Ratio & Proportion - Exercise 7(B)

Which of the following statements are true?

(i) 51 : 68 = 85 : 102

(ii) 1.5 : 2.5 = 3.6 : 6

(iii) 30 bags : 18 bags = ₹ 450 : ₹ 270

(iv) 32 kg : ₹ 36 = 8 kg : ₹ 9

Answer

(i) False
Reason —

First Ratio (51 : 68):

Let us find H.C.F. of 51 and 68:

$\begin{array}{r} 1 \\ 51 \overline{) 68} \\ \underline{-51} \\ 17 \overline{) 51} ( 3 \\ \underline{-51} \\ 0 \end{array}$

H.C.F. is 17.

Simplest form: $\dfrac{51 \div 17}{68 \div 17} = {\dfrac{3}{4}}$

Second Ratio (85 : 102):

Let us find H.C.F. of 85 and 102:

$\begin{array}{r} 1 \\ 85 \overline{) 102} \\ \underline{-85} \\ 17 \overline{) 85} ( 5 \\ \underline{-85} \\ 0 \end{array}$

H.C.F. is 17.

Simplest form: $\dfrac{85 \div 17}{102 \div 17} = {\dfrac{5}{6}}$

Since $3:4 \neq 5:6$, the statement is False.

(ii) True
Reason —

First Ratio = 1.5 : 2.5

Multiply by 10 to remove decimals = (1.5 x 10 : 2.5 x 10) = 15 : 25

Let us find H.C.F. of 15 and 25:

$\begin{array}{r} 1 \\ 15 \overline{) 25} \\ \underline{-15} \\ 10 \overline{) 15} ( 1 \\ \underline{-10} \\ 5 \overline{) 10} ( 2 \\ \underline{-10} \\ 0 \end{array}$

H.C.F = 5

Simplest form: $\dfrac{15 \div 5}{25 \div 5} = {\dfrac{3}{5}}$

Second Ratio = 3.6 : 6

Multiply by 10 to remove decimals = (3.6 x 10 : 6 x 10) = 36 : 60

Let us find H.C.F. of 36 and 60:

$\begin{array}{r} 1 \\ 36 \overline{) 60} \\ \underline{-36} \\ 24 \overline{) 36} ( 1 \\ \underline{-24} \\ 12 \overline{) 24} ( 2 \\ \underline{-24} \\ 0 \end{array}$

H.C.F. = 12.

Simplest form: $\dfrac{36 \div 12}{60 \div 12} = {\dfrac{3}{5}}$

Since both simplify to 3:5, the statement is True.

(iii) True
Reason —

First Ratio = 30 : 18

Let us find H.C.F. of 30 and 18:

$\begin{array}{r} 1 \\ 18 \overline{) 30} \\ \underline{-18} \\ 12 \overline{) 18} ( 1 \\ \underline{-12} \\ 6 \overline{) 12} ( 2 \\ \underline{-12} \\ 0 \end{array}$

H.C.F. = 6.

Simplest form: $\dfrac{30 \div 6}{18 \div 6} = {\dfrac{5}{3}}$

Second Ratio = 450 : 270

First, cancel the zeros: 45 : 27.

Let us find H.C.F. of 45 and 27:

$\begin{array}{r} 1 \\ 27 \overline{) 45} \\ \underline{-27} \\ 18 \overline{) 27} ( 1 \\ \underline{-18} \\ 9 \overline{) 18} ( 2 \\ \underline{-18} \\ 0 \end{array}$

H.C.F. = 9.

Simplest form: $\dfrac{45 \div 9}{27 \div 9} = {\dfrac{5}{3}}$

Since both simplify to 5:3, the statement is True.

(iv) False
Reason —

First Ratio = 32 : 36

Let us find H.C.F. of 32 and 36:

$\begin{array}{r} 1 \\ 32 \overline{) 36} \\ \underline{-32} \\ 4 \overline{) 32} ( 8 \\ \underline{-32} \\ 0 \end{array}$

H.C.F. = 4.

Simplest form: $\dfrac{32 \div 4}{36 \div 4} = {\dfrac{8}{9}}$

Second Ratio = 8 : 9

This is already in its simplest form: ${\dfrac{8}{9}}$

So numerically both ratios are equal.

