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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.

Also, find:

• Worksheet on Pictograph and Bar Graph
• Worksheet on Representing Data on Bar Graph
• Interpreting Bar Graph

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
Feb
Mar
Apr
May

(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.

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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

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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
English
Science
Social

where every circle represents 10 students. 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.

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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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Read More:

• Bar Graph on Graph Paper
• Reading Pictographs
• Represent Data on a Bar Graph

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) (ii) (iii)

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)

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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?

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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.

(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.

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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.

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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

(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.

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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.

Also, find:

• Problems on Cost Price, Selling Price and Rates of Profit and Loss
• Worksheet on Profit or Loss Percent

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%.

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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%.

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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.

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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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Read More:

• Worksheet on Calculating Profit or Loss
• Worksheet on Profit/Loss Involving Sales Tax

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

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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.

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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.

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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.

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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%.

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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%

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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%.

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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

• Properties of Subtraction
• Properties of Multiplication
• Properties of Division

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:

• Addition of Integers

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 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

• Worksheet on Profit or Loss Percent
• Worksheet on Calculating Profit or Loss
• Worksheet on Profit/Loss Involving Sales Tax

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%

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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

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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

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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

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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

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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

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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%

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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%

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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

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A successive discount is a discount that is given on the selling price of the product that has already had the discount on the marked price. While you cross a garment store, you come across the offers like 20% or 30% in blocks. The percentage we find on it is the discount offered by the shopkeeper to their customers.

For example, Suppose that a shopkeeper bought a shirt from the retailer at $500 and he decided to sell it at $800 and again put the discount of 20% where the final selling price will be $640. The customer feels that he got it for 20% off, but the shopkeeper sells it for a profit of $140 even after offering the discount. In short, we can say that, if the discount is again applied to the selling price, then it is determined as successive discounts. Successive discount is the amount of discount which is offered on the discount.

Successive Discounts Formula

To find the total discount in successive discounts case, Suppose that the first discount is x% and the second discount is y%, then the formula can be written as:
Total discount = x + y – \(\frac {xy}{100} \)%
There is another scenario like, Consider the original price of the shirt as ‘x’ and the first discount offered as ‘y’ and again the discount offered as ‘z’ on the new price. Then the selling price of the shirt is calculated as:
x-(y + z – \(\frac {yz}{100} \)) * x

Example:
Suppose that the online shopping website sells a product and it offers a discount of 10% on that product and again it offers more than 20% on the discounted value. Know the final value of the product.

In this case, let us suppose that the initial value is 100.
Given that the shopkeeper offered a discount of 10%, therefore it is (100-10) = 90
Then the shopkeeper again offered a discount of 20%, therefore it is 90 – 2(9) = 72
Hence the final value of the product is 72.
We can also calculate the final value by using the formula
Total discount = (x + y – xy / 100)%
x = 10% and y = 20%
Total discount = (10 + 20 – (10)(20) / 100)%
= (30 – 200/100)% = 28%

Successive Discounts Examples

Example 1:
Store is offering a t-shirt at the price of Rs.950. Successive discounts offered by the store are 30% and 50%. Calculate the selling price and total discount offered by the store?

Solution:
Given that, the price of the t-shirt = Rs. 950
Successive discounts are 30% and 50%
Total discounts = (x + y – xy / 100)%
= (30 + 50 – (30)(50) /100)% = 80 – 1500/100% = 65%
Discount = 65% of Rs. 950 = 65/100 * 950 = Rs. 617.5
Therefore, the selling price of the shirt = Rs. 950 – 617.5 = Rs. 332.5
Hence, the total discount offered is Rs. 617.5

The selling price is Rs. 332.5

Example 2:
Successive discounts on the product are 5%, 10%, and 15%. The price of the product in the store is $1000. Calculate the overall selling price and discount of the product?

Solution:
Given that, the price of the product = $1000
Successive discounts are 5%, 10% and 15%
Total discount of 5% and 10% are (x + y – xy / 100)% = 5 + 10 – (5)(10) / 100% = 15 – (50)/100% = 14.5%
Overall discount due to 14.5% and 15% = 14.5 + 15 – (14.5) * (15) /100%
= 29.5 – (217.5)/100% = 27.325%
Discount = 27.325 of $1000 = $273. 25
Total selling price of the product = Store price – overall discount
= $1000 – 273.25 = $726.75
Hence, the total discount offered = $273.25

The total selling price of the product = $726.75

Example 3:
The price of the product is $2250. The successive discounts are 10% and 20%. Find the selling price?

