5th Grade Data Handling Worksheet

5th Grade Data Handling Worksheet | Free Printable Data Handling Worksheets Grade 5

Use the 5th Grade Data Handling Worksheet to learn how the data represents with images. Worksheets on 5th Grade Data Handling provided here have different questions on bar graphs, pictographs, line graphs, etc. It is the best choice for every student to use our data handling exercise for class 5 pdf to crack the exam easily. The complete concept can easily understand by the students with our quizzes for grade 5 on data handling. Therefore, follow the step-by-step procedure given here to learn all the 5th Grade Data Handling problems.

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Data Handling Class 5 Questions PDF

1. Observe the below pictograph about how many cars were sold in five months of a year. One car represents 4 cars.

The month in a Year Number of Cars
Jan 5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet
Feb 5th grade handling worksheet5th grade handling worksheet
Mar 5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet
Apr 5th grade handling worksheet
May 5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet5th grade handling worksheet

(i) How many cars were sold in different months?
(ii) In which month was the maximum number of cars sold?
(iii) In which month was the minimum number of cars sold?
(iv) How many more cars were sold in May than Apr?
(v) How many total cars were sold in five months?

Solution:

We can get the information from the given pictograph. From the given information, every car is equal to the 4 number of cars.
(i) In Jan, the number of car images = 3.
January 3 x 4 = 12 cars.
In Feb, the number of car images = 2.
February 2 x 4 = 8 cars.
In Mar, the number of car images = 4.
March 4 x 4 = 16 cars.
In Apr, the number of car images = 1.
April 1 x 4 = 4 cars.
In May, the number of car images = 6.
May 6 x 4 = 24 cars.
(ii) In May month, the maximum number of cars (24) were sold.
(iii) In Apr month, the minimum number of cars (4) were sold.
(iv) 20 more cars were sold in May than April. 24 – 4 = 20 cars.
(v) 12 cars + 8 cars + 16 cars + 4 cars + 24 cars = 64 cars.


2. Choose the right answer to show the type of graph that will be used for the following information.
I. Number of Fans in the 5 rooms of an apartment.
(i) Double Bar graph (ii) Bar Graph
II. Favorite chocolate of the children.
(i) Bar graph (ii) Line Graph
III. A company’s product sale in the last 8 months.
(i) Pictograph (ii) Line Graph
IV. Marks obtained by Sam in different subjects in 3 terms of the year.
(i) Double Bar graph (ii) Bar Graph

Solution:

(i) From the given information, number of Fans in the 5 rooms of an apartment.
We can draw a Bar Graph to represent the given information.
Therefore, the answer is (ii) Bar Graph
(ii) From the given information, the Favorite chocolate of the children.
We can draw a Bar Graph to represent the given information.
Therefore, the answer is (i) Bar Graph
(iii) From the given information, A company’s product sale in the last 8 months.
We can draw a Line Graph to represent the given information.
Therefore, the answer is (ii) Line Graph
(iv) From the given information, marks were obtained by Sam in different subjects in 3 terms of the year.
We can draw a Double Bar Graph to represent the given information.
Therefore, the answer is (i) Double Bar graph


3. Check out an example, which will tell us how to interpret data using pictographs. From the given figure, the data of 150 students has been collected, who have different books. The data given was as follows:

No. of Students Books
30 Maths
20 English
50 Science
50 Social
Solution:

The above table data can be represented as a pictograph as follows:

Color No. of Students
Maths 5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets
English 5th grade handling worksheets5th grade handling worksheets
Science 5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets
Social 5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets5th grade handling worksheets

where every circle represents 10 students. 5th grade handling worksheets represents a book.

(i) What is the difference between the students who have English and Social books?

There are 5 students and 2 students having Social and English Books. So, the difference is 3 student books. 1 book means 10 students. So, the difference is 30 students.


4. Given below is the amount of money gained by selling cars from a company in the last 6 months. Tick the most suitable scale to be used for making a Bar Graph.

Months Jan Feb Mar Apr May Jun
Money Gained $2,00,000 $1,50,000 $4,50,000 $3,00,000 $1,00,000 $5,50,000

(i) $ 1,00,000 (ii) $ 25,000 (iii) $ 70,000 (iv) $ 75,000

Solution:

Given the amount of money gained by selling cars from a company in the last 6 months. The most suitable scale to be used for making a Bar Graph is $ 25,000 as every gained amount is divided by $ 25,000.

Therefore, the answer is (ii) $ 25,000


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5. Tick the right Pie Chart for the given information. Given information tells that the color and the number of students liked it.

Color Number of Students
Green 20
Blue 5
Red 2
Yellow 10

(i) 5th grade handling examples (ii)5th grade handling questions (iii) 5th grade handling solved examples

Solution:

Given information tells that the color and the number of students liked it. From the given data, Green is most liked by the more number of students. The second highest color liked by students is Yellow. The third highest color liked by students is Blue. Ans the fourth and last color liked by students is red. So, from the given images, the third image is satisfying the given data.

Therefore, the answer is (iii)


6. Answer the following questions.
(i) Which graph is used to compare data elements?
(ii) What is the collection of numbers gathered that provides some meaningful information or details?
(iii) What do we call when the data shows through pictures of objects?
(iv) On the scale of 1 unit length = 6 m, the bar of length 3 units will represent _______ m.
(v) Which graphs are used to represent the trend or change over time?

Solution:

(i) The Bar graph is used to compare data elements.
(ii) The Data is the collection of numbers gathered that provides some meaningful information or details.
(iii) We call it a Pictograph when the data shows through pictures of objects?
(iv) On the scale of 1 unit length = 6 m, the bar of length 3 units will represent 18 m.
(v) The Line graphs are used to represent the trend or change over time?


7. Observe the bar graph representing the number of ceiling fans sold in Mar month on different days of a week and answer the following.
5th grade handling questions and answers

(i) On which day were the maximum number of ceiling fans sold and how many?
(ii) On which day were the minimum number of ceiling fans sold and how many?
(iii) On which day were 100 ceiling fans sold?
(iv) Is there any equal number of ceiling fans sold?

Solution:

Given that a bar graph representing the number of ceiling fans sold in Mar month on different days of a week.
(i) On Sunday, the maximum number of ceiling fans is sold. 450 machines are sold on Thursday.
(ii) On Saturday, the minimum number of ceiling fans is sold. 50 machines are sold on Thursday.
(iii) On Thursday, 100 ceiling fans were sold.
(iv) No. There is no equal number of ceiling fans sold on any day of the week.


8. If ⊗ represents 2 pencils, what do the following represent?
(i) ⊗⊗
(ii) ⊗⊗⊗⊗⊗⊗
(iii) ⊗⊗⊗
(iv) ⊗⊗⊗⊗⊗
(v) ⊗⊗⊗⊗⊗⊗⊗⊗

Solution:

Given ⊗ represents 2 pencils.
(i) ⊗⊗
⊗ (one cross circle) = 2 pencils
⊗⊗ (two cross circles) = 2 × 2 pencils = 4 pencils.
Therefore, the answer is 4 pencils.
(ii) ⊗⊗⊗⊗⊗⊗
⊗ (one cross circle) = 2 pencils
⊗⊗⊗⊗⊗⊗ (six cross circles) = 6 × 2 pencils = 12 pencils
Therefore, the answer is 12 pencils.
(iii) ⊗⊗⊗
⊗ (one cross circle) = 2 pencils
⊗⊗⊗ (three cross circles) = 3 × 2 pencils = 6 pencils
Therefore, the answer is 6 pencils.
(iv) ⊗⊗⊗⊗⊗
⊗ (one cross circle) = 2 pencils
⊗⊗⊗⊗⊗ (five cross circles) = 5 × 2 pencils = 10 pencils
Therefore, the answer is 10 pencils.
(v) ⊗⊗⊗⊗⊗⊗⊗⊗
⊗ (one cross circle) = 2 pencils
⊗⊗⊗⊗⊗⊗⊗⊗ (eight cross circles) = 8 × 2 pencils = 16 pencils
Therefore, the answer is 16 pencils.


