## Factorize the Trinomial x^2 + px + q | How to Find Factorization of Trinomial of the Form x^2 + px + q?

Learn How to Factorize the Trinomial x2 + px +q? A Trinomial is a three-term algebraic expression. By Factoring the trinomial expression, we will get the product of two binomial terms. Here, the trinomial expression contains three terms which are combined with the operations like addition or subtraction. Here, we need to find the coefficient values, and based on the coefficient values, we can find out the binomial terms as products of trinomial expression.

## How to Factorize the Trinomial x2 + px + q?

To find x^2 + px + q, we have to find the two terms (m + n) = p and mn = q.
Substitute (m + n) = p and mn = q in x^2 + px + q.
x^2 + px + q = x^2 + (m + n)x + mn.
By expanding the above expression, we will get
x^2 + px + q = x^2 + mx + nx + mn.
separate the common terms from the above expression.
that is, x(x + m) + n(x + m).
factor out the common term.
that is, (x + m) (x + n).
So, x^2 + px + q = (x + m)(x + n).

### Factorization of Trinomial Steps

• Note down the given trinomial expression and compare the expression with the basic expression.
• Find out the product and sum of co-efficient values that is (m + n) and mn.
• Based on the above step, find out the two co-efficient values m and n.
• Finally, we will get the product of two terms which are equal to the trinomial expression.

### Examples on Factoring Trinomials of Form x2 + px + q

1. Resolve into factors

(i) a2 + 3a -28

Solution:
Given Expression is a2 + 3a -28.
Compare the a2 + 3a -28 with the x^2 + px +q
Here, p = m + n = 3 and q = mn = -28
q is the product of two co-efficient. That is, 7 *(- 4) = -28
p is the sum of two co-efficient. That is 7 + ( – 4) = 3.
So, a2 + 3a -28 = a2 + [7 + (-4)]a – 28.
= a2 + 7a – 4a – 28.
=a (a + 7) – 4(a + 7)
Factor out the common term.
That is, (a + 7) (a – 4).

Finally, the expression a2 + 3a -28 = (a + 7) (a – 4).

(ii) a2 + 8a + 15

Solution:
Given Expression is a2 + 8a + 15.
Compare the a2 + 8a + 15 with the x^2 + px +q.
Here, p = m + n = 8 and q = mn = 15.
q is the product of two co-efficient. That is, 5 * 3 = 15.
p is the sum of two co-efficient. That is 5 + 3 = 8.
So, a2 + 8a + 15 = a2 + (5 + 3)a + 15.
= a2 + 5a + 3a + 15.
=a (a + 5) + 3(a + 5).
Factor out the common term.
That is, (a + 5) (a + 3).

Finally, the expression a2 + 8a + 15= (a + 5) (a + 3).

2. Factorize the Trinomial

(i) a2 + 15a + 56

Solution:
Given Expression is a2 + 15a + 56.
Compare the a2 + 15a + 56 with the x^2 + px +q.
Here, p = m + n = 15 and q = mn = 56.
q is the product of two co-efficient. That is, 7 * 8 = 56.
p is the sum of two co-efficient. That is 7 + 8 = 15.
So, a2 + 15a + 56= a2 + (7 + 8) a + 56.
= a2 + 7a + 8a + 56.
=a (a + 7) + 8(a + 7).
Factor out the common term.
That is, (a + 7) (a + 8).

Finally, the expression a2 + 15a + 56= (a + 7) (a + 8).

(ii) a2 + a – 56

Solution:
Given Expression is a2 + a – 56.
Compare the a2 + a – 56 with the x^2 + px +q.
Here, p = m + n = 1 and q = mn = – 56.
q is the product of two co-efficient. That is, – 7 * 8 = – 56.
p is the sum of two co-efficient. That is – 7 + 8 = 1.
So, a2 + a – 56 = a2 + ( – 7 + 8)a – 56.
= a2 – 7a + 8a – 56.
=a (a – 7) + 8(a – 7).
Factor out the common term.
That is, (a – 7) (a + 8).

Finally, the expression a^2 + a – 56 = (a – 7) (a + 8).

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## Word Problems on Four-Digit Numbers | 4 Digit Numbers Problems with Solutions

If you are searching for four-digit number word problems, you have landed on the correct web page that gives the information regarding how to solve word problems on four-digit numbers. You can also see the solved examples of word problems on four-digit numbers. Try to practice using these 4 Digit Problems and test your knowledge and improvised on the area accordingly.

## Word Problems on Four-Digit Numbers

Here we will solve some of the word problems on 4-digit numbers. Apply the same method for solving the word problems on 4-digit numbers.

Example 1:

In a school, there are 4530 boys and 6890 girls. How many students are there in the school?

