\frac{\sqrt{5}}{\sqrt{3}}\)

\(\frac{\sqrt{45}}{\sqrt{27}}\) = \(\frac{\sqrt{9×5}}{\sqrt{9×3}}\)

= \(\frac{3\sqrt{5}}{3\sqrt{3}}\)

= \(\frac{\sqrt{5}}{\sqrt{3}}\)

∴ \(\frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}\)

9.2 × 7.1 = 65.32

5.1 x 3.4 = 17.34

Now, 65.32 > 17.34

∴ \(\frac{9.2}{5.1}>\frac{3.4}{7.1}\)

The ratio of ∠ A to ∠ B is 5 : 4.

Let the common multiple of this ratio be x.

Then ∠ A = 5x° and ∠ B = 4x°

∠ A + ∠ B = 180°         ... (Interior angles)

∴ 5x° + 4x° = 180°

∴ 5x + 4x = 180

∴ 9x = 180  ∴ x = 20

∠ B = 4x° = 4 × 20° = 80°

Answer is : The measure of ∠ B is 80°.

The ratio of the present ages of Albert and Salim is 5 : 9.

Let the common multiple of this ratio be x.

Then 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.

∴ \(\frac{5x + 5}{9x + 5}=\frac{3}{5}\)

5(5x + 5) = 3(9x + 5)

25x + 25 = 27x + 15

27x - 25x = 25 - 15

2x = 10

x = 5

∴ Present age of Albert = 5x = 5 x 5 = 25 years

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

Answer is : The present age of Albert is 25 years and the present age of Salim is 45 years.

Let the common multiple of the ratio 3 : 1 be x.

Then the length and the breadth of the rectangle are 3x cm and x cm respectively.

The perimeter of a rectangle = 2(l + b)

∴ 36 = 2(3x + x)                  ... (Given : the perimeter 36 cm)

∴ 36 = 2 x 4x

∴ x = \(\frac{9}{2}\)

3x = 3 x \(\frac{9}{2}\) = \(\frac{27}{2}\) = 13.5 cm;

x = \(\frac{9}{2}\) = 4.5 cm;

Answer is : The length of the rectangle is 13.5 cm and breadth is 4.5 cm.

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 x 4 = 124, Other number = 23x = 23 x 4 = 92

Answer is : The two numbers are 92 and 124.

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 x 2 = 20, Other number = 9x = 9 x 2 = 18

Answer is : The two numbers are 18 and 20.

a : b = 3 : 1. ∴ \(\frac{a}{b}=\frac{3}{1}\)

∴ a = 3b        ….(1)

b : c = 5 : 1.  ∴ \(\frac{b}{c}=\frac{5}{1}\)

∴ b = 5c        ….(2)

From (1) and (2),

a = 3b = 3 x 5c = 15c .. a = 15c   ….(3)

(i) \([\frac{a^3}{15b^2c}]^3\) = \([\frac{(15c)^3}{15(5c)^2c}]^3\) …[From (2) and (3)]

= \([\frac{15×15×15×c^3}{15×5×5×c^3}]^3\)

= \([\frac{15×15}{5×5}]^3\)

= (3 × 3)³ = (9)³ = 729

Answer is : The required value is 729.

(ii) \(\frac{a^2}{7bc}\) = \(\frac{(15c)^2}{7(5c)c}\)   …[From (2) and (3)]

= \(\frac{15×15×c^2}{7×5×c^2}\)

= \(\frac{45}{7}\)

Answer is : The required value is \(\frac{45}{7}\).