But a ratio should exist only between quantities of the same kind. Here the quantities are kg and ₹, which are different kinds. Therefore the statement is false.

Check whether the following numbers are in proportion or not :

(i) 30, 40, 45, 60

(ii) $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$

(iii) 0.8, 3, 2.4, 9

(iv) $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$

(v) $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$

Answer

(i) 30, 40, 45, 60

The given numbers are 30, 40, 45, 60.

We have:

$30 : 40 = \dfrac{30}{40} = \dfrac{3}{4}$

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

∴ (30 : 40) = (45 : 60)

Hence, 30, 40, 45 and 60 are in proportion.

(ii) $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$

The given numbers are $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$.

Convert mixed to improper fraction:

$2\dfrac{1}{2} = \dfrac{5}{2}$

$3\dfrac{1}{2} = \dfrac{7}{2}$

We have:

$2 : \dfrac{5}{2} = \dfrac{2}{5/2} = 2 \times \dfrac{2}{5} = \dfrac{4}{5}$

$3 : \dfrac{7}{2} = \dfrac{3}{7/2} = 3 \times \dfrac{2}{7} = \dfrac{6}{7}$

∴ $(2 : 2\dfrac{1}{2})$ ≠ $(3 : 3\dfrac{1}{2})$.

Hence, $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$ are not in proportion.

(iii) 0.8, 3, 2.4, 9

The given numbers are 0.8, 3, 2.4, 9.

Multiply by 10 to convert decimals to whole numbers:

We have:

$0.8 : 3 = \dfrac{0.8 \times 10}{3 \times 10} = \dfrac{8}{30} = \dfrac{4}{15}$

$2.4 : 9 = \dfrac{2.4 \times 10}{9 \times 10} = \dfrac{24}{90} = \dfrac{24 \div 6}{90 \div 6} = \dfrac{4}{15}$

∴ (0.8 : 3) = (2.4 : 9)

Hence, 0.8, 3, 2.4 and 9 are in proportion.

(iv) $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$

The given numbers are $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$.

We have:

$\dfrac{1}{5} : \dfrac{1}{8} = \dfrac{1}{5} \times \dfrac{8}{1} = \dfrac{8}{5}$

$\dfrac{1}{4} : \dfrac{1}{10} = \dfrac{1}{4} \times \dfrac{10}{1} = \dfrac{10}{4} = \dfrac{5}{2}$

$∴ \dfrac{1}{5} : \dfrac{1}{8}$ ≠ $\dfrac{1}{4} : \dfrac{1}{10}$

Hence, $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$ are not in proportion.

(v) $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$

The given numbers are $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$.

We have:

$\dfrac{1}{2} : \dfrac{1}{5} = \dfrac{1}{2} \times \dfrac{5}{1} = \dfrac{5}{2}$

$\dfrac{1}{6} : \dfrac{1}{15} = \dfrac{1}{6} \times \dfrac{15}{1} = \dfrac{15}{6} = \dfrac{5}{2}$

$∴ \dfrac{1}{2} : \dfrac{1}{5} = \dfrac{1}{6} : \dfrac{1}{15}$.

Hence, $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$ are in proportion.

Find the value of x in each of the following :

(i) 42 : 12 : : 7 : x

(ii) 1.8 : x : : 2.4 : 6.0

(iii) 6 : 0.8 : : x : 10

(iv) x : 1.6 : : 2.1 : 8.4

(v) $\dfrac{1}{9}$ : x : : $\dfrac{1}{3} : \dfrac{1}{4}$

(vi) 16 : x : : x : 25

Answer

(i) 42 : 12 : : 7 : x

In a proportion, we know that:

product of extremes = product of means

∴ 12 × 7 = 42 × x

⇒ 84 = 42x

⇒ x = $\dfrac{84}{42} = 2$

Hence, x = 2

(ii) 1.8 : x : : 2.4 : 6.0

In a proportion, we know that:

product of extremes = product of means

∴ x × 2.4 = 1.8 × 6.0

⇒ 2.4x = 10.8

⇒ x = $\dfrac{10.8}{2.4} = \dfrac{108}{24} = \dfrac{9}{2} = 4.5$

Hence, x = 4.5

(iii) 6 : 0.8 : : x : 10

In a proportion, we know that:

product of extremes = product of means

∴ 0.8 × x = 6 × 10

⇒ 0.8x = 60

⇒ x = $\dfrac{60}{0.8} = \dfrac{600}{8} = 75$.

