Direct & Inverse Variation

We have : Fewer toys, Less cost. [Direct variation]

Clearly, the two variables namely 'the number of toys' and 'the cost' have direct variation.

Now, cost of 40 toys = ₹ 1,024.

Cost of 1 toy = ₹ $\Big(\frac{1{,}024}{40}\Big)$

One dozen toys = 12 toys

Cost of 12 toys = ₹ $\Big(\frac{1{,}024}{40} \times 12\Big)$ = ₹ $\frac{12288}{40}$ = ₹ 307.20

Hence, the cost of one dozen such toys is ₹ 307.20.

We have : More money, More notebooks that can be purchased. [Direct variation]

Clearly, the two variables namely 'the money' and 'the number of notebooks' have direct variation.

Now, for ₹ 333, the number of notebooks purchased = 18.

For ₹ 1, the number of notebooks purchased = $\frac{18}{333}$.

For ₹ 425.50, the number of notebooks purchased = $\Big(\frac{18}{333} \times 425.50\Big)$ = $\frac{7659}{333}$ = 23.

Hence, 23 notebooks can be purchased for ₹ 425.50.

We have : Less money, Fewer pencils that can be bought. [Direct variation]

Clearly, the two variables namely 'the money' and 'the number of pencils' have direct variation.

Now, for ₹ 120, the number of pencils bought = 50.

For ₹ 1, the number of pencils bought = $\frac{50}{120}$.

For ₹ 108, the number of pencils bought = $\Big(\frac{50}{120} \times 108\Big)$ = 45.

Hence, 45 pencils can be bought for ₹ 108.

We have : Fewer rods, Less weight. [Direct variation]

Clearly, the two variables namely 'the number of rods' and 'the weight' have direct variation.

Now, weight of 45 rods = 12.6 kg.

Weight of 1 rod = $\frac{12.6}{45}$ kg.

Weight of 24 rods = $\Big(\frac{12.6}{45} \times 24\Big)$ kg = $\frac{302.4}{45}$ kg = 6.72 kg.

Hence, the weight of 24 such rods is 6.72 kg.

We have : More days, More earning. [Direct variation]

Clearly, the two variables namely 'the number of days' and 'the earning' have direct variation.

1 week = 7 days.

Now, earning of the labourer in 7 days = ₹ 784.

Earning of the labourer in 1 day = ₹ $\Big(\frac{784}{7}\Big)$ = ₹ 112.

Earning of the labourer in 15 days = ₹ (112 × 15) = ₹ 1,680.

Hence, the labourer will earn ₹ 1,680 in 15 days.

We have : More petrol, More distance covered. [Direct variation]

Clearly, the two variables namely 'the quantity of petrol' and 'the distance covered' have direct variation.

Now, distance covered in 4.5 litres of petrol = 56.7 km.

Distance covered in 1 litre of petrol = $\frac{56.7}{4.5}$ km = 12.6 km.

Distance covered in 26 litres of petrol = (12.6 × 26) km = 327.6 km.

Hence, the car will cover 327.6 km in 26 litres of petrol.

We have : Fewer hours, More men. [Inverse variation]

Clearly, the two variables namely 'the number of men' and 'the number of hours to do the work' have inverse variation.

Now, the work is done by 15 men in 36 hours.

The work is done in 1 hour by (15 × 36) men.

The work is done in 20 hours by $\Big(\frac{15 \times 36}{20}\Big)$ men = 27 men.

Hence, 27 men will be required to finish the work in 20 hours.

1 week = 7 days.

Now, 13 men can weave 117 baskets in 7 days.

1 man can weave 117 baskets in (7 × 13) days.

1 man can weave 1 basket in $\frac{7 \times 13}{117}$ days.

Let the required number of men be $x$.

$x$ men can weave 1 basket in $\frac{7 \times 13}{117 \times x}$ days.

$x$ men can weave 189 baskets in $\frac{7 \times 13 \times 189}{117 \times x}$ days.