\(\sqrt{0.04×0.4×a}\) = 0.4 × 0.04 × \(\sqrt{b}\)

\(\frac{\sqrt{a}}{\sqrt{b}}\) = \(\frac{0.4 × 0.04}{\sqrt{0.4 × 0.04}}\)

∴ \(\frac{a}{b}\) = \(\frac{(0.4 × 0.04)^2}{(0.4 × 0.04)}\)        …(squaring both sides)

= 0.4 × 0.04 = \(\frac{4}{10}×\frac{4}{100}\) = \(\frac{2}{5}×\frac{1}{25}\) = \(\frac{2}{125}\)

∴ \(\frac{a}{b}\) = \(\frac{2}{125}\)

Answer is : a : b = 2 : 125

(x + 3) : (x + 11) = (x - 2) : (x + 1)

∴ \(\frac{x + 3}{x + 11}\) = \(\frac{x - 2}{x + 1}\)

(x + 3)(x + 1) = (x - 2)(x + 11)

∴ x(x + 1) + 3(x + 1) = x(x + 11) - 2(x + 11)

∴ x² + x + 3x + 3 = x² + 11x - 2x -22

∴ x² + 4x + 3 = x² + 9x - 22

∴ 4x - 9x = - 22 - 3

∴ -5x = - 25

∴ x = -25/-5 = 5

Answer is : The value of x is 5.

\(\frac{a}{b}=\frac{7}{3}\) ….(Given)

Let the common multiple be k.

∴ a = 7k, b = 3k

∴ \(\frac{5a + 3b}{5a - 3b}\) = \(\frac{5(7k) + 3(3k)}{5(7k) - 3(3k)}\)

= \(\frac{35k + 9k}{35k - 9k}\)

= \(\frac{44k}{26k}\)

= \(\frac{22}{13}\)

Answer is : \(\frac{5a + 3b}{5a - 3b}\)  = 22 : 13

\(\frac{a}{b}=\frac{7}{3}\) ….(Given)

Let the common multiple be k.

∴ a = 7k, b = 3k

∴ \(\frac{2a^2 + 3b^2}{2a^2 - 3b^2}\) = \(\frac{2(7k)^2 + 3(3k)^2}{2(7k)^2 - 3(3k)^2}\)

= \(\frac{2×49k^2 + 3×9k^2}{2×49k^2 - 3×9k^2}\)

= \(\frac{98k^2 + 27k^2}{98k^2 - 27k^2}\)

= \(\frac{125k^2 }{71k^2 }\)

= \(\frac{125}{71}\)

Answer is : \(\frac{2a^2 + 3b^2}{2a^2 - 3b^2}\) = 125 : 71

\(\frac{a}{b}=\frac{7}{3}\) ….(Given)

Let the common multiple be k.

∴ a = 7k, b = 3k

∴ \(\frac{a^3 - b^3}{b^3}\) = \(\frac{(7k)^3 - (3k)^3}{(3k)^3}\)

= \(\frac{343k^3 - 27k^3}{27k^3}\)

= \(\frac{316k^3}{27k^3}\)

= \(\frac{316}{27}\)

Answer is : \(\frac{a^3 - b^3}{b^3}\) = 316 : 27

\(\frac{a}{b}=\frac{7}{3}\) ….(Given)

Let the common multiple be k.

∴ a = 7k, b = 3k

∴ \(\frac{7a + 9b}{7a - 9b}\) = \(\frac{7(7k) + 9(3k)}{7(7k) - 9(3k)}\)

= \(\frac{49k + 27k}{49k - 27k}\)

= \(\frac{76k}{22k}\)

= \(\frac{38}{11}\)

Answer is : \(\frac{7a + 9b}{7a - 9b}\) = 38 : 11

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

Applying componendo and dividendo, we get

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

\(\frac{30a^2}{8b^2}=\frac{54}{40}\)

\(\frac{a^2}{b^2}=\frac{54×8}{40×30}=\frac{54}{150}=\frac{9}{25}\)

∴ \(\frac{a}{b}=\frac{3}{5}\)       ... (Taking positive square root)

Answer is : \(\frac{a}{b}\) = 3 : 5

\(\frac{a}{b}=\frac{3}{5}\)      … [from Q.3.2(i)]

Now dividing each term of \(\frac{7a - 3b}{7a + 3b}\)  by b.

