Direct and Inverse Variations

$\frac{3}{18} = \frac{1}{6}$

$\frac{8}{48} = \frac{1}{6}$

So, we can say that x and y vary directly

$$\begin{aligned}\therefore \frac{3}{18} = \frac{5}{a} = \frac{b}{90} = \frac{8}{48}\\[1em] \Rightarrow \frac{3}{18} = \frac{5}{a} \text{ and } \frac{3}{18} = \frac{b}{90}\\[1em] \Rightarrow a = \frac{5 \times 18}{5} \text{ and } b = \frac{3 \times 90}{18}\\[1em] \Rightarrow a = \frac{90}{3} \text{ and } b = \frac{270}{18}\\[1em] \Rightarrow a = 30 \text{ and } b = 15\end{aligned}$$

Hence, option 2 is the correct option.

Let the machine make x screws.

Hours	Screws
5	45
12	x

Since it is a case of direct variation.

$$\begin{aligned}\therefore \frac{5}{45} = \frac{12}{x}\\[1em] \Rightarrow x = \frac{45 \times 12}{5}\\[1em] \Rightarrow x = \frac{540}{5}\\[1em] \Rightarrow x = 108\end{aligned}$$

Hence, option 4 is the correct option.

Let the cost of one dozen pens be x.

Pens	Cost
9	369
12	x

Since it is a case of direct variation.

$$\begin{aligned}\therefore \frac{9}{369} = \frac{12}{x}\\[1em] \Rightarrow x = \frac{369 \times 12}{9}\\[1em] \Rightarrow x = \frac{4,428}{9}\\[1em] \Rightarrow x = 492\end{aligned}$$

Hence, option 4 is the correct option.

Let the value of y when x = 25 be a

x	y
400	80
25	a

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{400}{80} = \frac{25}{a}\\[1em] \Rightarrow a = \frac{80 \times 25}{400}\\[1em] \Rightarrow a = \frac{2,000}{400}\\[1em] \Rightarrow a = 5\end{aligned}$$

Hence, option 1 is the correct option.

Let the man earn ₹ x.

Hours	Earing
48	₹ 2,400
36	₹ x

Since it is a case of direct variation.

$$\begin{aligned}\therefore \frac{48}{2400} = \frac{36}{x}\\[1em] \Rightarrow x = \frac{36 \times 2400}{48}\\[1em] \Rightarrow x = \frac{86,400}{48}\\[1em] \Rightarrow x = 1,800\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}\frac{x_1}{y_1} = \frac{3}{4.5} = \frac{2}{3},\\[1em] \frac{x_2}{y_2} = \frac{5}{7.5} = \frac{2}{3},\\[1em] \frac{x_3}{y_3} = \frac{8}{12} = \frac{2}{3},\\[1em] \frac{x_4}{y_4} = \frac{11}{16.5} = \frac{2}{3},\\[1em]\end{aligned}$$

$\Rightarrow \frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3} = \frac{x_4}{y_4}$ = a constant

Hence, x and y are in direct variation.

$$\begin{aligned}\frac{x_1}{y_1} = \frac{16}{32} = \frac{1}{2},\\[1em] \frac{x_2}{y_2} = \frac{30}{60} = \frac{1}{2},\\[1em] \frac{x_3}{y_3} = \frac{40}{80} = \frac{1}{2},\\[1em] \frac{x_4}{y_4} = \frac{56}{84} = \frac{2}{3},\\[1em]\end{aligned}$$

$\Rightarrow \frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3} ≠ \frac{x_4}{y_4}$ = not a constant

Hence, x and y are not in direct variation.

$$\begin{aligned}\frac{x_1}{y_1} = \frac{27}{81} = \frac{1}{3},\\[1em] \frac{x_2}{y_2} = \frac{45}{180} = \frac{1}{4},\\[1em] \frac{x_3}{y_3} = \frac{54}{216} = \frac{1}{4},\\[1em] \frac{x_4}{y_4} = \frac{75}{225} = \frac{1}{3},\\[1em]\end{aligned}$$

$\Rightarrow \frac{x_1}{y_1} ≠ \frac{x_2}{y_2} = \frac{x_3}{y_3} ≠ \frac{x_4}{y_4}$ = not a constant

Hence, x and y are not in direct variation.

