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

EXERCISE 10 B

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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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Q2 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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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)

Or

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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Q3 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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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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Q7 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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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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Q8 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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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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Q9 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper

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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

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