∴ \(\frac{7a - 3b}{7a + 3b}\) = \(\frac{7\frac{a}{b} - 3\frac{b}{b}}{7\frac{a}{b} + 3\frac{b}{b}}\)    …

= \(\frac{7\frac{3}{5} - 3}{7\frac{3}{5} + 3}\)    … \((\frac{a}{b}=\frac{3}{5})\)

= \(\frac{\frac{21-15}{5}}{\frac{21+15}{5}}\)

= \(\frac{6}{36}\) = \(\frac{1}{6}\)

Answer is : \(\frac{7a - 3b}{7a + 3b}\) = 1 : 6

\(\frac{a}{b}=\frac{3}{5}\)      … [from Q.3.2(i)]

∴ \(\frac{a^2}{b^2}=\frac{9}{25}\)

Now dividing each term of \(\frac{b^2 - 2a^2}{b^2 + 2a^2}\)  by b².

∴  \(\frac{b^2 - 2a^2}{b^2 + 2a^2}\) = \(\frac{(b^2)(b^2) - (2a^2)/(b^2)}{(b^2)/(b^2) + (2a^2)/(b^2)}\)

= \(\frac{1 - 2(\frac{a^2}{b^2})}{1 + 2(\frac{a^2}{b^2})}\)

= \(\frac{1 - 2(\frac{9}{25})}{1 + 2(\frac{9}{25})}\)

= \(\frac{1 - \frac{18}{25}}{1 + \frac{18}{25}}\)

= \(\frac{\frac{25-18}{25}}{\frac{25+18}{25}}\)

= \(\frac{7}{43}\)

Answer is : \(\frac{b^2 - 2a^2}{b^2 + 2a^2}\) = 7 : 43

\(\frac{a}{b}=\frac{3}{5}\)      … [from Q.3.2(i)]

∴ \(\frac{a^3}{b^3}=\frac{27}{125}\)

Now dividing each term of \(\frac{b^3 - 2a^3}{b^3 + 2a^3}\)  by b³.

∴  \(\frac{b^3 - 2a^3}{b^3 + 2a^3}\) = \(\frac{(b^3)(b^3) - (2a^3)/(b^3)}{(b^3)/(b^3) + (2a^3)/(b^3)}\)

= \(\frac{1 - 2(\frac{a^3}{b^3})}{1 + 2(\frac{a^3}{b^3})}\)

= \(\frac{1 - 2(\frac{27}{125})}{1 + 2(\frac{27}{125})}\)

= \(\frac{1 - \frac{54}{125}}{1 + \frac{54}{125}}\)

= \(\frac{\frac{125-54}{125}}{\frac{125+54}{125}}\)

= \(\frac{71}{179}\)

Answer is : \(\frac{b^3 - 2a^3}{b^3 + 2a^3}\)  = 71 : 179

\(\frac{3a + 7b}{3a - 7b}=\frac{4}{3}\)               ….(Given)

Applying componendo and dividendo, we get

∴ \(\frac{(3a + 7b)+(3a - 7b)}{(3a + 7b)-(3a - 7b)}=\frac{4+3}{4-3}\)

∴ \(\frac{6a}{14b}=\frac{7}{1}\)

∴ \(\frac{a}{b}=\frac{7×14}{1×6}=\frac{49}{3}\)

∴ \(\frac{a^2}{b^2}=\frac{49^2}{3^2}=\frac{2401}{9}\)

∴ \(\frac{a^2}{b^2}×\frac{3}{7}=\frac{49^2}{3^2}=\frac{2401}{9}×\frac{3}{7}\)                  ...(Multiplying both the sides by \(\frac{3}{7}\))

∴ \(\frac{3a^2}{7b^2}=\frac{343}{3}\)

∴ \(\frac{3a^2 + 7b^2}{3a^ - 7b^2}=\frac{343+3}{343-3} \)           ….... (By componendo-dividendo)

∴ \(\frac{3a^2 + 7b^2}{3a^ - 7b^2}=\frac{346}{340}=\frac{173}{170}\)

∴ \(\frac{3a^2 - 7b^2}{3a^ + 7b^2}=\frac{346}{340}=\frac{170}{173}\)     ….... (By invertendo)

Answer is : The value of the ratio \(\frac{3a^2 - 7b^2}{3a^ + 7b^2}\) is 170 : 173

Let us first decide whether x = 0 or x ≠ 0.

