RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

Other Exercises

Question 1.
Fill in the blanks: [NCERT]
(i) All points lying inside / outside a circle are called …….. points / ………. points.
(ii) Circles having the same centre and different radii are called …….. circles.
(iii) A point whose distance from the centre of a circle is greater than its radius lies in …….. of the circle.
(iv) A continuous piece of a circle is …….. of the circle.
(v) The longest chord of a circle is a ……… of the circle.
(vi) An arc is a …….. when its ends are the ends of a diameter.
(vii) Segment of a circle is the region between an are and ……..of the circle.
(viii)A circle divides the plane, on which it lies, in …….. parts.
Solution:
(i) All points lying inside / outside a circle are called interior points / exterior points.
(ii) Circles having the same centre and different radii are called concentric circles.
(iii) A point whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iv) A continuous piece of a circle is arc of the circle.
(v) The longest chord of a circle is a diameter of the circle.
(vi) An arc is a semi-circle when its ends are the ends of a diameter.
(vii) Segment of a circle is the region between an arc and centre of the circle.
(viii) A circle divides the plane, on which it lies, in three parts.

Question 2.
Write the truth value (T/F) of the following with suitable reasons: [NCERT]
(i) A circle is a plane figure.
(ii) Line segment joining the centre to any point on the circle is a radius of the circle.
(iii) If a circle is divided into three equal arcs each is a major arc.
(iv) A circle has only finite number of equal chords.
(v) A chord of a cirlce, which is twice as long is its radius is a diameter of the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure of an arc is the complement of the central angle containing the arc.
(viii)The degree measure of a semi-circle is 180°.
Solution:
(i) True.
(ii) True.
(iii) True.
(iv) False. As it has infinite number of equal chords.
(v) True.
(vi) False. It is a segment not sector.
(vii) False. As total degree measure of a circle is 360°.
(viii) True.

 

 

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1 are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.

Other Exercises

Question 1.
Solution:
x2 + 8x + 16
= (x)2 + 2 × x × 4 + (4)2
= (x + 4)2

Question 2.
Solution:
x2 + 14x + 49
= (x)2 + 2 × x × 7 + (7)2
= (x + 7)2 Ans.

Question 3.
Solution:
1 + 2x + x2
= (1)2 + 2 × 1 × x + (x)2
= (1 + x)2 Ans.

Question 4.
Solution:
9 + 6z + z2
= (3)2 + 2 x 3 x z + (z)2
= (3 + z)2 Ans.

Question 5.
Solution:
x2 + 6ax + 9a2
= (x)2 + 2 × x × 3a + (3a)2
= (x + 3a)2 Ans.

Question 6.
Solution:
4y2 + 20y + 25
= (2y)2 + 2 x 2y x 5 + (5)2
= (2y2 + 5)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 7.
Solution:
36a2 + 36a + 9
= 9 [4a2 + 4a + 1]
= 9 [(2a)2 + 2 x 2a x 1 + (1)2]
= 9 [2a + 1]2

Question 8.
Solution:
9m2 + 24m + 16
= (3m)2 + 2 x 3m x 4 + (4)2
= (3m + 4)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 9.
Solution:
z2 + z + \(\\ \frac { 1 }{ 4 } \)
= (z)2 + 2 x z x \(\\ \frac { 1 }{ 2 } \) + \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
= \({ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }\)

Question 10.
Solution:
49a2 + 84ab + 36b2
= (7a)2 + 2 x 7a x 6b + (6b)2
{ ∵ a2 + 2ab + b2 = (a + b)2}
= (7a + 6b)2

Question 11.
Solution:
p2 – 10p + 25
= (p)2 – 2 x p x 5 + (5)2
= (p – 5)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 12.
Solution:
121a2 – 88ab + 16b2
= (11a)2 – 2 x 11a x 4b + 4(b)2
= (11a – 4b)2

Question 13.
Solution:
1 – 6x + 9x2
= (1)2 – 2 x 1 x 3x + (3x)2
= (1 – 3x)2
{ ∵ a2 – 2ab + b2 = (a – b)2}

Question 14.
Solution:
9y2 – 12y + 4
= (3y)2 – 2 x 3y x 2 + (2)2
{ ∵ a2 – 2ab + b2 = (a – b)2}
= (3y – 2)2

Question 15.
Solution:
16x2 – 24x + 9
= (4x)2 – 2 x 4x x 3 + (3)2
= (4x – 3)2 Ans.

Question 16.
Solution:
m2 – 4mn + 4n2
= (m)2 -2 x m x 2n + (2n)2
= (m – 2n)2 Ans.

Question 17.
Solution:
a2b2 – 6abc + 9c2
= (ab)2 – 2 x ab x 3c + (3c)2
= (ab – 3c)2 Ans.

Question 18.
Solution:
m4 + 2m2n2 + n4
= (m2)2 + 2m2n2 + (n2)2
= (m2 + n2)2
{ ∵ a2 + 2ab + b2 = (a + b)2}

Question 19.
Solution:
(l + m)2 – 4lm
= l2 + m2 + 2lm – 4lm
= l2 + m2 – 2lm
= l2 – 2lm + m2
= (l – m)2
{ ∵ a2 – 2ab + b2 = (a – b)}

 

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.

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NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

MULTIPLE CHOICE QUESTIONS

Question 1.
A particle is moving in a circular path of radius r. The displacement after half a circle would be :
(a) Zero       (b) πr            (c) 2r                  (d) 2πr.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 1
Answer:
(c) Explanation : Particle is just opposite to the initial position on the circle.

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Question 2.
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g             (b) u2/2g                (c) u2/g                (d) u/2g.
Answer:
(b) Explanation : u2 – v2 = – 2gh or 0 – u2 = – 2gh (∴ u = 0 at highest point).
∴ h = u2/2g.

Question 3.
The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Answer:
(d) Explanation : Displacement ≤ distance.

Question 4.
If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 2

Question 5.
From the given u -t graph (Fig. 1), it can be inferred that the object is
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 3
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration.
Answer:
(a) Explanation : Velocity of object is constant.

Question 6.
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m s-1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity.
Answer:
(c) Explanation : Velocity of the boy changes continuously due to the change in direction of motion in circular path.

Question 7.
Area under a v — t graph represents a physical quantity which has the unit
(a) m2          (b) m                (c) m3              (d) m s-1.
Answer:
(b) Explanation : Area = v x t = (m/s) x s = m.

Question 8.
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 2 Choose the correct statement Q
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 4
(a) Car A is faster than car D
(b) Car B is the slowest
(c) Car D is faster than car C
(d) Car C is the slowest.
Answer:
(b) Explanation : Slope of distance – time graph = speed of object.

Question 9.
Which of the following figures represents uniform motion of a moving object correctly ?
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 5
Answer:
(a) Explanation : For uniform motion, distance ∞ time.

Question 10.
Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 6

Question 11.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal ?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun.
Answer:
(a)

SHORT ANSWER QUESTIONS

Question 12.
The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero ? Justify your answer.
Answer:
No. When object moves in a circular path of radius r, then displacement of object .after completing a circle is zero but distance travelled = 2πr.

