RD Sharma solution class 8 chapter 10 Direct and Inverse Variation Exercise 10.2

Exercise 10.2

Page-10.12

Question 1:

In which of the following tables x and y vary inversely:
(i)

x	4	3	12	1
y	6	8	2	24

(ii)

x	5	20	10	4
y	20	5	10	25

(iii)

x	4	3	6	1
y	9	12	8	36

(iv)

x	9	24	15	3
y	8	3	4	25

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Since} x \mathrm{and} y \mathrm{var}y \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ y = \frac{k}{x} \\ \Rightarrow xy = k \\ \therefore \mathrm{The} \mathrm{product} \mathrm{of} x \mathrm{and} y \mathrm{is} \mathrm{constant}. \\ \mathrm{In} \mathrm{all} \mathrm{cases}, \mathrm{the} \mathrm{product} xy \mathrm{is} \mathrm{constant} (i.e., 24). \\ \mathrm{Thus}, \mathrm{in} \mathrm{this} \mathrm{case}, x \mathrm{and} y \mathrm{var}y \mathrm{inversely}. \\ \left(\mathrm{ii}\right) \mathrm{In} \mathrm{all} \mathrm{ca}s\mathrm{es}, \mathrm{the} \mathrm{product} xy \mathrm{is} \mathrm{constant} \mathrm{for} \mathrm{any} \mathrm{two} \mathrm{pairs} \mathrm{of} \mathrm{values} \mathrm{for} x \mathrm{and} y. \\ \mathrm{Here}, xy = 100 \mathrm{for} \mathrm{all} \mathrm{cases} \\ \mathrm{Thus}, \mathrm{in} \mathrm{this} \mathrm{case}, x \mathrm{and} y \mathrm{var}y \mathrm{inversely}. \\ \left(\mathrm{iii}\right) \mathrm{If} x \mathrm{and} y \mathrm{var}y \mathrm{inversely}, \mathrm{the} \mathrm{product} xy \mathrm{should} \mathrm{be} \mathrm{constant}. \\ \mathrm{Here}, \mathrm{in} \mathrm{one} \mathrm{case}, \mathrm{product} = 6 \times 8 = 48 \mathrm{and} \mathrm{in} \mathrm{the} \mathrm{rest}, \mathrm{product} = 36 \\ \mathrm{Thus}, \mathrm{in} \mathrm{this} \mathrm{case}, x \mathrm{and} y \mathrm{do} \mathrm{not} \mathrm{var}y \mathrm{inversely}. \\ \left(\mathrm{iv}\right) \mathrm{If} x \mathrm{and} y \mathrm{var}y \mathrm{inversely}, \mathrm{the} \mathrm{produc}t xy \mathrm{should} \mathrm{be} \mathrm{constant}. \\ \mathrm{Here}, \mathrm{product} \mathrm{is} \mathrm{different} \mathrm{for} \mathrm{all} \mathrm{cases}. \\ \mathrm{Thus}, \mathrm{in} \mathrm{this} \mathrm{case}, x \mathrm{and} y \mathrm{do} \mathrm{not} \mathrm{var}y \mathrm{inversely}. \end{gathered}$$

Question 2:

It x and y vary inversely, fill in the following blanks:
(i)

x	12	16	...	8	...
y	...	6	4	...	0.25

(ii)

x	16	32	8	128
y	4	...	...	0.25

(iii)

x	9	...	81	243
y	27	9	...	1

Answer 2:

$$\begin{gathered} \left(i\right) \mathrm{Since} x \mathrm{and} y \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ xy = k \\ \mathrm{For} x = 16 \mathrm{and} y = 6, \mathrm{we} \mathrm{have}: \\ 16 \times 6 = k \\ \Rightarrow k = 96 \\ \mathrm{For} x = 12 \mathrm{and} k = 96, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 12y = 96 \\ \Rightarrow y = \frac{96}{12} \\ = 8 \\ \mathrm{For} y = 4 \mathrm{and} k = 96, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 4x = 96 \\ \Rightarrow x = \frac{96}{4} \\ = 24 \\ \mathrm{For} x = 8 \mathrm{and} k = 96, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 8y = 96 \\ \Rightarrow y = \frac{96}{8} \\ = 12 \\ \mathrm{For} y = 0.25 \mathrm{and} k = 96, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 0.25x = 96 \\ \Rightarrow x = \frac{96}{0.25} \\ = 384 \\ \left(\mathrm{ii}\right) \mathrm{Since} x \mathrm{and} y \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ xy = k \\ \mathrm{For} x = 16 \mathrm{and} y = 4, \mathrm{we} \mathrm{have}: \\ 16 \times 4 = k \\ \Rightarrow k = 64 \\ \mathrm{For} x = 32 \mathrm{and} k = 64, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 32y = 64 \\ \Rightarrow y = \frac{64}{32} \\ = 2 \\ \mathrm{For} x = 8 \mathrm{and} k = 64 \\ xy = k \\ \Rightarrow 8y = 64 \\ \Rightarrow y = \frac{64}{8} \\ = 8 \\ \left(\mathrm{iii}\right) \mathrm{Since} x \mathrm{and} y \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ xy = k \\ \mathrm{For} x = 9 \mathrm{and} y = 27 \\ 9 \times 27 = k \\ \Rightarrow k = 243 \\ \mathrm{For} y = 9 \mathrm{and} k = 243, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 9x = 243 \\ \Rightarrow y = \frac{243}{9} \\ = 27 \\ \mathrm{For} x = 81 \mathrm{and} k = 243, \mathrm{we} \mathrm{have}: \\ xy = k \\ \Rightarrow 81y = 243 \\ \Rightarrow y = \frac{243}{81} \\ = 3 \end{gathered}$$

Page-10.13

Question 3:

Which of the following quantities vary inversely as each other?
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.

Answer 3:

(i) If the number of men is more, the time taken to construct a wall will be less. Therefore, it is in inverse variation.
(ii) If the length of a journey is more, the price of the ticket will also be more. Therefore, it is not in inverse variation.
(iii) If the length of the journey is more, the amount of petrol consumed by the car will also be more. Therefore, it is not in inverse variation.
Thus, only (i) is in inverse variation.

Question 4:

It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table:

v (in cm3)	...	48	60	...	100	...	200
p (in atmospheres)	2	...	3/2	1	...	1/2	...

Answer 4:

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{volume} \mathrm{and} \mathrm{pressure} \mathrm{for} \mathrm{the} \mathrm{given} \mathrm{mass} \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ vp = k \\ \mathrm{For} v = 60 \mathrm{and} p = \frac{3}{2}, \mathrm{we} \mathrm{have}: \\ k = 60 \times \frac{3}{2} \\ = 90 \\ \mathrm{For} p = 2 \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ 2v = 90 \\ \Rightarrow v = \frac{90}{2} \\ = 45 \\ \mathrm{For} v = 48 \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ 48p = 90 \\ \Rightarrow p = \frac{90}{48} \\ = \frac{15}{8} \\ \mathrm{For} p = 1 \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ 1v = 90 \\ \Rightarrow v = \frac{90}{1} \\ = 90 \\ \mathrm{For} v = 100 \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ 100p = 90 \\ \Rightarrow v = \frac{90}{100} \\ = \frac{9}{10} \\ \mathrm{For} p = \frac{1}{2} \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ \frac{1}{2}v = 90 \\ \Rightarrow v = 90 \times 2 \\ = 180 \\ \mathrm{For} v = 200 \mathrm{and} k = 90, \mathrm{we} \mathrm{have}: \\ 200p = 90 \\ \Rightarrow p = \frac{90}{200} \\ = \frac{9}{20} \end{gathered}$$

Question 5:

If 36 men can do a piece of work in 25 days, in how many days will 15 men do it?

Answer 5:

Let x be the number of days in which 15 men can do a piece of work.