Since a and b vary directly,

$$\begin{aligned}\frac{3}{36} = \frac{x}{60} = \frac{y}{96} = \frac{10}{z}\\[1em] \Rightarrow \frac{3}{36} = \frac{x}{60}, \frac{3}{36} = \frac{y}{96} \text{ and } \frac{3}{36} = \frac{10}{z}\\[1em] \Rightarrow x = \frac{3 \times 60}{36}, y = \frac{3 \times 96}{36} \text{ and } z = \frac{36 \times 10}{3}\\[1em] \Rightarrow x = \frac{180}{36} , y = \frac{288}{36} \text{ and } z = \frac{360}{3}\\[1em] \Rightarrow x = 5, y = 8 \text{ and } z = 120\end{aligned}$$

Hence, x = 5, y = 8 and z = 120.

Let the truck consume x litres of diesel.

Diesel	Distance
28	448
64	x

Since it is a case of direct variation.

$$\begin{aligned}\therefore \frac{28}{448} = \frac{64}{x}\\[1em] \Rightarrow x = \frac{64 \times 448}{28}\\[1em] \Rightarrow x = \frac{28,672}{28}\\[1em] \Rightarrow x = 1,024\end{aligned}$$

Hence, the truck will cover 1,024 km distance.

Let the taxi charges ₹ x for 120 km.

Distance	Charge
100	₹ 1,800
120	₹ x

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{100}{1,800} = \frac{120}{x}\\[1em] \Rightarrow x = \frac{120 \times 1800}{100}\\[1em] \Rightarrow x = \frac{2,16,000}{100}\\[1em] \Rightarrow x = 2,160\end{aligned}$$

Hence, the taxi will charge ₹ 2,160.

Let x articles cost ₹ 1,750.

Articles	Cost
27	₹ 1,890
x	₹ 1,750

Since it is a case of direct variation.

$$\begin{aligned}\therefore \frac{27}{1,890} = \frac{x}{1,750}\\[1em] \Rightarrow x = \frac{27 \times 1,750}{1,890}\\[1em] \Rightarrow x = \frac{47,250}{1,890}\\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, the 25 articles can be bought for ₹ 1,750.

Let x kg rice cost ₹ 3,680.

Quantity	Cost
7	₹ 1,120
x	₹ 3,680

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{7}{1,120} = \frac{x}{3,680}\\[1em] \Rightarrow x = \frac{7 \times 3,680}{1,120}\\[1em] \Rightarrow x = \frac{25,760}{1,120}\\[1em] \Rightarrow x = 23\end{aligned}$$

Hence, the 23 kg of rice can be bought for ₹ 3,680.

Let 54 notebooks cost ₹ x.

Quantity	Cost
6	₹ 156
54	₹ x

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{6}{156} = \frac{54}{x}\\[1em] \Rightarrow x = \frac{54 \times 156}{6}\\[1em] \Rightarrow x = \frac{8,424}{6}\\[1em] \Rightarrow x = 1,404\end{aligned}$$

Hence, the cost of 54 notebooks is ₹ 1,404.

Let the x men be employed for digging 135 m long trench in one day.

Men	Length
22	27
x	135

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{22}{27} = \frac{x}{135}\\[1em] \Rightarrow x = \frac{22 \times 135}{27}\\[1em] \Rightarrow x = \frac{2,970}{27}\\[1em] \Rightarrow x = 110\end{aligned}$$

Hence, 110 men should be employed for digging 135 m long trench in one day.

Let x articles weigh 224 kg.

Articles	Weight
11	77
x	224

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{11}{77} = \frac{x}{224}\\[1em] \Rightarrow x = \frac{11 \times 224}{77}\\[1em] \Rightarrow x = \frac{2,464}{77}\\[1em] \Rightarrow x = 32\end{aligned}$$

Hence, 32 articles would weigh 224 kg.

(i) Let x km be distance travel in 36 minutes.

Distance	Time (in minutes)
120	60 (∵ 1 hour = 60 mins)
x	36

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{120}{60} = \frac{x}{36}\\[1em] \Rightarrow x = \frac{120 \times 36}{60}\\[1em] \Rightarrow x = \frac{4,320}{60}\\[1em] \Rightarrow x = 72\end{aligned}$$

Hence, the train will travel a distance of 72 km in 36 minutes.

(ii) Let the train take y minutes to cover 210 km.