9. 50 students were surveyed to find out about their favorite chocolate. The information collected has been tabulated. Observe the given data and draw a pie chart.

Favorite Chocolate Number of Students
Dairy Milk 10
Munch 5
KitKat 30
5-star 5

(i) Which is the most favorite chocolate of the students?
(ii) How many more students like Dairy Milk than Munch?
(iii) Which chocolates are liked by a similar number of students?

Solution:

Given 50 students were surveyed to find out about their favorite chocolate.
The pie chart for the given information is
5th grade handling worksheet problems
(i) KitKat is the favorite chocolate of the more students.
(ii) 5 more students like Dairy Milk than Munch.
(iii) Munch and 5-star chocolates are liked by a similar number of students.


Worksheet on Cost Price, Selling Price and Rates of Profit and Loss

Worksheet on Cost Price, Selling Price and Rates of Profit and Loss | Calculating C.P, S.P, Profit and Loss Worksheets

Worksheet on Cost Price, Selling Price, and Rates of Profit and Loss will help the students to learn different problems on C.P., S.P., Profit, and Loss. Check out every problem available in this article and solve them to know how to solve profit and loss problems. You can crack the exams easily by solving the problems in this article. We have explained every problem along with answers.

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Cost Price, Selling Price, and Rates of Profit and Loss Worksheet with Answers

1. A seller sells a keyboard at a price of Rs 1,500. If the cost price of the keyboard was Rs 2,500, find the profit/loss in which the shopkeeper is. Also, find the percent for the same.

Solution:

Given that a seller sells a keyboard at a price of Rs 1,500.
Selling Price = Rs 1,500
If the cost price of the keyboard was Rs 2,500,
Cost Price = Rs 2,500
Loss = Cost Price – Selling Price = Rs 2,500 – Rs 1,500 = Rs 1,000
Now, find the Loss%.
Loss Percentage = (LOSS/C. P) *100 = (1000/2500) *100 = 40%

Therefore, the loss is Rs 1,000 and the Loss% is 40%.


2. A man buys 15 TVs at a rate of Rs 12,000 per TV and sells them at a rate of Rs 13,500 per unit. Find the total profit/loss faced by the man. Also, find the percent for the same.

Solution:

Given that a man buys 15 TV at a rate of Rs 12,000 per TV.
The cost price of the TV = Rs 12,000
He sells them at a rate of Rs 13,500 per unit.
The Selling price of the TV = Rs 13,500
Since, SP>CP,
Profit = Selling price – Cost Price = Rs 13,500 – Rs 12,000 = RS 1500
Profit% = (Profit/Cost Price)*100% =(1500/12,000)*100% = 12.5%.

Therefore, the Profit = RS 1500 and Profit% = 12.5%.


3. A person bought two refrigerators at Rs 20,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.

Solution:

The Cost price of a refrigirator = Rs 20000
profit = 10%
Selling Price = [100 + p/100] × Cost Price = [100 + 10/100] × 20000
Selling Price = 110/100 × 20000 = 11 × 2000 = Rs. 22000
Loss = 10%
Selling price = [100 – L/100] × Cost Price = [100 – 10/100] × 20000
Selling Price = 90/100 × 20000 = 9 × 2000 = Rs. 18000
Total cost price = Rs 20000 + Rs 20000 = Rs 40000
Total selling price = Rs. 22000 + Rs. 18000 = Rs 40000
Here Cost Price = Selling Price

Therefore, there no profit or no loss.


4. An owner of a Hyundai car sells his car at a price of 5,50,000 with a loss present of 10.5%. Then find the price at which he purchased the Hyundai car and also find the loss suffered by the owner.

Solution:

Given that an owner of a Hyundai car sells his car at a price of RM 5,50,000 with a loss present of 10.5%.
The Selling Price SP = Rs. 5,50,000
Loss% = 10.5
Let the Cost Price CP is X.
Now, find the Cost price.
Selling Price = Cost Price – 10.5%
Rs. 5,50,000 = X – 10.5%
Rs. 5,50,000 = X*(100 – 10.5)/100
Rs. 5,50,000 = 0.895X
X = 5,50,000/0.895
X = 614525
Loss = 614525 – 5,50,000 = 64525

Therefore, the cost of the car is 614525. The loss suffered by the owner is Rs. 64525.


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5. A shop owner buys 20 television sets from a retailer at a rate of Rs 30,000 per set. He sells half of them at a profit of 25% and rests half at a loss of 20%. Find the overall profit/loss faced by the shopkeeper?

Solution:

Given that a shop owner buys 20 television sets from a retailer at a rate of Rs 30,000 per set. He sells half of them at a profit of 25% and rests half at a loss of 20%.
As he sells them at profit of 25%, (30,000 * 5 sets) * 25% = 37,500
If he sold the other half at a loss of 20%;, (30,000 * 5 sets) * 20% = 30,000

Profit (loss) = Rs. 37,500 – Rs. 30,000 = Rs. 7500


6. If the selling price of a commodity is Rs 8,000 with a profit of 15%, find the cost price of the commodity.

Solution:

Given that the selling price of a commodity is Rs 8,000.
Profit% = 15%.
The cost price of the commodity when the selling price and Profit% is
Cost price = (Selling Price × 100)/(100 + Profit percentage)
Cost price = (8000 * 100)/(100 + 15) = 80,000/115 = 6956.52

Therefore, the cost price of the commodity is Rs 6956.52.


7. If the selling price of a product is Rs 9,000 with a loss of 10%, find the cost price of the product?

Solution:

Given that the selling price of a commodity is Rs 9,000.
Loss% = 10%.
The cost price of the commodity when the selling price and Loss% is
Cost price = (Selling Price × 100)/(100 – Loss percentage)
Cost price = (9000 * 100)/(100 – 10) = 90,000/90 = 10000

Therefore, the cost price of the product is Rs 10000.


8. If a bike toy is bought at the cost price of Rs 10,000 and sold at a loss of 25%, find the selling price of the bike toy?

Solution:

Given that the cost price of a bike toy is Rs 10,000.
Loss% = 25%.
The selling price of the bike toy when the cost price and Loss% is
Selling Price = Cost Price [(100 – Loss Percentage)/100]
Selling Price = 10000 [(100 – 25)/100] = Rs 7500

Therefore, the Selling price of the bike toy is Rs 7500.


9. A ceiling fan is bought for Rs. 500 and sold for Rs. 600. Find the gain percent?

Solution:

Given that a ceiling fan is bought for Rs. 500 and sold for Rs. 600. From the given information, the cost price = Rs. 500 and Sale price = Rs. 600.
Now, find the Profit.
Profit or Gain = Selling Price – Cost Price = Rs. 600 – Rs. 500 = Rs. 100
Profit percent or gain percent = (Profit/Cost Price) x 100% = (100/600) x 100% = 16.66%

Therefore, the gain percent of the book is 16.66%.


