Chapter 9: Time & Speed - Practice Now 1 & 2 and Worked Example 1 & 2 - Solution

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Chapter 9: Time & Speed

FORMULAS:

• Arrival Time = Depart Time + Journey Time
• Depart Time = Arrival Time - Journey Time
• Journey Time = Arrival Time - Depart Time
• 1 hour = 60 minutes
• 1 hour = 3600 seconds
• 1 minute = 60 seconds
• 1 minute = ¹ / ₆₀ hours
• 1 second = ¹ / ₆₀ minutes
• 1 second = ¹ / ₃₆₀₀ hours

PRACTISED NOW

PRACTISE NOW 1:
A ship left port X at 22:45 on Friday and arrived at Port Y 7¹/₄ hours later. At what time and day the ship arrive at Port Y?
Solution:
Data:
Depart time: 22:45
Depart day: Friday
Journey time: 7¹/₄ hours
To change ¹/₄ hour into minute
¹/₄ x 60 = 15 minutes
Therefore
Journey time: 7 hours and 15 minutes = 7:15
Arrival time = ?
Arrival day = ?

Formula:
Arrival Time = Depart Time + Journey Time
Arrival Time = 22:45 + 7:15

Arrival time is 6:00 on next day 
Answer: The ship arrives at port Y, at 6:00 or 6:00 a.m. on Saturday.

PRACTISE NOW 2:
A bus left city P at 10:45 and arrived at city Q at 23:11 on the same day. How long was the bus journey?
Solution:
Data:
Depart time: 10:45
Arrival time = 23:11
Journey time: ?

Formula:
Journey Time = Arrival Time - Depart Time
Journey Time = 23:11 - 10:45 = 12:26

Answer: The bus journey took 12 hours and 26 minutes.

WORKED EXAMPLES

Worked Example 1:
A car leaves Town A at 21:15 on Wednesday and Town B 5 hours later. At what time and day does the car arrive at Town B?
Solution:
Data:
Departure time: 21:15
Departure day: Wednesday
Journey time: 5¹/₂ Hours.
To change ¹/₂ hour into minute
¹/₂ x 60 = 30 minutes
Therefore
Journey time: 5 hours and 30 minutes = 5:30
Arrival Time: ?
Arrival Day:?

Formula:
Arrival time = Departure time + Journey time
Arrival time = 21:15 + 5: 30

Arrival time is 2:45 on next day
Answer: The car arrives at Town B, at 2:45 or 2:45 a.m. on Thursday.

Worked Example 2:
A train left City A at 07:35 and arrived at City B at 13:05 on the same day. How long was the train journey?
Solution:
Data:
Departure time: 7:35
Arrival time: 13:05
Journey time: ?

Formula:
Journey time = Arrival time - Departure time
Journey time = 13:05 - 7:35

Answer: The train journey took 5 hours and 30 minutes.

Chapter 3: Profit & Loss - Exercise 3A - Solution

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Chapter 3.1: Profit & Loss
Exercise 3A

FORMULAS:

• Profit = Selling price - Cost price
• Loss = Cost Price - Selling price
• Selling Price (For Profit) = Cost Price + Profit
• Selling Price (For Loss) = Cost Price - Loss
• Cost Price (For Profit) = Selling Price - Profit
• Cost Price (For Loss) = Selling Price + Loss
• Profit % of Cost Price = ^Profit/_{Cost Price} x 100
• Profit % of Selling Price = ^Profit/_{Selling Price} x 100
• Loss % of Cost Price = ^Loss/_{Cost Price} x 100
• Loss % of Selling Price = ^Loss/_{Selling Price} x 100

 BASIC LEVEL 

1. Complete the table.

	Cost Price (PKR)	Selling Price (PKR)	Profit /Loss (PKR)	Profit/Loss As A Percentage Of The Cost Price
1.	40	45
2.	600	480
3.	88000			Profit of 4%
4.	5680			Loss of 22.5%
5.		28.14		Profit of 171/4%
6.		506.85		Loss of 7%

SOLUTION:
(a) Data:

• Cost Price = 40 PKR
• Selling Price = 45 PKR
• Profit = ? (Hint: Because selling price is greater than cost price)
• Profit% of the cost price = ?

For Profit:
Profit = Selling Price - Cost Price
⇒ Profit = 45 - 40
⇒ Profit = 5 PKR

Profit% of Cost Price

Answer: Profit is 5 PKR and Profit% of cost price is 12.5%.

(b) Data:

• Cost Price = 600 PKR
• Selling Price = 480 PKR
• Loss = ? (Hint: Because selling price is less than cost price)
• Loss% of the cost price = ?

For Loss:
Loss = Cost Price - Selling Price
⇒ Loss = 600 - 480
⇒ Loss = 120 PKR

Loss% of Cost Price

Answer: Loss is 120 PKR and Loss % of cost price is 20%.

