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S.chand books class 8 maths solution chapter 10 Direct and Inverse variation exercise 10 C

EXERCISE 10 C

Q1 | Ex-10C | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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Question 1

4 mens can make 4 cupboards in 4 days ; how many cupboards can 14 mens make in 14 days ?

Sol :

As number of men ∝ number of cupboards or Number of days

In 4 days 4 mens make 4 cupboards

∴ In 1 day 4 mens make $\dfrac{4}{4}$ cupboards [dividing 4 both sides]

∴ In 1 day 1 men make $\dfrac{4}{4\times 4}$ cupboards [dividing 4 both sides]

∴ In 1 day 14 mens make $\dfrac{4}{4\times 4}\times 14$ cupboards [multiplying 14 both sides]

∴ In 14 days 14 mens make $\dfrac{4}{4\times 4}\times 14 \times 14$ cupboards [multiplying 14 both sides]

$=\dfrac{4}{16}\times 14 \times 14$

= 49 cupboards

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Question 2

In a hostel it costs 1800 to keep 50 students for 8 weeks . For what length of time did the cost of keeping 90 students amount to 21060

Sol :

As cost ∝ Student or weeks

⇒To keep 50 students for 8 weeks it costs 1800

⇒To keep 50 students for 1 weeks it costs $\dfrac{1800}{8}$ [Dividing 8 both sides]

⇒To keep 1 students for 1 weeks it costs $\dfrac{1800}{8\times 50}$ [Dividing 50 both sides]

⇒To keep 90 students for 1 weeks it costs $\dfrac{1800}{8\times 50}\times 90$ = 405 [Multiplying 90 both sides]

⇒So with amount 21060 , 90 students stay $\dfrac{21060}{405}=52$ weeks

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Question 3

39 persons can repair a road in 12 days working 5 hours per day . In how many days will 30 persons working 6 hours per day complete the work ?

Sol :

$\text{Number of persons} \propto \dfrac{1}{\text{Number of working days}}$

Case 1:

39 persons working 5 hours per day to repair a road take 12 days which is equal to 2340

⇒39×5×12 = 2340..(i)

Case 2:

30 persons working 6 hours per day to repair a road take x days which is equal to 180x

⇒30×6 = 180x ..(ii)

From (i) and (ii) , we get

⇒2340 = 180x

⇒$x=\dfrac{2340}{180}$

⇒x = 13

⇒13 days

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Question 4

If 15 bottles of water are needed for seven men for two days , how many bottles are required for four mens for seven days ?

Sol :

Number of bottles ∝ Number of mens

⇒7 mens for 2 days need 15 bottles

⇒7 mens for 1 days need $\dfrac{15}{2}$ bottles [Dividing 2 both sides]

⇒1 men for 1 day need $\dfrac{15}{2\times 7}$ bottles [Dividing 7 both sides]

⇒1 men for 7 days need $\dfrac{15}{2\times 7}\times 7$ bottles [Multiplying 7 both sides]

⇒4 mens for 7 days need $\dfrac{15}{2\times 7}\times 7\times 4$ bottles [Multiplying 7 both sides]

⇒30 bottles are needed

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Question 5

10 cooks working for 8 hours each can prepare a meal for 536 people . How many cooks will be needed to prepare a meal for 737 people , if they are required to prepare a meal in 5 hours ?

Sol :

As cooks ∝ meal for peoples

⇒10 cooks working 8 hours to prepare meal for 536 peoples

⇒If 10 cooks work 1 hour, then they prepare meal for $\dfrac{536}{8}$ peoples

⇒If 1 cooks work 1 hour, then it prepare meal for $\dfrac{536}{8\times 10}$ peoples

⇒If 1 cooks work 5 hour, then it prepare meal for $\dfrac{536}{8\times 10}\times 5$ peoples = 33.5 ..(i)

⇒If we have to prepare meal for 737 people then cooks required is equal to

$=\dfrac{737}{33.5}$ [from (i)]

= 22 cooks

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Question 6

A garrison of 1200 men has sufficient rations for 25 days at the rate of 2400 g per man per day . If 300 men join them and the rations are reduced to 2000 g per man per day , how long will the food last all of them ?

