Ratio and Proportion — Exercise 9(B)

Is the following statement true?

27 : 36 = 4.5 : 6

Answer

27 : 36 = 4.5 : 6

We have :

27 = 3³ and 36 = 2² × 3²

∴ H.C.F. of 27 and 36 = 3² = 9.

So, $27 : 36 = \dfrac{27}{36} = \dfrac{27 \div 9}{36 \div 9} = \dfrac{3}{4} = 3 : 4$.

And, $4.5 : 6 = \dfrac{4.5}{6} = \dfrac{45}{60}$.

45 = 3² × 5 and 60 = 2² × 3 × 5

∴ H.C.F. of 45 and 60 = 3 × 5 = 15.

So, $\dfrac{45}{60} = \dfrac{45 \div 15}{60 \div 15} = \dfrac{3}{4} = 3 : 4$.

Since both the ratios are equal,

Hence, the given statement is true.

Is the following statement true?

$\dfrac{3}{4} : \dfrac{15}{16} = \dfrac{2}{3} : \dfrac{5}{6}$

Answer

$\dfrac{3}{4} : \dfrac{15}{16} = \dfrac{2}{3} : \dfrac{5}{6}$

We have :

$\dfrac{3}{4} : \dfrac{15}{16} = \dfrac{3}{4} \times \dfrac{16}{15} = \dfrac{48}{60}$

48 = 2⁴ × 3 and 60 = 2² × 3 × 5

∴ H.C.F. of 48 and 60 = 2² × 3 = 12.

So, $\dfrac{48}{60} = \dfrac{48 \div 12}{60 \div 12} = \dfrac{4}{5} = 4 : 5$.

And, $\dfrac{2}{3} : \dfrac{5}{6} = \dfrac{2}{3} \times \dfrac{6}{5} = \dfrac{12}{15}$.

12 = 2² × 3 and 15 = 3 × 5

∴ H.C.F. of 12 and 15 = 3.

So, $\dfrac{12}{15} = \dfrac{12 \div 3}{15 \div 3} = \dfrac{4}{5} = 4 : 5$.

Since both the ratios are equal,

Hence, the given statement is true.

Is the following statement true?

₹ 14 : ₹ 21 = 2 pens : 3 pens

Answer

₹ 14 : ₹ 21 = 2 pens : 3 pens

We have :

14 = 2 × 7 and 21 = 3 × 7

∴ H.C.F. of 14 and 21 = 7.

So, $₹\ 14 : ₹\ 21 = \dfrac{14}{21} = \dfrac{14 \div 7}{21 \div 7} = \dfrac{2}{3} = 2 : 3$.

And, $2 \text{ pens} : 3 \text{ pens} = \dfrac{2}{3} = 2 : 3$.

Since both the ratios are equal,

Hence, the given statement is true.

Is the following statement true?

6.5 km : 2.6 km = ₹ 60 : ₹ 24

Answer

6.5 km : 2.6 km = ₹ 60 : ₹ 24

We have :

$6.5 \text{ km} : 2.6 \text{ km} = \dfrac{6.5}{2.6} = \dfrac{65}{26}$

65 = 5 × 13 and 26 = 2 × 13

∴ H.C.F. of 65 and 26 = 13.

So, $\dfrac{65}{26} = \dfrac{65 \div 13}{26 \div 13} = \dfrac{5}{2} = 5 : 2$.

60 = 2² × 3 × 5 and 24 = 2³ × 3

∴ H.C.F. of 60 and 24 = 2² × 3 = 12.

So, $₹\ 60 : ₹\ 24 = \dfrac{60}{24} = \dfrac{60 \div 12}{24 \div 12} = \dfrac{5}{2} = 5 : 2$.

Since both the ratios are equal,

Hence, the given statement is true.

Check whether the following numbers are in proportion or not:

8, 12, 18, 24

Answer

We know that four numbers are in proportion if product of extremes = product of means.

8, 12, 18, 24

Product of extremes = 8 × 24 = 192.

Product of means = 12 × 18 = 216.

Since 192 ≠ 216, product of extremes ≠ product of means.

Hence, 8, 12, 18, 24 are not in proportion.

Check whether the following numbers are in proportion or not:

6.4, 3.6, 4.8, 2.7

Answer

We know that four numbers are in proportion if product of extremes = product of means.

6.4, 3.6, 4.8, 2.7

Product of extremes = 6.4 × 2.7 = 17.28.

Product of means = 3.6 × 4.8 = 17.28.