10. A seller buys chairs from a dealer at a rate of Rs 250 per chair. He sells them at a rate of Rs 325 per chair. He buys 3 chairs of the same type and at the same rate. Find the overall profit/loss. Also profit percent/ loss percent.

Solution:

Given that a seller buys a chair from a dealer at a rate of Rs 250 per battery.
The cost price rate = Rs 250 per chair.
Total cost price = Rs 250 x 3 = Rs 750
He sells them at a rate of Rs 325 per chair.
Selling price rate = Rs 325 per chair
Total selling price = Rs 3250
He buys 3 chairs of the same type and at the same rate.
Profit = total selling price – total cost price
= Rs 3250 – Rs 750 = Rs 2500
= Rs 2500

Profit percent = (2500/750) x 100 % = 333.33%


11. A person bought some car toys at the rate of 30 for Rs. 60 and sold them at 4 for Rs. 28. Find his gain or loss percent.

Solution:

Given that a person bought some car toys at the rate of 30 for Rs. 60 and sold them at 4 for Rs. 28.
Cost price of 30 pens = Rs. 60 → Cost price (CP) of 1 pen = Rs. 2.
Selling price of 4 pens = Rs. 28 → Selling price (SP) of 1 pen = Rs. 28/4 = Rs. 7
Therefore, Gain = 7 – 2 = 5.
Gain percent = 5/2 * 100 = 250%

Therefore, the answer is 250%.


Properties of Addition

Different Properties of Addition – Definition, Facts, Types, Examples

Properties of addition are different for different addition operations. It is necessary to know addition properties because we use every addition operation in our daily life activities. The numbers used for addition operations are called addends. All these properties of addition are used to reduce complex algebraic equations. We will also have properties for subtraction, multiplication, and division in our mathematics. Learn all the properties of addition in this article along with solved examples and practice questions.

The main properties of addition are
1. Commutative property
2. Associative Property
3. Distributive Property
4. Additive Identity Property
Check out all the properties with examples and explanation below.

Also, check

1. Commutative Property of Addition

When two numbers or integers are added the total remains the same even when the values of the numbers or integers changed it is considered as commutative property of addition

For example, let us consider A and B as two numbers or integers, then according to the commutative property of addition A+B = B+A.

Example:

Let us assume A = 15 and B = 10 then commutative property of addition becomes,
Firsly, add A and B: A + B = 15 + 10 = 25
Then, add B and A: B + A = 10 + 15 = 25.
Now, A + B  is equal to the B + A.

Here adding 15 and 10 & and adding 10 and 15 by interchanging the values gives the same result i.e 25.

2. Associative Property of Addition

When three numbers or integers are added the result remains the same even when the grouping or associating of the numbers or integers changed is considered as Associative Property of Addition.

For example, let us assume A, B, and C as three numbers or integers. According to Associative Property of Addition A + (B + C) = (A + B) + C. The associative property of addition is also applicable for multiplication. We use parenthesis to group the addends.

Example of Associative Property of Addition:

Let us assume A = 2, B = 3, and C = 4 then Associative property of addition becomes,
Firsly, add A with the B + C : A + (B + C) = 2 + (3 + 4) = 2 + 7 = 9.
Then, add (A + B) with C : (A + B) + C = (2 + 3) + 4 = 5 + 4 = 9.
Now, A + (B + C)  is equal to the (A + B) + C

Here adding 2, 3, and 4 & and adding 4, 3, and 2 by interchanging the values gives the same result i.e 9.

3. Distributive Property of Addition

Distributive Property of Addition says that When the addition of two numbers or integers is multiplied with the third number the result remains the same when the addition of each of the two numbers or integers is multiplied by the third number or integer.

For example, let us assume A, B, and C as three numbers or integers. According to Distributive Property of Addition, A × (B + C) = (A × B) + (A × C). The distributive property is the combination of both the multiplication operation and the addition operation.

Do Refer:

Example:

Let us assume A = 4, B = 5, and C = 6 then Distributive Property of Addition becomes,
Firsly, add  (B + C) and multiply with A : A × (B + C) = 4 × (5 + 6) = 4 × 11 = 44.
Then, multiply  (A × B) and Add it with the multiplication of (A × C) : (A × B) + (A × C) = (4 × 5) + (4 × 6) = 20 + 24 = 44.

Now, A × (B + C)  is equal to the (A × B) + (A × C).

4. Additive Identity Property of Addition

Additive Identity Property of Addition says that when any number gives the same number after adding it with another number i.e, zero. When you add a number with zero, then you will get the same number as the output. Therefore, the number zero is known as the identity element of addition.

For example, let us take a number A. According to Additive Identity Property of Addition, A + 0 = A or 0 + A = A.

Example:

Let us assume A = 3, then Additive Identity Property of Addition becomes,
Firsly, add  A + 0 = A
Then, add 0 + A = A

Therefore, A + 0 = A or 0 + A = A.

Some More Properties of Addition

Property of Opposites:

According to the Property of Opposites, if you add a number with its negative number, then you will get zero. As the addition of two numbers became zero, they are called additive inverses. This property is known as the inverse property of addition. Or if any number is added to its opposite number, the result becomes zero. Every real number will have its unique additive inverse value.

For example, let us take a number B. According to inverse property of addition, A + (-A) = 0 or (-A) + A = 0

Example:
Let us assume A = 8, then inverse property of addition becomes,
8 + (-8) = 8 – 8 = 0.
Hence Proved.

Sum of Opposite of Numbers:

Let us take two numbers C and D, then their opposites number will become -C and -D. The property becomes -(C + D) = (-C) + (-D)

Example:
Let us assume A = 6, B = 4 then the property to prove its equality becomes,
-(6 + 4) = (-6) + (-4)
-(6 + 4) = -6 -4
-10 = -10

Hence, the equality of this property is proved.

Properties of Addition Examples

Check out the below examples to understand the complete concepts of Properties of Addition.

Example 1:
Prove -(5 + 2) = (-5) + (-2)

Solution:
Given that -(5 + 2) = (-5) + (-2)
-(7) = -5 -2
-7 = -7
L.H.S = R.H.S

Example 2:
Prove 2, 3, and 4 obeys the Distributive Property of Addition.

Solution:
According to the Distributive Property of Addition, A × (B + C) = A × B + A × C
Let A = 2, B = 3, and C = 4.
2 × (3 + 4) = 2 × 3 + 2 × 4
2 × (7) = 6 + 8
14 = 14.
Hence the given numbers obey the Distributive Property of Addition.

Different Practice Questions on Properties of Addition

1. Use properties of addition: – (6+1) = _____.
(i) -7
(ii) -5
(iiii) -12
(iv) -9

Answer: (i) -7
Simplification: Given that – (6+1).
– (6+1) = -6 -1
– (6+1) = -7
Therefore, the answer is -7.

2. Which property describes the following problem? 5 + 4 = 4 + 5
(i) Identity Property
(ii) Commutative Property
(iii) Associative Property
(iv) Math Property

Answer: (ii) Commutative Property
Simplification: Given that 5 + 4 = 4 + 5.
9 = 9
According to the Commutative Property of addition A + B = B + A.
Therefore, the answer is the Commutative Property of addition.

3. Which property uses parentheses?
(i) identity
(ii) commutative
(iii) associative
(iv) all of the properties

Answer: (iv) all of the properties

4. Which property describes this problem? 9 + 0 = 9
(i) Commutative Property of Addition
(ii) Identity Property of Addition
(iii) Associative Property of Addition
(iv) This problem isn’t any of the properties.