Question 13.
How will the equations of motion for an object moving with a uniform velocity change ?
Answer:
Equations of motion of a uniformly accelerated motion of an object are

  1. V = u + at,
  2. S = ut + ½ at2,
  3. v2 – u2 = 2aS.

When object moves with a uniform velocity, its acceleration, a = 0. Hence equations of motion become

  1. V = u,
  2. S = ut,
  3. v2 = u2.

Question 14.
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in Fig. 4. Plot a velocity-time graph for the same.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 7
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 8

Question 15.
A car starts from rest and moves along the x-axis with constant acceleration 5 m s-2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 9
Velocity after 8 s,V = u + at = 0 + 5 x 8 = 40 ms-1.
Distance travelled in last 4 s moving with constant velocity, x2 = v t = 40 x 4 = 160 m
∴Total distance = x1 + x2 = 320 m.

Question 16.
A motorcyclist drives from A to B with a uniform speed of 30 km h-1 and returns back with a speed of 20 km h-1. Find its average speed.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 10
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 11

Question 17.
The velocity-time graph (Fig. 6) shows the motion of a cyclist. Find
(i) its acceleration
(ii) its velocity and
(iii) the distance covered by the cyclist in 15 seconds.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 12
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 13

Question 18.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height
Answer:
When stone is thrown vertically upwards, it has some initial velocity (u). The velocity of the stone goes on decreasing as it goes upwards and becomes zero at the maximum height. There after, stone begins to fall and its velocity goes on increasing but in opposite direction and becomes equal to the initial velocity (u) when it reaches the point of projection. v – t graph of a stone is shown in figure 7 A. The speed-time graph of the stone is shown in figure 7 B.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 14

Question 19.
An object is dropped from rest at the height of 150m and simultaneously another object is dropped from rest at the height of 100m.
What is the difference in their heights after 2 s if both the objects drop with same accelerations ? How does the difference in heights vary with time ?
Answer:
Distance moved by the object dropped from height of 150m in 2 seconds can be calculated using
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 15
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 16
Height of the object from ground after 2 seconds = 150 – 20 = 130 m Similarly, distance moved by the other object in 2 seconds = 20 m
Height of the second object from ground after 2 seconds = 100 – 20 = 80 m
∴ Difference in their heights after 2 seconds = 130 m – 80 m = 50 m
Difference in height of both objects remains the same with time.

Question 20.
An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 17

Question 21.
Using following data, draw time-displacement graph for a moving object :

Time (s)     0      2 4 6 8 10 12 14 16
Displacement (m) 0 2 4 4 4 6 4 2 0

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 18
Displacement-time graph is shown in figure 8.
Average velocity for first 4s = Slope of displacement-time graph
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 19

Question 22.
An electron moving with a velocity of 5 x 104 ms-1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms-2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 20

Question 23.
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Answer:
We know, distance travelled by a uniformly accelerated object in time t is given by
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 21

Question 24.
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12 : u22 (Assume upward acceleration is -g and downward acceleration to be +g).
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 22

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B

RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
\(\\ \frac { 3 }{ 5 } \)
= \(\\ \frac { 3X20 }{ 5X20 } \)
= \(\\ \frac { 60 }{ 100 } \)
= 60% (d)

Question 2.
Solution:
0.8%
= \(\\ \frac { 0.8 }{ 100 } \)
= \(\\ \frac { 8 }{ 10X100 } \)
= \(\\ \frac { 8 }{ 1000 } \)
= 0.008 (b)

Question 3.
Solution:
6 : 5
= \(\\ \frac { 6 }{ 5 } \)
= \(\\ \frac { 6X20 }{ 5X20 } \)
= \(\\ \frac { 120 }{ 100 } \)
= 120% (c)

Question 4.
Solution:
5% of a number is 9
Number = \(\\ \frac { 9X100 }{ 5 } \)
= 180 (d)

Question 5.
Solution:
Let x% of 90 = 120
=> \(\\ \frac { x }{ 100 } \) x 120 = 90
=> x% = \(\\ \frac { 120X100 }{ 90 } \)
= \(133\frac { 1 }{ 3 } %\) (c)

Question 6.
Solution:
Let x% 10 kg = 250 g
\(\\ \frac { x }{ 100 } \) x 10 kg
= \(\\ \frac { 250 }{ 1000 } \) kg
x = \(\\ \frac { 250X100 }{ 1000X10 } \)
= \(\\ \frac { 25 }{ 10 } \)%
= 2.5% (d)

Question 7.
Solution:
40% of x = 240
=> x = \(\\ \frac { 240 }{ 40 } \) x 100
= 600 (b)

Question 8.
Solution:
?% of 400 = 60
=> x% of 400 = 60
\(\\ \frac { x }{ 100 } \) x 400 = 60
x = \(\\ \frac { 60X100 }{ 400 } \)
= 15 (c)

Question 9.
Solution:
(180% of ?)÷2 = 504
\(\left( \frac { 180 }{ 100 } \times x \right) \div 2\) = 504
\(\frac { 180 }{ 100 } x\) = 504 x 2
\(x=\frac { 504\times 2\times 100 }{ 180 }\)
= 560 (d)

Question 10.
Solution:
20% of Rs. 800
= \(\\ \frac { 20 }{ 100 } \) x 800
= Rs 160 (a)

Question 11.
Solution:
Nitin gets = 98 marks
and it is 56% of total marks
Total-marks = \(\\ \frac { 98X100 }{ 56 } \)
= 175 (c)

Question 12.
Solution:
Let a number be = 1000
Then increase = 10%
Increased number = \(\\ \frac { 100X110 }{ 100 } \) = 110
Now decrease = 1%
Decreased number = \(\\ \frac { 110X90 }{ 100 } \) = 99
Difference = 100 – 99 = 1
% decrease = 1% (b)

Question 13.
Solution:
4 hours 30 min = \(4\frac { 1 }{ 2 } %\)
= \(\\ \frac { 9 }{ 2 } \) hours
% of a day = \(\\ \frac { 9 }{ 2 } \) x \(\\ \frac { 100 }{ 24 } \)%
= \(\\ \frac { 75 }{ 4 } \)%
= \(18\frac { 3 }{ 4 } %\) (a)

Question 14.
Solution:
Let total number of examinees = 100
Passed = 65
Failed = 100 – 65 = 35
Now 35% of total examinees = 420
Total examinees = \(\\ \frac { 420X100 }{ 35 } \)
= 1200 (c)

Question 15.
Solution:
Let number = x
Then x = \(\\ \frac { xX20 }{ 100 } \) = 40
=> 100x – 200x = 4000
80x = 4000
=> x = \(\\ \frac { 4000 }{ 80 } \) = 50
Number = 50 (a)

Question 16.
Solution:
Rate of decrease = \(27\frac { 1 }{ 2 } %\)
Let number = x
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B 16.1

Question 17.
Solution:
Let x% of 20 = 0.05
\(\\ \frac { x }{ 100 } \) x 20 = 0.05
x = \(\\ \frac { 0.05X100 }{ 20 } \)
= 0.25% (c)

Question 18.
Solution:
\(\\ \frac { 1 }{ 3 } \) = 1206 = 402
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9B 18.1

Question 19.
Solution:
x% of y is y% = \(\\ \frac { xy }{ 100 } \)
\(\\ \frac { y }{ 100 } \) × x
= y% of x (a)

Question 20.
Solution:
Let x% of \(\\ \frac { 2 }{ 7 } \) = \(\\ \frac { 1 }{ 35 } \)
=>\(\\ \frac { x }{ 100 } \) x \(\\ \frac { 2 }{ 7 } \) = \(\\ \frac { 1 }{ 35 } \)
x = \(\\ \frac { 1X100X7 }{ 35X2 } \)
= 10% (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Value Based Questions in Science for Class 9 Chapter 14 Natural Resources

Value Based Questions in Science for Class 9 Chapter 14 Natural Resources

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 14 Natural Resources

Question 1.
It has been made mandatory to install rain water harvesting system and solar water heater in all buildings in urban areas,

  1. What is the rationale when rain water already passes into drains ?
  2. Why are solar water heaters being installed when electric geijers are available ?

