Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 3 Data Handling Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

1. Find the mean of the following data :

Question (i).
3, 5, 7, 9, 11, 13, 15
Answer:
Mean = \(\frac{3+5+7+9+11+13+15}{7}\)
= \(\frac {63}{7}\)
= 9

Question (ii).
40, 30, 30, 0, 26, 60
Answer:
Mean = \(\frac{40+30+30+0+26+60}{6}\)
= \(\frac {183}{6}\)
= 9

2. Find the mean of the first five whole numbers.
Answer:
The first five whole numbers are : 0, 1, 2, 3, 4
Mean = \(\frac{0+1+2+3+4}{5}\)
= \(\frac {10}{5}\)
= 2
Hence, the mean of first five whole numbers = 2

3. A batsman scored the following number of runs in six innings :
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning.
Answer:
Mean runs
= \(\frac{36+35+50+46+60+55}{6}\)
= \(\frac {282}{6}\)
Hence, the mean runs scored by batsman in an innings = 47

4. The ages in years of 10 teachers of a school are :
32, 41, 28, 54, 35, 26, 23, 33, 38, 40
(i) What is the age of the oldest teacher and that of the youngest teacher ?
(ii) What is the range of the ages of the teachers ?
(iii) What is the mean age of these teachers ?
Answer:
Arranging the ages in ascending order, we get
23, 26, 28, 32, 33, 35, 38, 40, 41, 54
(i) Age of the oldest teacher = 54 years Age of the youngest teacher 23 years
(ii) Range of the ages = 54 – 23 = 31
(iii) Mean age of the teachers

= \(\frac {350}{10}\)
= 35 years.

5. The rain fall (in mm) ip a city on 7 days of a certain week was recorded as follows :

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) How many days had the rainfall less than the mean rainfall ?
Answer:
(i) Arranging the rainfall (in mm) in ascending order, we get
Highest Rainfall = 20.5 mm
Lowest Rainfall = 0.0 mm
Range of the rainfall = 20.5 mm – 0.0 mm
= 20.5 mm.

(ii) Mean Rainfall
= \(\frac{0.01+12.2+2.1+0.0+20.5+5.5+1.0}{7}\)
= \(\frac {41.31}{7}\)
= 5.9 mm.

(iii) Number of days having rainfall less than mean rainfall = 5 days.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Question 1.
Aftab tells his daughter, “Seven years ago I was seven times as old as you were then. Also, three years from now, I shall
be three times as old as you will be” (Isn’t this interesting ?). Represent this situation algebraically and graphically.
Solution:
Let Aftabs present age = x years
and Aftab’s daughter’s present age = y years
Algebraical-Situation
According to 1st condition,
x – 7 = 7(y – 7)
or x – 7 = 7y – 49
or x – 7y + 42 = 0
According to 2nd condition,
x + 3 = 3(y + 3)
or x + 3 = 3y + 9
or x – 3y – 6 = 0
∴ Pair of Line’ar Equation in two variables are
x – 7y + 42 = 0
and x – 3y – 6 = 0

Graphical – Situation:
x – 7y + 42 = 0
x – 3y – 6 = 0

x = 7y – 42 ………….(1)
x = 3y + 6 …………..(2)
Putting y = 5 in (1), we get
Putting y = 0 in (2), we get

x = 7 × 5 – 42 = 35 – 42
x = -7
Putting y = 6 in (1), we get
x = 7 × 6 – 42 = 42 – 42 = 0
Putting y = 7 in (1), we get:
x = 7 × 7 – 42 = 49 – 42 = 7

Table

Plotting the points A (-7, 5), B (0, 6), C (7, 7) and drawing a line joining them, we get the graph of the equation x – 7y + 42 = 0
x = 3 × 0 + 6
x = 0 + 6 = 6
Putting y = 3 in (2), we get:
x = 3 × 3 + 6
= 9 + 6 = 15
Putting y = -2 in (2), we get:
x = 3 × -2 + 6
= -6 + 6 = 0

Table:

Plotting the points D (6, 0), E (15, 3), F (0, -2) and drawing a line joining them, we get the graph of the equation x – 3y – 6 = 0