If x = 0, LHS = \(\frac{-20}{-5}\) = 4 and RHS = \(\frac{12}{3}\) = 4.

∴ LHS = RHS

∴ x = 0 is a solution of the given equation.

If x ≠ 0, then

\(\frac{x^2+12x-20}{3x - 5} × \frac{1}{4}=\frac{x^2+8x+12}{2x+3}× \frac{1}{4}\) …(Multiplying both the sides by \(\frac{1}{4}\))

∴ \(\frac{x^2+12x-20}{12x - 20} =\frac{x^2+8x+12}{8x+12}\)

∴ \(\frac{x^2+12x-20-(12x-20)}{12x - 20} =\frac{x^2+8x+12-(8x+12)}{8x+12}\)    ….... (By dividendo)

∴ \(\frac{x^2}{12x - 20} =\frac{x^2}{8x+12}\)

∴ (12x - 20) = (8x + 12)     ... (Dividing both the sides by x²)

∴ 12x - 8x = 12 + 20

∴ 4x = 32

∴ x = 8

Answer is : x = 0 or x = 8 is the solution of the given equation.

Let us first decide whether x = 0 or x ≠ 0.

If x = 0, LHS = \(\frac{63}{12}\)  = \(\frac{21}{4}\) and RHS = \(\frac{3}{-5}\)

∴ LHS ≠ RHS ∴ x ≠ 0

\(\frac{10x^2+15x+63}{5x^2-25x+12}=\frac{2x+3}{x-5}\)

\(\frac{5x(2x+3)+63}{5x(x-5)+12}=\frac{2x+3}{x-5}\)

Substituting a for 2x + 3 and b for x -5,

∴ \(\frac{5xa+63}{5xb+12}=\frac{a}{b}\)

∴ b(5xa + 63) = a(5xb + 12)

5xab + 63b = 5axb + 12a

∴  63b = 12a

∴  21b = 4a

Substituting the values of a and b,

21(x - 5) = 4(2x + 3)

∴  21x – 105 = 8x + 12

∴  21x – 8x = 12 + 105

∴ 13x = 117

∴ x = \(\frac{117}{13}\)  ∴ x = 9

Answer is : x = 9 is the solution of the given equation.

\(\frac{(2x+1)^2+(2x-1)^2}{(2x+1)^2-(2x-1)^2}=\frac{17}{8}\)

Applying componendo and dividendo, we get

\(\frac{(2x+1)^2+(2x-1)^2+[(2x+1)^2-(2x-1)^2]}{(2x+1)^2+(2x-1)^2-[(2x+1)^2-(2x-1)^2]}=\frac{17+8}{17-8}\)

∴ \(\frac{(2x+1)^2+(2x-1)^2+(2x+1)^2-(2x-1)^2}{(2x+1)^2+(2x-1)^2-(2x+1)^2+(2x-1)^2}=\frac{25}{9}\)

∴ \(\frac{2(2x+1)^2}{2(2x-1)^2}=\frac{25}{9}\)

∴ \(\frac{(2x+1)^2}{(2x-1)^2}=\frac{25}{9}\)

∴ \(\frac{2x+1}{2x-1}=\frac{5}{3}\)      …..(Taking square roots of both sides)

∴ 6x + 3 = 10x - 5

∴ 10x – 6x = 5 + 3

∴ 4x = 8

∴ x = 2

Answer is : x = 2 is the solution of the given equation.