Solution:

Number of boys in the school=4530

Number of girls in the school=6890

Therefore, the total number of students in the school=4530+6890=11,420.

Example 2:

In a garden, there are 5200 red roses, 2040 white roses, and 1000 orange roses. How many roses are there in the garden?

Solution:

Number of red roses=5200

Number of White roses=2040

Number of orange roses=1000

Total number of roses in the garden=5200 + 2040+1000=8,240.

Example 3:

In a village, there are 6050 males and 5678 females. What is the population in that village?

Solution:

Number of males in the village=6050

Number of females in the village=5678

The total population in that village=11,728.

Example 4:

In a library, there are 5420 computer books,3560 chemistry books, and 4289 English literature books. How many books are there in the library?

Solution:

Number of computers books=5420

Number of chemistry books=3560

Number of English books=4289

Therefore, the Total no of books in the library=5420+3560+4289=13,269.

Example 5:

In a School, there are 6000 children. If 3250 are boys, How many are girls?

Solution:

Total no of children in the school=6000

Total no of boys=3250

Therefore, the total no of girls=6000-3250=2750.

Example 6:

There are 8560 rice bags in the godown.5200 are taken out for distribution. How many bags are left in the godown?

Solution:

Total no of rice bags in the godown=8560

No of rice bags taken out for distribution=5200

Therefore, The total No of rice bags that are left=8560-5200=3,360.

Example 7:

There are 5000 rice bags and 6040 wheat bags in the godown.2000 rice bags are taken out for distribution. How many bags are there in the godown?

Solution:

No of rice bags in the godown=5000

No of rice bags taken out for distribution=2000

Total no of rice bags in the godown=5000-2000=3000

no of Wheat bags in the godown=6040

Total no of bags in the godown=3000+6040=9040.

Example 8:

There are two friends. one friend has 1050 rice bags and the other friend has 3050 wheat bags. Two friends contain how many bags?

Solution:

No of rice bags=1050

No of wheat bags=3050

Two friends contain bags=1050+3050=4,100.

Example 9:

If two four-digit numbers are added the sum is 4000. One number is 1000. Find out the other number?

Solution:

The sum of the two numbers = 4000.

one number=1000.

The Other number=4000-1000=3000.

Example 10:

In a village, there is a 5000 population.2030 are doing jobs and gone out of the village. How much population does the village have?

Solution:

no of population=5000

no of people going out of the village=2030

Total no of the population in the village=5000-2030=2970.

Example 11:

In an election, the number of votes polled is 8000.200 people are not voted due to various reasons. How many people are there in the village?

Solution:

People who are not voted=200

Total no of people in the village=8000+200=8200.

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## Factorization of Perfect Square Trinomials | How to Factor Perfect Square Trinomials?

If you are searching for help on Factorization of Perfect Square Trinomials you are the correct place. Check out all the problems explained in this article. We have given a detailed explanation for every individual problem along with correct answers. Also, we are providing all factorization concepts, worksheets on our website for free of cost. Therefore, learn every concept and improve your knowledge easily.

Learn to solve the given algebraic expressions using the below formulas.
(i) a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b)
(ii) a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b)

## Factoring Perfect Square Trinomials Examples

1. Factorization when the given expression is a perfect square

(i) m4 – 10m2n2 + 25n4

Solution:
Given expression is m4 – 10m2n2 + 25n4
The given expression m4 – 10m2n2 + 25n4 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = m2, b = 5n2
Apply the formula and substitute the a and b values.
m4 – 10m2n2 + 25n4
(m2)2 – 2 (m2) (5n2) + (5n2)2
(m2 – 5n2)2
(m2 – 5n2) (m2 – 5n2)

Factors of the m4 – 10m2n2 + 25n4 are (m2 – 5n2) (m2 – 5n2)

(ii) b2+ 6b + 9

Solution:
Given expression is b2+ 6b + 9
The given expression b2+ 6b + 9 is in the form a2 + 2ab + b2.
So find the factors of given expression using a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) where a = b, b = 3
Apply the formula and substitute the a and b values.
b2+ 6b + 9
(b)2 + 2 (b) (3) + (3)2
(b + 3)2
(b + 3) (b + 3)

Factors of the b2+ 6b + 9 are (b + 3) (b + 3)

(iii) p4 – 2p2 q2 + q4

Solution:
Given expression is p4 – 2p2 q2 + q4
The given expression p4 – 2p2 q2 + q4 is in the form a2 – 2ab + b2.
So find the factors of given expression using a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) where a = p2, b = q2
Apply the formula and substitute the a and b values.
p4 – 2p2 q2 + q4
(p2)2 – 2 (p2) (q2) + (q2)2
(p2 – q2)2
(p2 – q2) (p2 – q2)
From the formula (a2 – b2) = (a + b) (a – b), rewrite the above equation.
(p + q) (p – q) (p + q) (p – q)