From the graph it is clear that the two lines intersect at G (42, 12).
Hence, x = 42 and y = 12 is the solution of given pair of linear equations.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for 3900. Later, she buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically and geometrically. [Pb. 2019, Set-A, B, C]
Solution:
Let cost of one bat = x
Cost of one hail = y
Algehraical – Situation
According to 1st condition,
3x + 6y = 3900
or x + 2y = 1300
According to 2nd condition,
1x + 3y = 1300
∴ Pair of linear equations in two variables are:
x + 2y = 13001
and x + 3y = 1300

Graphical – Situation:
x + 2y = 1300
x = 1300 – 2y …………..(1)
Putting y = 0 in (1), we get
x = 1300 – 2 × 0
x= 1300
Putting y = 500 in (1), we get :
= 1300 – 2 × 500
= 1300 – 1000 = 300
Putting y = 650 in (1), we get :
x = 1300 – 2 × 650
1300 – 1300 = 0

Table:

Plotting the points A (1300, 0), B (300, 500) and C (0, 650) drawing a line joining them we get the graph of the equation x + 2y = 1300.
x + 3y = 1300
x = 1300 – 3y …………….(2)
Putting y = 0 in (2), we get :
x = 1300 – 3 × 0 = 1300
Putting y = 500 in (2), we get :
x = 1300 – 3 × 500
= 1300 – 1500 = – 200
Putting y = 300 in (2), we get :
x = 1300 – 3 × 300
= 1300 – 900 = 400

Table:

Plotting the points A (1300, 0), E (-200, 500), F (400, 300) and drawing a line joining them we get the graph of the
equation.
x + 3y = 1300

From the graph it is clear that the two lines intersect at A (1300, 0).
Hence x = 1300 and y = 0 is the solution of given pair of linear equations.

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let cost of 1 kg apples = ₹ x
Cost of 1 kg grapes = ₹ y
Algebraical – Situation
According to 1st condition,
2x + 1y = 160
According to 2nd condition,
4x + 2y = 300
∴ Pair of linear equations in two variables
2x + y = 160
and 4x + 2y = 300
Graphical – Situation
2x + y = 160
2x = 160 – y
x = \(\frac{160-y}{2}\) ……………(1)
Putting y = 0 in (1), we get :
x = \(\frac{160-0}{2}=\frac{160}{2}\) = 80
Putting y = 60 in (1), we get :
x = \(\frac{160-60}{2}=\frac{100}{2}\) = 50
Putting y = 160 in (1), we get :
x = \(\frac{160-160}{2}=\frac{0}{2}\) = 0

Plotting the points A (80, 0), B (50, 60), C (0, 160) and drawing a line joining them, we get the graph of the equation 2x + y = 160
Now 4x + 2y = 300
or 2x + y = 150
2x = 150 – y
x = \(\frac{160-y}{2}\) …………(2)
Putting y = 0 in (2), we get:
x = \(\frac{160-0}{2}=\frac{150}{2}\) = 75
Putting y = 50 in (2), we get:
x = \(\frac{150-50}{2}=\frac{100}{2}\) = 50
Putting y = 150 in (2), we get:
x = \(\frac{150-150}{2}=\frac{0}{2}\) = 0

Plotting the points D (75, 0), E (50, 50), F (0, 150) and drawing a line joining them, we get the graph of equation
4x + 2y = 300

From the graph, it is clear that the two lines do not intersect anywhere i.e. they are parallel.

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 10 Gravitation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 10 Gravitation

PSEB 9th Class Science Guide Gravitation Textbook Questions and Answers

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
Let m₁ and m₂ be the masses of the two objects A and B respectively and ‘r’ be the distance between their centres. Therefore, according to the law of Gravitation, the force of attraction between them is given ahead:

Therefore, force of attraction will become four times when the distance between the two objects is reduced to half.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Solution:
Suppose F is the gravitational force that acts on an object of mass’m’.
∴ F = G. \(\frac{\mathrm{M} m}{r^{2}}\) …………….. (i)
and F = mg ………………….. (ii)
From (i) and (ii)
F = \(\frac{\mathrm{GM} m}{r^{2}}\) = mg
It is clear that F ∝ m but acceleration due to gravity ‘g’ does not depend upon mass ‘m’. Hence all objects (light or heavy) fall with the same speed when there is no air resistance.