Solution:
Given that, the price of the product is $2250
The successive discounts are 10% and 20%
Total discount = (x + y – xy / 100)%
x = 10% and y = 20%
Total discount = 10 + 20 – (10)(20) 100% = (30 -200/100)%
Total discount = 28%
Discount = 28% of 2250 = (28/100) * 2250
Discount = 630
Therefore, the discount earned is 630
Selling Price = Marked Price – Discount = 2250 – 630 = 1620
Selling Price = 1620

Therefore, Selling Price = 1620
Discount = 630

Example 4:
Successive discounts of 30% and 20% are offered by the trader. Find the total discount offered?

Solution:
Given that, the successive discounts are 30% and 20%
Suppose that MRP = 100
Discount 1 = 30% of 100
Discount = 30
Hence, the price is 100-30 = 70
Now, the discount amount is 20%
Discount = 20% of 70
Discount = 14
Hence, the final price is 70-14 = 56
Total discount offered by the trader = MRP – Selling Price
= 100 – 56 = 44

Hence, the total discount offered = Rs. 44

Example 5:
3 successive discounts of 50%, 20%, and 10% are offered by a trader. Find the total discount percent?

Solution:
Suppose that MRP = 100
Given that, First discount = 50% of 100
Discount = 50
Hence, the price = 100-50 = 50
Second discount = 20% of 50 = 10
Hence, the price = 50 – 10 = 40
Final Discount = 10% of 40 = 4
Hence, the total price = 40 – 4 = 36
Therefore, total discount = MRP – Selling Price = 100 – 36 = 64
Total discount = 64/100 * 100 = 64%

Hence the total discount offered = 64%

We have mentioned all the tricks and tips to solve successive discount problems. Practice all the questions and improve your skills in solving the problems on successive discounts.

Frequently Asked Questions on Successive Discounts

1. How are successive discounts calculated?

Suppose that successive discounts are d1, d2 and d3, then the selling price is SP = (1-d1/100) * (1-d2/100) * (1-d3/100) * Marked Price

2. Is 30 or 40 successive discounts better?

The discount of 70% is better than the successive discounts of 30% and 40%.

3. How do you calculate two successive discount formulas?

The two discount formula is (x + y – xy / 100)%
Where x is the first discount and y is the second discount

4. What is the meaning of successive discounts?

Successive discount is the discounted price on the already given discount which is similar to compound interest.

If you are wondering what is the difference between discounts and markup then you have come the right way? Here we will give you insight on discount and markups definition, formulae, and solved examples explaining step by step on how to approach. Follow the concept and know the real-time examples of discount and markup by checking the below sections for detailed information.

Also, check:

• Worksheet on Calculating Discount on Marked Price
• Worksheet on Discount Percent

What is meant by Discount?

Discount is the reduction in the price/rate of some product/item. The main purposes of providing discounts are:

• Increase in sales
• Clear the old stock
• Encourage distributors
• Reward potential customers

Discount is considered the easiest way to increase the product demand and plays an important role in online products. We can check amazing offers while shopping for various products. These offers mainly concentrate to attract customers which are named as discounts. It is considered as the value/price of the total amount/quantity which is generally less than the original value. In general words, we can tell that the total amount is sold at a particular discount to attract customers.

Discount Formula

The discount formula is as follows:
Discount = Marked Price – Selling Price
where Marked Price (M.P) is the actual value of the product without the discount value.
Selling Price is what the customers pay for the product.
Discount is the percentage of the marked price.

Markup

Markup is considered as the total profit or gross on a particular service or commodity. It is defined as the percentage over cost price. For suppose, if the product cost is Rs. 100 and its selling price is Rs. 150, then the markup will be 50%. It is also defined as the difference between CP(cost price) and SP(selling price) of the product. It estimates the profit and loss of the business.

What is meant by Markup Price?

Markup is the difference between the retail price of a commodity and cost. The amount that is added to the cost determines the retail prices of particular items. Markup is combined with total C.P to meet the business costs and generation of profit. Markup represents the percentage or fixed amount of selling price or cost price.

Markup Formula

As mentioned above, markup value is the difference between the SP(selling price) and CP(cost price) of the product.
Markup = Retail – Cost

Markup Percentage

The formula to calculate the markup percentage is:
Sale Price = Cost * (1 + Markup) or Markup = 100 * (Sale Price – Cost Price)/Cost

Markup and Discount Examples

Example 1.
Lara wanted to gift her mom a dress. The cost of the dress was Rs. 450 and it is marked as 20% off on the dress. How much is the discount amount?

