Direct and Inverse Variation

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

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

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

⇒ $\frac{x}{y}$ is not a constant

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

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

$$\begin{aligned}\frac{3}{9} = \frac{1}{3}, \frac{5}{15} = \frac{1}{3}, \frac{7}{21} = \frac{1}{3},\\[1em] \frac{9}{27} = \frac{1}{3}, \frac{10}{30} = \frac{1}{3}\end{aligned}$$

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

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

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

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

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

$x$	$y$
3	45
5	$y_1$
$x_1$	90
$x_2$	120
10	$y_2$

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

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

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

$$\begin{aligned}\Rightarrow 3y_1 = 5 \times 45\\[1em] \Rightarrow y_1 = \frac{225}{3}\\[1em] \Rightarrow y_1 = 75\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 45x_1 = 3 \times 90\\[1em] \Rightarrow x_1 = \frac{270}{45}\\[1em] \Rightarrow x_1 = 6\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 45x_2 = 3 \times 120\\[1em] \Rightarrow x_2 = \frac{360}{45}\\[1em] \Rightarrow x_2 = 8\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 3y_2 = 10 \times 45\\[1em] \Rightarrow y_2 = \frac{450}{3}\\[1em] \Rightarrow y_2 = 150\end{aligned}$$

Thus, we have :

$x$	$y$
3	45
5	75
6	90
8	120
10	150

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

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

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

$x$	$y$
4	7
8	$y_1$
$x_1$	21
20	$y_2$
28	$y_3$

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

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

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

$$\begin{aligned}\Rightarrow 4y_1 = 8 \times 7\\[1em] \Rightarrow y_1 = \frac{56}{4}\\[1em] \Rightarrow y_1 = 14\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 7x_1 = 4 \times 21\\[1em] \Rightarrow x_1 = \frac{84}{7}\\[1em] \Rightarrow x_1 = 12\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 4y_2 = 20 \times 7\\[1em] \Rightarrow y_2 = \frac{140}{4}\\[1em] \Rightarrow y_2 = 35\end{aligned}$$

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

$$\begin{aligned}\Rightarrow 4y_3 = 28 \times 7\\[1em] \Rightarrow y_3 = \frac{196}{4}\\[1em] \Rightarrow y_3 = 49\end{aligned}$$

Thus, we have :

$x$	$y$
4	7
8	14
12	21
20	35
28	49

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

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

Length (in m)	Cost (in ₹)
8	250
5.8	$x$

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{8}{250} = \frac{5.8}{x}\\[1em] \Rightarrow x = \frac{250 \times 5.8}{8}\\[1em] \Rightarrow x = \frac{1{,}450}{8}\\[1em] \Rightarrow x = 181.25\end{aligned}$$

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

We know that,

1 week = 7 days

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

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

Days	Earning (in ₹)
7	3,360
18	$x$

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

Hence, it is a case of direct variation.

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

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

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

Dollars	Cost (in ₹)
175	14,700
$x$	48,048

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

Hence, it is a case of direct variation.

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

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

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

Distance (in km)	Petrol (in litres)
67.5	4.5
$x$	26.4

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{67.5}{4.5} = \frac{x}{26.4}\\[1em] \Rightarrow x = \frac{67.5 \times 26.4}{4.5}\\[1em] \Rightarrow x = \frac{1782}{4.5}\\[1em] \Rightarrow x = 396\end{aligned}$$

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

Converting the thickness into the same unit,

90 cm = 90 × 10 = 900 mm

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

Sheets	Thickness (in mm)
12	45
$x$	900

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{12}{45} = \frac{x}{900}\\[1em] \Rightarrow x = \frac{12 \times 900}{45}\\[1em] \Rightarrow x = \frac{10800}{45}\\[1em] \Rightarrow x = 240\end{aligned}$$

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

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.

Height of mast (in cm)	Length of ship (in cm)
6	$x$
900	3300

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{6}{x} = \frac{900}{3300}\\[1em] \Rightarrow x = \frac{6 \times 3300}{900}\\[1em] \Rightarrow x = \frac{19800}{900}\\[1em] \Rightarrow x = 22\end{aligned}$$

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

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

Length (in cm)	Mass (in g)
16	192
$x$	105

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{16}{192} = \frac{x}{105}\\[1em] \Rightarrow x = \frac{16 \times 105}{192}\\[1em] \Rightarrow x = \frac{1680}{192}\\[1em] \Rightarrow x = 8.75\end{aligned}$$

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

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

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

Map distance (in cm)	Actual distance (in km)
1	20
3.5	$x$

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

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{1}{20} = \frac{3.5}{x}\\[1em] \Rightarrow x = 3.5 \times 20\\[1em] \Rightarrow x = 70\end{aligned}$$

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

Converting the heights into metres,

Height of the water tank = 23 m 75 cm

= $23\frac{75}{100}$ 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

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

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

Height (in m)	Length of shadow (in m)
23.75	20
9.5	$x$

$$\begin{aligned}\Rightarrow \frac{23.75}{20} = \frac{9.5}{x}\\[1em] \Rightarrow x = \frac{9.5 \times 20}{23.75}\\[1em] \Rightarrow x = \frac{190}{23.75}\\[1em] \Rightarrow x = 8\end{aligned}$$

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

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

Height (in m)	Length of shadow (in m)
23.75	20
$y$	12

$$\begin{aligned}\Rightarrow \frac{23.75}{20} = \frac{y}{12}\\[1em] \Rightarrow y = \frac{23.75 \times 12}{20}\\[1em] \Rightarrow y = \frac{285}{20}\\[1em] \Rightarrow y = 14.25\end{aligned}$$

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.