Question 3.
What is magnitude of gravitational force between the earth and a 1 kg object on its surface? Take mass of earth to be 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m. G = 6.67 × 10^-11 Nm² kg^-2.
Solution:
Here, mass of the object (m) = 1 kg
Mass of the earth (M) = 6 × 10²⁴ kg
Radius of the earth (R) = 6.4 × 10⁶ m
The magnitude of force of gravitation between object of mass 1 kg and the earth

Question 4.
The earth and the moon are attracted to each other by gravitational force. Does the earth attracts the moon with a force that is greater than or smaller than or the same as the force with which the moon attracts the earth? Why?
Answer:
The earth attracts the moon with the same force as the force with which the moon attracts the earth. According to Newton’s third law., these two forces are equal and opposite.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
According to Newton’s third law, the moon also attracts earth with a force equal to that with which the earth attracts the moon. But the earth is much larger than the moon. So, the acceleration produced in the earth (a ∝ 1/m) is very less and is not noticeable.

Question 6.
What happens to the force between two objects, if

1. the mass of one object is doubled?
2. the distance between the objects is doubled and tripled?
3. the masses of both objects are doubled?

Answer:


i.e. the force becomes four times the original force.

Question 7.
What is the importance of universal law of gravitation?
Answer:
Importance of universal law of gravitation:

1. The gravitational force between the sun and the earth makes the earth move around the sun with a uniform speed.
2. The gravitational force between the earth and the moon makes the moon move around the earth with uniform speed.
3. The high and low tides are formed in sea due to the gravitational pull exerted by the sun and the moon on the surface of water.
4. It is the gravitational pull of the earth, which holds our atmosphere in place.
5. The gravitational pull of earth keeps us and other bodies firmly on the ground.

Question 8.
What is the acceleration of free fall?
Answer:
It is the acceleration produced when a body falls under the influence of the force of gravitation of the earth alone. Near the surface of the earth, its value is 9.8 m s^-2.

Question 9.
What do we call the gravitational force between the earth and an object?
Answer:
The gravitational force between the earth and an object is called weight of the object.

Question 10.
A person ‘A’ buys few grams of gold at poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why?
[Hint: The value of g is greater at the poles than at the equator.]

Answer:
The value of g at the equator is less than that at the poles. Hence, the few gm of gold at poles will measure less when taken to the equator. Therefore, the friend will not agree with the weight of the gold bought.

Question 11.
Why will a sheet of paper fall slower than one
Answer:
The sheet of paper will experience a larger air resistance due to its large surface area than that of its ball form.

Increased force of friction will reduce the forward driving force due to gravity. Hence sheet of paper falls slower than one that is crumbled into a ball.

Question 12.
Gravitational force on the surface of moon is 1/6th as strong as gravitational force on the earth. What is the weight in newton of a 10 kg object on moon and on the earth?
Solution:
Mass of the object on moon = 10 kg
Mass of the object on the earth = 10 kg
Acceleration due to gravity on the earth (g) = 9.8 m s^-2
Weight of the object on the earth (W) = m × g
= 10 × 9.8
= 98 N
Now weight of the object on moon’s surface = \(\frac {1}{6}\) × weight of the object on earth
= \(\frac {1}{6}\) × 98N
= 16.3 N

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m s^-1. Calculate :
1. The maximum height to which it rises
2. The total time it takes to return to the surface of earth.
Solution:
1. Here initial velocity of the ball (u) = 49 m s^-1
[At maximum height the ball comes to rest]
Final velocity of the ball (υ) = 0
Acceleration due to gravity (g) = – 9.8 m s² [in the upward direction]
Time to reach the maximum height (t) =?

∴ Total time taken to return to earth = Time for upward journey + Time for downward journey
= t + t
= 2 t
= 2 × t
= 2 × 5 s
= 10 s

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate the final velocity just before touching the ground.
Solution:
Here, the height of the tower, (h) = 19.6 m
Initial velocity of stone, (u) = 0
Acceleration due to gravity, (g) = + 9.8 m s^-1
Final velocity of the stone, (υ) = ?
Using equation of motion, υ² – u² = 2gh
υ² – (0)² = 2 × 9.8 × 19.6
υ² = 19.6 × 19.6
υ = \( \sqrt{{19.6 × 19.6}} \)
or υ = 19.6 ms^-1