But this time is given to be 3 days.

$$\begin{aligned}\Rightarrow \frac{7 \times 13 \times 189}{117 \times x} = 3 \\[1em] \Rightarrow \frac{17199}{117x} = 3 \\[1em] \Rightarrow 351x = 17199 \\[1em] \Rightarrow x = \frac{17199}{351} = 49\end{aligned}$$

Hence, 49 men will be needed to weave 189 baskets in 3 days.

We have : Less time, More speed. [Inverse variation]

Clearly, the two variables namely 'the speed of the car' and 'the time taken' have inverse variation.

Now, the journey is finished in 48 minutes at a speed of 65 kmph.

The journey is finished in 1 minute at a speed of (65 × 48) kmph.

The journey is finished in 40 minutes at a speed of $\Big(\frac{65 \times 48}{40}\Big)$ kmph = 78 kmph.

Hence, the car must go at a speed of 78 kmph to finish the journey in 40 minutes.

We have : Fewer parrots, Less consumption of nuts. [Direct variation]

Clearly, the two variables namely 'the number of parrots' and 'the quantity of nuts' have direct variation.

Number of parrots left in the zoo = (28 − 8) = 20.

Now, 28 parrots consume nuts = 7420 g.

1 parrot consumes nuts = $\frac{7420}{28}$ g = 265 g.

20 parrots consume nuts = (265 × 20) g = 5300 g.

Hence, 5300 g of nuts will be required in a day.

We have : More persons, More money earned in a day. [Direct variation]

In a day :

5 men can earn ₹ 700.

1 man can earn ₹ $\Big(\frac{700}{5}\Big)$ = ₹ 140.

7 men can earn ₹ (140 × 7) = ₹ 980.

7 women can earn ₹ 700.

1 woman can earn ₹ $\Big(\frac{700}{7}\Big)$ = ₹ 100.

11 women can earn ₹ (100 × 11) = ₹ 1,100.

∴ In a day, 7 men and 11 women can earn ₹ (980 + 1,100) = ₹ 2,080.

Hence, 7 men and 11 women would earn ₹ 2,080 per day.

We have : More labourers, More earning. [Direct variation]

More number of working days, More earning. [Direct variation]

Now, earning of 5 labourers in 15 days = ₹ 9,000.

Earning of 1 labourer in 15 days = ₹ $\Big(\frac{9{,}000}{5}\Big)$ = ₹ 1,800.

Earning of 1 labourer in 1 day = ₹ $\Big(\frac{1{,}800}{15}\Big)$ = ₹ 120.

Let the required number of labourers be $x$.

Earning of $x$ labourers in 1 day = ₹ $120x$.

Earning of $x$ labourers in 8 days = ₹ $(120x \times 8)$ = ₹ $960x$.

But this earning is given to be ₹ 6,720.

$$\begin{aligned}\Rightarrow 960x = 6720 \\[1em] \Rightarrow x = \frac{6720}{960} \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, 7 labourers can earn ₹ 6,720 in 8 days.

We have : More men, Less number of days the food will last. [Inverse variation]

After 10 days, the food is sufficient for 450 men for (80 − 10) days, i.e., 70 days.

Number of men in the fort after the new arrivals = (450 + 50) = 500.

For 450 men, the food is sufficient for 70 days.

For 1 man, the food is sufficient for (70 × 450) days.

For 500 men, the food is sufficient for $\Big(\frac{70 \times 450}{500}\Big)$ days = 63 days.

Hence, the remaining food will last for 63 days.

We have : Fewer typists, More working hours per day. [Inverse variation]

More working hours per day, Fewer days. [Inverse variation]

Now, 3 typists working 7 hours a day type the thesis in 10 days.

1 typist working 7 hours a day types the thesis in (10 × 3) days.

1 typist working 1 hour a day types the thesis in (10 × 3 × 7) days.

Let the required number of hours per day be $x$.

2 typists working 1 hour a day type the thesis in $\frac{10 \times 3 \times 7}{2}$ days.

2 typists working $x$ hours a day type the thesis in $\frac{10 \times 3 \times 7}{2 \times x}$ days = $\frac{105}{x}$ days.

But this time is given to be 21 days.

$$\begin{aligned}\Rightarrow \frac{105}{x} = 21 \\[1em] \Rightarrow 21x = 105 \\[1em] \Rightarrow x = \frac{105}{21} = 5\end{aligned}$$

Hence, 2 typists should work for 5 hours per day to finish the thesis in 21 days.