Answer: (ii) Identity Property of Addition
Simplification: Given that 9 + 0 = 9
According to the Additive Identity Property of Addition A + 0 = 0 + A = A.
Therefore, the answer is the Additive Identity Property of Addition.

5. The following problem is an example of which property? (6 + 5) + 3 = 6 + (5 + 3)
(i) Identity Property of Addition
(ii) Commutative Property of Addition
(iii) This problem isn’t any of the given properties.

Answer: (iii) This problem isn’t any of the properties.
Simplification: Given that (6 + 5) + 3 = 6 + (5 + 3)
According to the Associative Property of Addition (A + B) + C = A + (B + C).
Therefore, the answer is the (iii) This problem isn’t any of the given properties.

6. What number fills in the blank?
2 × ____ = (2 × 3) + (2 × 7).
(i) 10
(ii) 8
(iii) 6
(iv) 7

Answer: (i) 10
Simplification: Given that 2 × ____ = (2 × 3) + (2 × 7).
According to Distributive Property of Addition, A × (B + C) = (A × B) + (A × C).
A = 2, B = 3, and C = 7.
2 × ____ = (2 × 3) + (2 × 7)
2 × ____ = 6 + 14
2 × ____ = 20.
Therefore, the answer is (i) 10.

Frequently Asked Questions on Properties of Addition

1. Mention the different properties of addition?

Different properties of addition are
Commutative property
Associative property
Identity Property
Distributive property

2. What is the additive identity of 8?

The additive identity says that A + 0 = 0 + A = A.
Here the A = 8.
8 + 0 = 0 + 8 = 8.

3. Explain which property uses addition and multiplication operations at the same time?

The distributive property of addition uses both addition and multiplication operations at the same time. The distributive property of addition is A × (B + C) = A × B + A × C.

4. What does the commutative property of addition explains to us?

The commutative property of addition explains that the sum becomes the same even if the order of addends is changed in the process of addition.

Problems on Cost Price, Selling Price and Rates of Profit and Loss

Problems on Cost Price, Selling Price and Rates of Profit and Loss | Cost Price and Selling Price Problems with Solutions

Problems on Cost Price, Selling Price, and Rates of Profit and Loss along with solved examples are given in this article with a clear explanation. Students can easily understand the in-depth concepts of C.P., S.P., Profit, and Loss by solving various problems. Also, we have included shortcuts and different methods to solve the problems to help the students while solving questions. Furthermore, all cost price, selling price, profit, and loss formula, solved examples, and practice questions are included here for the best practice of the students.

Cost Price: The price for a product or goods bought by a retailer or merchant is known as the Cost Price.
Selling Price: The price for products or goods sold by a retailer or merchant is known as the Selling Price.
Profit: It is the difference between Selling Price and Cost Price.
Profit = Selling Price – Cost Price = S.P. – C.P.
Profit percent = [(S.P. – C.P)/C.P.] x 100%
Profit percent = (Profit/C.P.) x 100%
Loss: It is the difference between Cost Price and Selling Price.
Profit = Cost Price – Selling Price = C.P. – S.P.
Profit percent = [(C.P – S.P.)/C.P.] x 100%
Profit percent = (Loss/C.P.) x 100%

Also, find

Cost Price, Selling Price and Rates of Profit and Loss Questions with Answers

1. A chair is bought for Rs. 300 and sold for Rs. 700. Find the gain percent?
(i) 133.33%
(ii) 73.33%
(iii) 93.33%
(iv) 233.33%

Solution:

Given that a chair is bought for Rs. 300 and sold for Rs. 700. From the given information, the cost price = Rs. 300 and Sale price = Rs. 700.
Now, find the Profit.
Profit or Gain = Selling Price – Cost Price = Rs. 700 – Rs. 300 = Rs. 400
Profit percent or gain percent = (Profit/Cost Price) x 100% = (400/300) x 100% = 133.33%
The gain percent of the book is 133.33%

Therefore, the answer is (i) 133.33%


2. A retailer sells 65 m of cloth for Rs. 8,905 at the profit of Rs. 5/m of cloth. What is the cost price of 1 m of cloth?
(i) Rs. 72
(ii) Rs. 32
(iii) Rs. 132
(iv) Rs. 152

Solution:

Given that a retailer sells 65 m of cloth for Rs. 8,905 at the profit of Rs. 5/m of cloth.
Firstly, find out the Selling Price of the 1 m cloth.
Selling Price of 1m of cloth =  Rs. 8,905/65 = Rs. 137
Now, find the Cost Price of 1m of cloth.
Cost Price of 1m of cloth = Selling Price of 1m of cloth – profit on 1m of cloth
Cost Price of 1m of cloth = Rs. 137 – Rs. 5 = Rs. 132
The cost price of 1 m of cloth is Rs. 132

Therefore, the answer is (iii) Rs. 132


3. By selling a fan at Rs. 1600, a shopkeeper makes a profit of 25%. At what price should he sell the fan so as to make a loss of 25%?
(i) Rs. 690
(ii) Rs. 960
(iii) Rs. 540
(iv) Rs. 1200

Solution:

Given that by selling a fan at Rs. 1600, a shopkeeper makes a profit of 25%.
The selling price of a fan = Rs. 1600.
The shopkeeper makes a profit of 25%.
Firstly, find out the Cost Price.
Cost Price = (Selling Price) x [100/(100+Profit)]
Cost Price = (1600) x [100/(100+25)] = 1600 x [100/(125)]
Cost Price = 1280.
Now, find the Loss.
Loss = 25% = 25% of 1280 = Rs. 320.
Now, find the Selling Price.
Selling Price = Cost Price – Loss = 1280 – 320 = Rs. 960
The selling price of the fan to make a loss of 25% is Rs. 960

Therefore, the answer is (ii) Rs. 960


4. Alex bought 140 bottles at the rate of Rs. 200/bottle. The transport expenditure was Rs. 1,200. He paid an octroi at the rate of Rs. 1.55/bottle and labor charges were Rs. 300. What should be the selling price of 1 bottle, if he wants a profit of 20%?
(i) Rs. 247.125
(ii) Rs. 274.659
(iii) Rs. 245.687
(iv) Rs. 254.712

Solution:

Given that Alex bought 140 bottles at the rate of Rs. 200/bottle. The transport expenditure was Rs. 1,200. He paid an octroi at the rate of Rs. 1.55/bottle and labor charges were Rs. 300.
Total Cost Price per bottle = 200 + 1200/140 + 1.55 + 300/140 = 212.26
Selling Price = Cost Price[(100 + profit%)/100] = 212.26[(100 + 20%)/100] = 254.712
The selling price of 1 bottle, if he wants a profit of 20% is Rs. 254.712.

Therefore, the answer is (iv) Rs. 254.712


5. A man sold two cars for Rs. 5.8 lakhs each. On the one, he gained 10% and on the other, he lost 10%. What percent is the effect of the sale on the whole?
(i) 25% loss
(ii) 25% gain
(iii) 0.25% gain
(iv) 0.25% loss

Solution:

Given that a man sold two cars for Rs. 5.8 lakhs each. On the one, he gained 10% and on the other, he lost 10%.
Find out the loss%.
Loss% = (5/10)^2 = 1/4% = 0.25%.
The loss% that effect of the sale, on the whole, is 0.25%.