Answer:

  1. Rain water passed into drain does not enter the ground but is taken out of the urban area and poured into a water body at quite a distance with or without treatment. However, ground water is being withdrawn everywhere for drinking, industrial and irrigation purposes. As a result, the level of ground water is going down and arid conditions are being faced in many areas. In rain water harvesting, rain water from a building is passed directly into ground, dug wells or water pumps. This recharges ground water.
  2. Installation of solar water heaters is a method of saving electricity which is always in short supply due to rapid urbanisation, Industrialisation and intensive agriculture. Coal/gas based power plants are adding CO2 to the atmosphere causing global warming. Saving electricity is now an important social responsibility.

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Question 2.

  1. Sheela saw blue-green algae forming bloom in the village pond,
  2. Fish, which was previously abundant was no-where to be seen.
  3. The pond is giving a stink. Water of the pond is not even fit for cattle. Some of the cattle who were taken to the pond for drinking and bathing have fallen sick,
  4. What explanation will Sheela give for this to the villagers.

Answer:

  1. Bloom forming algae occur in a pond only when the quality of pond water has deteriorated due to pollution. Blue-green algae secrete toxins that are harmful to animals and humans,
  2. Fish must have died due to deficiency of oxygen in pond water. Oxygen deficiency occurs when there is excess of organic matter (organic loading). The aerobic decomposers consume the dissolved oxygen. This is followed by anaerobic decomposition of organic matter. It produces sulphides and other sludge producing substances. Blue-green algae can grow under such circumstances.
  3. Stink comes from anaerobic breakdown products of organic matter. The toxins released by blue-green algae further deteriorate the quality of water causing sickness and skin rashes in animals and humans.
  4. Sheela could explain to the villagers that deterioration of pond has been due to excess fertilizers used by them in their fields. Rain wash brought these fertilizers into the pond. There was initial spurt in the growth of plants due to this. The phenomenon is called eutrophication. Excess plant matter slowly caused organic loading of water that reduced its oxygen content, killing the fish and other aquatic animals. So fertilizers should be used very judiciously in the fields.

Question 3.

  1. Carbon dioxide concentration in the atmosphere has reached more than 390 ppm
  2. The antarctic and arctic waters are becoming acidic causing thinning of animal shells.
  3. Polar ice is melting,
  4. Some low lying islands have submerged in sea.

(a) What are the reasons behind all these changes occurring on earth ?
(b) Suggest a mechanism to control the same and if possible to reverse the trend.

Answer:

(a) The reason for these global changes is faulty recycling of carbon dioxide. More of carbon dioxide is being produced than its consumption in photosynthesis due to

  1. Excess combustion of fossil fuels in industries, power houses, automobiles, homes and other places,
  2. Reduced intake of CO2 by plants due to deforestation,
  3. Higher amounts of CO2 entering the atmosphere not only increases its atmospheric concentration but also the concentration of dissolved CO2 in water, especially in colder seas. This is making colder waters acidic. The acidity is thinning the shells of many marine animals.
  4. CO2 is a green house gas which is causting global warming. Rise in global temperature is melting snow over poles. This is raising the level of sea water slowly. It was already submerged some low lying islands.

(b) Suggestions :

  1. Large scale tree plantation in all vacant areas,
  2. Reduced felling of forest trees.
  3. Reducing consumption of fossil fuel by reducing dependence on thermal power plants, increasing efficiency of automobiles and switching over to alternate technologies that do not consume fossil fuels.

Question 4.
On a school trip to an industrial estate, students found that the marble used in the office buildings of most factories has lost its shine and become pitted.

  1. What is the reason of marble pitting
  2. Can this factor cause harm to vegetation as well ?
  3. Suggest way to prevent this.

Answer:

  1. Pitting and discolouring of marble is due to corrosive action of dry or wet acid rain, comprising sulphur dioxide and nitrogen oxides, emitted during combustion of fossil fuels in industries. The acid reacts with insoluble calcium carbonate of marble and converts it into soluble calcium sulphate and calcium nitrate.
  2. Yes, Acid rain can also cause harm to vegetation by
    1. Direct action of acid over plants causing death of leaves,
    2. Solubilisation of essential minerals and their leaching leaving only toxic minerals in the soil.
  3. Use of wet scrubbers to remove acidic gases from the industrial emissions.

Hope given Value Based Questions in Science for Class 9 Chapter 14 Natural Resources are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources

NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources

Question 1.
Rivers from land, add minerals to sea water. Discuss how.
Answer:
Rivers are formed from springs, melting snow and rain water flowing over land as run off. While, passing through and over the rocks, the flowing water picks up minerals. These dissolved minerals are taken by rivers to sea and make them available to marine organisms.

More Resources

Question 2.
How can we prevent loss of top soil ?
Answer:
Control of Soil Erosion: Soil erosion can be prevented by providing plant cover to soil. Roots of plants hold the soil firmly around them. They increase water percolation during rains and decrease runoff. Vegetation cover also reduces direct pounding of soil by rain drops. The various methods to control soil erosion are as follows :

  1. Vegetation Cover: Soil should not be left uncovered. Tree plantation should be undertaken where vegetation cover is poor or absent.
  2. Controlled Grazing: Grazing should be limited to certain seasons and for limited number of animals.
  3. Slope is divided into a number of flat fields for slowing down the flow of water.
  4. Contour Bunding: Small bunds are raised on the edges of fields to prevent loss of top soil through wind or water.
  5. Conservation Tillage: Instead of conventional tillage, reduced or no tillage can be practised. It prevents erosion.
  6. Wind Breaks: Wind erosion is reduced if rows of trees and shrubs are planted at right angles to the prevailing direction of wind.

Question 3.
How is the life of organisms living in water affected when water gets polluted ? (CCE 2012)
Answer:
Water gets polluted from domestic wastes, fertilizers, pesticides (from fields) and industrial wastes.

  1. Fertilizers and domestic wastes cause eutrophication or excessive growth and increased organic loading. This, reduces oxygen available to aquatic animals,
  2. Domestic waste carries pathogens of a number of diseases. Some of them can cause diseases in aquatic animals as well,
  3. Pesticides and industrial wastes carry toxins that directly harm the aquatic organisms.

Question 4.
During summer, if you go near the lake, you feel relief from heat. Why ?
Answer:
During day time, air over the land becomes heated up, rises upward and creates an area of low pressure. Lake water is not heated up so quickly. Evaporation of water from its surface also cools it up. Therefore, air over the surface of lake is cooler. It moves towards the land where low pressure exists. Therefore, during day dme a cool breeze flows from lake to the land nearby.