Distance	Time (in minutes)
120	60
210	y

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{120}{60} = \frac{210}{y}\\[1em] \Rightarrow y = \frac{210 \times 60}{120}\\[1em] \Rightarrow y = \frac{12,600}{120}\\[1em] \Rightarrow y = 105\end{aligned}$$

105 minutes = 60 minutes + 45 minutes

= 1 hours 45 minutes

Hence, the train will take 1 hours 45 minutes to cover 210 km.

$$\begin{aligned}a_1b_1 = a_2b_2 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 8 \times 12 = 6 \times b\\[1em] \Rightarrow 96 = 6 \times b\\[1em] \Rightarrow b = \frac{96}{6}\\[1em] \Rightarrow b = 16\end{aligned}$$

Hence, option 4 is the correct option.

Let 24 men take x days to make the same number of screws.

Men	Days
12	30
24	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 12 \times 30 = 24 \times x\\[1em] \Rightarrow 360 = 24x\\[1em] \Rightarrow x = \frac{360}{24}\\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, option 1 is the correct option.

Let the duration of each period be x minutes.

Periods	Duration(in min)
8	45
9	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 8 \times 45 = 9 \times x\\[1em] \Rightarrow 360 = 9x\\[1em] \Rightarrow x = \frac{360}{9}\\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}∵ x_1y_1 = 12 \times 5 = 60\\[1em] x_3y_3 = 10 \times 6 = 60\\[1em]\end{aligned}$$

So, we can say that this is inverse variation.

$$\begin{aligned}\Rightarrow a_1b_1 = a_2b_2 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 12 \times 5 = 4 \times a\\[1em] \Rightarrow 60 = 4 \times a\\[1em] \Rightarrow a = \frac{60}{4}\\[1em] \Rightarrow a = 15\end{aligned}$$

And, similarly

$$\begin{aligned}a_3b_3 = a_3b_3 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 10 \times 6 = 30 \times b\\[1em] \Rightarrow 60 = 30 \times b\\[1em] \Rightarrow b = \frac{60}{30}\\[1em] \Rightarrow b = 2\end{aligned}$$

Hence, option 4 is the correct option.

Original no. of students in hostel = 200

New no. of students in hostel = 200 + 100 = 300

For 300 students, let the provisions last for x days.

Students	Provision
200	30
300	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 200 \times 30 = 300 \times x\\[1em] \Rightarrow 6000 = 300x\\[1em] \Rightarrow x = \frac{6000}{300}\\[1em] \Rightarrow x = 20\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}∵ x_1y_1 = 4 \times 6 = 24,\\[1em] x_2y_2 = 3 \times 8 = 24,\\[1em] x_3y_3 = 12 \times 2 = 24,\\[1em] x_4y_4 = 1 \times 24 = 24,\\[1em]\end{aligned}$$

So, $x_1y_1 = x_2y_2 = x_3y_3 = x_4y_4$

Hence, x and y vary inversely.

$$\begin{aligned}∵ x_1y_1 = 30 \times 60 = 1,800,\\[1em] x_2y_2 = 120 \times 30 = 3,600,\\[1em] x_3y_3 = 60 \times 30 = 1,800,\\[1em] x_4y_4 = 24 \times 75 = 1,800,\\[1em]\end{aligned}$$

So, $x_1y_1 ≠ x_2y_2 ≠ x_3y_3 = x_4y_4$

Hence, x and y do not vary inversely.

$$\begin{aligned}∵ x_1y_1 = 10 \times 90 = 900,\\[1em] x_2y_2 = 30 \times 30 = 900,\\[1em] x_3y_3 = 60 \times 20 = 1,200,\\[1em] x_4y_4 = 10 \times 90 = 900,\\[1em]\end{aligned}$$

So, $x_1y_1 = x_2y_2 ≠ x_3y_3 ≠ x_4y_4$

Hence, x and y do not vary inversely.

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 4 \times 4 = 8 \times l = 2 \times m = 32 \times n\\[1em] \Rightarrow 16 = 8l = 2m = 32n\\[1em] \Rightarrow l = \frac{16}{8}, m = \frac{16}{2} \text{ and } n = \frac{16}{32}\\[1em] \Rightarrow l = 2, m = 8 \text{ and } n = 0.5\end{aligned}$$

Hence, l = 2, m = 8 and n = 0.5.

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 24 \times l = 32 \times 12 = m \times 8 = 16 \times n\\[1em] \Rightarrow 24l = 384 = 8m = 16n\\[1em] \Rightarrow l = \frac{384}{24}, m = \frac{384}{8} \text{ and } n = \frac{384}{16}\\[1em] \Rightarrow l = 16, m = 48 \text{ and } n = 24\end{aligned}$$

Hence, l = 16, m = 48 and n = 24.

