Ratio and Proportion — Exercise 7(B)

Check whether the following quantities form a proportion or not:

3x, 7x, 24 and 56

Answer

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 3x × 56 = 168x

Product of means = 7x × 24 = 168x

Since product of extremes = product of means, the quantities are in proportion.

Hence, 3x, 7x, 24 and 56 form a proportion.

Check whether the following quantities form a proportion or not:

0.8, 3, 2.4 and 9

Answer

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 0.8 × 9 = 7.2

Product of means = 3 × 2.4 = 7.2

Since product of extremes = product of means, the quantities are in proportion.

Hence, 0.8, 3, 2.4 and 9 form a proportion.

Check whether the following quantities form a proportion or not:

$1\dfrac{1}{2}, 3\dfrac{1}{4}, 4\dfrac{1}{2} \text{ and } 9\dfrac{3}{4}$

Answer

Four quantities form a proportion if product of extremes = product of means.

$1\dfrac{1}{2}, 3\dfrac{1}{4}, 4\dfrac{1}{2} \text{ and } 9\dfrac{3}{4} = \dfrac{3}{2}, \dfrac{13}{4}, \dfrac{9}{2} \text{ and } \dfrac{39}{4}$

Product of extremes = $\dfrac{3}{2} \times \dfrac{39}{4} = \dfrac{117}{8}$

Product of means = $\dfrac{13}{4} \times \dfrac{9}{2} = \dfrac{117}{8}$

Since product of extremes = product of means, the quantities are in proportion.

Hence, $1\dfrac{1}{2}, 3\dfrac{1}{4}, 4\dfrac{1}{2} \text{ and } 9\dfrac{3}{4}$ form a proportion.

Check whether the following quantities form a proportion or not:

0.4, 0.5, 2.9 and 3.5

Answer

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 0.4 × 3.5 = 1.4

Product of means = 0.5 × 2.9 = 1.45

Since product of extremes ≠ product of means, the quantities are not in proportion.

Hence, 0.4, 0.5, 2.9 and 3.5 do not form a proportion.

Check whether the following quantities form a proportion or not:

$2\dfrac{1}{2}, 5\dfrac{1}{2}, 3.0 \text{ and } 6.0$

Answer

Four quantities form a proportion if product of extremes = product of means.

$2\dfrac{1}{2}, 5\dfrac{1}{2}, 3.0 \text{ and } 6.0 = \dfrac{5}{2}, \dfrac{11}{2}, 3 \text{ and } 6$

Product of extremes = $\dfrac{5}{2} \times 6 = 15$

Product of means = $\dfrac{11}{2} \times 3 = \dfrac{33}{2} = 16.5$

Since product of extremes ≠ product of means, the quantities are not in proportion.

Hence, $2\dfrac{1}{2}, 5\dfrac{1}{2}, 3.0 \text{ and } 6.0$ do not form a proportion.

Find the fourth proportional of:

3, 12 and 4

Answer

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 3, b = 12, c = 4

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{x}\\[1em] \Rightarrow x = \dfrac{b \times c}{a}\\[1em] = \dfrac{12 \times 4}{3} \\[1em] = \dfrac{48}{3}\\[1em] = 16$

Hence, the fourth proportional is 16.

Find the fourth proportional of:

5, 9 and 45

Answer

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 5, b = 9, c = 45

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{x}\\[1em] \Rightarrow x = \dfrac{b \times c}{a}\\[1em] = \dfrac{9 \times 45}{5} \\[1em] = \dfrac{405}{5} \\[1em] = 81$

Hence, the fourth proportional is 81.