We have : Fewer workers, More days. [Inverse variation]

More working hours per day, Fewer days. [Inverse variation]

Now, 7 workers working 6 hours a day build the wall in 12 days.

1 worker working 6 hours a day builds the wall in (12 × 7) days.

1 worker working 1 hour a day builds the wall in (12 × 7 × 6) days.

3 workers working 1 hour a day build the wall in $\Big(\frac{12 \times 7 \times 6}{3}\Big)$ days.

3 workers working 8 hours a day build the wall in $\Big(\frac{12 \times 7 \times 6}{3 \times 8}\Big)$ days = 21 days.

Hence, 3 workers working 8 hours a day will take 21 days to build a similar wall.

We have : Fewer articles, Less cost. [Direct variation]

Cost of 20 articles = ₹ 90.

Cost of 1 article = ₹ $\Big(\frac{90}{20}\Big)$ = ₹ 4.50.

Cost of 9 articles = ₹ (4.50 × 9) = ₹ 40.50.

Hence, Option 2 is the correct option.

We have : More articles, More cost. [Direct variation]

Cost of 4 pairs of trousers = ₹ 1,000.

Cost of 1 pair of trousers = ₹ $\Big(\frac{1{,}000}{4}\Big)$ = ₹ 250.

Cost of 3 pairs of trousers = ₹ (250 × 3) = ₹ 750.

Cost of 8 shirts = ₹ 1,000.

Cost of 1 shirt = ₹ $\Big(\frac{1{,}000}{8}\Big)$ = ₹ 125.

Cost of 3 shirts = ₹ (125 × 3) = ₹ 375.

∴ Amount required to buy 3 pairs of trousers and 3 shirts = ₹ (750 + 375) = ₹ 1,125.

Hence, Option 4 is the correct option.

We have : Less length, Less weight. [Direct variation]

Weight of 22.5 metres of the rod = 85.5 kg.

Weight of 1 metre of the rod = $\frac{85.5}{22.5}$ kg = 3.8 kg.

Weight of 9 metres of the rod = (3.8 × 9) kg = 34.2 kg.

Hence, Option 1 is the correct option.

We have : More distance on the map, More actual distance. [Direct variation]

0.6 cm on the map represents 6.6 km.

1 cm on the map represents $\frac{6.6}{0.6}$ km = 11 km.

80.5 cm on the map represents (11 × 80.5) km = 885.5 km.

Hence, Option 4 is the correct option.

We have : More men, Fewer days. [Inverse variation]

15 men can reap the field in 35 days.

1 man can reap the field in (35 × 15) days.

21 men can reap the field in $\Big(\frac{35 \times 15}{21}\Big)$ days = 25 days.

Hence, Option 2 is the correct option.

We have : More days, Fewer persons. [Inverse variation]

The work is done by 20 persons in 7 days.

The work is done in 1 day by (20 × 7) persons.

The work is done in 28 days by $\Big(\frac{20 \times 7}{28}\Big)$ persons = 5 persons.

Hence, Option 1 is the correct option.

We have : Fewer pipes, More time. [Inverse variation]

Number of pipes remaining = (10 − 2) = 8.

10 pipes fill the tank in 24 minutes.

1 pipe fills the tank in (24 × 10) minutes.

8 pipes fill the tank in $\Big(\frac{24 \times 10}{8}\Big)$ minutes = 30 minutes.

Hence, Option 2 is the correct option.

We have : More radius, Fewer rounds. [Inverse variation]

For a cylinder of base radius 14 cm, the number of rounds = 140.

For a cylinder of base radius 1 cm, the number of rounds = (140 × 14).

For a cylinder of base radius 20 cm, the number of rounds = $\Big(\frac{140 \times 14}{20}\Big)$ = 98.

Hence, Option 2 is the correct option.

We have : More men, Less number of weeks the provisions will last. [Inverse variation]

After 4 weeks, the provisions are sufficient for 750 men for (20 − 4) weeks, i.e., 16 weeks.