Therefore, the answer is (iv) 0.25% loss


6. A bike is sold at 25% profit. If the CP and the SP of the bike are increased by Rs 80 and Rs 50 respectively, the profit% decreases by 10%. Find the cost price of the bike?
(i) 260
(ii) 240
(iii) 320
(iv) 220

Solution:

Given that a bike is sold at 25% profit. If the CP and the SP of the bike are increased by Rs 80 and Rs 50 respectively.
Let the Cost Price = x, then Selling Price = (125/100) × x  = 5x/4
New Cost Price (CP) = (x + 60), new Selling Price (SP) = (5x/4 + 30), new profit% = 25 – 15 = 10
So (5x/4 + 30) = (110/100) × (x + 60)
Solve, x = 240

Therefore, the answer is (ii) 240


7. A man bought some pens at the rate of 20 for Rs. 60 and sold them at 5 for Rs. 30. Find his gain or loss percent.

Solution:

Given that a man bought some pens at the rate of 20 for Rs. 60 and sold them at 5 for Rs. 30.
Cost price of 20 pens = Rs. 60 → Cost price (CP) of 1 pen = Rs. 3.
Selling price of 5 pens = Rs. 30 → Selling price (SP) of 1 pen = Rs. 30/5 = Rs. 6
Therefore, Gain = 6 – 3 = 3.
Gain percent = 3/3 * 100 = 100%

Therefore, the answer is 100%


8. A shopkeeper buys batteries from a dealer at a rate of Rs 350 per battery. He sells them at a rate of Rs 425 per battery. He buys 5 batteries of the same type and at the same rate. Find the overall profit/loss. Also profit percent/ loss percent.

Solution:

Given that a shopkeeper buys batteries from a dealer at a rate of Rs 350 per battery.
The cost price rate = Rs 350 per battery.
Total cost price = Rs 350 x 5 = Rs 1750
He sells them at a rate of Rs 425 per battery.
Selling price rate = Rs 425 per battery
Total selling price = Rs 4250
He buys 5 batteries of the same type and at the same rate.
Profit = total selling price – total cost price
= Rs 4250 – Rs 1750
= Rs 2500
Profit percent = (2500/1750) x 100 % = 142.85%


9. A shopkeeper sells a TV for Rs. 8,500 with a loss of Rs. 500. Find the price at which he had bought it from the dealer. Also, calculate the loss percent?

Solution:

Given that a shopkeeper sells a TV for Rs. 8,500 with a loss of Rs. 500. Find the price at which he had bought it from the dealer.
The selling price of the TV = Rs 8,500
The loss suffered by the shopkeeper = Rs 500
Now, find the Cost price.
We know that, Selling price = Cost Price – Loss
So, Cost Price = Selling Price + Loss
Cost Price = Rs 8,500 + Rs 500
= Rs 9,000
Loss percent = (Loss/Cost price) x 100% = (500/9000) x 100% = 5.55


Worksheet on Graph of Linear Relations in x, y

Worksheet on Graph of Linear Relations in x, y | Graphing Linear Relations in x, y Axis Worksheet with Answers

Worksheet on Graph of Linear Relations in x, y with step-wise solutions is available on this page. So, the students of grade 9 can go through the problems provided here and try to solve them and test how much they have learned from the chapter coordinate geometry. This page is the one-stop for the students who are lagging in graphing linear equations using x, y.

We will learn how to draw the graphs of linear relations in x, y coordinates. Hence practice the problems given below and score better grades in the exams. These concepts will be helpful for you in further classes.

See More:

Graphing Linear Relations in x, y Worksheet Answers

You can see different types of equations in the x-y plane in the graph of linear relations worksheet.

Example 1.
Draw the graph of the linear equation in two variables x – 3y + 1 = 0

Solution:

Given equation is x – 3y + 1 = 0
x + 1 = 3y
y = x + 1/3
In the given equation
If x = -1 then y = (-1) + 1/3 = -2/3
If x = 0 then y = (0) + 1/3= 1/3
If x = 1 then y = (1) + 1/3 = 4/3
If x = 2 then y = (2) + 1/3 = 7/3
Plot the graph using the points (-1,-2/3), (0,1/3), (1,4/3), (2,7/3).
Worksheet on Graph of Linear Relations in x, y_1


Example 2.
Draw the graph of the linear equation in two variables x + y = 1

Solution:

x + y = 1
– y = x – 1
y = -x + 1
If x = 0 then y = 0 + 1 = 1
If x = 1 then y = -1 + 1 = 0
If x = 2 then y = -2 + 1 = -1
If x = 4 then y = – 4 + 4 = 0
Then plot the following points in the graph (0,1), (1,0), (2,-2) and (4,0)
Worksheet on Graph of Linear Relations in x, y_2


Example 3.
Draw the graph of the linear equation in two variables y = 3x + 2.

Solution:

Given equation is y = 3x + 2
In the given equation
If x = -2 then y = 3(-2) + 2 = -6+2 = -4
If x = -1 then y = 3(-1) + 2 = -2
If x = 0 then y = 3(0) + 2 = 2
If x = 1 then y = 3(1) + 2 = 5
If x = 2 then y = 3(2) + 2 = 8
Plot the graph using the points (-2,-4), (-1,-2), (0,2), (1,5) and (2,8).
Worksheet on Graph of Linear Relations in x, y_3


Example 4.
Draw the graph of the linear equation in two variables x − y = 6

Solution:

x – y = 6
– y = 6 – x
y = x – 6
If x = 0 then y = 0 – 6 = -6
If x = 1 then y = 1 – 6 = -5
If x = 2 then y = 2 – 6 = -4
If x = 4 then y = 4 – 6 = -2
Then plot the graph using the points (0,-6), (1,-5), (2,-4), (4,-2).
Worksheet on Graph of Linear Relations in x, y_4


Example 5.
Draw the graph of the linear equation in two variables y = 3x + 1.

Solution:

Given equation is y = 3x + 1
In the given equation
If x = -5 then y = 3(-5) + 1 = -15+ 1 = -14
If x = -1 then y = 3(-1) + 1 = -2
If x = 0 then y = 3(0) + 1 = 1
If x = 1 then y = 3(1) + 1 = 4
If x = 2 then y = 3(2) + 1 = 7
Plot the graph using the points (-5,-14), (-1,-2), (0,1), (1,4) and (2,7).
Worksheet on Graph of Linear Relations in x, y_5


Example 6.
Draw the graph of the linear equation in two variables 4x + 3y + 1 = 0

Solution:

Given equation is 4x + 3y + 1 = 0
-3y = 4x + 1
y = -4x/3 -1/3
In the given equation
If x = -1 then y = -4(-1)/3 – 1/3 = 1
If x = 0 then y = (0)/3 – 1/3= -1/3
If x = 1 then y = -4(1)/3 – 1/3 = -5/3
If x = 2 then y = -4(2)/3 – 1/3 = -9/3
Plot the graph using the points (-1,1), (0,-1/3), (1,-5/3), (2,-9/3).
Worksheet on Graph of Linear Relations in x, y_6


Example 7.
Draw the graph of the linear equation in two variables 2x + y = 4

Solution:

2x + y = 4
– y = 2x – 4
y = -2x + 4
If x = 0 then y = -2(0) + 4 = 4
If x = 1 then y = -2(1) + 4 = 2
If x = 2 then y = -2(2) + 4 = 0
If x = 4 then y = -2(4) + 4 = -4
Then plot the following points in the graph (0,4), (1,2), (2,0) and (4,-4)
Worksheet on Graph of Linear Relations in x, y_7