Given, earning of 5 men = earning of 7 women

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

⇒ Earning of 10 men = earning of $\left(\frac{7}{5} \times 10\right)$ 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.

Number of women	Earning per day (in ₹)
7	2,625
27	$x$

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

Hence, it is a case of direct variation.

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

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

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.

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

$$\begin{aligned}90 \times 10 = 900, 60 \times 15 = 900, 45 \times 20 = 900,\\[1em] 30 \times 30 = 900, 20 \times 45 = 900\end{aligned}$$

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.

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

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

⇒ $xy$ is not a constant

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

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

Volume (in cm3)	Pressure (in mm of mercury)
630	360
720	$x$

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 630 \times 360 = 720 \times x\\[1em] \Rightarrow 226800 = 720x\\[1em] \Rightarrow x = \frac{226800}{720}\\[1em] \Rightarrow x = 315\end{aligned}$$

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

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

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

Number of children	Sweets for each child
20	4
16	$x$

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 20 \times 4 = 16 \times x\\[1em] \Rightarrow 80 = 16x\\[1em] \Rightarrow x = \frac{80}{16}\\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, each child will get 5 sweets.

Converting 90 paise into rupees,

90 paise = ₹ $\frac{90}{100} = 0.90$

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

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

Price per orange (in ₹)	Number of oranges
4.50	36
5.40	$x$

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.

$$\begin{aligned}\Rightarrow 4.50 \times 36 = 5.40 \times x\\[1em] \Rightarrow 162 = 5.40x\\[1em] \Rightarrow x = \frac{162}{5.40}\\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, Pooja can buy 30 oranges.

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

Number of men	Number of days
12	8
$x$	6

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 12 \times 8 = x \times 6\\[1em] \Rightarrow 96 = 6x\\[1em] \Rightarrow x = \frac{96}{6}\\[1em] \Rightarrow x = 16\end{aligned}$$

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

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

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

Number of taps	Time (in minutes)
8	27
6	$x$

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 8 \times 27 = 6 \times x\\[1em] \Rightarrow 216 = 6x\\[1em] \Rightarrow x = \frac{216}{6}\\[1em] \Rightarrow x = 36\end{aligned}$$

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

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

Number of persons	Number of months
560	9
$x$	5

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 560 \times 9 = x \times 5\\[1em] \Rightarrow 5040 = 5x\\[1em] \Rightarrow x = \frac{5040}{5}\\[1em] \Rightarrow x = 1008\end{aligned}$$

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

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

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

Bottles in each box	Number of boxes
10	30
12	$x$

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 10 \times 30 = 12 \times x\\[1em] \Rightarrow 300 = 12x\\[1em] \Rightarrow x = \frac{300}{12}\\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, 25 boxes would be filled.

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

Time (in minutes)	Speed (in km/h)
24	5
20	$x$

Less the time taken, more should be the speed.

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 24 \times 5 = 20 \times x\\[1em] \Rightarrow 120 = 20x\\[1em] \Rightarrow x = \frac{120}{20}\\[1em] \Rightarrow x = 6\end{aligned}$$

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

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.

Number of soldiers	Number of days
80	45
100	$x$

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.

$$\begin{aligned}\Rightarrow 80 \times 45 = 100 \times x\\[1em] \Rightarrow 3600 = 100x\\[1em] \Rightarrow x = \frac{3600}{100}\\[1em] \Rightarrow x = 36\end{aligned}$$

Hence, the food would last for 36 days.

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.

Number of soldiers	Number of days
1200	24
$x$	32

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.

$$\begin{aligned}\Rightarrow 1200 \times 24 = x \times 32\\[1em] \Rightarrow 28800 = 32x\\[1em] \Rightarrow x = \frac{28800}{32}\\[1em] \Rightarrow x = 900\end{aligned}$$

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

Hence, 300 soldiers left the fort.

We know that,

One day's work = $\frac{1}{\text{number of days to complete the work}}$

Farmer's one day's work = $\frac{1}{10}$

Wife's one day's work = $\frac{1}{8}$

$$\begin{aligned}\text{(Farmer + Wife)'s one day's work} = \frac{1}{10} + \frac{1}{8}\\[1em] = \frac{4 + 5}{40}\\[1em] = \frac{9}{40}\end{aligned}$$

Number of days taken by them working together = $\frac{1}{\frac{9}{40}} = \frac{40}{9} = 4\frac{4}{9}$

Hence, working together, the farmer and his wife can reap the field in $\mathbf{4\frac{4}{9}}$ days.