Since product of extremes = product of means,

Hence, 6.4, 3.6, 4.8, 2.7 are in proportion.

Check whether the following numbers are in proportion or not:

$11\dfrac{1}{3}, 9\dfrac{1}{3}, 8\dfrac{1}{2}$, 7

Answer

We know that four numbers are in proportion if product of extremes = product of means.

$11\dfrac{1}{3}, 9\dfrac{1}{3}, 8\dfrac{1}{2}$, 7

We have : $11\dfrac{1}{3} = \dfrac{34}{3}, 9\dfrac{1}{3} = \dfrac{28}{3}$ and $8\dfrac{1}{2} = \dfrac{17}{2}$.

Product of extremes = $\dfrac{34}{3} \times 7 = \dfrac{238}{3}$

Product of means = $\dfrac{28}{3} \times \dfrac{17}{2} = \dfrac{476}{6} = \dfrac{238}{3}$

Since product of extremes = product of means,

Hence, $\mathbf{11\dfrac{1}{3}, 9\dfrac{1}{3}, 8\dfrac{1}{2}}$, 7 are in proportion.

Check whether the following numbers are in proportion or not:

0.36, 1.8, 6.4, 32

Answer

We know that four numbers are in proportion if product of extremes = product of means.

0.36, 1.8, 6.4, 32

Product of extremes = 0.36 × 32 = 11.52.

Product of means = 1.8 × 6.4 = 11.52.

Since 11.52 = 11.52, product of extremes = product of means.

Hence, 0.36, 1.8, 6.4, 32 are in proportion.

Check whether the following numbers are in proportion or not:

$\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{9}{10}$

Answer

We know that four numbers are in proportion if product of extremes = product of means.

$\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{9}{10}$

Product of extremes = $\dfrac{3}{4} \times \dfrac{9}{10} = \dfrac{27}{40}$

Product of means = $\dfrac{5}{6} \times \dfrac{7}{8} = \dfrac{35}{48}$

Since $\dfrac{27}{40} \ne \dfrac{35}{48}$, product of extremes ≠ product of means.

Hence, $\mathbf{\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}, \dfrac{9}{10}}$ are not in proportion.

Find the value of $x$:

8 : $x$ :: 6 : 27

Answer

8 : $x$ :: 6 : 27

Product of extremes = product of means.

$\Rightarrow 8 \times 27 = x \times 6 \\[1em] \Rightarrow 6x = 216 \\[1em] \Rightarrow x = \dfrac{216}{6} = 36$

Hence, $\bm{x = 36}$.

Find the value of $x$:

5.6 : 3.5 :: $x$ : 1.25

Answer

5.6 : 3.5 :: $x$ : 1.25

Product of extremes = product of means.

$\Rightarrow 5.6 \times 1.25 = 3.5 \times x \\[1em] \Rightarrow 3.5x = 7 \\[1em] \Rightarrow x = \dfrac{7}{3.5} = 2$

Hence, $\bm{x = 2}$.

Find the value of $x$:

$1\dfrac{4}{5} : 2\dfrac{4}{5} :: x : 3\dfrac{1}{2}$

Answer

$1\dfrac{4}{5} : 2\dfrac{4}{5} :: x : 3\dfrac{1}{2}$

We have : $1\dfrac{4}{5} = \dfrac{9}{5}, 2\dfrac{4}{5} = \dfrac{14}{5}$ and $3\dfrac{1}{2} = \dfrac{7}{2}$.

Product of extremes = product of means.

$\Rightarrow \dfrac{9}{5} \times \dfrac{7}{2} = \dfrac{14}{5} \times x \\[1em] \Rightarrow \dfrac{63}{10} = \dfrac{14x}{5} \\[1em] \Rightarrow x = \left(\dfrac{63}{10} \times \dfrac{5}{14}\right) \\[1em] \Rightarrow x = \dfrac{9}{4} \\[1em] \Rightarrow x = 2\dfrac{1}{4}$

Hence, $\bm{x = 2\dfrac{1}{4}}$.

Find the value of $x$:

$\dfrac{2}{3} : \dfrac{4}{7} :: 1\dfrac{5}{6} : x$

Answer

$\dfrac{2}{3} : \dfrac{4}{7} :: 1\dfrac{5}{6} : x$

We have : $1\dfrac{5}{6} = \dfrac{11}{6}$.

Product of extremes = product of means.