Sol :

$\text{Number of men}\propto\dfrac{1}{\text{Number of Days}}$

Case 1:

⇒Total men 1200 has ration for 25 days at rate of 2400 g per man per day which is equal to 72000000

⇒1200×25×2400=72000000..(i)

Case 2:

⇒Total men 1200+300=1500 has ration for x days at rate of 2000 g per man per day which is equal to 3000000

⇒1500×x×2000 = 3000000x..(ii)

On dividing (i) by (ii) , we get number of days ration last for 1500 peoples at rate of 2000 g per man per day

⇒$x=\dfrac{72000000}{3000000}$

x = 24

24 days

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Question 7

If a man travels 65 km in 3 days by walking $7\dfrac{1}{2}$ hours a day , in how many days will he travel 156 km by walking 8 hours a day ?

Sol :

Case 1 :

⇒In 3 days a man walk $7\dfrac{1}{2}$ hours a day to cover 65 km

⇒In 3 days a man walk $\dfrac{15}{2}$ hours a day to cover 65 km

⇒In 1 days a man walk $\dfrac{15}{2}$ hours a day to cover $\dfrac{65}{3}$ km

⇒In 1 days a man walk 1 hour a day to cover $\dfrac{65}{3}\times \dfrac{2}{15}$ km

⇒In 1 days a man walk 8 hours a day to cover $\dfrac{65}{3}\times \dfrac{2}{15}\times 8$ km

⇒In x days a man walk 8 hours a day to cover $\dfrac{65}{3}\times \dfrac{2}{15}\times 8\times x$ km or $\dfrac{13}{3}\times \dfrac{2}{3}\times 8\times x$ or $\dfrac{208x}{9}$ km ..(i)

Case 2:

In x days a man walks 8 hours a day cover 156 km ..(ii)

Equation (ii) must be equal to (i)

⇒$\dfrac{208x}{9}=156$

⇒$x=\dfrac{156\times 9}{208}$

$=\dfrac{78\times 9}{104}=\dfrac{39\times 9}{52}$

$=\dfrac{351}{52}=\dfrac{27}{13}$

$=6\dfrac{3}{4}$ days

S.chand books class 8 maths solution chapter 10 Direct and Inverse variation exercise 10 B

EXERCISE 10 B

Q1 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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Question 1

The pressure P of an enclosed gas , held at a constant temperature , is inversely proportional to the volume V of the gas . Express P in terms of V and the constant of variations K . Calculate
(i) The value of K when P = 500 and V = 2

Sol :

Given : $\text{P}\propto \dfrac{1}{\text{V}}$ or $\text{P}= \dfrac{k}{\text{V}}$

⇒ PV = K

⇒ 500×2 = K

⇒K = 1000

(ii) The value of P when V = 5

Sol :

⇒$\text{P}= \dfrac{k}{\text{V}}$

⇒$\text{P}= \dfrac{1000}{5}$

⇒P = 200

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Question 2

How many days would it take 67 men to build a wall which 134 mens can build in 3 weeks ?

Sol :

Number of men (m)	134	67
Number of weeks(w)	3

As number of weeks (w) varies inversely with number of mens (m)

⇒$\text{No. of weeks}\propto \dfrac{1}{\text{No. of mens}}$ or

⇒$\text{No. of weeks}=\dfrac{k}{\text{No. of mens}}$

⇒k = w×m , where k is variable constant

⇒k = 3×134 = 402

Now , m = 67 , d = ? ⇒ $\text{No. of weeks}=\dfrac{k}{\text{No. of mens}}$

⇒$\text{w}=\dfrac{402}{67}$

⇒w = 6

= 6 weeks

ALTERNATE METHOD

134÷67=2

Simple explanation through logic:

If 2x the amount of people takes a certain time to complete a task, half the amount of people will take 2x the time to complete the task.

3 weeks × 2 = 6 weeks

6 weeks = 42 days

Some basic concept

Why we are not dividing 134 by 3 ?

Answer : Because No. of mens is inversely proportional to no . of weeks in other words on increasing first value(no. of mens) then other value decreases(no. of weeks)

Why we are multiplying 134 by 3 ?