Let x men do the same work in 42 days.

Men	Days
36	7
x	42

Since it is a case of inverse variation.

$$\begin{aligned}\therefore 36 \times 7 = x \times 42\\[1em] \Rightarrow 252 = 42x\\[1em] \Rightarrow x = \frac{252}{42}\\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, 6 men will do the same work in 42 days.

Let 21 pipes take x minutes to fill the tank.

Pipes	Time (in min)
12	42
21	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 12 \times 42 = 21 \times x\\[1em] \Rightarrow 504 = 21x\\[1em] \Rightarrow x = \frac{504}{21}\\[1em] \Rightarrow x = 24\end{aligned}$$

Hence, 21 pipes will take 24 minutes to fill the tank.

Original no. of men in the fort = 150

New no. of men in the fort = 150 - 25 = 125

For 125 men lets take food will last for x days.

The remaining food would last (45 - 10) = 35 days

Men	Provision
150	35
125	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 150 \times 35 = 125 \times x\\[1em] \Rightarrow 5,250 = 125x\\[1em] \Rightarrow x = \frac{5,250}{125}\\[1em] \Rightarrow x = 42\end{aligned}$$

Hence, the food last for 42 days.

Let 30 men do the same work in x days.

Men	Days
72	25
30	x

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 72 \times 25 = 30 \times x\\[1em] \Rightarrow 1,800 = 30x\\[1em] \Rightarrow x = \frac{1,800}{30}\\[1em] \Rightarrow x = 60\end{aligned}$$

Hence, 30 men will do the same work in 60 days.

Let x workers do the same work in 70 hours.

Workers	Time (in hours)
56	180
x	70

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 56 \times 180 = x \times 70\\[1em] \Rightarrow 10,080 = 70x\\[1em] \Rightarrow x = \frac{10,080}{70}\\[1em] \Rightarrow x = 144\end{aligned}$$

Hence, 144 workers will do the work in 70 hours.

Let the car take x hours at the speed of 75 km per hour.

Time (in hr)	Speed
6	50
x	75

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 6 \times 50 = x \times 75\\[1em] \Rightarrow 300 = 75x\\[1em] \Rightarrow x = \frac{300}{75}\\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, the car takes 4 hours at the speed of 75 km per hour.

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

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

(A + B)'s 1 day work = $\frac{1}{2} + \frac{1}{3}$

= $\frac{3 + 2}{6}$

= $\frac{5}{6}$

Hence, option 4 is the correct option.

Let work done in one day be $x$

Work done in 5 days = $\frac{1}{20}$

$$\begin{aligned}\therefore 5x = \frac{1}{20} \\[1em] \Rightarrow x = \frac{1}{20 \times 5} \\[1em] \Rightarrow x = \frac{1}{100}\end{aligned}$$

Hence, option 1 is the correct option.

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

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

(Ritu + Manish)'s 1 day work = $\frac{1}{4} + \frac{1}{6}$

= $\frac{3 + 2}{12}$

= $\frac{5}{12}$

No. days taken by Ritu & Manish together to knit the same sweater = $\frac{1}{\frac{5}{12}} = \frac{12}{5}$

Hence, option 3 is the correct option.

(A + B)'s 1 day work = $\frac{1}{4}$

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

B's 1 day work = $\frac{1}{4} - \frac{1}{6}$

= $\frac{3 - 2}{12}$

= $\frac{1}{12}$

B alone can do the work in 12 days

Hence, option 1 is the correct option.

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

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

(A + B)'s 1 day work = $\frac{1}{10} + \frac{1}{15}$

= $\frac{3 + 2}{30}$

= $\frac{5}{30}$

A and B can do the work in $\frac{30}{5} = 6$ days

Hence, A and B can do the work in 6 days.

A + B can do the work in $6\frac{2}{3} = \frac{20}{3}$ days.

(A + B)'s 1 day work = $\frac{3}{20}$

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

A's 1 day work = $\frac{3}{20} - \frac{1}{10}$

= $\frac{3 - 2}{20}$

= $\frac{1}{20}$

A can do the work in $\frac{20}{1} = 20$ days

Hence, A can do the work in 20 days.