Factors of the p4 – 2p2 q2 + q4 are (p + q) (p – q) (p + q) (p – q)

2. Factor using the identity

(i) 25 – a2 – 2ab – b2

Solution:
Given expression is 25 – a2 – 2ab – b2
Rearrange the given expression as 25 – (a2 + 2ab + b2)
a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b)
25 – (a + b)2
(5)2– (a + b)2
From the formula (a2 – b2) = (a + b) (a – b), rewrite the above equation.
[(5 + a + b)(5 – a – b)]

(ii) 1- 2mn – (m2 + n2)

Solution:
Given expression is 1- 2mn – (m2 + n2)
1- 2mn – m2 – n2
1 – (2mn + m2 + n2)
1 – (m + n)2
(1)2 – (m + n)2
From the formula (a2 – b2) = (a + b) (a – b), rewrite the above equation.
(1 + m + n) (1 – m + n)

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## Problems on Calculating Time | Time Word Problems with Solutions

Work with the Problems on Calculating Time and learn how to find Time when Speed and Distance are given. Know the Relationship between Speed Time and Distance with the formula provided. Check Worked out Examples for Distance with Solutions and cross-check your solutions while practicing. Check different types of Time Questions followed by illustrations for a better understanding of the concepts. Practice each of the Time Word Problems provided and score better grades in the exam.

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### Solved Time Questions with Answers

1. A car travel 70 km in 30 minutes. In how much time will it cover 120 km?

Solution:

Speed = Distance/Time

= 70 km/1/2 hr

= 140 kmph

Speed = Distance/Time

140 kmph = 120 km/Time

Time = 120 km/140 kmph

= 0.85 hr

car takes 0.85 hr to cover the distance 120 km.

2. Vinay covers 180 km by car at a speed of 60 km/hr. find the time taken to cover this distance?

Solution:

Speed = 60 km/hr

Distance = 180 km

Speed = Distance/Time

60 km/hr = 180 km/Time

Time = 180 km/60 km/hr

= 3 hr

Vinay takes 3 hrs to cover the distance 180 km at a speed of 60 km/hr.

3. A train covers a distance of 45 km in 20 minutes. Find the time taken by it to cover the same distance if its speed is decreased by 12 km/hr?

Solution:

Distance covered by train = 45 km

Time taken = 20 min = 20/60 = 1/3 hr

Speed of the train = Distance covered/Time taken

= 45km/1/3 hr

= 135 km/hr

Reduced Speed = 135 km/hr – 12 km/hr

= 123 km/hr

Time = Distance Traveled/Speed

= 45/123

= 0.365hr

= 0.365*60 min

= 21.95 min

4. A man is walking at a speed of 5 km per hour. After every km, he takes a rest for 3 minutes. How much time will it take to cover a distance of 6 km?

Solution:

Rest time = Number of rests * time of each rest

= 5*3 minutes

= 15 minutes

Total Time Taken = Distance/Speed +Rest Time

= (6/5)*60+15 minutes

= 72+15

= 87 minutes

5. A car takes 3 hours to cover a distance if it travels at a speed of 45 mph. What should be its speed to cover the same distance in 2 hours?

Solution:

Distance = Speed *Time

= 45mph*3 hr= 135 miles

Speed = Distance/Time

= 135 miles/2 hrs

= 67.5 miles per hour

Therefore, car needs to travel at a speed of 67.5 miles per hour in order to travel a distance of 135 miles in a time of 2 hrs.

Mcx Calculator

## Decimal Place Value Chart: Definition, How to Write, and Examples

Mathematics is the study of numbers, shapes, and patterns. It includes various complex and simple arithmetic topics that help people in their daily life routine. In Maths, Numbers play a major role and they can be of different types like Real Numbers, Whole Numbers, Natural Numbers, Decimal Numbers, Rational Numbers, etc. Today, we are going to discuss one of the main topics of Decimal Numbers. In Decimals, identifying the Decimal Place Values is a fundamental topic and everyone should know the techniques clearly. So, here we will be discussing elaborately the topic of Decimal Place Values Chart.

Let’s get into it.

## What is a Decimal in Math?

In algebra, a decimal number can be represented as a number whose whole number part and the fractional part is divided by a decimal point. The dot in a decimal number is called a decimal point. The digits following the decimal point show a value smaller than one.