Number of men	36	15
Number of days	25	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{hired} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{taken} \mathrm{to} \mathrm{do} \mathrm{a} \mathrm{piece} \mathrm{of} \mathrm{work} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 36 \times 25 = x \times 15 \\ \Rightarrow x = \frac{36 \times 25}{15} \\ = \frac{900}{15} \\ = 60 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{is} 60. \end{gathered}$$

Question 6:

A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men?

Answer 6:

Let x be the number days required to complete a piece of work by 125 men.
 

Number of men	50	125
Months	5	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{engaged} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{taken} \mathrm{to} \mathrm{do} \mathrm{a} \mathrm{piece} \mathrm{of} \mathrm{work} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, we have: \\ 50 \times 5 = 125x \\ \Rightarrow x = \frac{50 \times 5}{125} \\ = 2 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{months} \mathrm{is} 2. \end{gathered}$$

Question 7:

A work-force of 420 men with a contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?

Answer 7:

Let x be the extra number of men employed to complete the job in 7 months.
 

Number of men	420	x
Months	9	7

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{hired} \mathrm{and} \mathrm{the} \mathrm{time} \mathrm{required} \mathrm{to} \mathrm{finish} \mathrm{the} \mathrm{piece} \mathrm{of} \mathrm{work} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 420 \times 9 = 7x \\ \Rightarrow x = \frac{420 \times 9}{7} \\ = 540 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{extra} \mathrm{men} \mathrm{required} \mathrm{to} \mathrm{complete} \mathrm{the} \mathrm{job} \mathrm{in} 7 \mathrm{months} = 540 - 420 = 120 \end{gathered}$$

Question 8:

1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days?

Answer 8:

Number of men	1200	x
Days	35	25

Let x be the number of additional men required to finish the stock in 25 days.
$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{and} \mathrm{the} \mathrm{time} \mathrm{taken} \mathrm{to} \mathrm{finish} \mathrm{a} \mathrm{stock} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 1200 \times 35 = 25x \\ \Rightarrow x = \frac{1200 \times 35}{25} \\ = 1680 \\ \therefore \mathrm{Required} \mathrm{number} \mathrm{of} \mathrm{men} = 1680 - 1200 = 480 \\ \mathrm{Thus}, \mathrm{an} \mathrm{additional} 480 \mathrm{men} \mathrm{should} \mathrm{join} \mathrm{the} \mathrm{existing} 1200 \mathrm{men} \mathrm{to} \mathrm{finish} \mathrm{the} \mathrm{stock} \mathrm{in} 25 \mathrm{days}. \end{gathered}$$

Question 9:

In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel, how long will these provisions last?

Answer 9:

$\mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{with} \mathrm{food} \mathrm{provisions} \mathrm{for} 80 (\mathrm{i}.\mathrm{e}., 50+30) \mathrm{girls}.$

Number of girls	50	80
Number of days	40	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{girls} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{with} \mathrm{food} \mathrm{provisions} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 50 \times 40 = 80x \\ \Rightarrow x = \frac{50 \times 40}{80} \\ = \frac{2000}{80} \\ = 25 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{is} 25. \end{gathered}$$

Question 10:

A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance?

Answer 10:

Let the increased speed be x km/h.
 

Time (in h)	10	8
Speed (km/h)	48	x+48

$$\begin{gathered} \mathrm{Since} \mathrm{speed} \mathrm{and} \mathrm{time} \mathrm{taken} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{get}: \\ 10 \times 48 = 8\left(x + 48\right) \\ \Rightarrow 480 = 8x + 384 \\ \Rightarrow 8x = 480 - 384 \\ \Rightarrow 8x = 96 \\ = 12 \\ \mathrm{Thus}, \mathrm{the} \mathrm{speed} \mathrm{should} \mathrm{be} \mathrm{increased} \mathrm{by} 12\mathrm{km}/\mathrm{h}. \end{gathered}$$

Question 11:

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

Answer 11:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{after} 4 \mathrm{days}, \mathrm{out} \mathrm{of} 28 \mathrm{days}, \mathrm{the} \mathrm{fort} \mathrm{had} \mathrm{enough} \mathrm{food} \mathrm{for} 1200 \mathrm{soldiers} \mathrm{for} (28 - 4 = 24) \mathrm{days}. \\ \mathrm{Let} \mathrm{x} \mathrm{be} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{soldiers} \mathrm{who} \mathrm{left} \mathrm{the} \mathrm{fort} . \end{gathered}$$

Number of soldiers	1200	1200-x
Number of days for which food lasts	24	32

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{soldiers} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{for} \mathrm{which} \mathrm{the} \mathrm{food} \mathrm{lasts} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 1200 \times 24 = \left(1200 - x\right) \times 32 \\ \Rightarrow \frac{1200 \times 24}{32} = 1200 - x \\ \Rightarrow 900 = 1200 - x \\ \Rightarrow x = 1200 - 900 \\ = 300 \\ \mathrm{Thus}, 300 \mathrm{soldiers} \mathrm{left} \mathrm{the} \mathrm{fort}. \end{gathered}$$

Question 12:

Three spraying machines working together can finish painting a house in 60 minutes. How long will it take for 5 machines of the same capacity to do the same job?

Answer 12:

Let the time taken by 5 spraying machines to finish a painting job be x minutes.
 

Number of machines	3	5
Time (in minutes)	60	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{spraying} \mathrm{machines} \mathrm{and} \mathrm{the} \mathrm{time} \mathrm{taken} \mathrm{by} \mathrm{them} \mathrm{to} \mathrm{finish} \mathrm{a} \mathrm{painting} \mathrm{job} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 3 \times 60 = 5 \times x \\ \Rightarrow 180 = 5x \\ \Rightarrow x = \frac{180}{5} \\ = 36 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{time} \mathrm{will} \mathrm{be} 36 \mathrm{minutes}. \end{gathered}$$

Question 13:

A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How many new members are there in this group now?

Answer 13:

Let x be the number of new members in the group.
 

Number of members	3	x
Number of days	30	18

$$\begin{gathered} \mathrm{Since} \mathrm{more} \mathrm{members} \mathrm{can} \mathrm{finish} \mathrm{the} \mathrm{wheat} \mathrm{in} \mathrm{less} \mathrm{number} \mathrm{of} \mathrm{days}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{case} \mathrm{of} \mathrm{inverse} \mathrm{variation}. \\ \mathrm{Therefore}, \mathrm{we} \mathrm{get}: \\ 3 \times 30 = x \times 18 \\ \Rightarrow 90 = 18x \\ \Rightarrow x = \frac{90}{18} \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{new} \mathrm{members} \mathrm{in} \mathrm{the} \mathrm{group} = 5 - 3 = 2 \end{gathered}$$

Question 14:

55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days?

Answer 14:

Let x be the number of cows that can graze the field in 10 days .
 

Number of days	16	10
Number of cows	55	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{cows} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{taken} \mathrm{by} \mathrm{them} \mathrm{to} \mathrm{graze} \mathrm{the} \mathrm{field} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 16 \times 55 = 10 \times x \\ \Rightarrow x = \frac{16 \times 55}{10} \\ = 88 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{cows} \mathrm{is} 88 \end{gathered}$$

Question 15:

18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required?

Answer 15:

Let the number of men required to reap the field in 15 days be x.
 

Number of days	35	15
Number of men	18	x

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \mathrm{and} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{required} \mathrm{ro} \mathrm{reap} \mathrm{the} \mathrm{field} \mathrm{are} \mathrm{in} \mathrm{inverse} \mathrm{variation}, \mathrm{we} \mathrm{have}: \\ 35 \times 18 = 15 \times x \\ \Rightarrow x = \frac{35 \times 18}{15} \\ = 42 \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{men} \mathrm{is} 42. \end{gathered}$$

Question 16:

A person has money to buy 25 cycles worth Rs 500 each. How many cycles he will be able to buy if each cycle is costing Rs 125 more?