Hence, x = 75

(iv) x : 1.6 : : 2.1 : 8.4

In a proportion, we know that:

product of extremes = product of means

∴ 1.6 × 2.1 = x × 8.4

⇒ 3.36 = 8.4x

⇒ x = $\dfrac{3.36}{8.4} = \dfrac{336}{840} = 0.4$

Hence, x = 0.4

(v) $\dfrac{1}{9}$ : x : : $\dfrac{1}{3} : \dfrac{1}{4}$

In a proportion, we know that:

product of extremes = product of means

$∴ x \times \dfrac{1}{3} = \dfrac{1}{9} \times \dfrac{1}{4}$

⇒ $\dfrac{x}{3} = \dfrac{1}{36}$

⇒ x = $\dfrac{3}{36} = \dfrac{1}{12}$.

Hence, x = $\dfrac{1}{12}$

(vi) 16 : x : : x : 25

In a proportion, we know that:

product of extremes = product of means

∴ x × x = 16 × 25

⇒ x² = 400

⇒ x = $\sqrt{400} = 20$

Hence, x = 20

Find the fourth proportional to :

(i) 4, 9, 32

(ii) 15, 6, 7

(iii) 0.6, 1.5, 3

(iv) $\dfrac{1}{3}, \dfrac{2}{5}, 6$

(v) $2\dfrac{1}{2}, 2\dfrac{6}{7}, 3\dfrac{1}{2}$

(vi) 3 hrs 12 min, 24 min, 1 m 68 cm

Answer

(i) 4, 9, 32

Let the fourth proportional be x. Then 4, 9, 32, x are in proportion.

product of extremes = product of means

4 × x = 9 × 32

⇒ x = $\dfrac{9 \times 32}{4}$

⇒ x = 9 × 8 = 72

Hence, the fourth proportional is 72.

(ii) 15, 6, 7

Let the fourth proportional be x. Then 15, 6, 7, x are in proportion.

product of extremes = product of means

15 × x = 6 × 7

⇒ x = $\dfrac{42}{15} = \dfrac{14}{5} = 2.8$

Hence, the fourth proportional is 2.8

(iii) 0.6, 1.5, 3

Let the fourth proportional be x. Then 0.6, 1.5, 3, x are in proportion.

product of extremes = product of means

0.6 × x = 1.5 × 3

⇒ x = $\dfrac{4.5}{0.6} = \dfrac{4.5 \times 10}{0.6 \times 10} = \dfrac{45}{6} = 7.5$

Hence, the fourth proportional is 7.5

(iv) $\dfrac{1}{3}, \dfrac{2}{5}, 6$

Let the fourth proportional be x. Then $\dfrac{1}{3}, \dfrac{2}{5}, 6$, x are in proportion.

product of extremes = product of means

$\dfrac{1}{3} \times x = \dfrac{2}{5} \times 6 \\[1em] \Rightarrow \dfrac{x}{3} = \dfrac{12}{5} \\[1em] \Rightarrow x = \dfrac{12 \times 3}{5} \\[1em] \Rightarrow x = \dfrac{36}{5} \\[1em] \Rightarrow x = 7\dfrac{1}{5}$

Hence, the fourth proportional is $7\dfrac{1}{5}$

(v) $2\dfrac{1}{2}, 2\dfrac{6}{7}, 3\dfrac{1}{2}$

Convert to improper fractions: $\dfrac{5}{2}, \dfrac{20}{7}, \dfrac{7}{2}$. Let the fourth proportional be x.

Then $\dfrac{5}{2}, \dfrac{20}{7}, \dfrac{7}{2}$, x are in proportion.

product of extremes = product of means

$\dfrac{5}{2} \times x = \dfrac{20}{7} \times \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{5x}{2} = 10 \\[1em] \Rightarrow 5x = 10 \times 2 \\[1em] \Rightarrow 5x = 20 \\[1em] \Rightarrow x = \dfrac{20}{5} \\[1em] \Rightarrow x = 4$

Hence, the fourth proportional is 4

(vi) 3 hrs 12 min, 24 min, 1 m 68 cm

First, convert to the same units for each ratio:

1 hour = 60 min,

∴ 3 hrs = 3 x 60 min = 180 min

3 hrs 12 min = 180 min + 12 min = 192 min.