Number of men in the garrison after the reinforcement = (750 + 450) = 1200.

For 750 men, the provisions are sufficient for 16 weeks.

For 1 man, the provisions are sufficient for (16 × 750) weeks.

For 1200 men, the provisions are sufficient for $\Big(\frac{16 \times 750}{1200}\Big)$ weeks = 10 weeks.

Hence, Option 1 is the correct option.

We have : Fewer men, More number of days the food will last. [Inverse variation]

After 5 days, the food is sufficient for 120 men for (200 − 5) days, i.e., 195 days.

Number of men left = (120 − 30) = 90.

For 120 men, the food is sufficient for 195 days.

For 1 man, the food is sufficient for (195 × 120) days.

For 90 men, the food is sufficient for $\Big(\frac{195 \times 120}{90}\Big)$ days = 260 days.

Hence, Option 4 is the correct option.

We have : Less part of the cistern, Less time. [Direct variation]

Part of the cistern still to be filled = $\Big(1 - \frac{4}{5}\Big) = \frac{1}{5}$.

1 minute = 60 seconds.

$\frac{4}{5}$ of the cistern is filled in 60 seconds.

The whole cistern is filled in $\Big(60 \times \frac{5}{4}\Big)$ seconds = 75 seconds.

$\frac{1}{5}$ of the cistern is filled in $\Big(75 \times \frac{1}{5}\Big)$ seconds = 15 seconds.

Hence, Option 3 is the correct option.

(i) When $x$ and $y$ are in direct proportion, the ratio of the corresponding values remains the same.

Hence, when $\bm{x}$ and $\bm{y}$ are in direct proportion then $\bm{\frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3}} = \bm{k}$ (a constant) and so on.

(ii) The number of men and the number of days are in inverse proportion.

$x$ men can do the work in $y$ days.

1 man can do the work in ($y \times x$) days.

$y$ men can do the work in $\frac{y \times x}{y}$ days = $x$ days.

Hence, if $\bm{x}$ men can do a work in $\bm{y}$ days, then $\bm{y}$ men can do a work in $\bm{x}$ days.

(iii) The number of articles and the cost are in direct proportion.

Cost of $m$ articles = ₹ $n$.

Cost of 1 article = ₹ $\frac{n}{m}$.

Cost of $n$ articles = ₹ $\Big(\frac{n}{m} \times n\Big)$ = ₹ $\frac{n^2}{m}$.

Hence, if $\bm{m}$ articles cost ₹ $\bm{n}$ in all, then the total cost of $\bm{n}$ articles is ₹ $\bm{\frac{n^2}{m}}$.

(iv) The money and the number of oranges are in direct proportion.

For ₹ 52, the number of oranges bought = 8.

For ₹ 1, the number of oranges bought = $\frac{8}{52}$.

For ₹ 169, the number of oranges bought = $\Big(\frac{8}{52} \times 169\Big)$ = 26.

Hence, if 8 oranges cost ₹ 52, then the number of oranges that can be bought for ₹ 169 is 26.

(v) The number of pipes and the time taken are in inverse proportion.

6 pipes can fill the tank in 120 minutes.

1 pipe can fill the tank in (120 × 6) minutes.

5 pipes can fill the tank in $\Big(\frac{120 \times 6}{5}\Big)$ minutes = 144 minutes.

Hence, 5 pipes will fill the tank in 144 minutes.

(i) When $x$ and $y$ are in inverse proportion, the product of their corresponding values remains the same.

So, $x_1 y_1 = x_2 y_2$ and not $x_1 y_2 = x_2 y_1$.

Hence, the given statement is false (F).

(ii) The number of persons and the number of days are in inverse proportion.

3 persons can build the wall in 4 days.

1 person can build the wall in (4 × 3) days.

4 persons can build the wall in $\Big(\frac{4 \times 3}{4}\Big)$ days = 3 days.

So, 4 persons can build it in 3 days and not in $3\frac{1}{4}$ days.

Hence, the given statement is false (F).

(iii) In a fixed time, more the speed, more is the distance covered. So, the speed and the distance covered are in direct proportion.

Hence, the given statement is true (T).