Answer 16:

Let x be the number of cycles bought if each cycle costs Rs 125 more.
 

Cost of a cycle (in Rs)	500	625
Number of  cycles	25	x

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{in} \mathrm{inverse} \mathrm{variation}. \mathrm{Therefore}, \mathrm{we} \mathrm{get}: \\ 500 \times 25 = 625 \times x \\ \Rightarrow x = \frac{500 \times 25}{625} \\ = 20 \\ \therefore \mathrm{The} \mathrm{required} \mathrm{number} \mathrm{of} \mathrm{cycles} \mathrm{is} 20. \end{gathered}$$

Question 17:

Raghu has enough money to buy 75 machines worth Rs 200 each. How many machines can he buy if he gets a discount of Rs 50 on each machine?

Answer 17:

Let x be the number of machines he can buy if a discount of Rs. 50 is offered on each machine.
 

Number of machines	75	x
Price of each machine (in Rs)	200	150

$$\begin{gathered} \mathrm{Since} \mathrm{Raghu} \mathrm{is} \mathrm{getting} \mathrm{a} \mathrm{discount} \mathrm{of} \mathrm{Rs} 50 \mathrm{on} \mathrm{each} \mathrm{machine}, \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{each} \mathrm{machine} \mathrm{will} \mathrm{get} \mathrm{decreased} \mathrm{by} \mathrm{Rs} 50. \\ \mathrm{If} \mathrm{the} \mathrm{price} \mathrm{of} \mathrm{a} \mathrm{machine} \mathrm{is} \mathrm{less}, \mathrm{he} \mathrm{can} \mathrm{buy} \mathrm{more} \mathrm{number} \mathrm{of} \mathrm{machines}. \\ \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{case} \mathrm{of} \mathrm{inverse} \mathrm{variation}. \mathrm{Therefore}, \mathrm{we} \mathrm{have}: \\ 75 \times 200 = x \times 150 \\ \Rightarrow x = \frac{75 \times 200}{150} \\ = \frac{15000}{150} \\ = 100 \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{machines} \mathrm{he} \mathrm{can} \mathrm{buy} \mathrm{is} 100. \end{gathered}$$

Question 18:

If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
(iv) y = 35, find x when constant of variation = 7.

Answer 18:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Since} x \mathrm{and} y \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ xy = k \\ \mathrm{For} x = 3 \mathrm{and} y = 8, \mathrm{we} \mathrm{have}: \\ 3 \times 8 = k \\ \Rightarrow k = 24 \\ \mathrm{For} x = 4, \mathrm{we} \mathrm{have}: \\ 4y = 24 \\ \Rightarrow y = \frac{24}{4} \\ = 6 \\ \therefore y = 6 \\ \left(\mathrm{ii}\right) \mathrm{Since} x \mathrm{and} y \mathrm{vary} \mathrm{inversely}, \mathrm{we} \mathrm{have}: \\ xy = k \\ \mathrm{For} x = 5 \mathrm{and} y = 15, \mathrm{we} \mathrm{have}: \\ 5 \times 15 = k \\ \Rightarrow k = 75 \\ \mathrm{For} y = 12, \mathrm{we} \mathrm{have}: \\ 12x = 75 \\ \Rightarrow x = \frac{75}{12} \\ = \frac{25}{4} \\ \therefore x = \frac{25}{4} \\ \left(\mathrm{iii}\right) \mathrm{Given}: \\ x = 30 \mathrm{and} k = 900 \\ \therefore xy = k \\ \Rightarrow 30y = 900 \\ \Rightarrow y = \frac{900}{30} \\ = 30 \\ \therefore y = 30 \\ \left(\mathrm{iv}\right) \mathrm{Given}: \\ y = 35 \mathrm{and} k = 7 \\ \mathrm{Now}, xy = k \\ \Rightarrow 35x = 7 \\ \Rightarrow x = \frac{7}{35} \\ = \frac{1}{5} \\ \therefore x = \frac{1}{5} \end{gathered}$$