1 m = 100 cm,

∴ 1 m 68 cm = 100 cm + 68 cm = 168 cm.

Let the fourth proportional be x (in cm). We have:

192 : 24 :: 168 : x

Then 192 min, 24 min, 168 cm, x are in proportion.

product of extremes = product of means

192 x x = 24 x 168

⇒ x = $\dfrac{24 \times 168}{192}$

⇒ x = $\dfrac{1 \times 168}{8}$

⇒ x = 21

Hence, the fourth proportional is 21 cm.

Find the mean proportion between :

(i) 81 and 121

(ii) 1.8 and 0.2

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

(iv) 0.32 and 0.08

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

Answer

(i) 81 and 121

Mean proportion between 81 and 121

$= \sqrt{81 \times 121} \\[1em] = \sqrt{81} \times \sqrt{121} \\[1em] = 9 \times 11 \\[1em] = 99 \\[1em]$

Hence, the answer is 99

(ii) 1.8 and 0.2

Mean proportion between 1.8 and 0.2

$= \sqrt{1.8 \times 0.2} \\[1em] = \sqrt{0.36} \\[1em] = 0.6$

Hence, the answer is 0.6

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

Mean proportion between $\dfrac{2}{3}$ and $\dfrac{8}{27}$

$= \sqrt{\dfrac{2}{3} \times \dfrac{8}{27}} \\[1em] = \sqrt{\dfrac{16}{81}} \\[1em] = \dfrac{\sqrt{16}}{\sqrt{81}} \\[1em] = \dfrac{4}{9}$

Hence, the answer is $\dfrac{4}{9}$

(iv) 0.32 and 0.08

Mean proportion between 0.32 and 0.08

$= \sqrt{0.32 \times 0.08} \\[1em] = \sqrt{0.0256} \\[1em] = 0.16$

Hence, the answer is 0.16

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

Mean proportion between $\dfrac{1}{25}$ and $\dfrac{1}{16}$

$= \sqrt{\dfrac{1}{25} \times \dfrac{1}{16}} \\[1em] = \sqrt{\dfrac{1}{400}} \\[1em] = \dfrac{1}{\sqrt{400}} \\[1em] = \dfrac{1}{20}$

Hence, the answer is $\dfrac{1}{20}$

Find the third proportional to :

(i) 36, 12

(ii) 1.2, 0.6

(iii) $\dfrac{1}{9}, \dfrac{2}{3}$

(iv) 1m 60 cm, 40 cm

(v) 1 kg 250 g, 500 g

(vi) ₹ 2.40, ₹ 4.80

Answer

(i) 36, 12

Let the third proportional be x.

Then, 36 : 12 :: 12 : x.

product of extremes = product of means

36 × x = 12 × 12

⇒ x = $\dfrac{144}{36}$

⇒ x = 4

Hence, the third proportional is 4

(ii) 1.2, 0.6

Let the third proportional be x.

Then, 1.2 : 0.6 :: 0.6 : x.

product of extremes = product of means

1.2 × x = 0.6 × 0.6

⇒ x = $\dfrac{0.36}{1.2}$

⇒ x = $\dfrac{3.6}{12}$

⇒ x = 0.3

Hence, the third proportional is 0.3

(iii) $\dfrac{1}{9}, \dfrac{2}{3}$

Let the third proportional be x.

Then, $\dfrac{1}{9} : \dfrac{2}{3} :: \dfrac{2}{3} : x$.

product of extremes = product of means

$\dfrac{1}{9} \times x = \dfrac{2}{3} \times \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{x}{9} = \dfrac{4}{9} \\[1em] \Rightarrow x = \dfrac{4}{9} \times 9 \\[1em] \Rightarrow x = 4$

Hence, the third proportional is 4

(iv) 1 m 60 cm, 40 cm

First, convert to the same unit:

1 m = 100 cm

∴ 1 m 60 cm = 100 cm + 60 cm = 160 cm.

Let the third proportional be x.

Then, 160 : 40 :: 40 : x

product of extremes = product of means

160 × x = 40 × 40

⇒ x = $\dfrac{1600}{160}$

⇒ x = 10

Hence, the third proportional is 10 cm

(v) 1 kg 250 g, 500 g

First, convert to the same unit:

1 kg = 1000 g

∴ 1 kg 250 g = 1000 g + 250 g = 1250 g

Let the third proportional be x.