A can do $\frac{1}{5}$th of the work in 2 days.

$$\begin{aligned}\text{A's one day's work} = \frac{1}{5 \times 2}\\[1em] = \frac{1}{10}\end{aligned}$$

B can do $\frac{2}{3}$rd of the work in 8 days.

$$\begin{aligned}\text{B's one day's work} = \frac{2}{3 \times 8}\\[1em] = \frac{2}{24}\\[1em] = \frac{1}{12}\\[1em] \text{(A + B)'s one day's work} = \frac{1}{10} + \frac{1}{12}\\[1em] = \frac{6 + 5}{60}\\[1em] = \frac{11}{60}\end{aligned}$$

Number of days taken by A and B working together = $\frac{1}{\frac{11}{60}} = \frac{60}{11} = 5\frac{5}{11}$

Hence, A and B together can complete the work in $\mathbf{5\frac{5}{11}}$ days.

First tap's one minute work = $\frac{1}{20}$

Second tap's one minute work = $\frac{1}{12}$

$$\begin{aligned}\text{Both taps' one minute work} = \frac{1}{20} + \frac{1}{12}\\[1em] = \frac{3 + 5}{60}\\[1em] = \frac{8}{60}\\[1em] = \frac{2}{15}\end{aligned}$$

Number of minutes taken by both taps together = $\frac{1}{\frac{2}{15}} = \frac{15}{2} = 7\frac{1}{2}$

Hence, the tank will be full in $\mathbf{7\frac{1}{2}}$ minutes.

A's one day's work = $\frac{1}{6}$

B's one day's work = $\frac{1}{8}$

$$\begin{aligned}\text{(A + B)'s one day's work} = \frac{1}{6} + \frac{1}{8}\\[1em] = \frac{4 + 3}{24}\\[1em] = \frac{7}{24}\\[1em] \text{(A + B)'s 2 days' work} = \frac{7}{24} \times 2\\[1em] = \frac{7}{12}\\[1em] \text{Remaining work} = 1 - \frac{7}{12}\\[1em] = \frac{12 - 7}{12}\\[1em] = \frac{5}{12}\\[1em] \text{Number of days taken by A to finish the remaining work} = \frac{\text{Remaining work}}{\text{A's one day's work}} = \frac{\frac{5}{12}}{\frac{1}{6}}\\[1em] = \frac{5 \times 6}{12 \times 1}\\[1em] = \frac{30}{12}\\[1em] = \frac{5}{2}\\[1em] = 2\frac{1}{2}\end{aligned}$$

Hence, A will require $\mathbf{2\frac{1}{2}}$ days to finish the work.

A's one day's work = $\frac{1}{40}$

$$\begin{aligned}\text{A's 8 days' work} = \frac{1}{40} \times 8\\[1em] = \frac{1}{5}\\[1em] \text{Remaining work} = 1 - \frac{1}{5}\\[1em] = \frac{5 - 1}{5}\\[1em] = \frac{4}{5}\end{aligned}$$

B finishes $\frac{4}{5}$ of the work in 16 days.

$$\begin{aligned}\text{B's one day's work} = \frac{4}{5 \times 16}\\[1em] = \frac{4}{80}\\[1em] = \frac{1}{20}\\[1em] \text{(A + B)'s one day's work} = \frac{1}{40} + \frac{1}{20}\\[1em] = \frac{1 + 2}{40}\\[1em] = \frac{3}{40}\end{aligned}$$

Number of days taken by A and B working together = $\frac{1}{\frac{3}{40}} = \frac{40}{3} = 13\frac{1}{3}$

Hence, working together from the beginning, they would have taken $\mathbf{13\frac{1}{3}}$ days to complete the work.

A's one day's work = $\frac{1}{10}$

B's one day's work = $\frac{1}{15}$

$$\begin{aligned}\text{(A + B)'s one day's work} = \frac{1}{10} + \frac{1}{15}\\[1em] = \frac{3 + 2}{30}\\[1em] = \frac{5}{30}\\[1em] = \frac{1}{6}\\[1em] \text{A's 5 days' work} = \frac{1}{10} \times 5\\[1em] = \frac{1}{2}\end{aligned}$$

Let A and B work together for $x$ days.

Hence, (A + B)'s $x$ days' work + A's 5 days' work = 1

$$\begin{aligned}\Rightarrow \frac{1}{6} \times x + \frac{1}{2} = 1\\[1em] \Rightarrow \frac{x}{6} = 1 - \frac{1}{2}\\[1em] \Rightarrow \frac{x}{6} = \frac{1}{2}\\[1em] \Rightarrow x = \frac{1}{2} \times 6\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, A and B had worked together for 3 days.

