Direct and Inverse Variation — Exercise 9.1

Observe the following table and find if $x$ and $y$ are directly proportional:

Answer

Two quantities $x$ and $y$ are said to be in direct variation if $\dfrac{x}{y}$ is constant.

$\dfrac{5}{15} = \dfrac{1}{3}, \dfrac{8}{24} = \dfrac{1}{3}, \dfrac{12}{36} = \dfrac{1}{3},\\[1em] \dfrac{15}{60} = \dfrac{1}{4}, \dfrac{18}{72} = \dfrac{1}{4}, \dfrac{20}{100} = \dfrac{1}{5}$

Here, $\dfrac{5}{15} = \dfrac{8}{24} = \dfrac{12}{36} ≠ \dfrac{15}{60} = \dfrac{18}{72} ≠ \dfrac{20}{100}$

⇒ $\dfrac{x}{y}$ is not a constant

Hence, $\bm{x}$ and $\bm{y}$ are not directly proportional.

Observe the following table and find if $x$ and $y$ are directly proportional:

Answer

Two quantities $x$ and $y$ are said to be in direct variation if $\dfrac{x}{y}$ is constant.

$\dfrac{3}{9} = \dfrac{1}{3}, \dfrac{5}{15} = \dfrac{1}{3}, \dfrac{7}{21} = \dfrac{1}{3},\\[1em] \dfrac{9}{27} = \dfrac{1}{3}, \dfrac{10}{30} = \dfrac{1}{3}$

Here, $\dfrac{x}{y} = \dfrac{3}{9} = \dfrac{5}{15} = \dfrac{7}{21} = \dfrac{9}{27} = \dfrac{10}{30} = \dfrac{1}{3}$

⇒ $\dfrac{x}{y} = \dfrac{1}{3}$ (constant)

Hence, $\bm{x}$ and $\bm{y}$ are directly proportional.

If $x$ and $y$ are in direct variation, complete the following table:

Answer

Two quantities $x$ and $y$ are said to be in direct variation if $\dfrac{x}{y}$ is constant.

Let the missing entries be $y_1, x_1, x_2$ and $y_2$, then

As $x$ and $y$ are in direct variation, so we have $\dfrac{x}{y}$ = constant

$\Rightarrow \dfrac{3}{45} = \dfrac{5}{y_1} = \dfrac{x_1}{90} = \dfrac{x_2}{120} = \dfrac{10}{y_2}$

Taking $\dfrac{3}{45} = \dfrac{5}{y_1}$

$\Rightarrow 3y_1 = 5 \times 45\\[1em] \Rightarrow y_1 = \dfrac{225}{3}\\[1em] \Rightarrow y_1 = 75$

Taking $\dfrac{3}{45} = \dfrac{x_1}{90}$

$\Rightarrow 45x_1 = 3 \times 90\\[1em] \Rightarrow x_1 = \dfrac{270}{45}\\[1em] \Rightarrow x_1 = 6$

Taking $\dfrac{3}{45} = \dfrac{x_2}{120}$

$\Rightarrow 45x_2 = 3 \times 120\\[1em] \Rightarrow x_2 = \dfrac{360}{45}\\[1em] \Rightarrow x_2 = 8$

Taking $\dfrac{3}{45} = \dfrac{10}{y_2}$

$\Rightarrow 3y_2 = 10 \times 45\\[1em] \Rightarrow y_2 = \dfrac{450}{3}\\[1em] \Rightarrow y_2 = 150$

Thus, we have :

Hence, the missing entries are 75, 6, 8 and 150.

If $x$ and $y$ are in direct variation, complete the following table:

Answer

Two quantities $x$ and $y$ are said to be in direct variation if $\dfrac{x}{y}$ is constant.