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

B's 1 day work = $\frac{1}{20}$

(A + B)'s 1 day work = $\frac{1}{15} + \frac{1}{20}$

= $\frac{4 + 3}{60}$

= $\frac{7}{60}$

(A + B)'s 4 day work = $\frac{7}{60} \times 4$

= $\frac{7}{15}$

Remaining work = $1 - \frac{7}{15}$

= $\frac{15}{15} - \frac{7}{15}$

= $\frac{8}{15}$

Hence, $\frac{8}{15}$ work wil be left.

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

B's 1 day work = $\frac{1}{12}$

C's 1 day work = $\frac{1}{24}$

(A + B + C)'s 1 day work = $\frac{1}{6} + \frac{1}{12} + \frac{1}{24}$

= $\frac{4 + 2 + 1}{24}$

= $\frac{7}{24}$

A + B + C can do the work in $\frac{24}{7} = 3\frac{3}{7}$ days

Hence, A + B + C can do the work in $3\frac{3}{7}$ days.

(A + B)'s 1 day work = $\frac{1}{56}$

(A + B + C)'s 1 day work = $\frac{1}{42}$

C's 1 day work = $\frac{1}{42} - \frac{1}{56}$

= $\frac{4 - 3}{168}$

= $\frac{1}{168}$

C can do the work in $\frac{168}{1}$ days

Hence, C can do the work in 168 days.

A's 1 day work = $\frac{1}{24}$

(A + B)'s 1 day work = $\frac{1}{16}$

B's 1 day work = (A + B)'s 1 day work - A's 1 day work

= $\frac{1}{16} - \frac{1}{24}$

= $\frac{(3 - 2)}{48}$

= $\frac{1}{48}$

(A + B + C)'s 1 day work = $\frac{3}{32}$

(A + C)'s 1 day work = (A + B + C)'s 1 day work - B's 1 day work

= $\frac{3}{32} - \frac{1}{48}$

= $\frac{9 - 2}{96}$

= $\frac{7}{96}$

A + C can do the work in $\frac{96}{7} = 13\frac{5}{7}$ days

Hence, A + C can do the work in $13\frac{5}{7}$ days.

A's 1 day work = $\frac{1}{20}$

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

(A + B)'s 1 day work = $\frac{1}{20} + \frac{1}{15}$

= $\frac{3 + 4}{60}$

= $\frac{7}{60}$

(A + B)'s 6 day work = $\frac{7}{60} \times 6$

= $\frac{7}{10}$

Remaining work = $1 - \frac{7}{10}$

= $\frac{10}{10} - \frac{7}{10}$

= $\frac{3}{10}$

No. of days taken by B to finish the remaining work = $\frac{\text{Remaining work}}{\text{B's 1 day work}} = \frac{\frac{3}{10}}{\frac{1}{15}}$

= $\frac{3 \times 15}{10 \times 1}$

= $\frac{45}{10}$

= $\frac{9}{2}$

= $4\frac{1}{2}$

Hence, B will take $4\frac{1}{2}$ days to finish the remaining work.

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

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

(A + B)'s 1 day work = $\frac{1}{15} + \frac{1}{10}$

= $\frac{2 + 3}{30}$

= $\frac{5}{30}$

= $\frac{1}{6}$

(A + B)'s 2 day work = $\frac{1}{6} \times 2$

= $\frac{1}{3}$

Remaining work = $1 - \frac{1}{3}$

= $\frac{3}{3} - \frac{1}{3}$

= $\frac{2}{3}$

No. of days taken by A to finish the remaining work = $\frac{\text{Remaining work}}{\text{B's 1 day work}} = \frac{\frac{2}{3}}{\frac{1}{15}}$

= $\frac{2 \times 15}{3 \times 1}$

= $\frac{30}{3}$

= $10$

Hence, A will take 10 days to finish the remaining work.

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

B's 1 day work = $\frac{1}{18}$

(A + B + C)'s 1 day work = $\frac{1}{4}$

C's 1 day work = $\frac{1}{4} - \Big(\frac{1}{10} + \frac{1}{18}\Big)$

= $\frac{1}{4} - \Big(\frac{9 + 5}{90}\Big)$

= $\frac{1}{4} - \frac{14}{90}$

= $\frac{1}{4} - \frac{7}{45}$

= $\frac{45 - 28}{180}$

= $\frac{17}{180}$

C can do the work in $\frac{180}{17} = 10\frac{10}{17}$

Hence, C can do the work alone in $10\frac{10}{17}$ days.