Given, 3 women's work = 5 girls' work

⇒ 1 woman's work = $\frac{5}{3}$ girls' work

⇒ 7 women's work = $\left(\frac{5}{3} \times 7\right)$ girls' work = $\frac{35}{3}$ girls' work

⇒ 7 women and 11 girls' work = $\left(\frac{35}{3} + 11\right)$ girls' work

= $\frac{35 + 33}{3}$ girls' work

= $\frac{68}{3}$ girls' work

Since 5 girls take 17 days to complete the work,

Hence, 1 girl takes 5 × 17 = 85 days to complete the work.

$$\begin{aligned}\text{Number of days taken by }\frac{68}{3} \text{ girls} = \frac{85}{\frac{68}{3}}\\[1em] = \frac{85 \times 3}{68}\\[1em] = \frac{255}{68}\\[1em] = \frac{15}{4}\\[1em] = 3\frac{3}{4}\end{aligned}$$

Hence, 7 women and 11 girls working together will complete the work in $\mathbf{3\frac{3}{4}}$ days.

We know that,

Remuneration is in proportion of the work done.

A's one day's work = $\frac{1}{10}$

B's one day's work = $\frac{1}{15}$

Hence, A and B should share the money in the ratio $\frac{1}{10} : \frac{1}{15}$

Multiplying both the terms of the ratio by 30, we get :

$\frac{1}{10} \times 30 : \frac{1}{15} \times 30 = 3 : 2$

Sum of the terms of the ratio = 3 + 2 = 5

A's share = $\frac{3}{5}$ of ₹ 7,000 = ₹ $\left(\frac{3}{5} \times 7{,}000\right) = 4{,}200$

B's share = $\frac{2}{5}$ of ₹ 7,000 = ₹ $\left(\frac{2}{5} \times 7{,}000\right) = 2{,}800$

Hence, A gets ₹ 4,200 and B gets ₹ 2,800.

A's one day's work = $\frac{1}{2}$

B's one day's work = $\frac{1}{6}$

C's one day's work = $\frac{1}{3}$

$$\begin{aligned}\text{(A + B + C)'s one day's work} = \frac{1}{2} + \frac{1}{6} + \frac{1}{3}\\[1em] = \frac{3 + 1 + 2}{6}\\[1em] = \frac{6}{6} \\[1em] = 1.\end{aligned}$$

Number of days taken by A, B and C working together = $\frac{1}{1} = 1$

We know that,

Remuneration is in proportion of the work done.

Hence, they should divide the money in the ratio $\frac{1}{2} : \frac{1}{6} : \frac{1}{3}$

Multiplying all the terms of the ratio by 6, we get :

$\frac{1}{2} \times 6 : \frac{1}{6} \times 6 : \frac{1}{3} \times 6 = 3 : 1 : 2$

Sum of the terms of the ratio = 3 + 1 + 2 = 6

A's share = $\frac{3}{6}$ of ₹ 1,920 = ₹ $\left(\frac{3}{6} \times 1{,}920\right) = 960$

B's share = $\frac{1}{6}$ of ₹ 1,920 = ₹ $\left(\frac{1}{6} \times 1{,}920\right) = 320$

C's share = $\frac{2}{6}$ of ₹ 1,920 = ₹ $\left(\frac{2}{6} \times 1{,}920\right) = 640$

Hence, working together they require 1 day, and A gets ₹ 960, B gets ₹ 320 and C gets ₹ 640.

(A + B + C)'s one day's work = $\frac{1}{15}$

B's one day's work = $\frac{1}{30}$

C's one day's work = $\frac{1}{40}$

A's one day's work = (A + B + C)'s one day's work - B's one day's work - C's one day's work

$$\begin{aligned}=\frac{1}{15} - \frac{1}{30} - \frac{1}{40}\\[1em] = \frac{8 - 4 - 3}{120}\\[1em] = \frac{1}{120}\end{aligned}$$

Number of days taken by A alone = $\frac{1}{\frac{1}{120}} = 120$

Hence, A alone will do the work in 120 days.

Converting the mixed fraction into an improper fraction,

A, B and C together can plough the field in $4\frac{4}{5} = \frac{24}{5}$ days.

(A + B + C)'s one day's work = $\frac{1}{\frac{24}{5}} = \frac{5}{24}$

(A + C)'s one day's work = $\frac{1}{8}$

B's one day's work = (A + B + C)'s one day's work - (A + C)'s one day's work

$$\begin{aligned}= \frac{5}{24} - \frac{1}{8}\\[1em] = \frac{5 - 3}{24}\\[1em] = \frac{2}{24}\\[1em] = \frac{1}{12}\end{aligned}$$

Number of days taken by B alone = $\frac{1}{\frac{1}{12}} = 12$

Hence, B working alone will take 12 days to plough the field.

(A + B)'s one day's work = $\frac{1}{10}$

(B + C)'s one day's work = $\frac{1}{15}$

(C + A)'s one day's work = $\frac{1}{12}$

Adding these three, we get :

$$\begin{aligned}\text{2(A + B + C)'s one day's work} = \frac{1}{10} + \frac{1}{15} + \frac{1}{12}\\[1em] = \frac{6 + 4 + 5}{60}\\[1em] = \frac{15}{60}\\[1em] = \frac{1}{4}\\[1em] \text{(A + B + C)'s one day's work} = \frac{1}{4 \times 2}\\[1em] = \frac{1}{8}\end{aligned}$$

Number of days taken by A, B and C working together = $\frac{1}{\frac{1}{8}} = 8$