Question 5.
In coastal area, wind current moves from sea towards the land during day but during night it moves from land to the sea. Discuss the reason.
Answer:
Air Movement in Coastal Areas: In coastal areas, there is a regular flow of cooler air from sea towards the land during the daytime. It is because during the daytime land gets heated faster than water. Reradiation from land heats the air above it. The hot air rises and creates an area of low pressure. Sea water is not heated so rapidly. Air above sea remains comparatively cool. It has a higher pressure as compared to air over land. Therefore, cooler air present over sea moves towards land where low pressure exists.

Question 6.
Following are a few organisms :
(a) Lichens
(b) Mosses
(c) Mango tree
(d) Cactus.
Which among the above can grow on stones and also help in formation of soil ? Write the mode of their action for making soil.
Answer:
Lichens and Mosses.
Weathering by Living Organisms:
Living Organisms are cause biological weathering. Lichens growing on rock surface are able to extract minerals from the same. This produces small crevices where a thin layer of soil builds up. Mosses grow over these crevices. They cause deepening of crevices and building up of more soil in them. Deeper crevices form cracks.
The cracks become bigger when the roots of short lived herbs pass into them. With the passage of time roots of bigger plants pass into cracks. Cracks widen and cause slow fragmentation and later pulverisation of rocks.

Question 7.
Soil formation is done by both abiotic and biotic factors. List the name of these factors by classifying them as abiotic and biotic.
(CCE 2012)
Answer:
Abiotic Factors : Sun, water, wind.
Biotic Factors : Lichens, mosses, herbs, shrubs and trees.

Question 8.
All the living organisms are basically made up of C, N, S, P, H and O. How do they enter the living forms ? Discuss.
Answer:
Most of them first enter plants and become components of organic materials during the process of photosynthesis. They enter plants from air (CO2), water and ions (from soil). From plants the chemicals pass on to other organisms.

Question 9.
Why does the percentage of gases like oxygen, nitrogen and carbon dioxide remain almost the same in the atmosphere ?
Answer:
Through biogeochemical cycling where there is repeated circulation of biogenetic nutrients between abiotic and biotic components of the environment.

Question 10.
Why has moon very cold and very high temperature variations, e.g., from -190°C to 110°C even though it is at the same distance from the sun as the earth is ? (CCE 2012)
Answer:
Moon does not possess atomosphere. Atmosphere being a bad conductor acts as a temperature buffer on earth. This is not so on the moon. Therefore, moon gets heated up as the sun rays fall on its surface. It cools down drastically when there is no sunlight.

Question 11.
Why do people love to fly kites near the sea shore ? (CCE2012)
Answer:

  1. There is a regular unidirectional wind from sea to land.
  2. It helps in flying the kite high.
  3. The wind is cooler and provides comfort even in bright sun.

Question 12.
Wliy does Mathura refinery pose problems to the Taj Mahal ? (CCE 2014)
Answer:
Mathura refinery emits hydrocarbons and sulphur oxides. Hydrocarbons induce the formation of ozone which is highly oxidising. Sulphur dioxide produces acid rain which has a corroding effect on marble. Therefore, Mathura refinery does pose problems to Taj Mahal.

Question 13.
Why do not lichens occur in Delhi/Mumbai whereas they commonly grow in Manali/Ooty or Darjeeling ? (CCE 2011, 2012)
Answer:
Lichens are sensitive to sulphur dioxide which occurs in sufficient quantity in the atmosphere of Delhi due to large number of
vehicles, homes and factories using fossil fuels. Moreover, Delhi occurs in semi-arid area where, atmospheric moisture is low. In Manali and Darjeeling, the atmosphere is humid and sulphur dioxide pollution is comparatively low.

Question 14.
Wliy does water need conservation even though large oceans surround the land masses ? (CCE 2012)
Answer:
Marine water is unfit for consumption by humans, industries or land plants and animals. Therefore, terrestrial biota and human beings have to depend upon fresh water resources available on land (only about 0.5% of the total). The same has to be conserved.

Question 15.
There is a mass mortality of fish in a pond. What may be the reasons ?
Answer:

  1. Passage of pesticide rich water from crop fields. ,
  2. Pouring of industrial waste. .
  3. Pouring of hot water from an industry or thermal power plant.
  4. Release of waste rich in heavy metal and mercury.
  5. Blockage of fish gills by some suspended pollutant.
  6. Excessive eutrophication.

Question 16.
Lichens are called pioneer colonisers of bare rock. How can they help in formation of soil ?
Answer:
Ans. Lichens release small quantity of acids which corrode the surface of rocks creating small pores and releasing minerals. With the passage of time the porous surface changes into thin layer of soil.

Question 17.
“Soil is formed by water”. If you agree with this statement, then give reasons.
Answer:
Yes,  “action of water in weathering” (wetting and drying, frost action, abrasion).

  1. Wetting and Drying. Certain rock components can pick up and lose moisture. They undergo swelling and contraction resulting in fragmentation of rocks.
  2. Frost Action. Water seeping in cracks would swell up and exert a great pressure if it freezes due to low temperature. The rock would undergo fragmentation.
  3. Running water carrying rock fragments would break and grind rocks occurring in the pathway. Rain and hail also cause rock breaking.

Question 18.
Fertile soil has lots of hurpus. Why ? (CCE 2012)
Answer:
Fertile soils possess lots of humus because

  1. Humus is required for binding soil particles into crumbs. Crumb formation is needed for both hydration and aeration of soil,
  2. Humus makes the soil porous for easy passage of roots,
  3. It is source of minerals.
  4. Humus contains chemicals that promote growth of plants.

Question 19.
Why step farming is common in hills ?
Answer:
Step farming or terracing is carried out in the hills because it slows down the speed of rain water currents, checks soil erosion and increases water absorption by soil.

Question 20.
Why are root nodules useful for the plants ? (CCE 2012)
Answer:
Root nodules mostly occur in legume plants. These nodules contain nitrogen fixing bacteria named Rhizabium. The bacteria pick up nitrogen from soil atmosphere and convert it into organic compounds. The same pass into plant so that legumes become rich in proteins and other nitrogen compounds.

Question 21.
How do fossil fuels cause air pollution ? (CCE 2012)
Answer:
Fossil fuels (coal, petroleum) are burnt to obtain energy.
Combustion produces

  1. Suspended particles
  2. Sulphur oxides
  3. Nitrogen oxides
  4. Carbon monoxide
  5. Vapours of hydrocarbons.

All these are air pollutants. Hydrocarbons are a source of cancer. They also damage internal organs. Alongwith nitrogen oxides, they produce ozone and smog. Sulphur and nitrogen oxides give rise to acid rain. They are also irritants and damage eyes and nasal tract.

Question 22.
What are the causes of water pollution ? Discuss how you can contribute in reducing water pollution ?
Answer:
Causes of Water Pollution:

  1. 21% of all infectious diseases and 66% of all deaths of children below 5 years are caused by contamination of drinking and bathing water by pathogens present in sewage, e.g., cholera, typhoid, diarrhoea, dysentery, jaundice, etc.
  2. Toxic Chemicals: Industrial effluents pass down a number of toxic chemicals, g., mercury by paper industry (minamata disease), cadmium (itai-itai disease), lead (plumbism), nitrate (met-haemoglobinaemia).