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

Applying componendo and dividendo, we get

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

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

∴ \(\frac{2\sqrt{4x+1}}{2\sqrt{x+3}} = \frac{5}{3}\)

∴ \(\frac{\sqrt{4x+1}}{\sqrt{x+3}} = \frac{5}{3}\)

∴ \(\frac{4x+1}{x+3} = \frac{25}{9}\)         …..(Squaring both sides)

∴ 36x + 9 = 25x + 75

∴ 36x - 25x =  75 - 9

∴ 11x = 66

∴ x = 6

Answer is : x = 6 is the solution of the given equation

\(\frac{(4x+1)^2+(2x+3)^2}{4x^2+12x+9}=\frac{61}{36}\)

\(\frac{(4x+1)^2+(2x+3)^2}{(2x+3)^2}=\frac{61}{36}\)

Applying dividendo, we get

\(\frac{(4x+1)^2+(2x+3)^2-(2x+3)^2}{(2x+3)^2}=\frac{61-36}{36}\)

∴ \(\frac{(4x+1)^2}{(2x+3)^2}=\frac{25}{36}\)

∴ \(\frac{(4x+1)}{(2x+3)}=\frac{5}{6}\)           …..(Taking square roots of both sides)

∴ 24x + 6 = 10x + 15

∴ 24x - 10x  =  15 - 6

∴ 14x  =  9

∴ x = \(\frac{9}{14}\)

Answer is : x = \(\frac{9}{14}\) is the solution of the given equation

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

Applying componendo and dividendo, we get

\(\frac{(3x-4)^3-(x+1)^3+[(3x-4)^3+(x+1)^3]}{(3x-4)^3-(x+1)^3-[(3x-4)^3+(x+1)^3]}=\frac{61+189}{61-189}\)

∴ \(\frac{(3x-4)^3-(x+1)^3+(3x-4)^3+(x+1)^3}{(3x-4)^3-(x+1)^3-(3x-4)^3-(x+1)^3}=\frac{61+189}{61-189}\)

∴ \(\frac{2(3x-4)^3}{-2(x+1)^3}=\frac{250}{-128}\)

∴ \(\frac{(3x-4)^3}{(x+1)^3}=\frac{125}{64}\)

∴ \(\frac{(3x-4)}{(x+1)}=\frac{5}{4}\)            …..(Taking cube roots of both sides)

∴ 12x - 16 = 5x + 5

∴ 12x - 5x  =  5 + 16

∴ 7x  = 21

∴ x  = 3

Answer is : x = 3 is the solution of the given equation

\(\frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{....}=\frac{7x-9y}{....}\)

By the theorem on equal ratios;

\(\frac{3x+5y}{3(7)+5(3)}=\frac{7x-9y}{7(7)-9(3)}\)

\(\frac{3x+5y}{21+15}=\frac{7x-9y}{49-27}\)

\(\frac{3x+5y}{36}=\frac{7x-9y}{22}\)

\(\frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{36}=\frac{7x-9y}{22}\)

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2b+3c}{....}=\frac{....}{6-8+14}\)

By the theorem on equal ratios;

\(\frac{a-2b+3c}{3-2(4)+3(7}=\frac{2a-2b+2c}{6-8+14}\)

\(\frac{a-2b+3c}{3-8+21}=\frac{2a-2b+2c}{6-8+14}\)

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2b+3c}{16}=\frac{2a-2b+2c}{6-8+14}\)

5m – n = 3m + 4n

∴ 5m - 3m = 4n + n

∴ 2m = 5n

∴ \(\frac{m}{n}=\frac{5}{2}\)

∴ \(\frac{m^2}{n^2}=\frac{25}{4}\)              …..(Squaring both sides)

Applying componendo and dividendo, we get

\(\frac{m^2+n^2}{m^2-n^2}=\frac{25+4}{25-4}\)

\(\frac{m^2+n^2}{m^2-n^2}=\frac{29}{21}\)

Answer is : The value of the given expression is \(\frac{29}{21}\)

5m – n = 3m + 4n

∴ 5m - 3m = 4n + n

∴ 2m = 5n

∴ \(\frac{m}{n}=\frac{5}{2}\)

∴ \(\frac{m}{n}×\frac{3}{4}=\frac{5}{2}×\frac{3}{4}\)             …(Multiplying both the sides by \(\frac{3}{4} )

∴ \(\frac{3m}{4n}=\frac{15}{8}\)

Applying componendo and dividendo, we get

\(\frac{3m+4n}{3m-4n}=\frac{15+8}{15-8}\)