Answer : To understand this firstly lets discuss why we use multiplication

Suppose we have to give 2 pencil to every five person , then whats the Total no. of pencils

2 + 2 + 2 + 2 + 2 = 10 or 2×5 = 10

In the same way 134 mens work every week , total 3 weeks to complete work

No . of mens in 1st week + No. of mens work in 2nd week + No. of mens work in 3rd week = 402 or No. of mens ×3 = 402

402/67= 6 weeks

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Question 3

A car can complete a certain journey in 12 hours if it travels at 65 km/h . How much time will it take if it travels at 78 km/h ?

Sol :

Speed(S)	65 km/p	78 km/h
Time (T)	12 hours

As Speed (S) varies inversely with Time (T)

⇒$\text{S}\propto \dfrac{1}{\text{T}}$ or

⇒$\text{S}=\dfrac{k}{\text{T}}$

⇒k = S×T , where k is variable constant

⇒k = 65×12 = 780

Now , S = 78 , T = ? ⇒ $\text{S}=\dfrac{k}{T}$

⇒$\text{S}=\dfrac{780}{78}$

⇒S = 10

= 10 hours

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Question 4

A workforce of 210 men with a contractor can finish a work in 16 months . How many more man men should be employed so that the work is completed in 14 months ?

Sol :

Number of men (m)	210
Time(T)	16 months	14 months

As Number of mens (m) varies inversely with Time (T)

⇒$\text{m}= \dfrac{1}{\text{T}}$ or

⇒$\text{m}=\dfrac{k}{T}$

⇒k = T×m , where k is variable constant

⇒k = 16×210 = 3360

Now , T = 14 , m = ? ⇒ $\text{m}=\dfrac{k}{\text{T}}$

⇒$\text{m}=\dfrac{3360}{14}$

⇒m = 240

To complete work in 14 months , we need

= 240 - 210

= 30 more men

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Question 5

It is found that a book will contains 540 pages if 28 lines are allowed in a page . How many lines should be allowed in a page if the book has to contain 360 pages ?

Sol :

Number of pages (P)	540	360
Lines (L)	28

As Number of pages (P) varies inversely with Lines (L) ,

⇒$\text{P}\propto \dfrac{1}{\text{L}}$ or

⇒$\text{P}= \dfrac{k}{\text{L}}$ or

⇒k = P×L

⇒k = 540×28

⇒k = 15120 , where k is variable constant

Now , P = 360 , L = ? ⇒ , $\text{L}= \dfrac{k}{\text{P}}$ or

⇒$\text{L}= \dfrac{15120}{360}$

⇒L = 42

= 42 lines per page

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Question 6

Sreyash has enough money to buy 54 machines worth 200 each . How many machines can he buys if he gets a discount of 20 on each machine ?

Sol :

Number of Machines (M)	54
Cost of each (C)	200	200 - 20 = 180

As Number of Machines (M) varies inversely with Cost (C) ,

⇒$\text{M}\propto \dfrac{1}{\text{C}}$ or

⇒$\text{M}= \dfrac{k}{\text{C}}$ or

⇒k = M×C

⇒k = 54×200

⇒k = 10800 , where k is variable constant

Now , C = 180 , M = ? ⇒ , $\text{M}= \dfrac{k}{\text{C}}$ or

⇒$\text{M}= \dfrac{10800}{180}$

⇒M = 60

= 60 machines

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Question 7

If a ball moves at 135 m/s , it will strike an object in 4 seconds . If it moves at 120 m/s , how long will it takes to strike the same objects ?

Sol :

Speed (S)	135 m/s	120 m/s
Time (T)	4 seconds

As Speed (S) varies inversely with Time (T) ,

⇒$\text{S}\propto \dfrac{1}{\text{T}}$ or

⇒$\text{S}= \dfrac{k}{\text{T}}$ or

⇒k = S×T

⇒k = 135×4

⇒k = 540 , where k is variable constant

Now , S = 120 , T = ? ⇒ , $\text{T}= \dfrac{k}{\text{S}}$ or

⇒$\text{T}= \dfrac{540}{120}$

⇒T = 4.5 seconds

= 4.5 seconds

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Question 8

A man eats 200 g of rice a day and he has enough rice to last him 35 days . How long would the stock of rice last him if he were to eat 250 g of rice a day ?