A's 5 days work = $\frac{1}{4}$

A's 1 day work = $\frac{1}{4 \times 5}$

= $\frac{1}{20}$

B's 10 days work = $\frac{1}{3}$

B's 1 day work = $\frac{1}{3 \times 10}$

= $\frac{1}{30}$

(A + B)'s 1 day work = $\frac{1}{20} + \frac{1}{30}$

= $\frac{(3 + 2)}{60}$

= $\frac{5}{60}$

= $\frac{1}{12}$

A and B can do the work in $\frac{12}{1}$ days

Hence, A and B working together can complete the work in 12 days.

Tap's 1 hour work = $\frac{1}{3}$

Waste tap's 1 hour work = $-\frac{1}{5}$

Both tap's 1 hour work = $\frac{1}{3} - \frac{1}{5}$

= $\frac{5 - 3}{15}$

= $\frac{2}{15}$

Total time taken by both pipe = $\frac{15}{2}$ hours

Hence, with both taps kept open, the empty cistern will be full in $7\frac{1}{2}$ hours.

As $\frac{2}{6} = \frac{1}{3}$

And $\frac{9}{27} = \frac{1}{3}$

So, we can say x and y vary directly

$$\begin{aligned}\frac{2}{6} = \frac{a}{18}\\[1em] \Rightarrow a = \frac{2 \times 18}{6}\\[1em] \Rightarrow a = \frac{36}{6}\\[1em] \Rightarrow a = 6\end{aligned}$$

Hence, option 4 is the correct option.

3 x 12 = 36

12 x 3 = 36

So we can say x and y vary inversely

$$\begin{aligned}a_1b_1 = a_2b_2 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 3 \times 12 = b \times 9\\[1em] \Rightarrow 36 = b \times 9\\[1em] \Rightarrow b = \frac{36}{9}\\[1em] \Rightarrow b = 4\end{aligned}$$

Hence, option 1 is the correct option.

Let x notebooks cost ₹ 160.

Quantity	Cost
15	₹ 240
x	₹ 160

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{15}{240} = \frac{x}{160}\\[1em] \Rightarrow x = \frac{15 \times 160}{240}\\[1em] \Rightarrow x = \frac{2,400}{240}\\[1em] \Rightarrow x = 10\end{aligned}$$

Hence, option 2 is the correct option.

Let x men be required to complete the work in 10 days.

Men	Days
6	15
x	10

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 6 \times 15 = x \times 10\\[1em] \Rightarrow 90 = 10x\\[1em] \Rightarrow x = \frac{90}{10}\\[1em] \Rightarrow x = 9\end{aligned}$$

Hence, option 4 is the correct option.

Since it is a case of inverse variation,

$$\begin{aligned}\therefore 4 \times 6 = x \times 12\\[1em] \Rightarrow 24 = 12x\\[1em] \Rightarrow x = \frac{24}{12}\\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, option 1 is the correct option.

Since, Work done is constant.

Work done increases with more people working. Therefore, work done is directly proportional to the number of persons.

So, statement 1 is false.

Time taken decreases with more people working.

Therefore, time taken is inversely proportional to the number of persons.

So, statement 2 is false.

Hence, option 2 is the correct option.

If two numbers p and q are in direct variation, then $\frac{p}{q}$ will be constant.

So, reason is true.

For table,

$$\begin{aligned}\Rightarrow \frac{p_1}{q_1} = \frac{3}{8} \\[1em] \Rightarrow \frac{p_2}{q_2} = \frac{2}{12} = \frac{1}{6} \\[1em] \Rightarrow \frac{p_3}{q_3} = \frac{4}{6} = \frac{2}{3} \\[1em]\end{aligned}$$

∴ $\frac{p_1}{q_1} \ne; \frac{p_2}{q_2} \ne; \frac{p_3}{q_3}$

So, assertion (A) is false.

Hence, option 4 is the correct option.

If 3 pipes take 6 hours, then their combined rate of filling the tank is $\frac{1}{6}$ tank/hr.

Each pipe's rate = $\frac{1}{6} \times \frac{1}{3}$ = $\frac{1}{18}$ tank/hr

Thus, 6 pipes will fill = 6 x $\frac{1}{18} = \frac{1}{3}$ tank in an hr.

Time to fill 1 tank by 6 pipes = 3 hours

So, assertion (A) is true.

Time required to complete a certain work = $\frac{\text{Work to be completed}}{\text{One day's work}}$.

This is a standard formula used in time and work problems.