\(\frac{3m+4n}{3m-4n}=\frac{23}{7}\)

Answer is : The value of the given expression is \(\frac{23}{7}\)

No two of a, b, c are equal

∴ the value of each of (a-b), (b- c), (c - a) is not zero.

a(y + z) = b(z + x) = c(x + y)

∴ \(\frac{y + z}{bc} = \frac{z + x}{ca} = \frac{x + y}{ab}\)     ... (Dividing by abc)

By the theorem on equal ratios, each ratio

\(\frac{(x + y)-(z+x)}{ab-ca}=\frac{(y + z)-(x+y)}{bc-ab} = \frac{(z + x)-(y+z)}{ca-bc}\)

Also each ratio = \(\frac{(x + y)-(z+x)}{ab-ca}=\frac{x + y - z - x)}{a(b-c)}=\frac{y - z)}{a(b-c)}\)      …(i)

each ratio = \(\frac{(y + z)-(x+y)}{bc-ab} = \frac{y + z - x - y}{b(c-a)}=\frac{z - x}{b(c-a)}\)    …(ii)

each ratio = \(\frac{(z + x)-(y+z)}{ca-bc}=\frac{z + x - y -z}{c(a-b)}=\frac{x - y}{c(a-b)}\)       …(iii)

From (i), (ii) and (iii),

\(\frac{y - z)}{a(b-c)}\) = \(\frac{z - x}{b(c-a)}\) = \(\frac{x - y}{c(a-b)}\)

\(\frac{x}{3x-y-z} = \frac{y}{3y-z-x} = \frac{z}{3z-y-x}\)

By the theorem on equal ratios,

each ratio = \(\frac{x+y+z}{(3x-y-z)+(3y-z-x)+(3z-y-x)}\)

= \(\frac{x+y+z}{3x-y-z+3y-z-x+3z-y-x}\)

= \(\frac{x+y+z}{x+y+z}\) = 1     …( x + y + z ≠ 0)

\(\frac{ax+by}{x+y} = \frac{bx+az}{x+z} = \frac{ay+bz}{y+z}\)

By the theorem on equal ratios,

each ratio = \(\frac{(ax+by)+(bx+az)+(ay+bz)}{(x+y)+(x+z)+(y+z)}\)

= \(\frac{ax+by+bx+az+ay+bz}{x+y+x+z+y+z}\)

= \(\frac{x(a+b)+y(a+b)+z(a+b)}{2(x+y+z)}\)

= \(\frac{(a+b)(x+y+z)}{2(x+y+z)}\)

= \(\frac{a+b}{2}\)

\(\frac{y+z}{a} = \frac{z+x}{b} = \frac{x+y}{c}\)

By the theorem on equal ratios,

each ratio = \(\frac{z+x+x+y-y-z}{b+c-a}\)

= \(\frac{2}{b+c-a}\)…(i)

each ratio = \(\frac{x+y+y+z-z-x}{c+a-b}\)

= \(\frac{2y}{c+a-b}\)…(ii)

each ratio = \(\frac{y+z+z+x-x-y}{a+b-c}\)

= \(\frac{2z}{a+b-c}\)     …(iii)

From (i), (ii) and (iii),

\(\frac{2x}{b+c-a} = \frac{2y}{c+a-b} = \frac{2z}{a+b-c}\)

Multiplying each ratio by \(\frac{1}{2}\)

\(\frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c}\)

\(\frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{y-z}{x-z}\)

Multiplying numerator and denominator of third ratio by 5

∴ \(\frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{5(y-z)}{5(x-z)}\)

∴ \(\frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{5y-5z}{5x-5z}\)

By the theorem on equal ratios,

\(\frac{3x-5y+x+5z+5y-5z}{5z+3y+y-5x+5x-5z}\)

= \(\frac{4x}{4y}\) = \(\frac{x}{y}\)\)

Let us first decide whether x = 0 or x ≠ 0.