Sol :

Rice (R)	200 g	250 g
Number of Days (D)	35 days

As Rice (R) varies inversely with Number of Days (D) ,

⇒$\text{R}\propto \dfrac{1}{\text{D}}$ or

⇒$\text{R}= \dfrac{k}{\text{D}}$ or

⇒k = R×D

⇒k = 200×35

⇒k = 7000 , where k is variable constant

Now , R = 250 , D = ? ⇒ , $\text{D}= \dfrac{k}{\text{R}}$ or

⇒$\text{D}= \dfrac{7000}{250}$

⇒D = 28 days

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Question 9

A wheel of circumference 2.8 m revolve 765 times in traversing a certain distance . How many revolutions does a wheel of circumference 1.7 m make in traversing the same distance .

Sol :

Circumference of wheel (W)	2.8 m	1.7 m
Number of Revolution (R)	765

As Circumference of wheel (W) varies inversely with Number of Revolution (R)

⇒$\text{W}\propto \dfrac{1}{\text{R}}$ or

⇒$\text{W}= \dfrac{k}{\text{R}}$ or

⇒k = W×R

⇒k = 2.8×765

⇒k = 2142 , where k is variable constant

Now , W = 1.7 , R = ? ⇒ , $\text{R}= \dfrac{k}{\text{W}}$ or

⇒$\text{R}= \dfrac{2142}{1.7}$

⇒R = 1260 times

= 1260 times

S.chand mathematics solution class 8 chapter 10 Direct and inverse variation

EXERCISE 10 A

Q1 | Ex-10A | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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Question 1

Given that y is directly proportional to x and y = 40 , when x = 200 . Find the value of
(i) y , when x = 15

Sol: (i)

y ∝ x ⇒ $\dfrac{y}{x}=k$ , where k is the constant of variation

Given , $\dfrac{y}{x}=\dfrac{40}{200}=\dfrac{1}{5}$ ⇒ k = $\dfrac{1}{5}$

Also , $y = \dfrac{x}{5}$

$y=\dfrac{15}{5}$ [given: x=15]

= 3

(ii) x when y = 8

Sol: (ii)

y ∝ x ⇒ $\dfrac{y}{x}=k$ , where k is the constant of variation

Given , $\dfrac{y}{x}=\dfrac{40}{200}=\dfrac{1}{5}$ ⇒ k = $\dfrac{1}{5}$

Also , 5y = x [given: y=8]

x = 40

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Question 2

The length (in cm) stretched by a spring is directly proportional to the amount of force (in kg) applied . Given below are some observations about the force applied and the length stretched by a spring . Find the missing values in the table

Force (in kg)		20	25
Length stretched(in cm)	18	15		28

Sol:

As Force and Length vary directly . So $\dfrac{\text{Force}}{\text{Length}}$ is constant

$\dfrac{\text{Force}}{\text{Length}}=\dfrac{20}{15}=\dfrac{4}{3}$

∴The constant of variation $=\dfrac{4}{3}$

Now , $\dfrac{\text{Force}}{18}=\dfrac{4}{3}$⇒$\text{Force}=\dfrac{4\times 18}{3}$ = 24

$\dfrac{25}{\text{Length}}=\dfrac{4}{3}$⇒$\text{Length}=\dfrac{3\times 25}{4}=18\dfrac{3}{4}$

$\dfrac{\text{Force}}{28}=\dfrac{4}{3}$⇒$\text{Force}=\dfrac{4\times 28}{3}=37\dfrac{1}{3}$

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Question 3

Priya takes 4 hours in walking a distance of 20 km . What distance would she cover in 7 hours ?

Sol:

Distance	20 km
Time	4 hours	7 hours

As Distance ∝ Time ⇒$\dfrac{\text{Distance}}{\text{Time}}=k$

So $\dfrac{\text{Distance}}{\text{Time}}$ is constant

$\dfrac{\text{Distance}}{\text{Time}}=\dfrac{20}{4}=5$

∴The constant of variation = 5

Now , $\dfrac{\text{Distance}}{7}=5$

Distance = 7×5 = 35

ALTERNATE METHOD

As we know $\text{Speed}=\dfrac{\text{Distance}}{\text{time}}$

$=\dfrac{20}{4}=5$

= 5 km/hr

7 hours = 7×5 km

= 35 km

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Question 4

If 15 burners consume 90 cubic metre of gas in 2 hours , how much will 9 burners consume in the same time ?

