EXERCISE 10 B
Q1 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q2 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q3 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q4 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q5 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q6 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q7 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q8 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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
Q9 | Ex-10B | Class 8 | Direct and inverse variation | SChand Composite Maths | myhelper
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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