If x = 0, LHS = \(\frac{9}{21}\) = \(\frac{3}{7}\)  and RHS = \(\frac{-5}{3}\)

∴ LHS ≠ RHS ∴ x ≠ 0

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

Multiplying the numerator and the denominator of the second ratio by 4x

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

∴\(\frac{16x^2-20x+9}{8x^2+12x+21}\) = \(\frac{16x^2-20x)}{8x^2+12)}\)

By the theorem on equal ratios,

each ratio = \(\frac{16x^2-20x+9-(16x^2-20x)}{8x^2+12x+21-(8x^2+12)}\)

∴ \(\frac{16x^2-20x+9-16x^2+20x)}{8x^2+12x+21-8x^2-12)}\) = \(\frac{9}{21}\) = \(\frac{3}{7}\)

∴ \(\frac{4x-5}{2x+3}\) = \(\frac{3}{7}\)

∴ 7(4x - 5) = 3(2x + 3)

∴ 28x - 35 = 6x + 9

∴ 28x - 6x = 9 + 35

∴ 22x = 44

∴ x = 44/22 = 2

Answer is : 2 is the solution of the given equation.

Let us first decide whether y = 0 or y ≠ 0.

If y = 0, LHS = \(\frac{-12}{-4}\) = 3 and RHS = \(\frac{8}{1}\) = 8

∴ LHS ≠ RHS ∴ y ≠ 0

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

Multiplying the numerator and the denominator of the second ratio by 5y

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

∴ \(\frac{5y^2+40y-12}{5y+10y^2-4}\) = \(\frac{5y^2+40y}{5y+10y^2}\)

By the theorem on equal ratios,

each ratio = \(\frac{5y^2+40y-12-(5y^2+40y)}{5y+10y^2-4-(5y+10y^2)}\) = \(\frac{-12}{-4}\) = 3

∴ \(\frac{y+8}{1+2y}\) = 3

∴ y + 8 = 3(1 + 2y)

∴ y + 8 = 3 + 6y

∴ y - 6y = 3 - 8

∴ -5y = - 5

∴ y = -5/-5 = 1

Answer is : y = 1 is the solution of the given equation.

Let the number x to be subtracted to each of the given numbers.

Then the numbers (12 - x), (16 - x) and (21 - x) are in continued proportion.

\(\frac{12-x}{16-x}\) = \(\frac{16-x}{21-x}\) = k        ... (Say)

∴ k = \(\frac{12-x-(16-x)}{16-x-(21-x)}\) = \(\frac{12-x-16+x}{16-x-21+x)}\) = \(\frac{-4}{-5}\) = \(\frac{4}{5}\)

∴ \(\frac{12-x}{16-x}\) = \(\frac{4}{5}\)

∴ 5(12 - x) = 4(16 - x)

∴ 60 - 5x = 64 - 4x

∴ -5x + 4x = 64 - 60

∴ -x = 4

∴ x = -4

-4 should be subtracted to each of the given numbers.

i.e. 12 - (-4) = 16.

16 - (-4) = 20

21 - (-4) = 25

Check if 16, 20, and 25 are in continued proportion:

\(\frac{16}{20}\) = \(\frac{4}{5}\)

\(\frac{20}{25}\) = \(\frac{4}{5}\)

The number to be subtracted is −4.
This means adding 4 to each number results in 16, 20, and 25, which are in continued proportion with a common ratio of \(\frac{4}{5}\)

(28 - x) is the mean proportional of (23 - x) and (19 - x)

∴ \(\frac{23-x}{28-x}\) = \(\frac{28-x}{19-x}\)

By dividendo

\(\frac{23-x-(28-x)}{28-x}\) = \(\frac{28-x-(19-x)}{19-x}\)

∴ \(\frac{23-x-28+x}{28-x}\) = \(\frac{28-x-19+x}{19-x}\)

∴ \(\frac{-5}{28-x}\) = \(\frac{9}{19-x}\)

∴ -5(19 - x) = 9(28 - x)

∴ -95 + 5x = 252 - 9x

∴ 5x + 9x = 252 + 95

∴ 14x = 347

∴ x = \(\frac{347)}{14}\)

Let a, b and c be the three numbers in continued proportion.

