S.chand publication New Learning Composite mathematics solution of class 8 Chapter 9 Variation Exercise 9B

 Exercise 9B


Q1 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 1 

State the relationship between the given variables as an equation, using k for the constant of variation.

(a) The volume V a gas at a fixed temperature varies inversely as the pressure P.

Sol :

V=volume of gas

P=pressure

K=temperature(constant)

$V\propto \frac{1}{P}$

∴$V=\frac{K}{P}$


(b) The current l in an electrical circuit of fixed voltage varies inversely as the resistance R.

Sol :

I=current

R=ressistance

K=voltage(constant)

$I \propto \frac{1}{R}$

∴$I=\frac{K}{R}$


(c) The height h of a cylinder of fixed volume varies inversely as the area A of the base.

Sol :

h=height

A=Area

K=volume(constant)

$h\propto \frac{1}{A}$

∴$h=\frac{K}{A}$


(d) The frequency f of an electromagnetic wave is inversely proportional to the length l of the wave.

Sol :

f=frequency

l=length of wave

k=constant

$f\propto \frac{1}{l}$

∴$f=\frac{K}{l}$



Q2 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 2

Fill in the blanks in the following tables by determining first whether x and y very directly or inversely:

(a)

x 3 6 ? 27 ?
y 11 22 33 ? 880

Sol :

$\frac{y}{x}=\frac{11}{3}=k$

∴$\frac{y_4}{x_4}=k$

$x_4=33\times \frac{3}{11}$=9


$\frac{y_5}{x_5}=k$

$y_5=\frac{11}{5}\times 27$=99


$\frac{y_6}{x_6}=k$

$x_6=880\times \frac{3}{11}=240$


Ans 9,99,240  (Directly)


(b)

x 30 20 15 ? ?
y 6 4 ? 2 1
Sol :
y ∝ x

∴$\frac{y}{x}=k=\frac{6}{30}=\frac{1}{5}$

$\frac{y_4}{x_4}=k$
$y_4=15\times \frac{1}{5}$
=3


$\frac{y_5}{x_5}=k$
$x_5=\frac{y_5}{k}$
x5=2×5=10


$\frac{y_6}{x_6}=k$
$x_6=\frac{y_6}{k}$
x6=1×5=5

Ans 3,10,5 (Directly)

(c)

x 2 3 4 ? 8
y 48 ? 24 16 ?

Sol :

$y \propto \frac{1}{x}$

xy=k=48×2=96

y2x2=k

$y_2=\frac{x}{x_2}$

$=\frac{96}{3}=32$


y4x4=k

$x_4=\frac{k}{y_1}$

$x_4=\frac{96}{16}=6$


y5x5=k

$y_5=\frac{k}{x_5}$

$x_4=\frac{96}{8}=12$


Ans 32, 6, 12 (Inversly)


(d)

x 1 5 10 ? ?
y 125 ? 12.5 5 1


Sol :

$y\propto \frac{1}{x}$

∴xy=k=125×1=125


x2y2=k

$y=\frac{125}{5}=25$


x4y4=k

$x_4=\frac{125}{5}=25$

x5y5=k

$x_5=\frac{125}{1}=125$

Ans 25, 25, 125



Q3 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 3

 (a) If y varies inversely as x, and y = 9 when x = 2. find y when x = 3.

Sol :

y 9 ? ?
x 2 12 3


$y \propto \frac{1}{x}$ 

∴yx=k

Now , y=9 , x=2

∴k=yx

or k=9×2=18


Again, 

x=12

∴$y=\frac{18}{12}=\frac{3}{2}$


x=3

∴$y=\frac{k}{x}=\frac{18}{3}=6$


Ans $\frac{3}{2}$ ,3


(b) If u is inversely proportional to v, and is u = 12 when v = 3, find u when v = 9.

Sol :

$u \propto \frac{1}{v}$  

∴uv=k


Now u=12 ,v=3 ∴k=12×3=36

Agan , v=9  ∴$u=\frac{k}{v}=\frac{36}{9}=4$


(c) If c is inversely proportional to d, and if c = 18 when d = 2/3, find d when c = 6/7

Sol :

$c\propto \frac{1}{d}$ ∴cd=k

Now , 

c=18 ,$d=\frac{2}{3}$  ∴$k=18 \times \frac{2}{3}=12$

Again $c=\frac{6}{7}$  ∴$d=\frac{k}{c}12\times \frac{7}{6}=14$


(d) If m is inversely propotional to n, and if m = 0.02 when n = 5, find m when n = 0.2

Sol :

$m \propto \frac{1}{n}$ , ∴mn=k

Now ,m=0.02 , n=5  ∴k=0.02×5=0.10

mn=k

m×0.2=0.10

m=0.5



Q4 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 4

Navin cycles to his school at an average speed of 12 km/hr. It takes him 20 minutes to reach the school. If he wants to reach his school in 15 minutes, what should be his average speed?