(iv) More the number of men, less is the time taken by them to do a job. So, the number of men and the time taken are in inverse proportion.

Hence, the given statement is true (T).

If $x$ varies directly as $y^2$, then the ratio of $x$ and $y^2$ is a constant.

So, we can write $\frac{x}{y^2} = k$, i.e., $x = k y^2$, and not $x y^2 = k$.

∴ Assertion (A) is false.

Two quantities $x$ and $y$ are said to vary directly if $\frac{x}{y}$ is a constant quantity. This is the correct definition of direct variation.

∴ Reason (R) is true.

Hence, Option 4 is the correct option.

More the number of workers, less is the time taken to complete a job. Also, the product (number of workers × time taken) remains the same.

So, the number of workers and the time taken to complete a job is a case of indirect (inverse) variation.

∴ Assertion (A) is true.

Two quantities are said to be in inverse proportion, if the product of their corresponding values remains the same. This is the correct definition of inverse proportion, and it is exactly the property used to conclude the Assertion (A).

∴ Reason (R) is true and it is the correct explanation of Assertion (A).

Hence, Option 1 is the correct option.

Since $x$ and $y$ are in direct proportion,

$$\begin{aligned}\Rightarrow \frac{x}{y} = k \text{ (a constant)} \\[1em] \Rightarrow x = ky\end{aligned}$$

Taking the reciprocals of both the sides, we get :

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{ky} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{k} \times \frac{1}{y} \\[1em] \Rightarrow \frac{\Big(\frac{1}{x}\Big)}{\Big(\frac{1}{y}\Big)} = \frac{1}{k} \text{ (a constant)}\end{aligned}$$

As the ratio of $\frac{1}{x}$ and $\frac{1}{y}$ is a constant, they are in direct proportion.

Hence, Option 1 is the correct option.

We have : More time, Less speed. [Inverse variation]

The distance to the school is fixed. So, the speed and the time taken are in inverse proportion.

In 20 minutes, the required speed = 6 km/h.

In 1 minute, the required speed = (6 × 20) km/h.

In 24 minutes, the required speed = $\Big(\frac{6 \times 20}{24}\Big)$ km/h = 5 km/h.

Hence, Option 1 is the correct option.

Since $x$ is inversely proportional to $y$, the product $xy$ remains constant.

So, an increase in $x$ produces a decrease in $y$ of the same measure.

∴ If $x$ increases by $a$%, then $y$ decreases by $a$%.

Hence, Option 1 is the correct option.

Since $y$ is directly proportional to the cube root of $x$,

$y = k\sqrt[3]{x}$ .....(1)

where $k$ is a constant.

Substituting $y = 18$ and $x = 27$ in equation (1), we get :

$$\begin{aligned}\Rightarrow 18 = k\sqrt[3]{27} \\[1em] \Rightarrow 18 = k \times 3 \\[1em] \Rightarrow k = \frac{18}{3} = 6\end{aligned}$$

∴ Equation (1) becomes $y = 6\sqrt[3]{x}$.

Putting $x = 125$, we get :

$y = 6\sqrt[3]{125} = 6 \times 5 = 30$

Hence, Option 3 is the correct option.

We have : More children, Less number of days the food will last. [Inverse variation]

For 20 children, the food is sufficient for 15 days.

For 1 child, the food is sufficient for (15 × 20) days.

For 25 children, the food is sufficient for $\Big(\frac{15 \times 20}{25}\Big)$ days = 12 days.

Hence, Option 4 is the correct option.

Since $y$ is directly proportional to $x^2$,

$y = kx^2$ .....(1)

where $k$ is a constant.

Substituting $y = 8$ and $x = 4$ in equation (1), we get :

$$\begin{aligned}\Rightarrow 8 = k \times (4)^2 \\[1em] \Rightarrow 8 = 16k \\[1em] \Rightarrow k = \frac{8}{16} = \frac{1}{2}\end{aligned}$$

∴ Equation (1) becomes $y = \frac{1}{2}x^2$.

Putting $x = 5$, we get :

$y = \frac{1}{2} \times (5)^2 = \frac{25}{2} = 12.5$

Hence, Option 3 is the correct option.