Solution:

\(\displaystyle \frac{5}{7} = \displaystyle \frac{5 \times 4}{7 \times 4} = \displaystyle \frac{20}{28}\)

\(\displaystyle \frac{5}{7} = \displaystyle \frac{5 \times 7}{7 \times 7} = \displaystyle \frac{35}{49}\)

\(\displaystyle \frac{5}{7} = \displaystyle \frac{5 \times 0.5}{7 \times 0.5} = \displaystyle \frac{2.5}{3.5}\)

Solution:

\(\displaystyle \frac{9}{14} = \displaystyle \frac{9 \times 0.5}{14 \times 0.5} = \displaystyle \frac{4.5}{7}\)

\(\displaystyle \frac{9}{14} = \displaystyle \frac{9 \times 3}{14 \times 3} = \displaystyle \frac{27}{42}\)

\(\displaystyle \frac{9}{14} = \displaystyle \frac{9 \times 0.25}{14 \times 0.25} = \displaystyle \frac{2.25}{3.5}\)

Solution:

Let, the radius be r.

∴ Circumference of the circle = 2πr.

∴ \(\displaystyle \frac{\text {Radius}}{\text {Circumference}}\) = \(\displaystyle \frac{r}{2\pi r}\)

∴ \(\displaystyle \frac{\text {Radius}}{\text {Circumference}}\) = \(\displaystyle \frac{1}{2\pi}\)

∴ The ratio of radius to circumference of the circle is 1 ∶ 2π

Solution:

The radius is r.

∴ Area of the circle = πr².

∴ \(\displaystyle \frac{\text {Circumference}}{\text {Area}}\) = \(\displaystyle \frac{2\pi r}{\pi r^2}\)

∴ \(\displaystyle \frac{\text {Circumference}}{\text {Area}}\) = \(\displaystyle \frac{2}{r}\)

∴ The ratio of circumference to area of the circle is  2 ∶ r

Solution:

 The side is 7 cm.

∴ Diagonal of the square = 7\(\displaystyle \sqrt{2}\) cm.

∴ \(\displaystyle \frac{\text {Diagonal}}{\text {Side}}\) = \(\displaystyle \frac{7\sqrt{2} \text{ cm}}{7 \text{ cm}}\)

∴ \(\displaystyle \frac{\text {Diagonal}}{\text {Side}}\) = \(\displaystyle \frac{\sqrt{2}}{1}\)

∴ The ratio of diagonal of a square to its side is  \(\displaystyle \sqrt{2}\) ∶ 1

Solution:

 l = 5 cm and b = 3.5 cm.

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

And Area of the rectangle = l × b
= 5 × 3.5
= 17.5 ... (ii)

∴ \(\displaystyle \frac{\text {Perimeter}}{\text {Area}}\) = \(\displaystyle \frac{17}{\text {17.5}}\)

∴ \(\displaystyle \frac{\text {Perimeter}}{\text {Area}}\) = \(\displaystyle \frac{170}{\text {175}}\)

∴ \(\displaystyle \frac{\text {Perimeter}}{\text {Area}}\) = \(\displaystyle \frac{34}{35}\)

∴ The ratio of perimeter to area of that rectangle is  34 ∶ 35.

Solution:

\(\displaystyle \frac{\sqrt 5}{3}\), \(\displaystyle \frac{3}{\sqrt 7}\)

Now, \(\displaystyle \sqrt 5\) × \(\displaystyle \sqrt 7\)
= \(\displaystyle \sqrt{35}\) ... (i)

And 3 × 3
= 9 ... (ii)

Here, 35 < 81

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

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

Solution:

\(\displaystyle \frac{3\sqrt {5}}{5\sqrt {7}}\), \(\displaystyle \frac{\sqrt {63}}{\sqrt {125}}\)

Now, \(\displaystyle 3\sqrt {5} \times \sqrt {125}\)
= \(\displaystyle 3\sqrt {625} \)
= 3 × 25
= 75 ... (i)

And \(\displaystyle 5\sqrt {7} × \sqrt {63}\)
= \(\displaystyle 5\sqrt {441} \)
= 5 × 21
= 105 ... (ii)