So, reason (R) is true but reason (R) does not clearly explains assertion (A).

Hence, option 2 is the correct option.

We know that,

For two numbers to be in inverse variation,

p x q = constant

So, reason (R) is false.

From table,

$$\begin{aligned}\Rightarrow x_1y_1 = 5 \times 10 = 50, \\[1em] \Rightarrow x_2y_2 = 2 \times 25 = 50, \\[1em] \Rightarrow x_3y_3 = 4 \times 12.5 = 50. \\[1em]\end{aligned}$$

So, assertion (A) is true.

Hence, option 3 is the correct option.

Given,

The cost of 16 bulbs = ₹ 144

The cost of 1 bulb = $\frac{144}{16}$ = ₹ 9

Number of bulbs for ₹ 270 = $\frac{\text{Total money}}{\text{Price of each bulb}} = \frac{270}{9}$ = 30

So, assertion (A) is true.

In inverse variation,

$$\begin{aligned}\Rightarrow x_1y_1 = x_2y_2\\[1em] \Rightarrow \frac{x_1}{x_2} = \frac{y_2}{y_1}\\[1em]\end{aligned}$$

So, reason (R) is true but, reason (R) does not explains assertion(A).

Hence, option 2 is the correct option.

1. Since it is a case of direct variation,

$$\begin{aligned}\frac{150}{50} = \frac{x}{12.5}\\[1em] \Rightarrow \frac{3}{1} = \frac{x}{12.5}\\[1em] \Rightarrow x = 3 \times 12.5\\[1em] \Rightarrow x = 37.5\end{aligned}$$

Hence, when y = 12.5, then x = 37.5.

2. Since it is a case of direct variation,

$$\begin{aligned}\frac{150}{50} = \frac{75}{y}\\[1em] \Rightarrow \frac{3}{1} = \frac{75}{y}\\[1em] \Rightarrow y = \frac{75}{3}\\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, when x = 75, then y = 25.

1. Since it is a case of inverse variation,

$$\begin{aligned}60 \times 300 = x \times 90\\[1em] \Rightarrow 18000 = x \times 90\\[1em] \Rightarrow x = \frac{18000}{90}\\[1em] \Rightarrow x = 200\end{aligned}$$

Hence, when y = 90, then x = 200.

2. Since it is a case of inverse variation,

$$\begin{aligned}60 \times 300 = 300 \times y\\[1em] \Rightarrow 18000 = 300 \times y\\[1em] \Rightarrow y = \frac{18000}{300}\\[1em] \Rightarrow y = 60\end{aligned}$$

Hence, when x = 300, then y = 60.

Let x m be total length of 135 bars.

Iron bars	Length (in m)
153	680
135	x

Since it is a case of direct variation,

$$\begin{aligned}\therefore \frac{153}{680} = \frac{135}{x}\\[1em] \Rightarrow x = \frac{135 \times 680}{153}\\[1em] \Rightarrow x = \frac{91,800}{153}\\[1em] \Rightarrow x = 600\end{aligned}$$

Hence, 600 m is the total length of 135 bars.

Let 30 men take x days to repair the road.

Men	Days
12	25
30	x

Since it is a case of inverse variation,

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

Hence, 30 men will do the work in 10 days.

Cost of 1 dozen oranges = ₹ 90

Cost of 12 dozen oranges = ₹ 90 x 12 = ₹ 1,080

Price of oranges is increased by ₹ 30,

∴ New price of 1 dozen oranges = ₹ (90 + 30) = ₹ 120

Let Manoj buy x dozen oranges for ₹ 120 per dozen.

Oranges (in dozen)	Cost
12	90
x	120

As the price of oranges has increased, so now Manoj will be able to buy lesser number of oranges with the money he has.

∴ It is a case of inverse variation.

$$\begin{aligned}\therefore 12 \times 90 = x \times 120\\[1em] \Rightarrow 1080 = x \times 120\\[1em] \Rightarrow x = \frac{1080}{120}\\[1em] \Rightarrow x = 9\end{aligned}$$

Hence, Manoj can buy 9 dozen oranges for ₹ 1,080.

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

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

(C + A)'s 1 day work = $\frac{1}{16}$

2(A + B + C)'s 1 day work = $\frac{1}{8} + \frac{1}{12} + \frac{1}{16}$

= $\frac{6 + 4 + 3}{48}$

= $\frac{13}{48}$

(A + B + C)'s 1 day work = $\frac{13}{48 \times 2}$

= $\frac{13}{96}$

No. of days to complete the work when A, B and C are working together = $\frac{96}{13} = 7\frac{5}{13}$

Hence, A, B and C working together can complete the work in $7\frac{5}{13}$ days.