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m s^-1. Taking g = 10 m s^-2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Solution:
Initial velocity of the stone, (w) = 40 ms^-1,
Final velocity of the stone on reaching maximum height (υ) = 0 [At rest]
Acceleration due to gravity, (g) = – 10 m s² [upward direction]
Maximum height reached, (h) = ?
We know υ² – u² = 2gh
(0)² – (40)² = 2 × (- 10) × h
– 40 × 40 = – 2 × 10 × h
∴ h = \(\frac {-40×40}{-2×10}\)
= 80 m
Since stone goes 80 m upwards and then returns to the point of throw by moving 80 m downward.
∴ Total distance travelled by stone = h + h
= 2 h
= 2 × 80 m
= 160 m
As the stone returns to the initial point of throw, therefore, net displacement is zero (0)

Question 16.
Calculate the force of gravitation between the earth and the sun, given the mass of earth = 6 × 10²⁴ kg and of the sun = 2 × 10³⁰ kg. Average distance between the two is 1.5 × 10¹¹ m.
Solution:
Given, mass of the earth (m₁) = 6 × 10²⁴ kg
Mass of the sun, (m₂) = 2 × 10³⁰ kg
Average distance between the earth and the sun (d) = 1.5 × 10^-11 m
G = 6.7 × 10^-11 N – m² /kg²
Force of gravitation (F) = ?
According to the universal law of gravitation,

Question 17.
A stone is allowed to fall from the top of the tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m s^-1. Calculate when and where the two stones will meet? (g = 10 ms^-2)
Solution:

Height of the tower = 100 m
Suppose a stone is allowed to fall from point A at the top of tower and another stone is projected vertically upward from point C. Let us suppose that these two stones meet at point B after ‘t’ seconds.
Distance covered by first stone (AB) = x
∴ Distance covered by second stone (CB) = (100 – x)
Downward Journey of first stone
u = 0
g = + 10 m s^-2
(S) = x metres
using S = ut + \(\frac {1}{2}\)gt²
x = 0 × t + \(\frac {1}{2}\) × 10 × t²
x = 0 + 5 × t²
⇒ t² = \(\frac {x}{5}\) …………..(1)
Upward journey of second stone
u = 25 ms^-1
(S) = (100 – x) metres
g = – 10 m s^-2
using S = ut + \(\frac {1}{2}\)gt²
(100 – x) = 25 × t + \(\frac {1}{2}\)(-10) × t²
(100 – x) = 25t – 5t²
or 5t² = 25t – 100 + x
From (1) and (2)
\(\frac {x}{5}\) = \(\frac {25t-100+x}{5}\)
or x = 25t – 100 + x
0 = 25t – 100
25t = 100
∴ t = \(\frac {100}{25}\) = 4s
Now substituting the value of t = 4s in (1)
(4)² = \(\frac {x}{5}\)
16 = \(\frac {x}{5}\)
∴ x = 16 × 5 = 80 m
i.e. the first stone will cover a distance of 80 m in the downward direction, and second stone will cover upward distance = 100 – x
= 100 – 80
= 20 m

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find

1. Velocity with which it was thrown up,
2. the maximum height it reached; and
3. its position after 4 s.

Solution:
Total time taken (t) = 6 s
Time taken by the ball for upward joumey= Time taken by the ball for downward journey
= \(\frac {6s}{2}\)
= 3 s
(i) Suppose the ball is thrown upwards with initial velocity u
g = – 9.8 m/s²
t = 3 s
υ = 0 [the ball stops on reaching the maximum height]
Maximum height(S) = h
using υ = u + gt
0 = u + (-9.8) × 3
0 = u – 29.4
∴ u = 29.4 ms^-1

∴ Height of the ball from the thrower = (44.1 – 4.9) m
= 39.2 m

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act.
Answer:
If an object is immersed in a liquid then the buoyant force due to displaced liquid acts on the object in vertically upward direction.

Question 20.
Why does a block of plastic immersed under water come to the surface of water?
Or
Give reason why, a block of plastic when immersed underwater comes up to the surface of water.
Answer:
As density of plastic is less than the density of water. The upward thrust applied by displaced water on the plastic will be more than the weight of the plastic. So plastic block will float on water.

Question 21.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink?
Solution:
Here, density of water, p_w = 1gm cm^-3
Mass of substance, m = 50g
Volume of substance, V = 20cm³
We know, density of substance, ρ = \(\frac {m}{v}\)
= \(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^{3}}\)
= 2.5 g cm^-3
As the density of the substance is greater than the density of water, the given substance will sink in water.