Solution:
Given that, Cost of the dress = Rs. 450
Discount Percentage = 20%
Discount Amount = Original Price * Discount Rate
Discount Amount = 450 * 20% = 450 * (20/100) = 90
Therefore, the discount amount is Rs. 90
Hence the final cost of the dress is Cost Price – Discount Amount
= 450 – 90 = 340

The final cost is 340.

Example 2.
Soheal bought a watch which was originally Rs. 5000. He got a discount of 50% on the total amount. How much is the discount amount?

Solution:
Given that, the cost of the watch = Rs. 5000
Discount he got = 50%
Discount Amount = Original Price * Discount Rate
Discount Amount = 5000 * 50%
= 5000 * (50/100)
= 2500
Therefore, the discount amount is 2500
Hence the final cost of the watch is Cost Price – Discount Amount
= 5000 – 2500
= 2500

The final cost of the watch is 2500.

Example 3.
The original price of the book is Rs. 400. I got 20% on the total amount. How much is the discount amount?

Solution:
Given that, the cost of the book is Rs. 400
The discount I got = 20%
Discount Amount = Original Price * Discount Rate
Discount Amount = 400 * 20% = 400 * (20/100) = 80
Therefore, the discount amount is Rs. 80
Hence the final cost of the book is Cost Price – Discount Amount
= 400 – 80 = 320

The final cost of the book is Rs. 320

Example 4.
A car dealer advertises a 7% markup over cost. Find the selling price of the car that cost the dealer $13,000.

Solution:
Given that, Markup = 7% of 13,000 = (7/100) * 13000 = 910
The markup price is 910
Selling Price = Cost Price + Markup Price
=13,000 + 900
=13,900

The selling price of car = 13,900

Example 5.
A ring that costs the jeweler $360 sells for $630. Find the markup rate?

Solution:
Given that, Cost of the ring = $360
Selling Price = $630
Markup Price = Selling Price – Cost Price
= 630 – 360 = 270
The markup price = $270
Markup Percentage = Markup Price / Cost Price
= 270/360 = 75%

Therefore, the markup percentage = 75%

Example 6.
A person got a loan from a bank at a rate of 3% per year for some period. In how much period of time his loan of Rs. 65,000 will become Rs. 68,000.

Solution:
Given that, Principal amount = Rs. 65,000
Profit for bank = Rs. 68,000 – 65000 = 3900
Markup rate = 3/100per year
Profit per year = 650 * 3 = Rs. 1,950
The total amount of time period to clear a loan is 3900 – 1950 = 1950

Therefore, it takes 2 years to become Rs. 68,000 from Rs. 65,000 is 2 years.

Hence, the complete information is given on discount and markup. Check our page for more new concepts and information. Discount and Markup are used in day-to-day life and solve various problems of business.

Overhead expenses are business and other costs which are not related to direct materials, labor, and production. Overhead expenses are some of the indirect costs which are not related to particular business activities. Calculating the overhead expenses is not only important for budgeting but also to determine the charge or investment for a product or service. Suppose that you have a good business that is service-based. Apart from the direct investments or costs, indirect costs like insurance, rent, utilities are considered as overheads expenses.

To understand it better we will consider another example, ie., Suppose a person bought a TV at the cost price of Rs 12,000. Now, he took cable connection for the TV. He has to pay the cable bill every month which is considered as an overhead expense.

What are Overhead Expenses?

Overhead Expenses support the business but they do not generate any revenue. These expenses are mandatory and you have to pay them irrespective of your revenue. The main examples of overhead expenses are property taxes, utilities, office supplies, insurance, rent, accounting and legal expenses, advertising expenses, government licenses and fees, depreciation, and property taxes.

Types of Overhead Expenses

Among the overhead expenses, not all the expenses are the same or equal. These expenses are divided into 3 categories. Know the different expenses and their types which can create a meaningful budget for the business. The different types of overhead expenses are:

• Fixed Overhead
• Variable Overhead
• Semi-Variable Overhead

1. Fixed Overhead Expense

These expenses are something which won’t change from month to month. If a fixed overhead has to change then it changes only annually during the renewal period. Examples of fixed overhead are insurance, salaries, rent. These overhead expenses are easy to budget and plan. These fixed overheads are tough to reduce or restrict the cash flow.