Here, 75 < 105

∴ \(\displaystyle 3\sqrt {5} \times \sqrt {125}\) < \(\displaystyle 5\sqrt {7} × \sqrt {63}\)

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

Solution:

\(\displaystyle \frac{5}{18}\), \(\displaystyle \frac{17}{121}\)

Now, \(\displaystyle 5 \times 121\)
= 605 ... (i)

And \(\displaystyle 17 \times 18\)
= 306 ... (ii)

Here, 605 > 306

∴ \(\displaystyle \frac{5}{18}\) > \(\displaystyle \frac{17}{121}\)

Solution:

\(\displaystyle \frac{\sqrt{80}}{\sqrt{48}}\), \(\displaystyle \frac{\sqrt {45}}{\sqrt{27}}\)

Now, \(\displaystyle \sqrt{80} \times \sqrt{27}\)

= \(\displaystyle \sqrt{16\times 5} \times\sqrt{9 \times 3}\)

= \(\displaystyle 4 \sqrt{5} \times 3\sqrt{3}\)

= \(\displaystyle 12 \sqrt{15}\) ... (i)

And \(\displaystyle \sqrt{45} \times \sqrt{48}\)

= \(\displaystyle \sqrt{9\times 5} \times\sqrt{16 \times 3}\)

= \(\displaystyle 3 \sqrt{5} \times 4\sqrt{3}\)

= \(\displaystyle 12 \sqrt{15}\) ... (ii)

Here, \(\displaystyle 12 \sqrt{15}\) = \(\displaystyle 12 \sqrt{15}\)

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

Solution:

\(\displaystyle \frac{9.2}{5.1}\), \(\displaystyle \frac{3.4}{7.1}\)

Now, \(\displaystyle \frac{9.2}{5.1}\) > 1

And, \(\displaystyle \frac{3.4}{7.1}\) < 1

∴ \(\displaystyle \frac{9.2}{5.1}\) > \(\displaystyle \frac{3.4}{7.1}\)

Solution:

The ratio of \(\angle\)A and \(\angle\)B is 5 ∶ 4.

Let, \(\angle\)A = 5x and \(\angle\)B = 4x

Now, the adjacent angles of a parallelogram are supplementary.

∴ \(\angle\)A + \(\angle\)B = 180°

∴ 5x + 4x = 180°

∴ 9x = 180°

∴ \(\displaystyle x = \frac{180}{9}\)

∴ x = 20° ... (i)

∴ \(\angle\)B = 4x = 4 × 20° = 80°

∴ The measure of \(\angle\)B is 80°

Solution:

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

Let the present ages of Albert and Salim be 5x years and 9x years respectively.

Five years hence,
Albert’s age
= 5x + 5 years
And Salim’s age
= 9x + 5 years

Five years hence, the ratio of their ages will be 3 ∶ 5.

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

∴ 5(5x + 5) = 3(9x + 5)

∴ 25x + 25 = 27x + 15

∴ 25x − 27x = 15 − 25

∴ − 2x = − 10

i.e. 2x = 10

∴ \(\displaystyle x = \frac{10}{2}\)

∴ x = 5 ... (i)

∴ Albert’s present age = 5 × 5 = 25 years.

And Salim’s present age = 9 × 5 = 45 years.

∴ The present ages of Albert and Salim are 25 years and 45 years respectively.

Solution:

The ratio of length and breadth of a rectangle is 3 ∶ 1.

Let the length of the rectangle be 3x cm and the breadth be x cm.

Now,
The Perimeter of the rectangle
= 2(length + breadth)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x

∴ \(\displaystyle x = \frac{36}{8}\)

∴ \(\displaystyle x = \frac{9}{2}\)

∴ x = 4.5 cm... (i)

∴ Length = 3x = 3 × 4.5 = 13.5 cm

And Breadth = x = 4.5 cm

∴ The length and breadth of the rectangle are 13.5 cm and 4.5 cm respectively.

Solution:

The ratio of two numbers is 31 ∶ 23.