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

(B + C)'s 1 day work = $\frac{1}{36}$

(A + B + C)'s 1 day work = $\frac{1}{18}$

A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work

= $\frac{1}{18} - \frac{1}{36}$

= $\frac{2 - 1}{36}$

= $\frac{1}{36}$

Number of days required by A alone = 36 days

Hence, A requires 36 days to complete the work alone.

(ii) C's 1 day work = (A + B + C)'s 1 day work - (A + B)'s 1 day work

= $\frac{1}{18} - \frac{1}{24}$

= $\frac{4 - 3}{72}$

= $\frac{1}{72}$

Number of days required by C alone = 72 days

Hence, C requires 72 days to complete the work alone.

(iii) (A + C)'s 1 day work = $\frac{1}{36} + \frac{1}{72}$

= $\frac{2 + 1}{72}$

= $\frac{3}{72}$

No. of days required to complete the work with A and C working together = $\frac{72}{3} = 24$ days

Hence, A and C together will complete the work in 24 days.

(A + B)'s 1 day work = $\frac{1}{40}$

(B + C)'s 1 day work = $\frac{1}{30}$

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

2(A + B + C)'s 1 day work = $\frac{1}{40} + \frac{1}{30} + \frac{1}{24}$

= $\frac{(3 + 4 + 5)}{120}$

= $\frac{12}{120}$

= $\frac{1}{10}$

(A + B + C)'s 1 day work = $\frac{1}{2 \times 10}$

= $\frac{1}{20}$

No. of days required to complete the work when A, B and C are working together = 20 days

Hence, A, B and C working together can complete the work in 20 days.

(ii) A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work

= $\frac{1}{20} - \frac{1}{30}$

= $\frac{(3 - 2)}{60}$

= $\frac{1}{60}$

∴ A alone will complete the work in 60 days.

B's 1 day work = (A + B + C)'s 1 day work - (C + A)'s 1 day work

= $\frac{1}{20} - \frac{1}{24}$

= $\frac{(6 - 5)}{120}$

= $\frac{1}{120}$

∴ B alone will complete the work in 120 days.

C's 1 day work = (A + B + C)'s 1 day work - (A + B)'s 1 day work

= $\frac{1}{20} - \frac{1}{40}$

= $\frac{(2 - 1)}{40}$

= $\frac{1}{40}$

∴ C alone will complete the work in 40 days.

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

B's 1 day work = $\frac{1}{12}$

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

Let the work get completed in x days.

∴ A's 2 days work + B's (x - 3) days work + C's x days work = 1

$$\begin{aligned}\Rightarrow \frac{1}{10} \times 2 + \frac{1}{12} \times (x - 3) + \frac{1}{15} \times x = 1\\[1em] \Rightarrow \frac{1}{5} + \frac{(x - 3)}{12} + \frac{x}{15} = 1\\[1em] \Rightarrow \frac{[12 + 5(x - 3) + 4x]}{60} = 1\\[1em] \Rightarrow \frac{(12 + 5x - 15 + 4x)}{60} = 1\\[1em] \Rightarrow\frac{(9x - 3)}{60} = 1\\[1em] \Rightarrow 9x - 3 = 1 \times 60\\[1em] \Rightarrow 9x - 3 = 60\\[1em] \Rightarrow 9x = 60 + 3\\[1em] \Rightarrow 9x = 63\\[1em] \Rightarrow x = \frac{63}{9}\\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the work lasts for 7 days.

Work done by P in 1 minute = $\frac{1}{24}$

Work done by Q in 1 minute = $\frac{1}{32}$

Let the first pipe be turned off after x minutes.

Then, P's x minutes work + Q's 16 minutes work = 1

$$\begin{aligned}\Rightarrow \frac{1}{24} \times x + \frac{1}{32} \times 16 = 1\\[1em] \Rightarrow \frac{x}{24} + \frac{1}{2} = 1\\[1em] \Rightarrow \frac{x}{24} = 1 - \frac{1}{2}\\[1em] \Rightarrow \frac{x}{24} = \frac{1}{2}\\[1em] \Rightarrow x = \frac{1}{2} \times 24\\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, pipe P must be closed after 12 minutes.