2. Variable Overhead Expense

These expenses are mostly affected by business activities and not by sales. Some of the examples of variable overhead expenses are office supplies, legal expenses, repairs, advertising expenses, and maintenance expenses. It is no guarantee that office supplies will not change according to sales volume. In the same way, advertising expenses may increase during peak sales. The drawback of variable expense is that it is difficult to predict while budgeting.

3. Semi-Variable Expenses

These expenses are also not the same from month to month. These semi-variable expenses are also not completely unpredictable and some examples of these expenses are many utilities, hourly wages, some commissions, and vehicle expenses.

Also, See:

• Worksheet on Calculating Profit or Loss
• Worksheet on Calculating Overhead Charges

Calculating Overhead Rate

Calculating overhead rate is an important factor in the business. It determines the exact amount of sales that goes into overhead expenses. To calculate it, we have to add all the overhead expenses and divide that number with your sales. The formula of overhead rate is:

Overhead Rate = Overhead Expenses / Sales

Overhead Charges Examples

Example 1.
An industry estimated the factory overhead for the period of 10 years at 1,60,000. The estimation of materials produced for 40,000 units is 200,000. Production requires 40,000 hours of man work at the estimated wage cost of 80,000. Machines will run for 25,000 hours approximately. Calculate the overhead rate on each of the following bases:
i. Direct labor cost
ii. Machine hours
iii. Prime Cost

Solution:

To find the direct labour cost, machine hours and prime cost we have to calculate the overhead rate.
(i) Direct Labour Cost
= (Estimated Factory OverHead / Estimated Direct Labour Cost) * 100
= (1,60,000 / 80,000) * 100
= 200%
(ii) Machine hours
= (Estimated Factory Overhead / Estimated Machine hours)
= 1,60,000 / 25,000
= 6.40 per machine hour
(iii) Prime Cost Basis
= Estimated Factory Overhead / Estimated prime cost
= 1,60,000 / (2,00,000 + 80,000)) * 100
= 89%

Example 2.
A shopkeeper purchased a second hand car for Rs. 1,40,000. He spent Rs. 15,000 on its repair and painting and then sold it for Rs. 17,000. Find his profit or loss?

Solution:
Cost Price = Rs. 1,40,000
Overhead Charges = Rs. 15,000
C.P.N = (1,40,000 + 15,000) = Rs. 1,55,000
S.P = Rs. 17,000
Therefore, S.P > C.P
Hence, it is profit
Profit Percent = S.P – C.P
P = 170000 – 155000
P = Rs. 15000
P% = P / C.P.N * 100
P% = 15000/155000 * 100
P% = 300/31%
Therefore, the profit percentage is 300/31%

Example 3.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300. Determine his profit percentage?

Solution:
Cost Price of radio = Rs. 225
Overhead expenses = Rs. 15
Selling Price of radio = Rs. 300
Net Cost Price = Rs. 225 + 15 = 240
Profit % = Selling Price – Cost Price / Cost Price * 100
P% = 300 – 240 / 240 * 100
P% = 60/240 * 100
P% = 25%
Therefore, the profit percentage = 25%

Frequently Asked Questions on Overhead Costs

1. What are the examples of overhead expenses?

The examples of overhead expenses are interest, labor, advertising, insurance, accounting fees, travel expenditure, telephone bills, supplies, utilities, taxes, legal fees, repairs, legal fees, etc.

2. What are the types of overhead?

The types of overheads are fixed overhead expense, semi-variable overhead expense, and variable overhead expense.

3. What is the minimum percentage for overhead?

The minimal percentage is that it should not exceed 35% of the total revenue. In growing or small businesses, the overhead percentage factor is usually considered as the critical figure which is of concern.

4. How will overhead affect profit?

Overhead represents the supporting costs of production or service delivery. If there is an increase in overhead, it reduces profits by the exactly same amount.

Successor and Predecessor are the terms that are mentioned just before and after the number or term. Learn the definition, examples, and differences of successor and predecessor. Know the logic to find the successor and predecessor of the number. We are explaining the concept with images and solved examples to make it more clear and easy. Check it out!!

Do Refer: Successor and Predecessor of a Whole Number

Successor and Predecessor in Maths

Check out the What is the Successor and Predecessor from the below details. Also, find out different examples to understand deep about Predecessor and Successor.

What is Predecessor?

The predecessor is the value that comes immediately before/right before the particular value. Suppose that the particular value is x, then the predecessor value is before the value of that particular value i.e., x-1. Therefore, to find the predecessor value of any number, we have to subtract 1 from the given value.