Reduction in Water Pollution :

  1. Sewage: The authorities can be prevailed upon not to pass untreated sewage into water bodies.
  2. Garbage: Domestic waste or garbage should not be thrown on the banks of water body.
  3. Industrial Effluents. Asking the industrial houses not to pass the untreated effluents into water body.
  4. Washing Clothes: Washing clothes should be discouraged on the banks of water bodies as detergents are a source of eutrophication.
  5. Vegetation: Plant trees, shrubs and herbs on the banks. This checks soil erosion and siltation of water bodies. It also reduces run off from fields.

Question 23.
A motor car, with its glass totally closed, is parked directly under the sun. The inside temperature of the car rises very high. Explain why? (CCE 2012)
Answer:
Glass is transparent to sunlight, allowing it to pass into the interior of the car. It heats up the interior. But heat waves do not escape from the car due to glass being opaque to the same. As a result, the interior of the car placed in the sun will become very hot. It is similar to glass house effect.

Question 24.
Justify “Dust is a pollutant.” (CCE 2012)
Answer:

  1. Dust consists of suspended particles. The latter pass into nasal tract and cause a lot of discomfort including allergic asthma, bronchitis, cough and cold,
  2. Dust reduces light intensity,
  3. Its particles are eye irritants,
  4. Dust settles over plant foliage and reduces photosynthetic activity,
  5. Dust particles block stomata reducing gaseous exchange,
  6. Dust particles can pick up toxic metals and chemicals being emitted by industries.

Question 25.
Explain the role of the sun in the formation of soil.
Answer:

Paedogenesis: It is the process of formation of soil from rocky crust of earth. Paedogenesis occurs in two stages, weathering and humification.

  1. Weathering
    It is the process of breaking down of rock present on the surface of earth into fine particles. Weathering is caused by three types of processes — physical, chemical and biological. Physical weathering is pulverisation of rocky matter caused by physical phenomena like atmospheric changes (heating, cooling, wetting-drying, frost action) and mechanical forces (abrasion by rain and hail, rolling stones carried by water, wave action, sand blast by wind). Chemical weathering is pulverisation of rocks by chemical action in wet weather like hydration, oxidation, reduction, carbonation and solubilisation. Biological weathering is the breaking down of rocks by the action of living organisms.
    Sun: It heats up the rocks during the day. Heating causes the rocks to expand. During night, the rocks cool and contract. Different parts of a rock expand and contract at different rates resulting in its cracking and breaking up into smaller pieces or fragments.
  2. Humification
    It is the addition of partially decomposed organic matter or humus to the weathered rock particles to form soil. Humus binds the finely ground rock particles into aggregates called crumbs. Crumb formation is needed for both hydration and aeration of soil.

Question 26.
Carbon dioxide is necessary for plants. Why do we consider it as a pollutant ? (CCE 2014)
Answer:
Besides being essential for photosynthesis, carbon dioxide is a green house gas. Upto 350 ppm, it is both a good raw material as well as essential for keeping the earth warm. However, when its concentration rises (as presently it is 400 ppm), it becomes pollutant because it results in global warming. The latter is quite dangerous because it will melt polar ice, raise the water level of oceans and submerge several coastal areas and islands.

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources are helpful to complete your science homework.

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HOTS Questions for Class 9 Science Chapter 8 Motion

HOTS Questions for Class 9 Science Chapter 8 Motion

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 8 Motion

Question 1.
Under what condition, the average speed is equal to the magnitude of the average velocity. Justify your answer.
(CBSE 2010, 2012)
Answer:
HOTS Questions for Class 9 Science Chapter 8 Motion image - 1
Now, the average speed = magnitude of average velocity if the total distance travelled by an object in a given time interval is equal to the magnitude of the displacement of the object in the same time interval. This is possible, when an object moves in one direction along a straight line. Thus, average speed is equal to the magnitude of the average velocity if an object moves in one direction along a straight line.

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Question 2.
A body can have zero average velocity but not zero average speed. Explain. (CBSE 2013)

Or

Give an example of a motion in which average velocity is zero, but the average speed is not zero. (CBSE 2015)
Answer:
HOTS Questions for Class 9 Science Chapter 8 Motion 2
Since, the total distance travelled by a moving body increases with the time, so average speed of a moving body can not be zero.
HOTS Questions for Class 9 Science Chapter 8 Motion image - 2
Since total displacement of a moving body can be zero, so the average velocity of a moving body can be zero.
Example. When an athlete completes one round in a circular track in the given time interval, then his total displacement is zero but total distance travelled is equal to the length of the circular track. Hence, his average velocity is zero but his average speed is not zero.

Question 3.
An object P is moving with a constant velocity for 5 minutes. Another object Q is moving with changing velocity for 5 minutes. Out of these two objects, which one has acceleration. Explain. (CBSE 2012)
Change in velocity
Answer:
HOTS Questions for Class 9 Science Chapter 8 Motion image - 4
Since, the velocity of object P is not changing or change in velocity of the object is zero, therefore, object P has no acceleration. On the other hand, there is change in velocity of the object Q, so it has acceleration.

Question 4.
Can an object be accelerated if it is moving with constant speed ? Justify your answer with an example.

Or

Can a body have constant speed and still be accelerating ? Give an example. (CBSE 2010)

Or

Explain how is it possible for an object to move with a constant speed but with uniform acceleration.
(CBSE 2012)
Answer:
An object moving with constant speed can be accelerated if its direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.

Question 5.
Two trains A and B start moving at the same time. The distances travelled by them in given intervals of time are given below. State which train has uniform motion and which train has non-uniform motion.

Time Distance travelled by train A (in km) Distance travelled by train B (in km)
6.00 pm 0 0
6.15 pm 10 15
6.30 pm 20 24
6.45 pm 30 32
7.00 pm 40 38
7.15 pm 50 42
7.30 pm 60 47

Answer:
Since train A travels equal distances in equal intervals of time (i.e. in every 15 minutes), so the motion of train A is uniform motion. On the other hand, train B travels unequal distances in equal intervals of time (i.e. in every 15 minutes), so the motion of train B is non-uniform motion.

Hope given HOTS Questions for Class 9 Science Chapter 8 Motion are helpful to complete your science homework.

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Value Based Questions in Science for Class 9 Chapter 8 Motion

Value Based Questions in Science for Class 9 Chapter 8 Motion

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Class 9 Chapter 8 Motion

VALUE BASED QUESTIONS

Question 1.
Ram hired a three wheeler from bus stand to his house. The driver A’ of three wheeler followed the shortest path and reached the destination in half an hour. He charged Rs. 50 from Ram. On the same day, Ram’s sister also came to her house after attending a function in other city. She also hired a three wheeler. The driver ‘B’ of the three wheeler took an hour to reach her house and charged Rs. 150 from her.

  1. Is the displacement of both the three wheelers from bus stand to Ram’s house same or different ?
  2. What values are shown by driver A ?