$\Rightarrow \dfrac{2}{3} \times x = \dfrac{4}{7} \times \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{2x}{3} = \dfrac{44}{42} = \dfrac{22}{21} \\[1em] \Rightarrow x = \left(\dfrac{22}{21} \times \dfrac{3}{2}\right) \\[1em] \Rightarrow x = \dfrac{11}{7} \\[1em] \Rightarrow x = 1\dfrac{4}{7}$

Hence, $\bm{x = 1\dfrac{4}{7}}$.

Find the fourth proportional to:

2.8, 14 and 3.5

Answer

2.8, 14 and 3.5

Let the fourth proportional to 2.8, 14 and 3.5 be $x$. Then,

$\Rightarrow 2.8 : 14 :: 3.5 : x \\[1em] \Rightarrow 2.8 \times x = 14 \times 3.5 \\[1em] \Rightarrow 2.8x = 49 \\[1em] \Rightarrow x = \dfrac{49}{2.8} = 17.5$

Hence, the required fourth proportional is 17.5.

Find the fourth proportional to:

$3\dfrac{1}{3}, 1\dfrac{2}{3}$ and $2\dfrac{1}{2}$

Answer

$3\dfrac{1}{3}, 1\dfrac{2}{3}$ and $2\dfrac{1}{2}$

We have : $3\dfrac{1}{3} = \dfrac{10}{3}, 1\dfrac{2}{3} = \dfrac{5}{3}$ and $2\dfrac{1}{2} = \dfrac{5}{2}$.

Let the fourth proportional to $\dfrac{10}{3}, \dfrac{5}{3}$ and $\dfrac{5}{2}$ be $x$. Then,

$\Rightarrow \dfrac{10}{3} : \dfrac{5}{3} :: \dfrac{5}{2} : x \\[1em] \Rightarrow \dfrac{10}{3} \times x = \dfrac{5}{3} \times \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{10x}{3} = \dfrac{25}{6} \\[1em] \Rightarrow x = \left(\dfrac{25}{6} \times \dfrac{3}{10}\right) \\[1em] \Rightarrow x = \dfrac{5}{4} \\[1em] \Rightarrow x = 1\dfrac{1}{4}$

Hence, the required fourth proportional is $\mathbf{1\dfrac{1}{4}}$.

Find the fourth proportional to:

$1\dfrac{5}{7}, 2\dfrac{3}{14}$ and $3\dfrac{3}{5}$

Answer

$1\dfrac{5}{7}, 2\dfrac{3}{14}$ and $3\dfrac{3}{5}$

We have : $1\dfrac{5}{7} = \dfrac{12}{7}, 2\dfrac{3}{14} = \dfrac{31}{14}$ and $3\dfrac{3}{5} = \dfrac{18}{5}$.

Let the fourth proportional to $\dfrac{12}{7}, \dfrac{31}{14}$ and $\dfrac{18}{5}$ be $x$. Then,

$\Rightarrow \dfrac{12}{7} : \dfrac{31}{14} :: \dfrac{18}{5} : x \\[1em] \Rightarrow \dfrac{12}{7} \times x = \dfrac{31}{14} \times \dfrac{18}{5} \\[1em] \Rightarrow \dfrac{12x}{7} = \dfrac{558}{70} \\[1em] \Rightarrow x = \left(\dfrac{558}{70} \times \dfrac{7}{12}\right) \\[1em] \Rightarrow x = \dfrac{93}{20} \\[1em] \Rightarrow x = 4\dfrac{13}{20}$

Hence, the required fourth proportional is $\mathbf{4\dfrac{13}{20}}$.

Find the fourth proportional to:

$1\dfrac{1}{5}, 1\dfrac{3}{5}$ and 2.1

Answer

$1\dfrac{1}{5}, 1\dfrac{3}{5}$ and 2.1

We have : $1\dfrac{1}{5} = \dfrac{6}{5}, 1\dfrac{3}{5} = \dfrac{8}{5}$ and $2.1 = \dfrac{21}{10}$.

Let the fourth proportional to $\dfrac{6}{5}, \dfrac{8}{5}$ and $\dfrac{21}{10}$ be $x$. Then,

$\Rightarrow \dfrac{6}{5} : \dfrac{8}{5} :: \dfrac{21}{10} : x \\[1em] \Rightarrow \dfrac{6}{5} \times x = \dfrac{8}{5} \times \dfrac{21}{10} \\[1em] \Rightarrow \dfrac{6x}{5} = \dfrac{168}{50} \\[1em] \Rightarrow x = \left(\dfrac{168}{50} \times \dfrac{5}{6}\right) \\[1em] \Rightarrow x = \dfrac{28}{10} \\[1em] \Rightarrow x = 2.8$

Hence, the required fourth proportional is 2.8.