Sol :

Speed=s=12 km/hr , time=t=20min

∴$s\propto \frac{1}{t}$  ∴st=k

s1=12 , t1=20 , k=12×20=240

t2=15min  ∴$S_2=\frac{k}{t_2}=\frac{240}{15}=16$km/hr



Q5 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 5

The time needed to travel from one place to another is inversely proportional to the speed. A person travelling 72 km/hr can go from Dehradun to Lucknow in 10 hours. How fast must the person travel to make the trip in 9 hours?

Sol :

$T \propto \frac{1}{S}$  ∴TS=K

S1=72km/hr  ,T1=10 h  ∴K=S1T1=72×10=720

∴T1=9 h  , $S_1=\frac{K}{T_1}=\frac{720}{9}=80$km/h



Q6 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 6

28 pumps can empty a reservoir in 18 hours. In how many hours can 42 such pumps do the same work?

Sol :

$P\propto \frac{1}{T}$  ∴PT=K

P1=28  ,T1=18 h  ∴K=P1T1=18×28=504

∴P2=42  , $T_2=\frac{K}{P_2}=\frac{504}{42}=12$h



Q7 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 7

A stock of food grains is enough for 400 persons in 9 days. How long will the same stock last for 300 persons?

Sol :

$M\propto \frac{1}{D}$  MD=K

∴M1=400 per  D1=9  ∴K=M1D1=400×9=3600      

∴M2=300 per  $D_2=\frac{K}{M_2}=\frac{3600}{300}=12$days



Q8 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 8

A contractor, who had a workforce of 630 persons, undertook to complete a portion of a stadium in 14 months. He was asked to complete the job in 9 months. How many extra persons had he to employ?

Sol :

$M\propto \frac{1}{D}$  ∴MD=K

∴M1=630 per  D1=14 months  ∴K=M1D1=630×14=8820      

∴D2=9 months  $M_2=\frac{K}{D_2}=\frac{8820}{9}=980$days

∴Extra person=980-630=350



Q9 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 9

Working 4 hours a day, Savita can type a manuscript in 15 days. How many hours a day should she work so as to finish the work in 10 days?

Sol :

$T\propto \frac{1}{D}$   ∴TD=K

∴T1=4 hour  D1=15 days  ∴K=T1D1=4×15=60      

D2=10 days  $T_2=\frac{k}{D_2}=\frac{60}{10}$=6 hour  



Q10 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 10

A train moving at a speed of 60 km/hr covers a certain distance in 7.5 hours. What should be the speed of the train to cover the same distance in 6 hours?

Sol :

$S \propto \frac{1}{T}$   ∴ST=K

∴S1=60 km/hour  T1=7.5 hour  ∴K=S1T1=60×7.5=450      

T2=6 hour  $S_2=\frac{450}{6}$=75 km/hour  



Q11 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 11

A garrison of 800 men had provisions for 39 days, However a reinforcement of 500 men arrived. For how many days will the food last now?

Sol :

$M \propto \frac{1}{D}$  ∴DM=K

∴M1=800 min  D1=39 days  ∴K=M1D1=800×39=31200      

M2=500 min  $D_2=\frac{31200}{500}$=62.4 days

∴The food last now=(62.4-39)=23.4≈24days



Q12 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 12

A beseieged town has provisions to last for 3 weeks. Its population is 22400. How many people must be sent away in order that the provisions may last for 7 weeks?

Sol :

$P \propto \frac{1}{W}$  ∴PW=K

∴P1=22400 min  W1=3 days  ∴K=P1W1=22400×3=67200      

W2=7 $P_2=\frac{67200}{7}$=9600 

∴Number of provisions=22400-9600

=12800



Q13 | Ex-9B | Class 8 | S.Chand | New Learning Composite maths | Variation | myhelper

Question 13

A hostel had rations for 60 days for 500 students. After 12 days, 300 more students joined the hostel. How long will the remaining rations last?

Sol :

(60-12)=48 days for (500+300)=800

students 500 800
days 48 x

∴$x=\frac{48\times 500}{800}$

=30 days

No comments:

Post a Comment

Contact Form

Name

Email *

Message *