∴ Let, the greater number be 31x and the smaller number be 23x.

From the given information,
 31x + 23x = 216
∴ 54x = 216

∴ \(\displaystyle x = \frac{216}{54}\)

∴ x = 4 ... (i)

∴ The greater number = 31 × 4 = 124

And the smaller number = 23 × 4 = 92

∴ The numbers are 124 and 92.

Solution:

The ratio of two numbers is 10 ∶ 9.

∴ Let, the greater number be 10x and the smaller number be 9x.

From the given information,
 10x × 9x = 360
∴ 90x² = 360

∴ \(\displaystyle x^2 = \frac{360}{90}\)

∴ x² = 4 ... (i)

∴ \(\displaystyle x = \sqrt{4}\)

∴ \(\displaystyle x = 2\)

∴ The greater number
= 10x
= 10 × 2
= 20

And the smaller number
= 9x
= 9 × 2
= 18

∴ The numbers are 20 and 18.

Solution:

a ∶ b = 3 ∶ 1 ... (Given)

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

∴ a = 3b ... (i)

Also, b ∶ c = 5 ∶ 1 ... (Given)

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

∴ b = 5c ... (ii)

From (i) and (ii),

 a = 3b = 3 × 5c

∴ a = 15c... (iii)

Now, we have to find the value of \(\displaystyle \left(\frac{a^3}{15b^2c}\right)^3\)

From (ii) and (iii),

 \(\displaystyle \left(\frac{a^3}{15b^2c}\right)^3\)

= \(\displaystyle \left(\frac{(15c)^3}{15 \times (5c)^2 \times c}\right)^3\)

= \(\displaystyle \left(\frac{15\times 15\times 15 \times c^3}{15 \times 5 \times 5 \times c^2 \times c}\right)^3\)

= 9³

= 729

∴ \(\displaystyle \left(\frac{a^3}{15b^2c}\right)^3\) = 729 ... (iv)

Solution:

 \(\displaystyle \frac {a^2}{7bc}\)

From (ii) and (iii),

= \(\displaystyle \frac {(15c)^2}{7 \times 5c \times c}\)

= \(\displaystyle \frac {15 \times 15 \times c^2}{7 \times 5 \times c^2}\)

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

∴ \(\displaystyle \frac {a^2}{7bc} = \frac {45}{7}\) ... (iv)

Solution:

\(\sqrt {0.04 \times 0.4 \times a}\) = \(0.04 \times 0.4 \times \sqrt {b}\) ... (Given)

Squaring both sides,

 \(\displaystyle 0.04 \times 0.4 \times a\) = \(\displaystyle (0.04)^2 \times (0.4)^2 \times b\)

∴ \(\displaystyle 0.04 \times 0.4 \times a\) = \(\displaystyle (0.04)^2 \times (0.4)^2 \times b\)

∴ \(\displaystyle \frac {a}{b}\) = \(\displaystyle \frac {(0.04)^2 \times (0.4)^2}{0.04 \times 0.4}\)

∴ \(\displaystyle \frac {a}{b}\) = \(\displaystyle \frac {0.04 \times 0.04 \times 0.4 \times 0.4}{0.04 \times 0.4}\)

∴ \(\displaystyle \frac {a}{b}\) = \(\displaystyle 0.04 \times 0.4\)

∴ \(\displaystyle \frac {a}{b}\) = \(\displaystyle \frac {4}{100} \times \frac {4}{10}\)

∴ \(\displaystyle \frac {a}{b}\) = \(\displaystyle \frac {2}{125}\)

∴ a ∶ b = 2 ∶ 125

Solution:

(x + 3) ∶ (x + 11) = (x − 2) ∶ (x + 1) ... (Given)

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

∴ (x + 3)(x + 1) = (x − 2)(x + 11)

∴ x² + x + 3x + 3 = x² + 11x − 2x − 22

∴ 4x + 3 = 9x − 22

∴ 4x − 9x = − 22 − 3

∴ − 5x = − 25

i.e. 5x = 25

∴ \(\displaystyle x = \frac{25}{5}\)

∴ x = 5