Let the missing entries be $y_1, x_1, y_2$ and $y_3$, then

As $x$ and $y$ are in direct variation, so we have $\dfrac{x}{y}$ = constant

$\Rightarrow \dfrac{4}{7} = \dfrac{8}{y_1} = \dfrac{x_1}{21} = \dfrac{20}{y_2} = \dfrac{28}{y_3}$

Taking $\dfrac{4}{7} = \dfrac{8}{y_1}$

$\Rightarrow 4y_1 = 8 \times 7\\[1em] \Rightarrow y_1 = \dfrac{56}{4}\\[1em] \Rightarrow y_1 = 14$

Taking $\dfrac{4}{7} = \dfrac{x_1}{21}$

$\Rightarrow 7x_1 = 4 \times 21\\[1em] \Rightarrow x_1 = \dfrac{84}{7}\\[1em] \Rightarrow x_1 = 12$

Taking $\dfrac{4}{7} = \dfrac{20}{y_2}$

$\Rightarrow 4y_2 = 20 \times 7\\[1em] \Rightarrow y_2 = \dfrac{140}{4}\\[1em] \Rightarrow y_2 = 35$

Taking $\dfrac{4}{7} = \dfrac{28}{y_3}$

$\Rightarrow 4y_3 = 28 \times 7\\[1em] \Rightarrow y_3 = \dfrac{196}{4}\\[1em] \Rightarrow y_3 = 49$

Thus, we have :

Hence, the missing entries are 14, 12, 35 and 49.

If 8 metres cloth costs ₹250, find the cost of 5.8 metres of the same cloth.

Answer

Let the cost of 5.8 metres of cloth be ₹ $x$.

More the length of the cloth, more will be its cost.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{8}{250} = \dfrac{5.8}{x}\\[1em] \Rightarrow x = \dfrac{250 \times 5.8}{8}\\[1em] \Rightarrow x = \dfrac{1{,}450}{8}\\[1em] \Rightarrow x = 181.25$

Hence, the cost of 5.8 metres of the same cloth is ₹ 181.25.

If a labourer earns ₹3360 per week, how much will he earn in 18 days?

Answer

We know that,

1 week = 7 days

So, the labourer earns ₹ 3,360 in 7 days.

Let the labourer earn ₹ $x$ in 18 days.

More the number of days worked, more will be the earning.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{7}{3{,}360} = \dfrac{18}{x}\\[1em] \Rightarrow x = \dfrac{3{,}360 \times 18}{7}\\[1em] \Rightarrow x = \dfrac{60{,}480}{7}\\[1em] \Rightarrow x = 8{,}640$

Hence, the labourer will earn ₹ 8,640 in 18 days.

If 175 U.S. dollars cost ₹14700, how many dollars can be purchased in ₹48048?

Answer

Let $x$ U.S. dollars be purchased in ₹ 48,048.

More the money spent, more will be the number of dollars purchased.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{175}{14{,}700} = \dfrac{x}{48{,}048}\\[1em] \Rightarrow x = \dfrac{175 \times 48{,}048}{14{,}700}\\[1em] \Rightarrow x = \dfrac{48{,}048}{84}\\[1em] \Rightarrow x = 572$

Hence, 572 U.S. dollars can be purchased in ₹ 48,048.

If a car travels 67.5 km in 4.5 litres of petrol, how many kilometres will it travel in 26.4 litres of petrol?

Answer

Let the car travel $x$ km in 26.4 litres of petrol.

More the quantity of petrol, more will be the distance travelled.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{67.5}{4.5} = \dfrac{x}{26.4}\\[1em] \Rightarrow x = \dfrac{67.5 \times 26.4}{4.5}\\[1em] \Rightarrow x = \dfrac{1782}{4.5}\\[1em] \Rightarrow x = 396$

Hence, the car will travel 396 km in 26.4 litres of petrol.

If the thickness of a pile of 12 cardboard sheets is 45 mm, then how many sheets of the same cardboard would be 90 cm thick?

Answer

Converting the thickness into the same unit,

90 cm = 90 × 10 = 900 mm

Let $x$ sheets of the same cardboard be 900 mm thick.

More the number of sheets, more will be the thickness of the pile.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{12}{45} = \dfrac{x}{900}\\[1em] \Rightarrow x = \dfrac{12 \times 900}{45}\\[1em] \Rightarrow x = \dfrac{10800}{45}\\[1em] \Rightarrow x = 240$

Hence, 240 sheets of the same cardboard would be 90 cm thick.

In a model of a ship, the mast (flagstaff) is 6 cm high, while the mast of the actual ship is 9 m high. If the length of the ship is 33 m, how long is the model of the ship?