Question 22.
The volume of 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^-3 ? What will be the mass of the water displaced by this packet?
Solution:
Here, mass of the packet (m) = 500 g
Volume of packet (V) = 350 cm³
∴ Density of sealed packet ρ = \(\frac {m}{v}\)
= \(\frac{500 \mathrm{~g}}{350 \mathrm{~cm}^{3}}\)
= 1.43 gcm^-3
But Density of water ρ_w = 1 g cm^-3
As density of sealed packet is more than that of water, the sealed packet will sink in water.
∴ Volume of sealed packet immersed in water = V = 350 cm³
Weight of water displaced by the packet = Vρ_w
= 350 × 1
= 350g.

Science Guide for Class 9 PSEB Gravitation InText Questions and Answers

Question 1.
State the universal law of gravitation.
Answer:
Newton’s universal law of gravitation. This law states that every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force always acts along the line joining their centres.

If m₁ and m₂ are the masses of two objects lying distance d apart, then force F between them is:
F = \(\frac{\mathrm{Gm}_{1} m_{2}}{d^{2}}\)
where G is a constant, called universal gravitational constant.

Question 2.
Write the formula to find the magnitude of gravitational force between the earth and an object on the surface of the earth.
Answer:

Let ‘m’ be the mass of object on the earth and the mass of earth be ‘M’. If ‘R’ is the radius of the earth, then the formula for gravitational force between earth and object is:
F = \(\frac{\mathrm{Gm} M}{R^{2}}\)
Since the size of the object is very small as compared to that of the earth, therefore distance between centre of object and centre of the earth is taken to be equal to radius of the earth.

Question 3.
What is meant by Free Fall?
Answer:
Free Fall: An object is said to be in a state of free fall when it falls towards the earth under the influence of gravitational force between the object and the earth. There is no change in the direction of motion of the body but value of velocity keeps changing due to attraction of earth.
It falls towards earth with an acceleration of 9.8 m s^-2.

Question 4.
What is meant by acceleration due to gravity?
Answer:
Acceleration due to gravity: The acceleration produced in the motion of a body falling under the force of gravity is called acceleration due to gravity. It is denoted by ‘g’.

Question 5.
What is the difference between the mass of an object and its weight?
Answer:
Difference between mass and weight:

Question 6.
Why is the weight of the object on moon -th of its weight on the earth?
Answer:
We know that, Mass of earth (M_e) = 100 × Mass of moon (M_m)
Radius of earth (R_e) = 4 × Radius of moon (R_m)
Since the mass and radius of moon is less than that of the earth therefore, moon exerts lesser \(\frac {1}{6}\)th force of attraction on the object. Hence the weight of the object on moon is \(\frac {1}{6}\)th of the weight of the same object on earth.

Question 7.
Why is it difficult to hold a school bag having a strap made of thin and strong string? (Imp.)
Answer:
We know force per unit area is called pressure i.e. P = \(\frac {F}{A}\). Now for the constant force, the pressure experienced is inversely proportional to area. Now, when the string is thin, it has less area of cross-section and hence, exerts greater pressure on the hand for the given weight of school bag. Thus, it becomes difficult to hold the school bag.

Question 8.
What do you mean by buoyancy?
Answer:
Buoyancy means upward thrust acting in a body when the body is completely or partly immersed in a fluid (i.e. liquid or a gas).

Question 9.
Why does an object float or sink when placed on the surface of water?
Answer:
When the object has density less than the density of water i.e. 1 gm/cm³ then it, floats on the surface of water, because, it displaces more weight of water than its own weight. The upward force applied by displaced water is called buoyant force. As buoyant force is more than its own weight, therefore, it floats.

When the object has a density of more than 1 gem^-3, then it sinks in water, because, it always displaces less weight of water than its own weight. As buoyant force is less than its own weight, therefore, it sinks.

Question 10.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
With a weighing machine, we find weight and not mass, Your weight as noted by the machine is 42 kg f (or 42 kg wt) and not 42 kg. The actual weight is more than 42 kg. since you have displaced some air when weighed in it. However, the mass will remain the same in all cases.