Sol:

Gas (m³)	90 m³
Burner	15	9

As Gas consume ∝ Burner ⇒$\dfrac{\text{Gas}}{\text{Burner}}=k$

So $\dfrac{\text{Gas}}{\text{Burner}}$ is constant

$\dfrac{\text{Gas}}{\text{Burner}}=\dfrac{90}{15}=6$

∴The constant of variation = 6

Now , $\dfrac{\text{Gas}}{9}=6$

Distance = 6×9 = 54

= 54 cubic metre or 54 m³

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Question 5

The railway charges 5600 to carry a certain amount of luggage for 350 km . What should the charge be carry the same amount of luggage for 425 km ?

Sol :

Charges	5600
Distance (Km)	350 km	425 km

As Charges ∝ Distance ⇒$\dfrac{\text{Charges}}{\text{Distance}}=k$

So $\dfrac{\text{Charges}}{\text{Distance}}$ is constant

$\dfrac{\text{Charges}}{\text{Distance}}=\dfrac{5600}{350}=16$

∴The constant of variation = 16

Now , $\dfrac{\text{Charges}}{425}=16$

Distance = 16×425 = 6800

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Question 6

89 litres of oil cost 2091.50 . What is the cost of 15 litres ?

Sol :

Oil Cost	2091.50
Litres	89	15

As Cost ∝ Litres ⇒$\dfrac{\text{Cost}}{\text{Litre}}=k$

So $\dfrac{\text{Cost}}{\text{Litre}}$ is constant

$\dfrac{\text{Cost}}{\text{Litre}}=\dfrac{2091.50}{89}=23.50$

∴The constant of variation = 23.50

Now , $\dfrac{\text{Cost}}{15}=23.50$

Cost = 23.50×15 = 352.50

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Question 7

68 packets weigh 1 kg 632 g . What will be weight of 70 packets ?

Sol :

Weigh	1 kg 632 g (1632g)
Packets	68	70

As Weigh ∝ Packets ⇒$\dfrac{\text{Weigh}}{\text{Packets}}=k$

So $\dfrac{\text{Weigh}}{\text{Packets}}$ is constant

$\dfrac{\text{Weigh}}{\text{Packets}}=\dfrac{1632}{68}=24$

∴The constant of variation = 24

Now , $\dfrac{\text{Weigh}}{70}=24$

Cost = 24×70 = 1680 g

1680 g = 1 kg 680 g

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Question 8

If a man working for 49 hours earns 1715 , how much will he earn working for 27 hours ?

Sol :

Earns	1715
Hours (hours)	49	27

As Earns ∝ Hours ⇒$\dfrac{\text{Earns}}{\text{Hours}}=k$

So $\dfrac{\text{Earns}}{\text{Hours}}$ is constant

$\dfrac{\text{Earns}}{\text{Hours}}=\dfrac{1715}{49}=35$

∴The constant of variation = 35

Now , $\dfrac{\text{Earns}}{27}=35$

Cost = 35×27 = 945

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Question 9

The distance travelled by a ball dropped from an airplane is directly proportional to the square of time t . Given that t=2 seconds when d = 24 meters , find the distance the ball drops in 10 seconds .

Sol:

Distance (m)	24 m
Time (sec)	2 sec	10 sec

As Distance ∝ ${Time}^2$

⇒$\dfrac{\text{Distance}}{\text{Time}^2}=k$

So $\dfrac{\text{Distance}}{\text{Time}^2}$ is constant

$\dfrac{\text{Distance}}{\text{Time}}=\dfrac{24}{2^2}=\frac{24}{4}=6$

∴The constant of variation = 6

Now , $\dfrac{\text{Distance}}{{10}^2}=\frac{D}{100}=6$

Distance = 60×10 = 600

= 600 metres

EXERCISE 10 C

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EXERCISE 10 B

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EXERCISE 10 A

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