A's one day's work = (A + B + C)'s one day's work - (B + C)'s one day's work

$$\begin{aligned}= \frac{1}{8} - \frac{1}{15}\\[1em] = \frac{15 - 8}{120}\\[1em] = \frac{7}{120}\end{aligned}$$

Number of days taken by A alone = $\frac{120}{7} = 17\frac{1}{7}$

B's one day's work = (A + B + C)'s one day's work - (C + A)'s one day's work

$$\begin{aligned}= \frac{1}{8} - \frac{1}{12}\\[1em] = \frac{3 - 2}{24}\\[1em] = \frac{1}{24}\end{aligned}$$

Number of days taken by B alone = $\frac{1}{\frac{1}{24}} = 24$

C's one day's work = (A + B + C)'s one day's work - (A + B)'s one day's work

$$\begin{aligned}= \frac{1}{8} - \frac{1}{10}\\[1em] = \frac{5 - 4}{40}\\[1em] = \frac{1}{40}\end{aligned}$$

Number of days taken by C alone = $\frac{1}{\frac{1}{40}} = 40$

Hence, working all together they will finish the work in 8 days, and working alone A takes $\mathbf{17\frac{1}{7}}$ days, B takes 24 days and C takes 40 days.

Portion of the tank filled by the pipe in one hour = $\frac{1}{12}$

Portion of the tank filled in one hour when the waste pipe is also open = $\frac{1}{16}$

$$\begin{aligned}\text{Portion of the tank emptied by the waste pipe in one hour} = \frac{1}{12} - \frac{1}{16}\\[1em] = \frac{4 - 3}{48}\\[1em] = \frac{1}{48}\end{aligned}$$

Time taken by the waste pipe to empty the full tank = $\frac{1}{\frac{1}{48}} = 48$

Hence, the waste pipe will take 48 hours to empty the full tank.

(i) Two quantities are said to be in direct variation if an increase (or decrease) in one quantity causes increase (or decrease) in other quantity.

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

(iii) The total cost of articles varies directly to the number of articles purchased.

(iv) More work is done in more time.

(v) The time taken to finish a work varies inversely to the number of men at work.

(vi) The speed of a moving object varies inversely to the time taken to cover a certain distance.

(vii) The number of articles varies inversely with the cost per article, if a fixed amount is available.

(viii) Remuneration is in proportion of work done.

(i) Two quantities $x$ and $y$ are said to be in inverse variation if $xy$ is constant and not if $\frac{x}{y}$ is constant.

Hence, the given statement is false.

(ii) Number of days needed to complete the work = $\frac{1}{\text{one day's work}}$

Hence, the given statement is true.

(iii) Two quantities $x$ and $y$ are in direct variation if $\frac{x}{y}$ is constant, i.e. if $x = ky$, where $k$ is constant of variation.

Hence, the given statement is true.

(iv) More the number of men at work, more will be the work done. So, the work done varies directly and not inversely to the number of men at work.

Hence, the given statement is false.

(v) In the given time, more the speed of a moving object, more will be the distance covered by it.

Hence, the given statement is true.

(vi) More the money deposited in a bank, more will be the interest earned on it.

Hence, the given statement is true.

(vii) More the number of articles purchased, more will be the total cost. So, the total cost increases and does not decrease.

Hence, the given statement is false.

(viii) At the same time, more the length of the object, more will be the length of its shadow.

Hence, the given statement is true.

(ix) More the consumption of petrol, more will be the distance covered. So, the distance covered varies directly and not inversely to the consumption of petrol.

Hence, the given statement is false.

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

$$\begin{aligned}\Rightarrow xy = k\\[1em] \Rightarrow x = \frac{k}{y}\\[1em] \Rightarrow x \propto \frac{1}{y}\end{aligned}$$

So, all the three given relations represent inverse variation.

Hence, Option 4 is the correct option.

More the length of the wire, more will be its cost, i.e. length and cost are in direct variation.

Let the cost of 8 metre wire be ₹ $x$.

Length (in m)	Cost (in ₹)
12	240
8	$x$

$$\begin{aligned}\Rightarrow \frac{12}{240} = \frac{8}{x}\\[1em] \Rightarrow x = \frac{240 \times 8}{12}\\[1em] \Rightarrow x = \frac{1{,}920}{12}\\[1em] \Rightarrow x = 160\end{aligned}$$

Hence, Option 1 is the correct option.

More the quantity of wheat, more will be its cost, i.e. quantity and cost are in direct variation.

Let the cost of 20 kg wheat be ₹ $x$.

Wheat (in kg)	Cost (in ₹)
5	120
20	$x$

$$\begin{aligned}\Rightarrow \frac{5}{120} = \frac{20}{x}\\[1em] \Rightarrow x = \frac{120 \times 20}{5}\\[1em] \Rightarrow x = \frac{2{,}400}{5}\\[1em] \Rightarrow x = 480\end{aligned}$$

Hence, Option 4 is the correct option.

More the number of men at work, less will be the number of days taken, i.e. number of men and number of days are in inverse variation.

Let 30 men complete the work in $x$ days.