Answer:

  1. Displacement of both the three wheelers is same.
  2. Driver A is honest. He is concerned for others.

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Question 2.
Two boys A and B were driving cars on a road. There was a sign board on the side of road reading “speed limit” : 50 km/h. Boy A applied brakes and started driving at 45 km/h. On the other hand boy B did not bother about the speed limit. He continued to drive his car at 70 km/h. A police man stopped the car of boy B and fined him If Rs. 500.00. What values are shown by boy A ?
Answer:
boy A is good observer and being a good citizen, he follows the traffic rules.

Hope given Value Based Questions in Science for Class 9 Chapter 8 Motion are helpful to complete your science homework.

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RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A

RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A.

Other Exercises

Solve :

Question 1.
Solution:
8x + 3 = 27 + 2x
=> 8x – 2x
=> 27 – 3
=> 6x = 24
=> x = \(\\ \frac { 24 }{ 6 } \) = 4
x = 4

Question 2.
Solution:
5x + 7 = 2x – 8
=> 5x – 2x = – 8 – 7
=> 3x = – 15
=> x = \(\\ \frac { -15 }{ 3 } \) = – 5
x = – 5

Question 3.
Solution:
2z – 1 = 14 – z
=> 2z + z = 14 + 1
=> 3z = 15
=> z = \(\\ \frac { 15 }{ 3 } \) = 5
z = 5

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 4.
Solution:
9x + 5 = 4(x – 2) +8
=> 9x + 5 = 4x – 8 + 8
=> 9x – 4x = – 8 + 8 – 5
=> 5x = – 5
=> x = \(\\ \frac { -5 }{ 5 } \) = – 1
x = – 1

Question 5.
Solution:
\(\\ \frac { 7y }{ 5 } \) = y – 4
Multiplying both sides by 5,
\(\\ \frac { 7y }{ 5 } \) x 5 = 5(y – 4)
=> 5 (y-4)
=> 7y = 5y – 20
=> 7y – 5y = – 20
=> 2y = – 20
=> y = \(\\ \frac { -20 }{ 2 } \) = – 10
Hence y = – 10 Ans.

Question 6.
Solution:
3x + \(\\ \frac { 2 }{ 3 } \) = 2x + 1
=> 3x – 2x = 1 – \(\\ \frac { 2 }{ 3 } \)
=> x = \(\\ \frac { 3-2 }{ 3 } \) = \(\\ \frac { 1 }{ 3 } \)
Hence x = \(\\ \frac { 1 }{ 3 } \) Ans.

Question 7.
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
=> 15y – 2y + 5y = 60 – 18 – 30
=> 18y = 12
=> y = \(\\ \frac { 12 }{ 18 } \) = \(\\ \frac { 2 }{ 3 } \)
=> y = \(\\ \frac { 2 }{ 3 } \)

Question 8.
Solution:
3(5x – 7) – 2(9x – 11) = 4(8x – 13) – 17
=> 15x – 21 – 18x + 22 = 32x – 52 – 17
=> 15x – 18x – 32x = – 52 – 17 + 21 – 22
=> 15x – 50x = – 70
=> – 35x = – 70
=> x = \(\\ \frac { -70 }{ -35 } \) = 2
x = 2

Question 9.
Solution:
\(\\ \frac { x-5 }{ 2 } \) – \(\\ \frac { x-3 }{ 5 } \) = \(\\ \frac { 1 }{ 2 } \)
Multiplying each term by 10, the L.C.M. of 2 and 5
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 9.1

Question 10.
Solution:
\(\\ \frac { 3t-2 }{ 4 } \) – \(\\ \frac { 2t+3 }{ 3 } \) = \(\\ \frac { 2 }{ 3 } \) – t
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 10.1

Question 11.
Solution:
\(\frac { 2x+7 }{ 5 } +\frac { 3x+11 }{ 2 } =\frac { 2x+8 }{ 3 } -5\)
Multiplying by 30, the L.C.M. of 5, 2 and 3.
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 11.1

Question 12.
Solution:
\(\frac { 5x-4 }{ 6 } =4x+1-\frac { 3x+10 }{ 2 } \)
Multiplying by 6, the L.C.M. of 6 and 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 12.1

Question 13.
Solution:
\(5x-\frac { 1 }{ 3 } \left( x+1 \right) =6\left( x+\frac { 1 }{ 30 } \right) \)
=> 5x – \(\\ \frac { x+1 }{ 3 } \) = 6x + \(\\ \frac { 1 }{ 5 } \)
Multiplying by 15, the L.C.M. of 3 and 5
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 13.1

Question 14.
Solution:
\(4-\frac { 2\left( z-4 \right) }{ 3 } =\frac { 1 }{ 2 } \left( 2z+5 \right)\)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 14.1

Question 15.
Solution:
\(\frac { 3\left( y-5 \right) }{ 4 } -4y=3-\frac { \left( y-3 \right) }{ 2 } \)
Multiplying by 4, the L.C.M. of 4 and 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 15.1

Question 16.
Solution:
\(\\ \frac { 8x-3 }{ 3x } \) = \(\\ \frac { 2 }{ 1 } \)
By cross multiplication,
8x – 3 = 6x
=> 8x – 6x = 3
=> 2x = 3
=> x = \(\\ \frac { 3 }{ 2 } \)
x = \(\\ \frac { 3 }{ 2 } \)

Question 17.
Solution:
\(\\ \frac { 9x }{ 7-6x } \) = \(\\ \frac { 15 }{ 1 } \)
By cross multiplication,
9x = 105 – 90x
=> 9x + 90x = 105
=> 99x = 105
=> x = \(\\ \frac { 105 }{ 99 } \) = \(\\ \frac { 35 }{ 33 } \)
x = \(\\ \frac { 35 }{ 33 } \)

Question 18.
Solution:
\(\\ \frac { 3x }{ 5x+2 } \) = \(\\ \frac { -4 }{ 1 } \)
By cross multiplication,
3x × 1 = – 4×(5x + 2)
=> 3x = – 20x – 8
=> 3x + 20x = – 8
=> 23x = – 8
=> x = \(\\ \frac { -8 }{ 23 } \)
Hence x = \(\\ \frac { -8 }{ 23 } \)

Question 19.
Solution:
\(\\ \frac { 6y-5 }{ 2y } \) = \(\\ \frac { 7 }{ 9 } \)
By cross multiplication,
9(6y – 5) = 7 × 2y
=> 54y – 45 = 14y
=> 54y – 14y = 45
=> 40y = 45
=> y = \(\\ \frac { 45 }{ 40 } \) = \(\\ \frac { 9 }{ 8 } \)
Hence y = \(\\ \frac { 9 }{ 8 } \) Ans.

Question 20.
Solution:
\(\\ \frac { 2-9z }{ 17-4z } \) = \(\\ \frac { 4 }{ 5 } \)
By cross multiplication,
5 (2 – 9z) = 4(17 – 4z)
=> 10 – 45z = 68 – 16z
=> – 45z + 16z = 68 – 10
=> – 29 = 58
=> z = \(\\ \frac { 58 }{ -29 } \) = – 2
Hence z = – 2 Ans.

Question 21.
Solution:
\(\\ \frac { 4x+7}{ 9-3x } \) = \(\\ \frac { 1 }{ 4 } \)
By cross multiplication,
4(4x + 7) = 1 (9 – 3x)
=> 16x + 28 = 9 – 3x
=> 16x + 3x = 9 -28
=> 19x = – 19
=> x = \(\\ \frac { -19 }{ 19 } \) = – 1
Hence x = – 1 Ans.