Find the fourth proportional of:

2.1, 1.5 and 8.4

Answer

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 2.1, b = 1.5, c = 8.4

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{x}\\[1em] \Rightarrow x = \dfrac{b \times c}{a}\\[1em] = \dfrac{1.5 \times 8.4}{2.1} \\[1em] = \dfrac{12.6}{2.1} \\[1em] = 6.0$

Hence, the fourth proportional is 6.0

Find the fourth proportional of:

$\dfrac{1}{3}, \dfrac{2}{5} \text{ and } 8.4$

Answer

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where $a = \dfrac{1}{3}, b = \dfrac{2}{5}, c = 8.4$

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{x}\\[1em] \Rightarrow x = \dfrac{b \times c}{a}\\[1em] = \dfrac{\dfrac{2}{5} \times 8.4}{\dfrac{1}{3}} \\[1em] = \dfrac{2}{5} \times 8.4 \times 3 \\[1em] = \dfrac{2 \times 8.4 \times 3}{5} \\[1em] = \dfrac{50.4}{5} \\[1em] = 10.08$

Hence, the fourth proportional is 10.08.

Find the fourth proportional of:

4 hours 40 minutes, 1 hour 10 minutes and 16 hours

Answer

Convert into minutes:

4 hours 40 minutes = (4 × 60) + 40 = 280 minutes

1 hour 10 minutes = (1 × 60) + 10 = 70 minutes

16 hours = 16 × 60 = 960 minutes

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 280, b = 70, c = 960

$\Rightarrow \dfrac{a}{b} = \dfrac{c}{x}\\[1em] \Rightarrow x = \dfrac{b \times c}{a}\\[1em] = \dfrac{70 \times 960}{280} \\[1em] = \dfrac{67200}{280} \\[1em] = 240 \text{ minutes} = 4 \text{ hours}$

Hence, the fourth proportional is 4 hours.

Find the third proportional of:

27 and 9

Answer

Let the third proportional to a and b be x, then a : b :: b : x, where a = 27, b = 9

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{x}\\[1em] \Rightarrow x = \dfrac{b \times b}{a} = \dfrac{b^2}{a}\\[1em] = \dfrac{9^2}{27} \\[1em] = \dfrac{81}{27} \\[1em] = 3$

Hence, the third proportional is 3.

Find the third proportional of:

2 m 40 cm and 40 cm

Answer

2 m 40 cm = 240 cm.

Let the third proportional to a and b be x, then a : b :: b : x, where a = 240, b = 40

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{x}\\[1em] \Rightarrow x = \dfrac{b \times b}{a} = \dfrac{b^2}{a}\\[1em] = \dfrac{40^2}{240} \\[1em] = \dfrac{1600}{240} \\[1em] = \dfrac{20}{3} = 6\dfrac{2}{3} \text{ cm}$

Hence, the third proportional is $6\dfrac{2}{3}$ cm.

Find the third proportional of:

1.8 and 0.6

Answer

Let the third proportional to a and b be x, then a : b :: b : x, where a = 1.8, b = 0.6

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{x}\\[1em] \Rightarrow x = \dfrac{b \times b}{a} = \dfrac{b^2}{a}\\[1em] = \dfrac{0.6^2}{1.8} \\[1em] = \dfrac{0.36}{1.8} \\[1em] = 0.2$

Hence, the third proportional is 0.2.

Find the third proportional of:

$\dfrac{1}{7} \text{ and } \dfrac{3}{14}$

Answer

Let the third proportional to a and b be x, then a : b :: b : x, where $a = \dfrac{1}{7}, b = \dfrac{3}{14}$

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{x}\\[1em] \Rightarrow x = \dfrac{b \times b}{a} = \dfrac{b^2}{a}\\[1em] = \dfrac{\left(\dfrac{3}{14}\right)^2}{\dfrac{1}{7}} \\[1em] = \dfrac{9}{196} \times 7 \\[1em] = \dfrac{63}{196} \\[1em] = \dfrac{9}{28}$

Hence, the third proportional is $\dfrac{9}{28}$.

Find the third proportional of:

1.6 and 0.8

Answer

Let the third proportional to a and b be x, then a : b :: b : x, where a = 1.6, b = 0.8

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{x}\\[1em] \Rightarrow x = \dfrac{b \times b}{a} = \dfrac{b^2}{a}\\[1em] = \dfrac{0.8^2}{1.6} \\[1em] = \dfrac{0.64}{1.6} \\[1em] = 0.4$

Hence, the third proportional is 0.4.

Find the mean proportional between:

16 and 4

Answer

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 16, b = 4

$\Rightarrow \dfrac{a}{x} = \dfrac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{16 \times 4} \\[1em] = \sqrt{64} \\[1em] = 8$

Hence, the mean proportional is 8.