(c) Data:

• Cost Price = 88 000 PKR
• Profit% of the cost price = 4%
• Selling Price = ?
• Profit = ?

There are Two Methods To Solve It:
METHOD 1:
For Profit:

For Selling Price:
Selling Price = Cost Pice + Profit
⇒ Selling Price = 88 000 + 3520 = 91 520 PKR

METHOD 2:
For Selling Price:
First we will find selling price on profit percentage of cost price i.e. 4%.
(Hint: We have given profit percentage of the cost price = 4%, which means if the cost price is 100 PKR than the profit is 4 PKR.)
If
⇒ Cost price = 100 PKR
⇒ Profit = 4 PKR
Then
⇒ Selling price = Cost Price + Profit
⇒ Selling Price = 100 + 4 = 104 PKR

Now We will find Selling Price, If the cost Price is 88 000 PKR, as:

For Profit:
Profit = Selling Price - Cost Price
⇒ Profit = 91520 - 88000
⇒ Profit = 3520 PKR

Answer: Profit is 3520 PKR and selling price is 91 520 PKR.

(d) Data:

• Cost Price = 5680 PKR
• Loss% of the cost price = 22.5%
• Selling Price = ?
• Loss = ?

There are Two Methods To Solve It:
METHOD 1:
For Loss:

For Selling Price:
Selling Price = Cost Pice - Loss
⇒ Selling Price = 5680 - 1278 = 4402 PKR

METHOD 2:
For Selling Price:
First we will find selling price on Loss percentage of cost price i.e. 22.5%.
(Hint: We have given loss percentage of the cost price = 22.5%, which means if the cost price is 100 PKR than the loss is 22.5 PKR.)
If
⇒ Cost price = 100 PKR
⇒ Loss = 22.5 PKR
Then
⇒ Selling price = Cost Price - loss
⇒ Selling Price = 100 + 22.5 = 77.5 PKR

Now We will find Selling Price, If the cost Price is 5680 PKR, as:

For Loss:
Loss = Cost Price - Selling Price
⇒ Loss = 5680 - 4402
⇒ Loss = 1278 PKR

Answer: Loss is 1278 PKR and selling price is 4402 PKR.

(e) Data:

• Selling Price = 28.14 PKR
• Profit% of the cost price = 17¹/₄% = 17.25%
• Cost Price = ?
• Profit = ?

For Cost Price:
First we will find cost price on profit percentage of cost price i.e. 17.25%.
(Hint: We have given profit percentage of the cost price = 17.25%, which means if the cost price is 100 PKR than the profit is 17.25 PKR.)
If
⇒ Cost price = 100 PKR
⇒ Profit = 17.25 PKR
Then
⇒ Selling price = Cost Price + Profit
⇒ Selling Price = 100 + 17.25 = 117.25 PKR

Now We will find cost Price, If the selling Price is 28.14 PKR, as:

For Profit:
Profit = Selling Price - Cost Price
⇒ Profit = 28.14 - 24
⇒ Profit = 4.14 PKR

Answer: Profit is 4.14 PKR and cost price is 24 PKR.

(f) Data:

• Selling Price = 506.85 PKR
• Loss% of the cost price = 7%
• Cost Price = ?
• Loss = ?

For Cost Price:
First we will find cost price on loss percentage of cost price i.e. 7%.
(Hint: We have given loss percentage of the cost price = 7%, which means if the cost price is 100 PKR than the loss is 7 PKR.)
If
⇒ Cost price = 100 PKR
⇒ loss = 7 PKR
Then
⇒ Selling price = Cost Price - loss
⇒ Selling Price = 100 - 7 = 93 PKR

Now We will find cost Price, If the selling Price is 506.85 PKR, as:

For Loss:
Loss = Cost Price - Selling Price
⇒ Loss = 545 - 506.85
⇒ Loss = 38.15 PKR

Answer: Loss is 38.15 PKR and cost price is 545 PKR.

	Cost Price (PKR)	Selling Price (PKR)	Profit /Loss (PKR)	Profit/Loss As A Percentage Of The Cost Price
1.	40	45	Profit = 5	Profit of 12.5%
2.	600	480	Loss = 120	Loss of 20
3.	88000	91520	Profit = 3520	Profit of 4%
4.	5680	4402	Loss = 1278	Loss of 22.5%
5.	24	28.14	Profit = 4.14	Profit of 171/4%
6.	545	506.85	Loss =38.15	Loss of 7%

Chapter 8 - Linear Equation & Coordinate geometry - Worked Examples

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Unit 8.1: Linear Equation In Two Variables

An equation is said to be linear equation in two variables if it is written in the form of:

ax + by + c = 0

where a, b & c are real numbers and the coefficients of x and y, i.e., a and b respectively, are not equal to zero.