Answer

Converting the measurements of the actual ship into centimetres,

Height of the mast of the actual ship = 9 m = 9 × 100 = 900 cm

Length of the actual ship = 33 m = 33 × 100 = 3300 cm

Let the length of the model of the ship be $x$ cm.

Less the height of the mast, less will be the length of the ship.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{6}{x} = \dfrac{900}{3300}\\[1em] \Rightarrow x = \dfrac{6 \times 3300}{900}\\[1em] \Rightarrow x = \dfrac{19800}{900}\\[1em] \Rightarrow x = 22$

Hence, the length of the model of the ship is 22 cm.

The mass of an aluminium rod varies directly with its length. If a 16 cm long rod has a mass of 192 g, find the length of the rod whose mass is 105 g.

Answer

Let the length of the rod whose mass is 105 g be $x$ cm.

Less the mass of the rod, less will be its length.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{16}{192} = \dfrac{x}{105}\\[1em] \Rightarrow x = \dfrac{16 \times 105}{192}\\[1em] \Rightarrow x = \dfrac{1680}{192}\\[1em] \Rightarrow x = 8.75$

Hence, the length of the rod whose mass is 105 g is 8.75 cm.

Anita has to drive from village A to village B. She measures a distance of 3.5 cm between these villages on the map. What is the actual distance between the villages if the map scale is 1 cm = 20 km?

Answer

Given, the map scale is 1 cm = 20 km.

Let the actual distance between the two villages be $x$ km.

More the distance on the map, more will be the actual distance.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{1}{20} = \dfrac{3.5}{x}\\[1em] \Rightarrow x = 3.5 \times 20\\[1em] \Rightarrow x = 70$

Hence, the actual distance between the villages is 70 km.

A 23 m 75 cm high water tank casts a shadow 20 m long. Find at the same time:

(i) the length of the shadow cast by a tree 9 m 50 cm high.

(ii) the height of the tree if the length of the shadow is 12 m.

Answer

Converting the heights into metres,

Height of the water tank = 23 m 75 cm

= $23\dfrac{75}{100}$ m = 23.75 m

At the same time of the day, more the height of an object, more will be the length of its shadow.

Hence, it is a case of direct variation.

(i) Height of the tree = 9 m 50 cm

= $9\dfrac{50}{100}$ m = 9.5 m

Let the length of the shadow cast by the tree be $x$ m.

$\Rightarrow \dfrac{23.75}{20} = \dfrac{9.5}{x}\\[1em] \Rightarrow x = \dfrac{9.5 \times 20}{23.75}\\[1em] \Rightarrow x = \dfrac{190}{23.75}\\[1em] \Rightarrow x = 8$

Hence, the length of the shadow cast by the tree is 8 m.

(ii) Let the height of the tree be $y$ m.

$\Rightarrow \dfrac{23.75}{20} = \dfrac{y}{12}\\[1em] \Rightarrow y = \dfrac{23.75 \times 12}{20}\\[1em] \Rightarrow y = \dfrac{285}{20}\\[1em] \Rightarrow y = 14.25$

Converting the height into metres and centimetres,

14.25 m = 14 m + 0.25 × 100 cm = 14 m 25 cm

Hence, the height of the tree is 14 m 25 cm.

If 5 men or 7 women can earn ₹2625 per day, how much would 10 men and 13 women will earn per day?

Answer

Given, earning of 5 men = earning of 7 women

⇒ Earning of 1 man = earning of $\dfrac{7}{5}$ women

⇒ Earning of 10 men = earning of $\left(\dfrac{7}{5} \times 10\right)$ women = earning of 14 women

⇒ Earning of 10 men and 13 women = earning of (14 + 13) women = earning of 27 women

Let 27 women earn ₹ $x$ per day.

More the number of women, more will be the earning per day.

Hence, it is a case of direct variation.

$\Rightarrow \dfrac{7}{2{,}625} = \dfrac{27}{x}\\[1em] \Rightarrow x = \dfrac{2{,}625 \times 27}{7}\\[1em] \Rightarrow x = \dfrac{70{,}875}{7}\\[1em] \Rightarrow x = 10{,}125$

Hence, 10 men and 13 women will earn ₹ 10,125 per day.