Example 8.
Draw the graph of the linear equation in two variables 2x − y = 6

Solution:

2x – y = 6
– y = 6 – 2x
y = 2x – 6
If x = 0 then y = 2(0) – 6 = -6
If x = 1 then y = 2(1) – 6 = -4
If x = 2 then y = 2(2) – 6 = -2
If x = 4 then y = 2(4) – 6 = 2
Then plot the graph using the points (0,-6), (1,-4), (2,-2), (4,2).
Worksheet on Graph of Linear Relations in x, y_8


Example 9.
Draw the graph of the linear equation in two variables x – y + 2 = 0

Solution:

Given equation is x – y + 2 = 0
x + 2 = y
In the given equation
If x = -1 then y = (-1) + 2 = 1
If x = 0 then y = (0) + 2 = 2
If x = 1 then y = (1) + 2 = 3
If x = 2 then y = (2) + 2 = 4
Plot the graph using the points (-1,1), (0,2), (1,3), (2,4).
Worksheet on Graph of Linear Relations in x, y_9


Example 10.
Draw the graph of the linear equation in two variables x + y = 6

Solution:

x + y = 6
– y = x – 6
y = -x + 6
If x = 0 then y = 0 + 6 = 6
If x = 1 then y = -1 + 6 = 5
If x = 2 then y = -2 + 6 = 4
If x = 4 then y = – 4 + 6 = 2
Then plot the following points in the graph (0,6), (1,5), (2,4) and (4,2)
Worksheet on Graph of Linear Relations in x, y_10


Co-ordinate Geometry

Co-ordinate Geometry – Introduction, Definition, Formulas, Theorems | How do you Solve Coordinate Geometry?

In maths, coordinate geometry is one of the most important and interesting concepts at the secondary level. Coordinate geometry is the part of the geometry that uses two or more to specify the point. The position of the point or figure can be determined in a line or three-dimensional space. The concept of coordinate geometry is explained with definitions, formulas, examples here. In addition to this, we also provide worksheets on coordinate geometry to test your knowledge of this chapter.

Topics in Co-ordinate Geometry Grade 9 | Coordinate Geometry List of Contents

Introduction to Coordinate Geometry

Coordinate geometry is determined as the study of geometry using coordinate points. With the help of the points, we can find the distance between the two points, divide the lines into m:n ratio, find the midpoint, calculate the area of a triangle in the cartesian plane, and so on.

Definition of Coordinate geometry It is a part of the geometry where the position of the point is determined using coordinates.
What is meant by coordinate? Coordinates are nothing but the set of values that helps to show the exact point in the coordinate plane.
Distance Formula The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).
What is meant by Coordinate Plane? A coordinate plane is a two-dimensional plane that is formed by the intersection of two perpendicular lines that is x-axis and y-axis.
Section Formula Section formula is used to divide any line into two parts m:n ratio
Mid-point theorem Mid-point theorem is used to find the coordinates at which a line is divided into two parts.

What is a Co-ordinate and a Co-ordinate Plane?

Coordinate and coordinate plane sounds the same but they are different. A coordinate is the set of values that is used to show the exact point in the coordinate plane. Whereas a coordinate plane is a 2D plane that is formed by the intersection of two perpendicular lines that is x-axis and y-axis. The point where the two axes intersect is known as the origin. The location of the point (x, y) is known as the coordinates.
There are four quadrants in the graph they are,
Quadrant 1: (+x, +y)
Quadrant 2 : (-x, +y)
Quadrant 3 : (-x, -y)
Quadrant 4 : (+x, -y)
coordinate geometry

Equation of a Line in Cartesian Plane

Equation of the line can be represented as follows,
i. General Form:
The general form of a line is given as Ax + By + C = 0.
ii. Slope intercept form:
Let x, y be the coordinate of a point that passes through the line, m be the slope of the line, and c be the y-intercept.
y=mx + c
iii. Intercept Form of a line:
Consider a and b be the x-intercept and y-intercept of a line, then the equation of a line is represented as
y=mx + c

Slope of a Line

The general form of the line is Ax + By + C = 0.
The slope can be found by converting this form to the slope-intercept form.
Ax + By + C = 0
By = − Ax – C
By = − Ax – C
y = -A/B x – C/B
Comparing the above equation with y = mx + c,
m = -A/B
By using this formula we can find the slope of a line from the general equation of a line.

Coordinate Geometry Formulas and Theorems

Distance Formula:
The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).
d = √(x2 – x1)² + (y2 – y1)²
where (x1, y1) and (x2, y2) are the coordinates of the plane.
Mid-point Theorem:
The mid-point theorem is used to find the coordinates at which a line is divided into two parts. Consider two points A and B having the coordinates (x1, y1) and (x2, y2)
M(x, y) = [(x1+x2)/2, (y1+y2)/2]
Angle Formula:
Consider two lines A and B, having their slopes to be m1 and m2 respectively.
Let “θ” be the angle between these two lines, then the angle between them.
tanθ = (m1-m2)/1+m1m2
Case 1: When the two lines are parallel to each other,
m1 = m2 = m
tanθ = (m1-m2)/1+m1m2 = 0
Thus θ = 0
Case 2: When the two lines are perpendicular to each other,
m1.m2 = -1
tanθ = (m1-m2)/1+m1m2
tanθ = (m1-m2)/1+(-1) = (m1-m2)/0 = undefined
θ = 90°

Coordinate Geometry Example Questions

Example 1.
Find the distance of the point P (7, 8) from the x-axis.
Solution:
We know that,
(x, y) = (7, 8) is a point on the Cartesian plane in the first quadrant.
x = Perpendicular distance from the y-axis
y = Perpendicular distance from x-axis
Therefore, the perpendicular distance from x-axis = y coordinate = 8

Example 2.
Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 4) and B (3, 6).
Solution:
Let P (x, y) be equidistant from the points A (2, 4) and B (3, 7).
Therefore AP = BP …[Given]
AP² = BP² …[Squaring both sides]
(x – 2)² + (y – 4)² = (x – 3)² + (y – 6)²
x² – 4x + 4 + y² – 8y + 16 = x² – 6x + 9 + y² -12y + 36
-4x -8y + 20 = -6x – 12y + 45
-4x -8y + 20 + 6x + 12y – 45 = 0
2x + 4y – 25 = 0
2x + 4y = 25
2x – 25 = -4y
2x – 25 = -4y is the required relation.

Example 3.
Three vertices of a parallelogram taken in order are (1,0) (2,1) and (2,3) respectively. Find the coordinates of the fourth vertex
Solution:
Consider D(x,y) as the fourth vertex.
Let A(1,0) B(2,1) C(2,3) and D(x,y) be the vertices of a parallelogram.
ABCD took each other.
Since the diagonal of a parallelogram bisects each other.
Coordinates of the mid point of AC = coordinates of the midpoint of BD.
(1 + 2/2 , 0 + 3/2) = (2 + x/2 , 1 + y/2)
(3/2 , 3/2) = (2 + x/2 , 1 + y/2)
2 + x/2 + 3/2
= 2 + x = 3
x = 3 – 2
x = 1
And
1 + y/2 = 3/2
1 + y = 3
y = 3 – 1
y = 2
Hence coordinates of the fourth vertex D(1,2)

Example 4.
Find that value(s) of x for which the distance between the points P(x, 3) and Q(7, 10) is 10 units.
Solution:
Given that
PQ = 10
PQ² = 10²
PQ² = 100
(7 – x)² + (10 – 3)² = 100
Using distance formula
(7 – x)² + (10 – 3)² = 100
(7 – x)² = 100 – 49
(7 – x)² = 51
(7 – x) = √51
7 – x = + or – 7.1
7 – x = 7.1 or 7 – x = -7.1
x = 7.1 – 7 or x = -7.1 – 7
x = 0.1 or x = -14.1

Example 5.
Find the centroid of the triangle whose vertices are (1,2) (3,4) (5,6).
Solution:
The centroid of the triangle whose vertices are (1,2) (3,4) and (5,6)
The centroid of the triangle ABC = (x1 + x2 + x3/2, y1 + y2 + y3/2)
The centroid of the triangle whose vertices A(1,2) B(3,4), C(5,6)
= (1+3+5/2 ,2+ 4 + 6/2)
(9/2, 12/2)
(9/2, 6).