Find the mean proportional between:

3 and 27

Answer

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 3, b = 27

$\Rightarrow \dfrac{a}{x} = \dfrac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{3 \times 27} \\[1em] = \sqrt{81} \\[1em] = 9$

Hence, the mean proportional is 9.

Find the mean proportional between:

0.9 and 2.5

Answer

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 0.9, b = 2.5

$\Rightarrow \dfrac{a}{x} = \dfrac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{0.9 \times 2.5} \\[1em] = \sqrt{2.25} \\[1em] = 1.5$

Hence, the mean proportional is 1.5.

Find the mean proportional between:

0.6 and 9.6

Answer

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 0.6, b = 9.6

$\Rightarrow \dfrac{a}{x} = \dfrac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{0.6 \times 9.6} \\[1em] = \sqrt{5.76} \\[1em] = 2.4$

Hence, the mean proportional is 2.4.

Find the mean proportional between:

$\dfrac{1}{4} \text{ and } \dfrac{1}{16}$

Answer

Let the mean proportional between a and b be x, then a : x :: x : b, where $a = \dfrac{1}{4}, b = \dfrac{1}{16}$

$\Rightarrow \dfrac{a}{x} = \dfrac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{\dfrac{1}{4} \times \dfrac{1}{16}} \\[1em] = \sqrt{\dfrac{1}{64}} \\[1em] = \dfrac{1}{8}$

Hence, the mean proportional is $\dfrac{1}{8}$.

If A : B = 3 : 5 and B : C = 4 : 7, find A : B : C.

Answer

A : B = 3 : 5 and B : C = 4 : 7

To combine, make the value of B the same. B appears as 5 and 4; their L.C.M. is 20.

A : B = 3 : 5 = (3 × 4) : (5 × 4) = 12 : 20

B : C = 4 : 7 = (4 × 5) : (7 × 5) = 20 : 35

∴ A : B : C = 12 : 20 : 35

Hence, A : B : C = 12 : 20 : 35.

If x : y = 2 : 3 and y : z = 5 : 7, find x : y : z.

Answer

x : y = 2 : 3 and y : z = 5 : 7

Make the value of y the same. y appears as 3 and 5; their L.C.M. is 15.

x : y = 2 : 3 = (2 × 5) : (3 × 5) = 10 : 15

y : z = 5 : 7 = (5 × 3) : (7 × 3) = 15 : 21

∴ x : y : z = 10 : 15 : 21

Hence, x : y : z = 10 : 15 : 21.

If m : n = 4 : 9 and n : s = 3 : 7, find m : s.

Answer

$\text{Since, } m : n = 4 : 9 \Rightarrow \dfrac{m}{n} = \dfrac{4}{9} \text{ and } n : s = 3 : 7 \Rightarrow \dfrac{n}{s} = \dfrac{3}{7} \\[1em] \Rightarrow \dfrac{m}{s} = \dfrac{m}{n} \times \dfrac{n}{s} \\[1em] = \dfrac{4}{9} \times \dfrac{3}{7} \\[1em] = \dfrac{12}{63} \\[1em] = \dfrac{4}{21}$

i.e., m : s = 4 : 21

Hence, m : s = 4 : 21.

If P : Q = $\dfrac{1}{2} : \dfrac{1}{3} \text{ and }$Q : R = $1\dfrac{1}{2} : 1\dfrac{1}{3}$, find P : R.

Answer

P : Q = $\dfrac{1}{2} : \dfrac{1}{3}$ = (multiplying both terms by 6) = 3 : 2

Q : R = $1\dfrac{1}{2} : 1\dfrac{1}{3} = \dfrac{3}{2} : \dfrac{4}{3}$ = (multiplying both terms by 6) = 9 : 8

$\text{Since, } \text{ P : Q }= 3 : 2 \Rightarrow \dfrac{\text{ P }}{\text{ Q }} = \dfrac{3}{2} \text{ and } \text{ Q : R }= 9 : 8 \Rightarrow \dfrac{\text{ Q }}{\text{ R }} = \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{\text{ P }}{\text{ R }} = \dfrac{\text{ P }}{\text{ Q }} \times \dfrac{\text{ Q }}{\text{ R }} \\[1em] = \dfrac{3}{2} \times \dfrac{9}{8} \\[1em] = \dfrac{27}{16}$

i.e., P : R = 27 : 16

Hence, P : R = 27 : 16.