Examples:

If x = 15, then the predecessor value of 15 is 15 – 1 = 14
If x = 21, then the predecessor value of 21 is 21 – 1 = 20
If x = 49, then the predecessor value of 49 is 49 – 1 = 48
If x = 90, then the predecessor value of 90 is 90 – 1 = 90
If x = 115, then the predecessor value of 115 is 115 – 1 = 114

Therefore, the predecessor of any number is one less than the original whole number.

What is Successor?

The successor is the value that comes immediately after/right after the particular value. Suppose that the particular value is x, then the successor value is after the value of that particular value i.e., x + 1. Therefore, to find the successor value of any number, we have to add 1 to the given value.

Examples:
If x = 15, then the successor value of 15 is 15 + 1 = 16
If x = 21, then the successor value of 21 is 21 + 1 = 22
If x = 49, then the successor value of 49 is 49 + 1 = 50
If x = 90, then the successor value of 90 is 90 + 1 = 91
If x = 115, then the successor value of 115 is 115 + 1 = 116

Therefore, the successor of any number is one greater than the original whole number.

How to find the Successor and Predecessor of a Number?

To find the predecessor and successor of any value, we apply the basic subtraction and addition methods.

To evaluate the successor and predecessor of any value, we have to apply the basic method of addition and subtraction, respectively. For the successor, we need to add 1 to the given number whereas for the predecessor we have to subtract 1 from the given number. Finding a successor and predecessor is very easy and quick.

• Successor = Given number + 1
• Predecessor = Given number – 1

Let us see some solved examples here to understand better.

Successor and Predecessor Examples

Example 1.
Find the successor of the following numbers:
(i) 15
(ii) -11
(iii) -85
(iv) 91
(v) 149
(vi) 44
(vii) 87
(viii) 78

Solution:
The successor values of the numbers are as follows:
(i) 15 + 1 = 16
The successor value of 15 is 16
(ii) -11 + 1 = -10
The successor value of -11 is – 10
(iii) -85 + 1 = -84
The successor value of -85 is -84
(iv) 91 + 1 = 92
The successor value of 91 is 92
(v) 149 + 1 = 150
The successor value of 149 is 150
(vi) 44 + 1 = 45
The successor value of 44 is 45
(vii) 87 + 1 = 88
The successor value of 87 is 88
(viii) 78 + 1 = 79
The successor value of 78 is 79

Example 2.
Find the predecessor of the following numbers:
(i) -15
(ii) -81
(iii) 65
(iv) -9
(v) 22
(vi) 198
(vii) 55

Solution:
The predecessor values of the numbers are as follows:
(i) -15 – 1 = -16
The predecessor value of -15 is -16
(ii) -81 – 1 = -82
The predecessor value of -81 is -82
(iii) 65 – 1 = 64
The predecessor value of 65 is 64
(iv) – 9 – 1 = -10
The predecessor value of -9 is -10
(v) 22 – 1 = 21
The predecessor value of 22 is 21
(vi) 198 – 1 = 197
The predecessor value of 198 is 197
(vii) 55 – 1 = 54
The predecessor value of 55 is 54

Example 3.
Write the successor and predecessor of the following numbers:
(i) 94
(ii) 114
(iii) 32
(iv) 65
(v) 78

Solution:
The successor and predecessor of the numbers are as follows:
(i) The number is 94
Successor value of 94 is 94 + 1 = 95
Predecessor value of 94 is 94 – 1 = 93
(ii) The number is 114
Successor value of 114 is 114 + 1 = 115
Predecessor value of 114 is 114 – 1 = 113
(iii)The number is 32
Successor value of 32 is 32 + 1 = 33
Predecessor value of 32 is 32 – 1 = 31
(iv)The number is 65
Successor value of 65 is 65 + 1 = 66
Predecessor value of 65 is 65 – 1 = 64
(v)The number is 78
Successor value of 78 is 78 + 1 = 79
Predecessor value of 78 is 78 – 1 = 77

FAQs on Successor and Predecessor

1. What is the main difference between successor and predecessor

The successor is the number that comes after the original number. The predecessor is the number that comes before the original number.

2. Is there any natural number available that has no predecessor value?

Yes, there is a natural number that has no predecessor value i.e., 1. The reason for this is that natural numbers start from 1.

3. Is there any natural number available that has no successor value?

No, there is no natural number that has no successor value. As the natural numbers are infinite and it has no last value.

4. What is the formula for successor and predecessor numbers?

The formula for successor and predecessor are

• Successor = Original number + 1
• Predecessor = Original number – 1