FAQs on Coordinate Geometry

1. How can you prepare and complete coordinate geometry?

Pay attention to the classes and practice the problems thoroughly. Make use of the links given on our page and clarify your doubts.

2. How Can You Define the Position of an Object on a Floor?

The students of 9th grade can define the position of an object on the floor by considering two adjacent walls as two coordinate axes.

3. What are the signs of the coordinates in the four quadrants of a cartesian plane?

The signs of the coordinates in the four quadrants of a cartesian plane are (+, +) in the first quadrant, (-, +) in the second quadrant, (-, -) in the third quadrant of a cartesian plane, and (+, +) in the fourth quadrant of a cartesian plane.

Slope of Graph y = mx+c

Slope of Graph y = mx+c – Introduction, Formula, Examples | How do you find the Slope of y = mx + c?

In the previous articles, we have discussed about slope and y-intercept, plotting points in the XY plane, independent variables and dependent variables, etc. Let us learn how to solve the slope of the graph y = mx + c from this page. In addition that you can find the problems on the slope of the graph y = mx + c in the below section.

Also Read:

Introduction to Slope-Intercept Form | y=mx+c Equation of Straight Line Slope

y = mx + c is an equation of the line having a slope and y-intercept of the line in coordinate geometry. m represents the slope or gradient of the line and c is the y-intercept that cuts the point on the y-axis. The line cuts the y-axis at the point (0, c) which is the distance from the origin to the point c.
m = (y – c)/(x – 0)
m = (y – c)/x
mx = y – c
mx + c = y
y = mx + c
y = mx+c

y = mx + c Straight Line Graphs Slope Examples

Example 1.
What is the slope of the equation 6x – 4y + 5 = 0?
Solution:
Given that the equation is
6x – 4y + 5 = 0
4y = 6x + 5
y = 6/4x + 5/4.
y = 3/2x + 5/4
Comparing the equation with y = mx + c,
we have m = 3/2.
Therefore, the slope of the line is 3/2
Slope of the graph y = mx+c_1

Example 2.
Find the equation of a line in the form of y = mx + c, having a slope of 4 units and an intercept of -6 units.
Solution:
Given that
The slope of the line, m = 4 and The y-intercept of the line, c = -6.
We know that
The slope-intercept form of the equation of a line is y = mx + c.
From the equation
y = 4x – 6
Therefore the required equation of the line is y = 4x – 6

Example 3.
Convert the equation 5x + 4y = 12 into y = mx + c and find its y-intercept.
Solution:
Given that the equation is
5x + 6y + 12 = 0
6y = 5x + 12
y = 5/6x + 12/6.
y = 5/6x + 2
Comparing the equation with y = mx + c,
we have m = 5/6 and c = 2
Therefore, the slope of the line is 5/6. y-intercept is c = 2

Example 4.
What is the slope of the equation 2x – 4y + 10 = 0?
Solution:
Given that the equation is
2x – 4y + 10 = 0
4y = 2x + 10
y = 2/4x + 10/4.
y = 1/2x + 5/2
Comparing the equation with y = mx + c,
we have m = 1/2.
Therefore, the slope of the line is 1/2.

Example 5.
What is the slope and y-intercept of the line 3x – 6y + 12 = 0?
Solution:
Given that the equation is
3x – 8y + 12 = 0
8y = 3x + 12
y = 3/8x + 12/8.
y = 3/8x + 3/2
Comparing the equation with y = mx + c,
we have m = 3/8 and c = 3/2
Therefore, the slope of the line is ⅜ and y intercept c = 3/2

FAQs on Slope of the graph y = mx + c

1. What is the equation of a straight line?

The equation of the straight line is y = mx +c
where,
m is the slope and c is the y-intercept.

2. How do I find the slope in a graph?

Take two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points. Find the difference in x-coordinates for these two points. Divide the difference in y-coordinates by the difference in x-coordinates.

3. What is mx in the slope formula?

In the equation of a straight line y = mx + c, the slope is m which is multiplied by x and c is the y-intercept.

y-intercept of the Graph of y = mx + c

Finding y-intercept of the Graph y = mx + c | Tricks to find Slope Intercept Form of Line y=mx+c

y = mx + c is called the slope-intercept form equation of the straight line. In this article, we are going to learn more details about the y = mx + c graph of the straight line. In y = mx + c, m represents the slope value of the equation and c is the y-intercept. The graph y = mx + c cuts the y-axis at the point p the OP is the y-intercept of the graph and O is the origin.

Related Topics:

y-intercept of y = mx + c

The general form of the equation of the straight line is y = mx + c where m is slope or gradient and c is the y-intercept that cuts the y-axis at some point c. It is a linear equation and the variables x and y are the coordinates on the plane. x, y are independent and dependent variables respectively. The y-intercept of the line that is on the graph y = mx + c is c.

y = mx+c

Slope y = mx+c Formula

The equation y = mx + c is derived from the slope formula.
m = (y-c)/(x-0)
m = (y-c)/x
mx = y – c
mx + c = y
y = mx + c
Thus the slope-intercept form of the equation of the line is derived.

How to find the y-intercept of Equation of Straight Line y = mx + c?

  1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
  2. Substitute x = 0 into the equation to find the y-intercept.
  3. Declare the coefficient of x.

Slope Intercept Form of Equation y = mx + c Examples

Example 1.
What is the y-intercept of the graph of 6x + 2y = 3?
Solution:
Given that the equation is 6x + 2y = 3
2y = -6x + 3
y = – 6/2x + 3/2
y = -3x + 3/2
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3/2. So, the y-intercept = 3/2

Example 2.
What is the y-intercept of the graph of x + y = 3?
Solution:
Given that the equation is 7x + 2y = 3
y = -x + 3
y = – x + 3
y = -x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3

Example 3.
What is the y intercept of the graph of 5y = 2x + 3
Solution:
Given that the equation is 5y = 2x + 3
y = 2/5x + 3/5
We know that
y = mx + c
Comparing the equation with y = mx + c
We get c = 3/5
Therefore the y-intercept = 3/5

Example 4.
What is the y-intercept of the graph of 4x + 2y = 6?
Solution:
Given that the equation is 4x + 2y = 6
4x + 2y – 6 = 0
2y = -4x + 6
y = – 4/2x + 6/2
y = -2x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3

Example 5.
What is the y-intercept of the graph of 7x + 8y = 26?
Solution:
Given that the equation is 7x + 8y = 26
8y = -7x + 26
y = – 7/8x + 26/8
y = -7/8x + 13/4
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 13/4. So, the y-intercept = 13/4

FAQs on y-intercept of the Graph of y = mx + c

1. What is y = mx + c?

The expression y = mx + c is an equation of a line
Here m is the slope of the line and the y-intercept of c.
This equation is formed by knowing the slope of the line and the intercept which the line cuts on the y-axis. This equation y = mx + c is the basic equation of the line.