Question 22.
Solution:
\(\\ \frac { 7y+4}{ y+2 } \) = \(\\ \frac { -4 }{ 3 } \)
By cross multiplication,
3 (7y + 4) = – 4 (y + 2)
=> 21y + 12 = – 4y – 8
=> 21y + 4y = – 8 – 12
=> 25y = – 20
=> y = \(\\ \frac { -20 }{ 25 } \) = \(\\ \frac { -4 }{ 5 } \)
y = \(\\ \frac { -4 }{ 5 } \)

Question 23.
Solution:
\(\\ \frac { 15(2-y)-5(y+6) }{ 1-3y } \) = \(\\ \frac { 10 }{ 1 } \)
By cross multiplication,
15 (2 – y) – 5(y + 6) = 10 (1 – 3y)
=> 30 – 15y – 5y – 30 = 10 – 30y
=> – 15y – 5y + 30y = 10 – 30 + 30
=> 30y – 20y = 10
=> 10y = 10
y = \(\\ \frac { 10 }{ 10 } \) = 1
Hence y = 1 Ans.

Question 24.
Solution:
\(\\ \frac { 2x-(7-5x) }{ 9x-(3x+4x) } \) = \(\\ \frac { 7 }{ 6 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 24.1

Question 25.
Solution:
\(m-\frac { \left( m-1 \right) }{ 2 } =1-\frac { \left( m-2 \right) }{ 3 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 25.1

Question 26.
Solution:
\(\\ \frac { 3x+5 }{ 4x+2 } \) = \(\\ \frac { 3x+4 }{ 4x+7 } \)
By cross multiplication,
(3x + 5)(4x + 7) = (3x + 4)(4x + 2)
=> 12x² + 21x + 20x + 35
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 26.1

Question 27.
Solution:
\(\\ \frac { 9x-7 }{ 3x+5 } \) = \(\\ \frac { 3x-4 }{ x+6 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 27.1

Question 28.
Solution:
\(\\ \frac { 2-7x }{ 1-5x } \) = \(\\ \frac { 3+7x }{ 4+5x } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A 28.1

 

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B

RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B.

Other Exercises

Question 1.
Solution:
Marked price of cooler = Rs 4650
Rate of discount = 18%
Selling price
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 1.1

Question 2.
Solution:
Marked price = Rs 960
Selling price = Rs 816
Total Discount = M.P. – S.P.
= Rs 960 – Rs 816
= Rs 144
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 2.1

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
S.P. of shirt = Rs 1092
Discount = Rs 208
M.P. of shirt = S.P. + discount
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 3.1

Question 4.
Solution:
S.P. of toy = Rs 216.20
Discount = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 4.4

Question 5.
Solution:
S.P. of tea set = Rs 528
Rate of discount = 12%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 5.1

Question 6.
Solution:
Let C.P. of goods = Rs 100
Marked price = Rs 100 + 35
= Rs 135
Rate of discount = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 6.1

Question 7.
Solution:
Let C.P. of phone = Rs 100
.’. Marked price = Rs 100 + 40
= Rs 140
Rate of discount = 30%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 7.1

Question 8.
Solution:
C.P. of fan = Rs. 1080
Gain = 25%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 8.1

Question 9.
Solution:
C.P of refrigerator = Rs. 11515
and gain % = 20%
S.P. of refrigerator
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 9.1

Question 10.
Solution:
C.P. of ring = Rs. 1190
Gain = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 10.1

Question 11.
Solution:
Let Marked price = Rs. 100
Discount = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 11.1

Question 12.
Solution:
Let C.P. = Rs. 100
Gain = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 12.1

Question 13.
Solution:
Marked price of TV = Rs. 18500
Series of two successive discounts = 20% and 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 13.1

Question 14.
Solution:
Let M.P. = Rs. 100
First discount = 20%
and second discount = 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 14.1

 

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7D

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D.

Other Exercises

Question 1.
Solution:
x2 + 5x + 6
= x2 + 2x + 3x + 6
{6 = 2 x 3, 5 = 2 + 3}
= x (x + 2) + 3 (x + 2)
= (x + 2) (x + 3) Ans.

Question 2.
Solution:
y2 + 10y + 24
= y2 + 6y + 4y + 24
{24 = 6 x 4, 10 = 6 + 4}
= y (y + 6) + 4 (y + 6)
= (y + 6) (y + 4)

Question 3.
Solution:
z2 + 12x + 27
{27 = 9 x 3, 12 = 9 + 3}
= z2 + 9z + 3z + 27
= z (z + 9) + 3 (z + 9)
= (z + 9) (z + 3)

Question 4.
Solution:
p2 + 6p + 8
= p2 + 4p + 2p + 8
{ 8 = 4 x 2, 6 = 4 + 2}
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question 5.
Solution:
x2 + 15x + 56
= x2 + 8x + 7x + 56
{56 = 8 x 7, 15 = 8 x 7}
= x(x + 8) + 7(x + 8)
= (x + 8) (x + 7) Ans.

Question 6.
Solution:
y2 + 19y + 60
= y2 + 15y + 4y + 60
{60 = 15 x 4, 19 = 15 + 4}
= y (y + 15) + 4 (y + 15)
= (y + 15) (y + 4)

Question 7.
Solution:
x2 + 13x + 40
= x2 + 5x + 8x + 40
{40 = 5 x 8, 13 = 5 + 8}
= x(x + 5) + 8(x + 5)
= (x + 5) (x + 8) Ans.

Question 8.
Solution:
q2 – 10q + 21
= q2 – 7q – 3q + 21
{21 = ( – 7)x( – 3), – 10 = – 7 – 3}
= q {q – 7) – 3 (q – 7)
= (q – 7)(q – 3)

Question 9.
Solution:
p2 + 6p – 16
= p2 + 8p – 2p – 16
{ – 16 = 8x( – 2),6 = 8 – 2}
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Question 10.
Solution:
x2 – 10x + 24
= x2 – 6x – 4x + 24
{24 = ( – 6) x ( – 4), – 10 = – 6 – 4 }
= x(x – 6) – 4(x – 6)
= (x – 6) (x – 4) Ans.

Question 11.
Solution:
x2 – 23x + 42
= x2 – 2x – 21x + 42
{42 = ( – 2) x ( – 21), – 23 = – 2 – 21}
= x(x – 2) – 21(x – 2)
= (x – 2) (x – 21) Ans.

Question 12.
Solution:
x2 – 17x + 16
= x2 – x – 16x + 16
{ 16 = ( – 1) x ( – 16), 17 = – 1 – 16}
= x (x – 1) – 16(x – 1)
= (x – 1) (x – 16) Ans.

Question 13.
Solution:
y2 – 21y + 90
= y2 – 15y – 6y + 90
{90 = ( – 15)x ( – 6), – 21 = – 15 – 6}
= y(y – 15) – 6 (y – 15)
= (y – 15) (y – 6)

Question 14.
Solution:
x2 – 22x + 117
= x2 – 13x – 9x + 117
{117 = ( – 13) x ( – 9), – 22 = – 13 – 9}
= x(x – 13) – 9(x – 13)
= (x – 9) (x – 13) Ans.