Find the third proportional to:

12, 16

Answer

12, 16

Let the third proportional to 12 and 16 be $x$. Then,

$\Rightarrow 12 : 16 :: 16 : x \\[1em] \Rightarrow 12 \times x = 16 \times 16 \\[1em] \Rightarrow 12x = 256 \\[1em] \Rightarrow x = \dfrac{256}{12} \\[1em] \Rightarrow x = \dfrac{64}{3} \\[1em] \Rightarrow x = 21\dfrac{1}{3}$

Hence, the required third proportional is $\mathbf{21\dfrac{1}{3}}$.

Find the third proportional to:

4.5, 6

Answer

4.5, 6

Let the third proportional to 4.5 and 6 be $x$. Then,

$\Rightarrow 4.5 : 6 :: 6 : x \\[1em] \Rightarrow 4.5 \times x = 6 \times 6 \\[1em] \Rightarrow 4.5x = 36 \\[1em] \Rightarrow x = \dfrac{36}{4.5} = 8$

Hence, the required third proportional is 8.

Find the third proportional to:

$5\dfrac{1}{2}, 16\dfrac{1}{2}$

Answer

$5\dfrac{1}{2}, 16\dfrac{1}{2}$

We have : $5\dfrac{1}{2} = \dfrac{11}{2}$ and $16\dfrac{1}{2} = \dfrac{33}{2}$.

Let the third proportional to $\dfrac{11}{2}$ and $\dfrac{33}{2}$ be $x$. Then,

$\Rightarrow \dfrac{11}{2} : \dfrac{33}{2} :: \dfrac{33}{2} : x \\[1em] \Rightarrow \dfrac{11}{2} \times x = \dfrac{33}{2} \times \dfrac{33}{2} \\[1em] \Rightarrow \dfrac{11x}{2} = \dfrac{1089}{4} \\[1em] \Rightarrow x = \left(\dfrac{1089}{4} \times \dfrac{2}{11}\right) \\[1em] \Rightarrow x = \dfrac{99}{2} \\[1em] \Rightarrow x = 49\dfrac{1}{2}$

Hence, the required third proportional is $\mathbf{49\dfrac{1}{2}}$.

Find the third proportional to:

$3\dfrac{1}{2}, 8\dfrac{3}{4}$

Answer

$3\dfrac{1}{2}, 8\dfrac{3}{4}$

We have : $3\dfrac{1}{2} = \dfrac{7}{2}$ and $8\dfrac{3}{4} = \dfrac{35}{4}$.

Let the third proportional to $\dfrac{7}{2}$ and $\dfrac{35}{4}$ be $x$. Then,

$\Rightarrow \dfrac{7}{2} : \dfrac{35}{4} :: \dfrac{35}{4} : x \\[1em] \Rightarrow \dfrac{7}{2} \times x = \dfrac{35}{4} \times \dfrac{35}{4} \\[1em] \Rightarrow \dfrac{7x}{2} = \dfrac{1225}{16} \\[1em] \Rightarrow x = \left(\dfrac{1225}{16} \times \dfrac{2}{7}\right) \\[1em] \Rightarrow x = \dfrac{175}{8} \\[1em] \Rightarrow x = 21\dfrac{7}{8}$

Hence, the required third proportional is $\mathbf{21\dfrac{7}{8}}$.

Find the mean proportion between:

8 and 18

Answer

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

∴ Mean proportion between 8 and 18 = $\sqrt{8 \times 18} = \sqrt{144} = 12$

Hence, the mean proportion between 8 and 18 is 12.

Find the mean proportion between:

0.3 and 2.7

Answer

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

0.3 and 2.7

Mean proportion between 0.3 and 2.7 = $\sqrt{0.3 \times 2.7} = \sqrt{0.81} = 0.9$

Hence, the mean proportion between 0.3 and 2.7 is 0.9.

Find the mean proportion between:

$66\dfrac{2}{3}$ and 6

Answer

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

$66\dfrac{2}{3}$ and 6

We have : $66\dfrac{2}{3} = \dfrac{200}{3}$.

Mean proportion = $\sqrt{\dfrac{200}{3} \times 6} = \sqrt{400} = 20$

Hence, the mean proportion between $\mathbf{66\dfrac{2}{3}}$ and 6 is 20.