2. How to use y = mx + c?

  1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
  2. Substitute x = 0 into the equation to find the y-intercept.
  3. Declare the coefficient of x.

3. What is the y-intercept of the graph 5x + 2y = 5?

5x + 2y = 5
2y = -5x + 5
y = -5/2 x + 5/2
Now compare the equation with y = mx + c
c = 5/2 so, the y-intercept = 5/2.

Plotting a Point in Cartesian Plane

Plotting a Point in Cartesian Plane and Determining the Quadrants | How to draw a Cartesian Plane and Label Points on Graph?

Plotting a Point in Cartesian Plane is very easy if you know how to draw a cartesian plane in a coordinate geometry graph. A cartesian graph is an intersection of two points at O. They are represented by two numbers in parentheses, separated by a comma that is called coordinates. Learn more about the cartesian plane from here and know how to draw a cartesian plane and then plot points on the cartesian graph. Scroll down this page to learn plotting points on a cartesian plane with solved examples.

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What is Cartesian Plane?

The cartesian plane is a representation of the position of a point by referring to it in terms of x-axis horizontal line and y-axis vertical line respectively. The point of intersection of x-axes and y-axes is known as the origin. The origin is represented by the letter ‘O’. The coordinates have two values the first number represents the x-coordinate and the second number represents the y-coordinate.
Cartesian Plane_1

How to Plot Points in Cartesian Plan?

  1. First, check the signs of the coordinates of a point and then identify the quadrant in which the point is to be plotted.
  2. Take the rectangular cartesian frames XOX’ and YOY’ intersecting at the right angles at origin ‘O’.
  3. Take a point P on the x-axis on the side of the involved quadrant such that the distance of the point P from the origin O equals the value of the x-coordinate.
  4. Draw a perpendicular line PQ on the x-axis. Take a point M on this perpendicular such that PM is equal to the value of the y-coordinate and M in the relevant quadrant. Thus the point M is plotted as per the given coordinate points.

Plotting a Point in Cartesian Plane Examples

Plot the following points in the cartesian plane then identify which quadrant or axis it belongs

Example 1.
Plot the points (3,4) in the Cartesian plane.
Solution:
Given that the points are (3,4)
Here the x-coordinate is 3 and the y-coordinate is 4.
(3, 4) is a point in the first quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (3,4)
Cartesian Plane_2

Example 2.
Plot the points (5,-6) in the Cartesian plane.
Solution:
Given that the points are (5,-6)
Here the x-coordinate is 5 and the y-coordinate is -6.
(5, -6) is a point in the fourth quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (5,-6)
Cartesian Plane_3

Example 3.
Plot the points (-7,-8) in the Cartesian plane.
Solution:
Given that the points are (-7,-8)
Here the x-coordinate is -7 and the y-coordinate is -8.
(-7, -8) is a point in the third quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (-7,-8)
Cartesian Plane_4

Example 4.
Plot the points (-9,2) in the Cartesian plane.
Solution:
Given that the points are (-9,2)
Here the x-coordinate is -9 and the y-coordinate is 2.
(-9, 2) is a point in the second quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (-9,2)
Cartesian Plane_5

Example 5.
Plot the points (1,1) in the Cartesian plane.
Solution:
Given that the points are (1,1)
Here the x-coordinate is 1 and the y-coordinate is 1.
(1, 1) is a point in the first quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (1,1).
Cartesian Plane_6

Invariant Points for Reflection in a Line

Invariant Points for Reflection in a Line – Definition, Facts, Examples | How do you find Invariant Points and Lines?

Invariant Points for Reflection in a Line is nothing a point that remains unvaried after the transformation applied to it. Any point on a graph on the line of reflection is an invariant point. Learn more about Invariant Points Under Reflection in a Line from this article. By making this page as a reference you can know about the Invariant Points on the graph on the line of reflection which helps you to score good marks in the exams.

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Invariant Points for Reflection in a Line – Definition

Invariant points means the points which lie on the line and when reflected in the line. So, only those points are invariant which lie on the y-axis.
The invariant points must have x-coordinate = 0.
Therefore, only (0, 4) is the invariant point. No points lying outside the line will be an invariant point.

Invariant Points in Reflection in a Line Examples with Answers

Example 1.
Which of the following points (-2, 0), (0, -5), (3, -3) are invariant points when reflected in the x-axis?
Solution:
Given points are (-2,0), (0,-5) and (-3,3)
We know that
The points lying on the line are invariant points. when reflected in the line.
So, only those points are invariant which lie on the x-axis.
Hence, the invariant points must have y-coordinate = 0.
Therefore, only (-2, 0) is the invariant point.

Example 2.
Which of the following points (-2, 3), (0, 6), (4, -3), (-3, 6) are invariant points when reflected in the line parallel to the x-axis at a distance of 4 on the positive side of the y-axis?
Solution:
Given points are (-2,3), (0,6), (4,-3) and (-3,6)
We know that,
The points lying on the line are invariant points when reflected in the line.
So, only those points are invariant which are on the line parallel to the x-axis at a distance of 4 on the positive side of the y-axis.
Hence, the invariant points must have a y-coordinate = 4.
Therefore, (0, 6) and (-3, 6) are the invariant points.

Example 3.
Points A and B have coordinates (4,6) respectively. Find the reflection of A’ of A under reflection in the x-axis and A” of A under reflection in the y axis.
Solution:
Given that the coordinates are (4,6)
A’ = image of A under reflection in the x axis = (4,-6).
A” = image of A under reflection in the y axis = (-4,6).

Example 4.
A point P(a,b) is reflected in the x-axis to P'(6,-5). Write down the values of a and b. P” is the image of P reflected in the y axis. Write down the coordinates of P”. Find the coordinates of P”‘, when P is reflected in the line parallel to y-axis such that x = 3.
Solution:
A point P(a,b) is reflected in the x axis P'(6,-5)
We know that Mx(x,y) = (x,-y)
Thus, the coordinates of P are (6,5)
Hence a = 6, and b = 5.
P” = image of P reflected in the y axis = (-6,5)
P”‘ = reflection of P in the line (x = 3) = (-9,5).

Example 5.
Which of the following points (-1, 0), (0, -8), (2, -7) are invariant points when reflected in the x-axis?
Solution:
Given points are (-1,0), (0,-8) and (2,-7)
We know that
The points lying on the line are invariant points. when reflected in the line.
So, only those points are invariant which lie on the x-axis.
Hence, the invariant points must have y-coordinate = 0.
Therefore, only (-1, 0) is the invariant point.

FAQs on Invariant Points Under Reflection in a Line

1. Is a line of invariant points an invariant line?

In linear transformation maps the origin to the origin.
So the origin is always an invariant point under a linear transformation. Every point on the invariant line maps to a point on the line itself.

2. What are invariant points?

Invariant points are points on a line or a shape which do not move when the specific transformation is applied.

3. How do you find invariant points on a graph?

Sketch each graph by using the key points, including invariant points, Determine the image points on the graph by square rooting the points. And locate invariant points on y = f(x) and y = g(x). When graphing the square root of a function, invariant points occur at y = 0 and y = 1.