Number of men	Number of days
10	6
30	$x$

$$\begin{aligned}\Rightarrow 10 \times 6 = 30 \times x\\[1em] \Rightarrow 60 = 30x\\[1em] \Rightarrow x = \frac{60}{30}\\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, Option 1 is the correct option.

More the quantity of petrol, more will be the distance covered, i.e. petrol and distance are in direct variation.

Let the distance covered in 15 litres of petrol be $x$ km.

Distance (in km)	Petrol (in litres)
80	5
$x$	15

$$\begin{aligned}\Rightarrow \frac{80}{5} = \frac{x}{15}\\[1em] \Rightarrow x = \frac{80 \times 15}{5}\\[1em] \Rightarrow x = \frac{1200}{5}\\[1em] \Rightarrow x = 240\end{aligned}$$

Hence, Option 2 is the correct option.

Number of students after 50 new students joined = 200 + 50 = 250

More the number of students, less will be the number of days for which the food lasts, i.e. number of students and number of days are in inverse variation.

Let the food last for $x$ days for 250 students.

Number of students	Number of days
200	20
250	$x$

$$\begin{aligned}\Rightarrow 200 \times 20 = 250 \times x\\[1em] \Rightarrow 4000 = 250x\\[1em] \Rightarrow x = \frac{4000}{250}\\[1em] \Rightarrow x = 16\end{aligned}$$

Hence, Option 2 is the correct option.

More the number of persons at work, less will be the number of days taken, i.e. number of persons and number of days are in inverse variation.

Let 4 persons paint the house in $x$ days.

Number of persons	Number of days
3	8
4	$x$

$$\begin{aligned}\Rightarrow 3 \times 8 = 4 \times x\\[1em] \Rightarrow 24 = 4x\\[1em] \Rightarrow x = \frac{24}{4}\\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, Option 2 is the correct option.

The photograph is enlarged 100000 times, so the actual length of the bacteria is $\frac{1}{100000}$ times the length in the photograph.

Actual length of the bacteria = $\frac{5}{100000}$ cm

= 0.00005 cm

= 5 × 10^-5 cm

Converting the length into metres,

5 × 10^-5 cm = 5 × 10^-5 × $\frac{1}{100}$ m = 5 × 10^-7 m

So, all the three given values represent the actual length of the bacteria.

Hence, Option 4 is the correct option.

At the same time, more the height of an object, more will be the length of its shadow, i.e. height and length of shadow are in direct variation.

Let the height of the tree be $x$ m.

Height (in m)	Length of shadow (in m)
12	8
$x$	6

$$\begin{aligned}\Rightarrow \frac{12}{8} = \frac{x}{6}\\[1em] \Rightarrow x = \frac{12 \times 6}{8}\\[1em] \Rightarrow x = \frac{72}{8}\\[1em] \Rightarrow x = 9\end{aligned}$$

Hence, Option 2 is the correct option.

Converting the time into minutes,

1 hour = 60 minutes

Less the number of pipes, more will be the time taken to fill the tank, i.e. number of pipes and time are in inverse variation.

Let 4 pipes fill the tank in $x$ minutes.

Number of pipes	Time (in minutes)
5	60
4	$x$

$$\begin{aligned}\Rightarrow 5 \times 60 = 4 \times x\\[1em] \Rightarrow 300 = 4x\\[1em] \Rightarrow x = \frac{300}{4}\\[1em] \Rightarrow x = 75\end{aligned}$$

Hence, Option 1 is the correct option.

Portion of the tank filled by the first tap in one hour = $\frac{1}{8}$

Portion of the tank emptied by the second tap in one hour = $\frac{1}{10}$

$$\begin{aligned}\text{Portion of the tank filled in one hour when both taps work together} = \frac{1}{8} - \frac{1}{10}\\[1em] = \frac{5 - 4}{40}\\[1em] = \frac{1}{40}\end{aligned}$$

Time taken to fill the tank = $\frac{1}{\frac{1}{40}} = 40$ hours

Hence, Option 4 is the correct option.

If on increasing a quantity $x$, another quantity $y$ also increases, then $x$ and $y$ vary directly and not inversely to each other.

So, Statement I is false.

Two quantities $x$ and $y$ vary inversely if $xy = k$, i.e. if $x = \frac{k}{y}$, where $k$ is a constant.

So, Statement II is true.

Hence, Option 2 is the correct option.

Less the time in which the tank is to be filled, more will be the number of pipes required, i.e. number of pipes and time are in inverse variation.

Half of the time = $\frac{10}{2} = 5$ hours

Let $x$ pipes be needed to fill the tank in 5 hours.

Number of pipes	Time (in hours)
5	10
$x$	5

$$\begin{aligned}\Rightarrow 5 \times 10 = x \times 5\\[1em] \Rightarrow 50 = 5x\\[1em] \Rightarrow x = \frac{50}{5}\\[1em] \Rightarrow x = 10\end{aligned}$$

So, Statement I is true.

Time required to do a certain work = $\frac{\text{work to be done}}{\text{one day's work}}$

So, Statement II is true.

Hence, Option 3 is the correct option.