Find the mean proportion between:

1.25 and 0.45

Answer

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

1.25 and 0.45

Mean proportion between 1.25 and 0.45 = $\sqrt{1.25 \times 0.45} = \sqrt{0.5625} = 0.75$

Hence, the mean proportion between 1.25 and 0.45 is 0.75.

Find the mean proportion between:

$\dfrac{1}{7}$ and $\dfrac{4}{63}$

Answer

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

$\dfrac{1}{7}$ and $\dfrac{4}{63}$

Mean proportion = $\sqrt{\dfrac{1}{7} \times \dfrac{4}{63}} = \sqrt{\dfrac{4}{441}} = \dfrac{2}{21}$

Hence, the mean proportion between $\mathbf{\dfrac{1}{7}}$ and $\mathbf{\dfrac{4}{63}}$ is $\mathbf{\dfrac{2}{21}}$.

If 28 is the third proportional to 7 and $x$, find the value of $x$.

Answer

Since 28 is the third proportional to 7 and $x$, we have :

$\Rightarrow 7 : x = x : 28 \\[1em] \Rightarrow x \times x = 7 \times 28 \\[1em] \Rightarrow x^2 = 196 \\[1em] \Rightarrow x = \sqrt{196} = 14$

Hence, $\bm{x = 14}$.

If 18, $x$, 50 are in continued proportion, find the value of $x$.

Answer

Since 18, $x$, 50 are in continued proportion, we have :

$\Rightarrow 18 : x = x : 50 \\[1em] \Rightarrow x \times x = 18 \times 50 \\[1em] \Rightarrow x^2 = 900 \\[1em] \Rightarrow x = \sqrt{900} = 30$

Hence, $\bm{x = 30}$.

A rod was cut into two pieces in the ratio 7 : 5. If the length of the smaller piece was 45.5 cm, then find the length of the longer piece.

Answer

Let the lengths of the two pieces be $7x$ cm and $5x$ cm.

Clearly, the smaller piece is of length $5x$ cm.

$\Rightarrow 5x = 45.5 \\[1em] \Rightarrow x = \dfrac{45.5}{5} = 9.1$

∴ Length of the longer piece = (7 × 9.1) cm = 63.7 cm.

Hence, the length of the longer piece is 63.7 cm.

The area of two rectangular fields are in the ratio 5 : 9. Find the area of the smaller field if that of the larger field is 2331 sq. metres.

Answer

Let the areas of the two rectangular fields be $5x$ sq. metres and $9x$ sq. metres.

Clearly, the area of the larger field is $9x$ sq. metres.

$\Rightarrow 9x = 2331 \\[1em] \Rightarrow x = \dfrac{2331}{9} = 259$

∴ Area of the smaller field = (5 × 259) sq. metres = 1295 sq. metres.

Hence, the area of the smaller field is 1295 sq. metres.

What number must be subtracted from each of the numbers 41, 55, 36, 48, so that the differences are proportional?

Answer

Let $x$ be the required number. Then,

$\Rightarrow \dfrac{41 - x}{55 - x} = \dfrac{36 - x}{48 - x} \\[1em] \Rightarrow (41 - x)(48 - x) = (55 - x)(36 - x) \\[1em] \Rightarrow 1968 - 41x - 48x + x^2 = 1980 - 55x - 36x + x^2 \\[1em] \Rightarrow 1968 - 89x = 1980 - 91x \\[1em] \Rightarrow 91x - 89x = 1980 - 1968 \\[1em] \Rightarrow 2x = 12 \\[1em] \Rightarrow x = 6$

Hence, the required number is 6.

An alloy is to contain copper and zinc in the ratio 9 : 4. Find the quantity of zinc to be melted with $2\dfrac{2}{5}$ kg of copper, to get the desired alloy.

Answer

Let the required quantity of zinc be $x$ kg.

We have : $2\dfrac{2}{5} = \dfrac{12}{5}$.

Then,

$\Rightarrow 9 : 4 = \dfrac{12}{5} : x \\[1em] \Rightarrow 9 \times x = 4 \times \dfrac{12}{5} \\[1em] \Rightarrow 9x = \dfrac{48}{5} \\[1em] \Rightarrow x = \left(\dfrac{48}{5} \times \dfrac{1}{9}\right) \\[1em] \Rightarrow x = \dfrac{16}{15} \\[1em] \Rightarrow x = 1\dfrac{1}{15}$

Hence, the required quantity of zinc is $\mathbf{1\dfrac{1}{15}}$ kg.