Question 15.
Solution:
x2 – 9x + 20
= x2 – 5x – 4x + 20
{ 20 = ( – 5) x ( – 4), – 9 = – 5 – 4}
= x(x – 5) – 4(x – 5)
= (x – 5) (x – 4) Ans.

Question 16.
Solution:
x2 + x – 132
= x2 + 12x – 11x – 132
{ – 132 = 12 x ( – 11), 1 = 12 – 11}
= x(x + 12) – 11(x + 12)
= (x + 12) (x – 11) Ans.

Question 17.
Solution:
x2 + 5x – 104
= x2 + 13x – 8x – 104
{ – 104 = 13 x ( – 8), 5 = 13 – 8}
= x(x + 13) – 8(x + 13)
= (x + 13) (x – 8) Ans.

Question 18.
Solution:
y2 + 7y – 144
= y2 + 16y – 9y – 144
{144 = – 16 x 9, 7 = 16 – 9}
= y(y +16) – 9(y + 16)
= (y + 16) (y – 9) Ans.

Question 19.
Solution:
z2 + 19z – 150
= z2 + 25z – 6z – 150
{ – 150 = 25 x ( – 6), 19 = 25 – 6}
= z(z + 25) – 6(z + 25)
= (z + 25) (z – 6) Ans.

Question 20.
Solution:
y2 + y – 72
= y2 + 9y – 8y – 72
{ – 72 = 9x( – 8), 1 = 9 – 8}
= y(y + 9) – 8(y + 9)
= (y + 9) (y – 8) Ans

Question 21.
Solution:
a2 + 6a – 91
= a2 + 13a – 7a – 91
{ – 91 = 13x( – 7), 6 = 13 – 7}
= a (a + 13) – 7 (a + 13)
= (a + 13) (a – 7)

Question 22.
Solution:
p2 – 4p – 11
= p2 – 11p + 7p – 77
{ – 77 = – 11 x 7, – 4 = – 11 + 7}
= p(p – 11) + 7 (p – 11)
= (p – 11)(p + 7)

Question 23.
Solution:
x2 – 7x – 30
= x2 – 10x + 3x – 30
{ – 30 = – 10 x 3, – 7 = – 10 + 3}
= x(x – 10) + 3(x – 10)
= (x – 10) (x + 3) Ans.

Question 24.
Solution:
x2 – 14 x + 3 x – 42
{ – 11 = – 14 + 3], – 42 = – 14 x 3}
= x (x – 14) + 3 (x – 14)
= (x – 14) (x + 3) Ans.

Question 25.
Solution:
x2 – 5x – 24
= x2 – 8x + 3x – 24
{ – 24 = – 8 x 3, – 5 = – 8 + 3}
= x(x – 8) + 3(x – 8)
= (x – 8) (x + 3) Ans.

Question 26.
Solution:
y2 – 6y – 135
= y2 – 15y + 9y – 135
{ – 135 = – 15 x 9, – 6 = – 15 + 9}
= y (y – 15) + 9 (y – 15)
= (y – 15) (y + 9)

Question 27.
Solution:
z2 – 12z – 45
= z2 – 15z + 3z – 45
= z (z – 15) + 3 (z – 15)
{ – 45 = – 15 x 3, – 12 = – 15 + 3}
= (z – 15)(z + 3)

Question 28.
Solution:
x2 – 4x – 12
= x2 – 6x + 2x – 12
{ – 12 = – 6 x 2, – 4 = – 6 + 2}
= x (x – 6) + 2 (x – 6)
= (x – 6) (x + 2)

Question 29.
Solution:
3x2 + 10x + 8
= 3x2 + 6x + 4x + 8
{ 3 x 8 = 24, 24 = 6 x 4,10 = 6 + 4}
= 3x(x + 2) + 4(x + 2)
= (x + 2) (3x + 4) Ans.

Question 30.
Solution:
3y2 + 14y + 8
= 3y2 + 12y + 2y + 8
{3 x 8 = 24, 24 = 12 x 2, 14 = 12 + 2}
= 3y (y + 4) + 2 (y + 4)
= (y + 4) (3y + 2)

Question 31.
Solution:
3z2 – 10z + 8
= 3z2 – 6z – 4z + 8
{ 3 x 8 = 24, 24 = ( – 6)x( – 4), – 10 = – 6 – 4}
= 3z (z – 2) – 4 (z – 2)
= (z – 2) (3z – 4)

Question 32.
Solution:
2x2 + x – 45
= 2x2 + 10x – 9x – 45
{2 x ( – 45) = – 90,- 90 = (10)x( – 9), 1 = 10 – 9}
= 2x (x – 5) + 9 (x – 5)
= (x – 5) (2x + 9)

Question 33.
Solution:
6p2 + 11p – 10
= 6x2 + 15x – 4x – 10
{6x ( – 10) = 60, – 60 = 15 x ( – 4),11 = 15 – 4}
= 3x (2x + 5) – 2 (2x + 5)
= (3x – 2) (2x + 5) Ans.

Question 34.
Solution:
2x2 – 20x + 3x – 30
{2 x ( – 30) = – 60, – 17 = – 20 + 3, – 60 = – 20 x 3}
= 2x (x – 10) + 3 (x – 10)
= (x – 10) (2x + 3) Ans.

Question 35.
Solution:
7y2 -19y – 6
= 7y2 – 21 y + 2y – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7y(y – 3) + 2 (y – 3)
= (y – 3) (7y + 2)

Question 36.
Solution:
28 – 31x – 5x2
= 28 – 35x + 4x – 5x2
{28 x ( – 5)= – 140, -140 = – 35 x 4, – 31 = – 35 + 4}
= 7 (4 – 5x) + x (4 – 5x)
= (4 – 5x) (7 + x)

Question 37.
Solution:
3 + 23z – 8z2
= 3 + 24z – z – 8z2
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3 (1 + 8z) – z (1 + 8z)
= (1 + 8z) (3 – z)

Question 38.
Solution:
6x2 – 5x – 6
= 6x2 – 9x + 4x – 6
{ 6 x ( – 6) = – 36,- 36 = – 9 x 4, – 5 = – 9 + 4}
= 3x (2x – 3) + 2 (2x – 3)
= (2x – 3) (3x + 2)

Question 39.
Solution:
3m2 + 24m + 36
= 3 {m2 + 8m + 12}
= 3 {m2 + 6m + 2m + 12)
{12 = 6 x 2, 8 = 6 + 2}
= 3 {m (m + 6) + 2 (m + 6)}
= 3 (m + 6) (m + 2)

Question 40.
Solution:
4n2 – 8n + 3
= 4n2 – 6n – 2n + 3
{4 x 3 = 12, 12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2n (2n – 3) – 1 (2n- 3)
= (2n – 3) (2n – 1)

Question 41.
Solution:
6x2 – 17x – 3
= 6x2 – 18x + x – 3
{6 x ( – 3)= – 18, – 18 = – 18 x 1, – 17 = – 18 + 1}
= 6x (x – 3) + 1 (x – 3)
= (x – 3) (6x + 1)

Question 42.
Solution:
7x2 – 19x – 6
7x2 – 19x – 6
= 7x2 – 21x + 2x – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7x (x – 3) + 2 (x – 3)
= (x – 3) (7x + 2)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.