Rohit's one day's work = $\frac{1}{1} = 1$

Virat's one day's work = $\frac{1}{2}$

Rishabh's one day's work = $\frac{1}{4}$

$$\begin{aligned}\text{(Rohit + Virat + Rishabh)'s one day's work} = 1 + \frac{1}{2} + \frac{1}{4}\\[1em] = \frac{4 + 2 + 1}{4}\\[1em] = \frac{7}{4}\end{aligned}$$

Number of days taken by them working together = $\frac{1}{\frac{7}{4}} = \frac{4}{7}$

As $\frac{4}{7}$ is less than 1, they will take less than 1 day to build the wall.

So, Statement I is true.

Remuneration is in proportion of the work done. Since Rohit, Virat and Rishabh do different amounts of work, they should divide the money in the ratio $1 : \frac{1}{2} : \frac{1}{4}$, i.e. 4 : 2 : 1 and not equally among themselves.

So, Statement II is false.

Hence, Option 1 is the correct option.

(i) Two quantities $x$ and $y$ vary inversely to each other if $xy$ is constant.

Hence, (i) → (c).

(ii) Two quantities $m$ and $n$ vary directly if $\frac{m}{n}$ is constant, i.e. $m \propto n$.

Hence, (ii) → (f).

(iii) Two quantities $p$ and $q$ vary inversely if $pq$ is constant, i.e. $p \propto \frac{1}{q}$.

Hence, (iii) → (d).

(iv) If $x$ and $y$ vary inversely, then $xy$ is constant. So, on decreasing $x$, the value of $y$ will increase in proportion.

Hence, (iv) → (b).

(v) Two quantities $x$ and $y$ vary directly to each other if $\frac{x}{y}$ is constant.

Hence, (v) → (a).

(vi) If $x$ and $y$ vary directly, then $\frac{x}{y}$ is constant. So, on decreasing $x$, the value of $y$ will decrease in proportion.

Hence, (vi) → (e).

Thus, we have :

Column I	Column II
(i) $x$ and $y$ vary inversely to each other	(c) $xy$ = constant
(ii) Mathematical representation of direct variation of quantities $m$ and $n$	(f) $m \propto n$
(iii) Mathematical representation of inverse variation of quantities $p$ and $q$	(d) $p \propto \frac{1}{q}$
(iv) If $x$ and $y$ vary inversely then on decreasing $x$	(b) $y$ will increase in proportion
(v) $x$ and $y$ vary directly to each other	(a) $\frac{x}{y}$ = constant
(vi) If $x$ and $y$ vary directly then on decreasing $x$	(e) $y$ will decrease in proportion

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

$$\begin{aligned}\Rightarrow \frac{6}{18} = \frac{p}{27} = \frac{15}{q}\\[1em] \Rightarrow \frac{1}{3} = \frac{p}{27} \text{ and } \frac{1}{3} = \frac{15}{q}\\[1em] \Rightarrow p = \frac{27}{3} \text{ and } q = 15 \times 3\\[1em] \Rightarrow p = 9 \text{ and } q = 45\end{aligned}$$

Hence, $\bm{p} = 9$ and $\bm{q} = 45$.

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

$$\begin{aligned}\Rightarrow 2 \times 45 = a \times 12.5 = 6 \times b\\[1em] \Rightarrow 90 = 12.5a = 6b\\[1em] \Rightarrow a = \frac{90}{12.5} \text{ and } b = \frac{90}{6}\\[1em] \Rightarrow a = 7.2 \text{ and } b = 15\end{aligned}$$

Hence, $\bm{a = 7.2}$ and $\bm{b = 15}$.

The rainfall in a month does not increase or decrease in a fixed ratio with the number of days. It may rain heavily on some days and may not rain at all on the other days.

So, the rainfall and the number of days are neither in direct variation nor in inverse variation. Hence, the total rainfall in April cannot be found from the rainfall of the first 20 days.

Hence, we can't say what the total rainfall in April would be.

Number of hours for which Mamta worked = 48 - 6 = 42

Let Mamta earn ₹ $x$ for 42 hours of work.

Working hours	Earning (in ₹)
48	3,240
42	$x$

Less the number of hours worked, less will be the earning.

Hence, it is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{48}{3{,}240} = \frac{42}{x}\\[1em] \Rightarrow x = \frac{3{,}240 \times 42}{48}\\[1em] \Rightarrow x = \frac{1{,}36{,}080}{48}\\[1em] \Rightarrow x = 2{,}835\end{aligned}$$

Hence, Mamta earned ₹ 2,835.

New working hours per day = 10 + 2 = 12

Let Navjot do the same work in $x$ days working 12 hours per day.

Working hours per day	Number of days
10	6
12	$x$

More the working hours per day, less will be the number of days required.

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 10 \times 6 = 12 \times x\\[1em] \Rightarrow 60 = 12x\\[1em] \Rightarrow x = \frac{60}{12}\\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, Navjot can do the same work in 5 days.

Converting 30 paise into rupees,

30 paise = ₹ $\frac{30}{100} = 0.30$

Hence, the new price of each banana = ₹ (1.50 - 0.30) = ₹ 1.20

Let Sharmila buy $x$ bananas at ₹ 1.20 per banana.

