Direct and Inverse Variation — Exercise 9.2

Which of the following are in inverse variation?

(i) Number of students in a hostel and consumption of food.

(ii) Time taken by a train to cover a fixed distance and the speed of the train.

(iii) Area of land and its cost.

(iv) The number of people working and the time to complete the work.

(v) The quantity of rice and its cost.

Answer

Two quantities are said to vary inversely if an increase (or decrease) in one quantity causes a decrease (or increase) in the other quantity.

(i) More the number of students in a hostel, more will be the consumption of food.

Hence, it is a case of direct variation.

(ii) For a fixed distance, more the speed of the train, less will be the time taken by it to cover that distance.

Hence, it is a case of inverse variation.

(iii) More the area of land, more will be its cost.

Hence, it is a case of direct variation.

(iv) More the number of people working, less will be the time taken to complete the work.

Hence, it is a case of inverse variation.

(v) More the quantity of rice, more will be its cost.

Hence, it is a case of direct variation.

Hence, (ii) and (iv) are in inverse variation.

Observe the following table and find whether the variables (here $x$ and $y$) are in inverse variation:

Answer

Two quantities $x$ and $y$ are said to be in inverse variation if $xy$ is constant.

$90 \times 10 = 900, 60 \times 15 = 900, 45 \times 20 = 900,\\[1em] 30 \times 30 = 900, 20 \times 45 = 900$

Here, $xy = 90 \times 10 = 60 \times 15 = 45 \times 20 = 30 \times 30 = 20 \times 45 = 900$

⇒ $xy = 900$ (constant)

Hence, $\bm{x}$ and $\bm{y}$ are in inverse variation.

Observe the following table and find whether the variables (here $x$ and $y$) are in inverse variation:

Answer

Two quantities $x$ and $y$ are said to be in inverse variation if $xy$ is constant.

$75 \times 10 = 750, 45 \times 30 = 1350, 30 \times 25 = 750,\\[1em] 20 \times 35 = 700, 10 \times 65 = 650$

⇒ $xy$ is not a constant

Hence, $\bm{x}$ and $\bm{y}$ are not in inverse variation.

Under the condition that the temperature remains constant, the volume of gas is inversely proportional to its pressure. If the volume of gas is 630 cubic centimetres at a pressure of 360 mm of mercury, then what will be the pressure of the gas, if its volume is 720 cubic centimetres at the same temperature?

Answer

Let the pressure of the gas be $x$ mm of mercury when its volume is 720 cubic centimetres.

Given, the volume of the gas is inversely proportional to its pressure.

Hence, it is a case of inverse variation.

$\Rightarrow 630 \times 360 = 720 \times x\\[1em] \Rightarrow 226800 = 720x\\[1em] \Rightarrow x = \dfrac{226800}{720}\\[1em] \Rightarrow x = 315$

Hence, the pressure of the gas will be 315 mm of mercury.

A packet of sweets was distributed among 20 children and each of them received 4 sweets. How many sweets will each child get, if the number of children is reduced by 4?

Answer

Number of children after the reduction = 20 - 4 = 16

Let each child get $x$ sweets when there are 16 children.

Less the number of children, more will be the number of sweets each child gets.

Hence, it is a case of inverse variation.

$\Rightarrow 20 \times 4 = 16 \times x\\[1em] \Rightarrow 80 = 16x\\[1em] \Rightarrow x = \dfrac{80}{16}\\[1em] \Rightarrow x = 5$

Hence, each child will get 5 sweets.

Pooja has enough money to buy 36 oranges at the rate of ₹4.50 per orange. How many oranges she can buy if the price of each orange is increased by 90 paise?

Answer

Converting 90 paise into rupees,

90 paise = ₹ $\dfrac{90}{100} = 0.90$

Hence, the new price of each orange = ₹ (4.50 + 0.90) = ₹ 5.40

Let Pooja buy $x$ oranges at ₹ 5.40 per orange.

For the same amount of money, more the price of each orange, less will be the number of oranges bought.

Hence, it is a case of inverse variation.

$\Rightarrow 4.50 \times 36 = 5.40 \times x\\[1em] \Rightarrow 162 = 5.40x\\[1em] \Rightarrow x = \dfrac{162}{5.40}\\[1em] \Rightarrow x = 30$

Hence, Pooja can buy 30 oranges.