Worked Example 1: Three times of one number is added to four times of other number The result obtained is 53. Represent this statement in the form of a linear equation.
Solution:
Let x be one number and y be another number.
Three times of x = 3x
Four times of y = 4y
Hence, 3x + 4y = 53 is a linear equation with two variables x and y.

MORE EXAMPLES:

Construct a linear equation with two variables for the given statements.
Example 1:
10x + 4y = 3 and
-x + 51y = 2
are linear equations in two variables.

Example 2:
Ahmed runs a stationary shop. He sold 25 notebooks and 30 registers for PKR 2400, in the month of January.
Solution:
We can express the above statement in a generalised form as given below.
Let x stand for one notebook. 25 notebooks mean 25x
Let y stand for one register. 30 registers mean 30y.
Hence, 25x + 30y = PKR 2400 is a linear equation with two variables x and y.


Example 3:
Six chairs and four tables cost PKR 25 000.
Solution:
The equation will be:
⇒ 6a + 4b = PKR 25 000,
where a stands for one chair,
and b stands for one table.

Example 4:
The difference of one-fourth of a number and one- fifth of another number is 1.
Solution:
The equation will be:

Example 5:
A number when added to 32 makes 100.
Solution:
⇒ x + 32 = 100 Ans.

Example 6:
A number multiplied by 4 and 3 taken away from the product gives the answer 9.
Solution:
⇒ 4x - 3 = 9 Ans.

Unit 8.3: Graph Of Linear Equation

Linear Equation in One Variable:

A linear equation in one variable is of the form:

• ax + b = 0.
• a + by = 0.

Here, a & b are numbers and x & y are variable. a & b are coefficient of x & y respectively.

Line Graph:
A linear equation can be represented as a line graph.
In order to draw the line graph we require several pairs of coordinates. These coordinates represent the relationship given in the equation.

Graph of ax + b = 0 Or ay + b = 0
The solution of ax + b = 0 is unchanged if any number is added, subracted, multiplied, or divided on both side of the equation. It means there is only one solution of the equation. On a graph, it appears to be a straight line either horizontal or vertical.

1) Graph of ax + b = 0
Worked Example 2:
(i) Let us draw a line of linear equation x + 6 = 10 on the graph.
Solution:
⇒ x + 6 = 10
Then x = 10 - 6 = 4
or x = 4
The graph of x = 4 is a vertical line.

2) Graph of a + by = 0
(ii) Let us draw a line of linear equation 7 + y = 9 on the graph.
Solution:
⇒ 7 + y = 9
Then y = 9 - 7 = 2
or y = 2
The graph of y = 2 is a horizontal line.

Linear Equation in OneTwo Variable:

If an ordered pair (x, y) is a solution to a linear equation in two variables, then it lies on the graph of the equation.
[Note: Ordered pair  is always written in small bracket & first value is always 'x' & second value is always 'y' i.e.. (x, y)]
A graph of a linear equation in two variable (ax + by = c) is a straight line.

Example 1:
For y = 3x, find its ordered pairs.
Solution:
⇒ y = 3x (The y value is always equal to '3 times of the x value).
Let x = 1
⇒ y = 3 x 1 = 3
∴ Ordered Pair = (1, 3)

Let x = 2
⇒ y = 3 x 2 = 6
∴ Ordered Pair = (2, 6)

Let x = 5
⇒ y = 3 x 5 = 15
∴ Ordered Pair = (5, 15)

Answer: (1, 3), (2, 6) and (5, 15) are all coordinates on the line y = 3x.

Example 2:
Find the coordinate or ordered pairs of the equation y = 3x + 1 and plot these coordinates on graph paper.
Solution:
We use substitution to calculate the values.
For y = 3x + 1
We replace the value of x for different numbers and record the y value.
Let x = 0
⇒ y = (3 x 0) + 1
⇒ y = 0 + 1 = 1
∴ Ordered Pair = (0, 1)

Let x = 1
⇒ y = (3 x 1) + 1
⇒ y = 3 + 1 = 4
∴ Ordered Pair = (1, 4)

Let x = 2
⇒ y = (3 x 2) + 1
⇒ y = 6 + 1 = 7
∴ Ordered Pair = (2, 7)

For convenience, a table of values is used to create the coordinates.

x	0	1	2
y	1	4	7

We have coordinates of the equation as (0, 1), (1, 4), (2, 7). Now we plot these coordinates on the graph paper.

Worked Example 3: Find the coordinates of point P located on the graph.
Solution:

• Begin at the point P and draw a dotted line either up or down the x-axis. The line to the point on x-axis is the x-coordinate.
• Similarly, begin at point P and draw a touching d form P dotted line touching to the point on y-axis is y-coordinate.
• In the given graph vertical line from P touches x-axis at x = 2
• Horizontal line from P touches y-axis at y = 4

∴ The coordinates of P(x, y) = (2, 4) Answer.