Price per banana (in ₹)	Number of bananas
1.50	24
1.20	$x$

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

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 1.50 \times 24 = 1.20 \times x\\[1em] \Rightarrow 36 = 1.20x\\[1em] \Rightarrow x = \frac{36}{1.20}\\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, Sharmila can buy 30 bananas.

After 10 days, the remaining food is sufficient for 180 soldiers for (40 - 10) days, i.e. 30 days.

After 10 days, 30 soldiers leave the fort.

Hence, the number of soldiers in the fort = 180 - 30 = 150

Let the remaining food last for $x$ days when there are 150 soldiers.

Number of soldiers	Number of days
180	30
150	$x$

Less the number of soldiers, more will be the number of days for which the food lasts.

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 180 \times 30 = 150 \times x\\[1em] \Rightarrow 5400 = 150x\\[1em] \Rightarrow x = \frac{5400}{150}\\[1em] \Rightarrow x = 36\end{aligned}$$

Hence, the total number of days for which the food will last = 10 + 36 = 46

Hence, the food will last for a total of 46 days.

Given, 4 goats' work = 6 sheep's work

⇒ 4 goats and 14 sheep's work = (6 + 14) sheep's work = 20 sheep's work

Let 20 sheep graze the same field in $x$ days.

Number of sheep	Number of days
6	40
20	$x$

More the number of sheep, less will be the number of days taken to graze the field.

Hence, it is a case of inverse variation.

$$\begin{aligned}\Rightarrow 6 \times 40 = 20 \times x\\[1em] \Rightarrow 240 = 20x\\[1em] \Rightarrow x = \frac{240}{20}\\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, 4 goats and 14 sheep will take 12 days to graze the same field.

Portion of the tank filled by the first tap in one hour = $\frac{1}{20}$

Portion of the tank emptied by the second tap in one hour = $\frac{1}{30}$

$$\begin{aligned}\text{Portion of the tank filled in one hour when both taps are opened together} = \frac{1}{20} - \frac{1}{30}\\[1em] = \frac{3 - 2}{60}\\[1em] = \frac{1}{60}\end{aligned}$$

Time taken to fill the tank completely = $\frac{1}{\frac{1}{60}} = 60$ hours

Hence, the time taken for the tank to be half full = $\frac{60}{2} = 30$ hours

Hence, it will take 30 hours for the tank to be half full.

First ant's one hour work = $\frac{1}{3}$

Second ant's one hour work = $\frac{1}{4}$

Third ant's one hour work = $\frac{1}{6}$

$$\begin{aligned}\text{All three ants' one hour work} = \frac{1}{3} + \frac{1}{4} + \frac{1}{6}\\[1em] = \frac{4 + 3 + 2}{12}\\[1em] = \frac{9}{12}\\[1em] = \frac{3}{4}\end{aligned}$$

Time taken by the three ants together = $\frac{1}{\frac{3}{4}} = \frac{4}{3} = 1\frac{1}{3}$ hours

The share of each ant is in proportion of the work done by it.

Hence, they should share the grasshopper in the ratio $\frac{1}{3} : \frac{1}{4} : \frac{1}{6}$

Multiplying all the terms of the ratio by 12, we get :

$\frac{1}{3} \times 12 : \frac{1}{4} \times 12 : \frac{1}{6} \times 12 = 4 : 3 : 2$

Sum of the terms of the ratio = 4 + 3 + 2 = 9

First ant's share = $\frac{4}{9}$ of 63 g = $\left(\frac{4}{9} \times 63\right)$ g = 28 g

Second ant's share = $\frac{3}{9}$ of 63 g = $\left(\frac{3}{9} \times 63\right)$ g = 21 g

Third ant's share = $\frac{2}{9}$ of 63 g = $\left(\frac{2}{9} \times 63\right)$ g = 14 g

Hence, the three ants together take $\mathbf{1\frac{1}{3}}$ hours, and their shares are 28 g, 21 g and 14 g respectively.

(A + B)'s one day's work = $\frac{1}{12}$

(B + C)'s one day's work = $\frac{1}{15}$

Subtracting the second from the first, we get :

$$\begin{aligned}\text{A's one day's work - C's one day's work} = \frac{1}{12} - \frac{1}{15}\\[1em] = \frac{5 - 4}{60}\end{aligned}$$

= $\frac{1}{60}$ ....(1)

Since A is twice as good a workman as C,

Hence, A's one day's work = 2 × C's one day's work ....(2)

Substituting the value from (2) in (1), we get :

$$\begin{aligned}\Rightarrow 2 \times \text{C's one day's work} - \text{C's one day's work} = \frac{1}{60}\\[1em] \Rightarrow \text{C's one day's work} = \frac{1}{60}\\[1em] \text{Hence, A's one day's work} = 2 \times \frac{1}{60}\\[1em] = \frac{2}{60}\\[1em] = \frac{1}{30}\end{aligned}$$

Number of days taken by A alone = $\frac{1}{\frac{1}{30}} = 30$

Hence, A alone will do the same work in 30 days.