If a : b = 1.5 : 3.5 and b : c = 5 : 6, find a : c.

Answer

a : b = 1.5 : 3.5 = 15 : 35 = 3 : 7

$\text{Since, } a : b = 3 : 7 \Rightarrow \dfrac{a}{b} = \dfrac{3}{7} \text{ and } b : c = 5 : 6 \Rightarrow \dfrac{b}{c} = \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{a}{b} \times \dfrac{b}{c} \\[1em] = \dfrac{3}{7} \times \dfrac{5}{6} \\[1em] = \dfrac{15}{42} \\[1em] = \dfrac{5}{14}$

i.e., a : c = 5 : 14

Hence, a : c = 5 : 14.

If $1\dfrac{1}{4} : 2\dfrac{1}{3} = p : q \text{ and } q : r = 4\dfrac{1}{2} : 5\dfrac{1}{4}$, find p : r.

Answer

p : q = $1\dfrac{1}{4} : 2\dfrac{1}{3} = \dfrac{5}{4} : \dfrac{7}{3}$ = (multiplying both terms by 12) = 15 : 28

q : r = $4\dfrac{1}{2} : 5\dfrac{1}{4} = \dfrac{9}{2} : \dfrac{21}{4}$ = (multiplying both terms by 4) = 18 : 21 = 6 : 7

$\text{Since, } p : q = 15 : 28 \Rightarrow \dfrac{p}{q} = \dfrac{15}{28} \text{ and } q : r = 6 : 7 \Rightarrow \dfrac{q}{r} = \dfrac{6}{7} \\[1em] \Rightarrow \dfrac{p}{r} = \dfrac{p}{q} \times \dfrac{q}{r} \\[1em] = \dfrac{15}{28} \times \dfrac{6}{7} \\[1em] = \dfrac{90}{196} \\[1em] = \dfrac{45}{98}$

i.e., p : r = 45 : 98

Hence, p : r = 45 : 98.

If x : y = 5 : 4 and 2 : x = 3 : 8, find the value of y.

Answer

Given:

2 : x = 3 : 8

Product of extremes = product of means:

2 × 8 = 3 × x

x = $\dfrac{16}{3}$

Now, x : y = 5 : 4

$\Rightarrow \dfrac{x}{y} = \dfrac{5}{4} \\[1em] \Rightarrow y = \dfrac{4x}{5}\\[1em] \Rightarrow y = \dfrac{4}{5} \times \dfrac{16}{3} \\[1em] \Rightarrow y = \dfrac{64}{15} = 4\dfrac{4}{15}$

Hence, the value of y is $\dfrac{64}{15}, \text { i.e. } 4\dfrac{4}{15}$.

Find the value of x, when 2.5 : 4 = x : 7.5.

Answer

Given:

2.5 : 4 = x : 7.5

$\Rightarrow \dfrac{2.5}{4} = \dfrac{x}{7.5}$

Product of extremes = Product of means

⇒ 2.5 × 7.5 = 4 × x

$\Rightarrow 4x = 18.75 \\[1em] \Rightarrow x = \dfrac{18.75}{4} \\[1em] = \dfrac{1875}{400} \\[1em] = \dfrac{75}{16} \\[1em] = 4\dfrac{11}{16}$

Hence, the value of x is $4\dfrac{11}{16}$.

Show that 2, 12 and 72 are in continued proportion.

Answer

Three quantities a, b and c are in continued proportion if a : b :: b : c, i.e. if b² = a × c.

Here, a = 2, b = 12 and c = 72.

b² = 12² = 144

a × c = 2 × 72 = 144

Since b² = a × c (144 = 144), the quantities are in continued proportion.

Hence, 2, 12 and 72 are in continued proportion.