Question 11.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Answer:
The bag of cotton will actually be heavier than the iron bar. Cotton is bulky and has more area as compared to the area of the iron bar. Due to more area occupied by cotton bags, it experiences more upthrust because of the displaced volume of air. This upthrust reduces the downward pull and hence its weight as recorded by the weighing machine will be lesser.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

1. Write the first step that you will use to separate the variable and then solve the equation.

Question (i).
x + 1 = 0
Answer:
Given equation x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = -1
or x = – 1

Question (ii).
x – 1 = 5
Answer:
Given equation is x – 1 = 5
Adding 1 to both sides we get
x – 1 + 1 = 5 + 1
or x = 6
Thus x = 6 is the solution of the given equation

Question (iii).
x + 6 = 2
Answer:
Given equation is x + 6 = 2
Subtracting 6 from both sides, we get:
x + 6 – 6 = 2 – 6
or x = – 4
Thus, x = – 4 is the solution of the given equation.

Question (iv).
y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y + 4 – 4 = 4 – 4
or y = 0
Thus, y = 0 is the solution of the given equation.

Question (v).
y – 3 = 3
Answer:
Given equation is y – 3 = 3
Adding 3 to both sides we get
y – 3 + 3 = 3 + 3
or y = 6
Thus, y = 6 is the solution of the given equation.

2. Write the first step that you will use to separate the variable and then sotye the equation :

Question (i).
3x = 15
Answer:
Given equation is 3x = 15
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{15}{3}\)
or x = 5

Question (ii).
\(\frac{P}{7}\) = 4
Answer:
Given equation is \(\frac{P}{7}\) = 4
Multiplying both sides by 7, we get
7 × \(\frac{P}{7}\) = 7 × 4
or p = 28
Thus, p = 28 is the solution of the given equation.

Question (iii).
8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
or y = \(\frac {9}{2}\)

Question (iv).
20x = – 10
Answer:
Given equation is
20x = – 10
Dividing both sides by 20
\(\frac{20 x}{20}=\frac{-10}{20}\)
or x = \(\frac {-1}{2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

Question (i).
5x + 7 = 17
Answer:
Given equation is 5x + 7 = 17
Subtracting 7 from both sides, we get
5x + 7 – 7 = 17 – 7
or 5x = 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{10}{5}\)
or x = 2

Question (ii).
\(\frac{20 x}{3}\) = 40
Answer:
Given equation is \(\frac{20 x}{3}\) = 40
Multiplying both sides by 3, we get
3 × \(\frac{20 x}{3}\) = 3 × 40
or 20x = 3 × 40
Dividing both sides by 20, we get
\(\frac{20 x}{20}\) = \(\frac{3 \times 40}{20}\)
or x = 6

Question (iii).
3p – 2 = 46
Answer:
Given equation is 3p – 2 = 46
Adding 2 to both sides, we get
3p – 2 + 2 = 46 + 2
or 3 p = 48
Dividing both sides by 3, we get:
\(\frac{3 p}{3}=\frac{48}{3}\)
or p = 16

4. Solve the following equations :

Question (i).
10x + 10 = 100
Answer:
Given equation is 10x + 10 = 100
Subtracting 10 from both sides, we get
10x + 10 – 10 = 100 – 10
or 10x = 90
Dividing both sides by 10, we get
\(\frac{10 x}{10}=\frac{90}{10}\)
or x = 9
Thus x = 9 is the solution of the given equation.

Question (ii).
\(\frac{-p}{3}\) = 5
Answer:
Given equation is \(\frac{-p}{3}\) = 5
Multiplying both sides by – 3, we get
– 3 × \(\frac{-p}{3}\) = -3 × 5
or p = -15
Thus p = – 15 is the solution of the given equation.

Question (iii).
3x + 12 = 0
Answer:
Given equation is 3x + 12 = 0
Subtracting 12 from both sides, we get
3x + 12 – 12 = – 12
or 3x = – 12
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{-12}{3}\)
or x = -4
Thus x = – 4 is the solution of the given equation.

Question (iv).
2q – 6 = 0
Answer:
The given equation is 2q – 6 = 0
Adding 6 to both sides, we get
2q – 6 + 6 = 0 + 6
or 2q = 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{6}{2}\)
or q = 3
Thus, q = 3 is the solution of the given equation.

Question (v).
3p = 0
Answer:
The given equation is 3p = 0
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{0}{3}\)
or p = 0
Thus, p = 0 is the solution of the given equation.

Question (vi).
3s = -9
Answer:
The given equation is
3s = -9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=-\frac{9}{3}\)
or s = – 3
Thus, s = – 3 is the solution of the given equation.