Then, 1250 : 500 :: 500 : x

product of extremes = product of means

1250 x x = 500 x 500

⇒ x = $\dfrac{250000}{1250}$

⇒ x = $\dfrac{25000}{125}$

⇒ x = 200

Hence, the third proportional is 200 g

(vi) ₹ 2.40, ₹ 4.80

Let the third proportional be x.

Then, 2.40 : 4.80 :: 4.80 : x.

product of extremes = product of means

2.40 × x = 4.80 × 4.80

⇒ x = $\dfrac{4.80 \times 4.80}{2.40}$

⇒ x = 2 x 4.80

⇒ x = 9.60

Hence, the third proportional is ₹ 9.60

Show that 6, 36, 216 are in continued proportion.

Answer

If 6, 36, 216 are in continued proportion, then it should satisfy the condition b² = a x c

Here a = 6, b = 36, c = 216

∴ 36² = 6 x 216

1296 = 1296

Since it satisfies the condition b² = a x c, the numbers 6, 36, and 216 are in continued proportion.

∴ 6, 36, 216 are in continued proportion

If 8 pens cost ₹ 356, what is the cost of 14 pens?

Answer

Given:

Cost of 8 pens = ₹ 356

Cost of 14 pens = ?

Let the cost of 14 pens = ₹ x

Ratio of pens : Ratio of costs

8 : 14 :: 356 : x

By the rule: Product of Extremes = Product of Means:

8 × x = 14 × 356

⇒ x = $\dfrac{14 \times 356}{8}$

⇒ x = 14 x 44.5

⇒ x = 623

Hence, the cost of 14 pens is ₹ 623.

A uniform iron bar of length 7 m weighs 22.4 kg. How much does the same bar of length 13 m weigh ?

Answer

Given:

Weight of 7 m iron bar = 22.4 kg

Weight of 13 m iron bar = ?

Let the weight of 13 m iron bar = x kg

Ratio of lengths : Ratio of weights

7 : 13 :: 22.4 : x

By the rule: Product of Extremes = Product of Means:

7 × x = 13 × 22.4

⇒ x = $\dfrac{13 \times 22.4}{7}$

⇒ x = 13 x 3.2

⇒ x = 41.6

Hence, the bar of length 13 m weighs 41.6 kg.

A distance of 68 km is represented on a map by 1.7 cm. What distance is represented by 8.5 cm on the same map?

Answer

Given:

Map distance 1.7 cm = Actual distance 68 km

Map distance 8.5 cm = ?

Let actual distance for 8.5 cm = x km

Ratio of map distances : Ratio of actual distances

1.7 : 8.5 :: 68 : x

By the rule: Product of Extremes = Product of Means:

1.7 x x = 8.5 x 68

⇒ x = $\dfrac{8.5 \times 68}{1.7}$

⇒ x = 5 x 68

⇒ x = 340

Hence, the distance represented is 340 km.

A bus is running at a uniform speed. It covers a distance of 435 km in 6 hours. How much distance will it cover in 8 hours ?

Answer

Given:

Distance covered in 6 hours = 435 km

Distance covered in 8 hours = ?

Let distance covered in 8 hours = x km

Ratio of times : Ratio of distances

6 : 8 :: 435 : x

By the rule: Product of Extremes = Product of Means:

6 × x = 8 × 435

⇒ x = $\dfrac{8 \times 435}{6}$

⇒ x = 8 x 72.5

⇒ x = 580

Hence, the bus will cover 580 km in 8 hours.

If 15 men can dig a trench 35 m long in 1 day, then how many men can dig a similar trench 84 m long in 1 day ?

Answer

Given:

Men required for 35 m trench = 15 men

Men required for 84 m trench = ?

Let men required for 84 m trench = x men

Length of trench : Number of men

35 : 84 :: 15 : x

By the rule: Product of Extremes = Product of Means:

35 × x = 84 × 15

x = $\dfrac{84 \times 15}{35}$

x = $\dfrac{84 \times 3}{7}$

x = 12 x 3

x = 36

Hence, 36 men are required to dig the 84 m trench.