It takes 8 days for 12 men to construct a wall. How many men should be put on the job if it is required to be constructed in 6 days?

Answer

Let $x$ men be put on the job to construct the wall in 6 days.

Less the number of days, more will be the number of men required.

Hence, it is a case of inverse variation.

$\Rightarrow 12 \times 8 = x \times 6\\[1em] \Rightarrow 96 = 6x\\[1em] \Rightarrow x = \dfrac{96}{6}\\[1em] \Rightarrow x = 16$

Hence, 16 men should be put on the job to construct the wall in 6 days.

Eight taps through which water flows at the same rate can fill a tank in 27 minutes. If two taps go out of order, how long will the remaining taps take to fill the tank?

Answer

Number of taps in working condition = 8 - 2 = 6

Let 6 taps take $x$ minutes to fill the tank.

Less the number of taps, more will be the time taken to fill the tank.

Hence, it is a case of inverse variation.

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

Hence, the remaining taps will take 36 minutes to fill the tank.

A contractor undertook a contract to complete a part of stadium in 9 months with a team of 560 persons. Later on, it was required to complete the job in 5 months. How many extra persons should he employ to complete the work?

Answer

Let $x$ persons be required to complete the job in 5 months.

Less the number of months, more will be the number of persons required.

Hence, it is a case of inverse variation.

$\Rightarrow 560 \times 9 = x \times 5\\[1em] \Rightarrow 5040 = 5x\\[1em] \Rightarrow x = \dfrac{5040}{5}\\[1em] \Rightarrow x = 1008$

Hence, the extra persons to be employed = 1008 - 560 = 448

Hence, the contractor should employ 448 extra persons to complete the work.

A batch of bottles were packed in 30 boxes with 10 bottles in each box. If the same batch is packed using 12 bottles in each box, how many boxes would be filled?

Answer

Let $x$ boxes be filled when each box has 12 bottles.

More the number of bottles in each box, less will be the number of boxes filled.

Hence, it is a case of inverse variation.

$\Rightarrow 10 \times 30 = 12 \times x\\[1em] \Rightarrow 300 = 12x\\[1em] \Rightarrow x = \dfrac{300}{12}\\[1em] \Rightarrow x = 25$

Hence, 25 boxes would be filled.

Vandana takes 24 minutes to reach her school if she goes at a speed of 5 km/h. If she wants to reach school in 20 minutes, what should be her speed?

Answer

Let Vandana's speed be $x$ km/h to reach the school in 20 minutes.

Less the time taken, more should be the speed.

Hence, it is a case of inverse variation.

$\Rightarrow 24 \times 5 = 20 \times x\\[1em] \Rightarrow 120 = 20x\\[1em] \Rightarrow x = \dfrac{120}{20}\\[1em] \Rightarrow x = 6$

Hence, Vandana's speed should be 6 km/h.

A fort is provided with food for 80 soldiers to last for 60 days. Find how long would the food last if 20 additional soldiers join after 15 days?

Answer

After 15 days, the remaining food is sufficient for 80 soldiers for (60 - 15) days, i.e. 45 days.

After 15 days, 20 additional soldiers join the fort.

Hence, the number of soldiers in the fort = 80 + 20 = 100

Let the remaining food last for $x$ days when there are 100 soldiers.

More the number of soldiers, less will be the number of days for which the food lasts.

Hence, it is a case of inverse variation.

$\Rightarrow 80 \times 45 = 100 \times x\\[1em] \Rightarrow 3600 = 100x\\[1em] \Rightarrow x = \dfrac{3600}{100}\\[1em] \Rightarrow x = 36$

Hence, the food would last for 36 days.

1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were sent to another fort and thus, the food lasted for 32 more days. How many soldiers left the fort?

Answer

After 4 days, the remaining food is sufficient for 1200 soldiers for (28 - 4) days, i.e. 24 days.

The remaining food actually lasted for 32 more days.

Let $x$ soldiers be left in the fort after some of them were sent to another fort.

More the number of days for which the food lasts, less will be the number of soldiers.

Hence, it is a case of inverse variation.

$\Rightarrow 1200 \times 24 = x \times 32\\[1em] \Rightarrow 28800 = 32x\\[1em] \Rightarrow x = \dfrac{28800}{32}\\[1em] \Rightarrow x = 900$

Hence, the number of soldiers who left the fort = 1200 - 900 = 300

Hence, 300 soldiers left the fort.