Then b = 12 and a + c = 26  ... (1)

b² = ac

∴ (12)² = ac

∴ ac = 144 ∴ a = \(\frac{144}{c}\)    ... (2)

Substituting a =  from (2) in a + c = 26 in (1),

∴ \(\frac{144}{c}\) + c = 26

∴ 144 + c² = 26c         ... (Multiplying both the sides by c)

∴ c² - 26c + 144 = 0

∴ c² - 18c - 8c + 144 = 0

∴ c(c - 18) - 8(c - 18) = 0

∴ (c - 18) (c - 8) = 0

∴ c – 18 = 0 or c – 8 = 0

c = 18 or c = 8

Now, a + c = 26

∴ a + 18 = 26

∴ a = 26 - 18

∴ a = 8

OR a + c = 26

∴ a + 8 = 26

∴ a = 26 - 8

∴ a = 18

Answer is : The required numbers are 8, 12 and 18 or 18, 12 and 8.

(a + b + c)(a – b + c) = a² + b² + c²

∴ [(a + c) + b][(a + c) - b] = a² + b² + c²

∴ (a + c) ² - b² = a² + b² + c²

∴ [(a + b)(a - b) = a² - b²]

∴ a² + 2ac + c² - b² = d² + b² + c²

∴ 2ac = b² + b²

∴ 2ac = 2b²

∴ b² = ac

∴ a, b, c are in the continued proportion.

\(\frac{a)}{b}\) = \(\frac{b)}{c}\)

∴ b² = ac

LHS = (a + b + c)(b - c)

=ab + b²+ bc - ca - bc - c²

= ab - c² + b² - ca

=ab - c² + ca - ca

= ab - c²

= RHS

Note : This question can be solved using k-method.

Using k-method.

\(\frac{a)}{b}\) = \(\frac{b)}{c}\) = k

∴ a = bk, b = ck

∴ a = ck.k = ck²

LHS = (a² + b²)(b² + c²)

= [(ck²)² + (ck)²][(ck)² + c²]

= [c²k⁴ + c²k²][c²k² + c²]

= [c²k²(k² + 1)][c²(k² + 1)]

= c⁴k²(k² + 1)²             …..(i)

RHS = (ab + bc)²

= [b(a + c)]²

= b²(a + c)²

= c²k²(ck² + c)²

= c²k²c²(k² + 1)²

= c⁴k²(k² + 1)²             …..(ii)

From (i) and (ii),

(a² + b²)(b² + c²) = (ab + bc)².

Using k-method.

\(\frac{a)}{b}\) = \(\frac{b)}{c}\) = k

∴ a = bk, b = ck

∴ a = ck.k = ck²

LHS = \(\frac{a^2+b^2)}{ab}\) = \(\frac{(ck^2)^2+(ck)^2)}{(ck^2)(ck)}\)

= \(\frac{c^2k^4+(c^2k^2)}{(c^2k^3)}\)

= \(\frac{c^2k^2(k^2+1)}{c^2k^3}\)

= \(\frac{k^2+1}{k}\)               …..(i)

RHS = \(\frac{a+c)}{b}\) = \(\frac{ck^2+c)}{ck}\)

= \(\frac{c(k^2+1))}{ck}\)

= \(\frac{k^2+1}{k}\)              …..(ii)

From (i) and (ii),

\(\frac{a^2+b^2}{ab}\) = \(\frac{a+c}{b}\)

Let the mean proportional of \(\frac{x+y)}{x-y}\) and \(\frac{x^2-y^2}{x^2y^2}\) be m.

Then, m² = \(\frac{x+y)}{x-y}\) x \(\frac{x^2-y^2}{x^2y^2}\)

= \(\frac{x+y)}{x-y}\) x \(\frac{(x+y)(x-y)}{x^2y^2}\)

∴ m² = \(\frac{(x+y)^2}{x^2y^2}\)

∴ m = \(\frac{x+y}{xy}\)  ... (Taking square root of both the sides)

The mean proportional is \(\frac{x+y}{xy}\)