### What is the Place Value of Decimals?

Place value is a positional notation system where the position of a digit in a number, determines its value. The place value for decimal numbers is arranged exactly the identical form of treating whole numbers, but in this case, it is reverse. On the basis of the preceding exponential of 10, the place value in decimals can be decided.

### Decimal Place Value Chart

On the place value chart, the numbers on the left of the decimal point are multiplied with increasing positive powers of 10, whereas the digits on the right of the decimal point are multiplied with increasing negative powers of 10 from left to right.

• The first digit after the decimal represents the tenths place.
• The second digit after the decimal represents the hundredths place.
• The third digit after the decimal represents the thousands place.
• The rest of the digits proceed to fill in the place values until there are no digits left.

### How to write the place value of decimals for the number 132.76?

• The place of 6 in the decimal 132.76 is 6/100
• The place of 7 in the decimal 132.76 is 7/10
• The place of 2 in the decimal 132.76 is 2
• The place of 3 in the decimal 132.76 is 30
• The place of 1 in the decimal 132.76 is 100.

#### Examples:

1.  Write the place value of digit 7 in the following decimal number: 5.47?

The number 7 is in the place of hundredths, and its place value is 7 x 10 -2 = 7/100 = 0.07.

2. Identify the place value of the 6 in the given number: 689.87?

Given number is 689.87

The place of 6 in the decimal 689.87 is 600 or 6 hundreds.

3. Write the following numbers in the decimal place value chart.

(i) 4532.079

(ii) 490.7042

Solutions:

(i) 4532.079

4532.079 in the decimal place value chart.

(ii) 490.7042

490.7042 in the decimal place value chart.

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## Understanding Overheads Expenses – Definition, Types, Examples | How do you Calculate Overhead Expenses?

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.

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.

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:

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.

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:

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:

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
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
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%

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?

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.

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## 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.

Also, find:

## 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.

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%

## Dividend and Rate of Dividend – Definition, Formula, Examples | How to Calculate Dividend and Dividend Rate?

Students who are in search of the Dividend and Rate of Dividend examples can get them on this page. The profit which a shareholder gets for its investment from the organization is known as a dividend. Know what is dividend and what is rate of dividend from here. Let us discuss in detail the dividend and rate of dividend like definitions, formulas, procedures on how to calculate dividend, dividend rate, etc. So, refer to our page to improve your math skills and also to gain good marks in the exams.

## Dividend and Rate of Dividend – Definitions

Dividend plays an important role in starting a business or company. The share of the annual profit of a company shared among its shareholders is known as a dividend. The rate of dividend is expressed as a percentage of the face value is called the rate of dividend.

### Difference Between Dividend Rate Vs. Dividend yield

Dividend and dividend rates both are not the same. The dividend rate is the amount of share that is obtained from the shareholders such as mutual funds, stock market, etc. Whereas the dividend is the share of the profit among the shareholders.

### Dividend and Rate of Dividend Formulas

• Dividend Rate = Divided per share/current price
• Dividend Yield = Dividend per share/ Market value per share

Refer More:

### How to Calculate Dividend Rate?

The estimation of the dividend rate of a venture, asset, or portfolio includes duplicating the latest intermittent dividend installments by the number of installment periods in a single year.

### Dividend and Rate of Dividend Question and Answers

Example 1.
130 shares of Rs 30 each paying 10% dividend.
Solution:
Number of shares = 130
Price of each share = 30
Therefore, total investment = Rs(30 × 130) = 1300
Dividend = 10%
Hence annual income = 10×1300/100 = 130

Example 2.
40 shares of Rs 200 each available at Rs 35 and playing 6% dividend.
Solution:
Number of shares = 50
Price of each share = 200
Face value of 40 shares = Rs( 200 × 40) = 8000
Dividend = 6%
Hence annual income = 6 × 8000/100 = 480

Example 3.
180 shares of Rs 50 each paying 15% dividend.
Solution:
Number of shares = 180
Price of each share = 50
Therefore, total investment = Rs(50 × 180) = 9000
Dividend = 15%
Hence annual income = 15×9000/100 = 1350

Example 4.
70 shares of Rs 60 each available at Rs 65 and playing 4% dividend.
Solution:
Number of shares = 50
Price of each share = 60
Face value of 70 shares = Rs( 60 × 70) = 4200
Dividend = 4%
Hence annual income = 4 × 4200/100 = 168

Example 5.
260 shares of Rs 60 each paying 2% dividend
Solution:
Number of shares = 260
Price of each share = 60
Therefore, total investment = Rs(60 × 260) = 15,600
Dividend = 2%
Hence annual income = 2×15,600/100 = 312

### FAQs on Dividend and Rate of Dividend

1. What is the rate of dividend?

The dividend rate is expressed as the percentage or yield, which is a financial ratio that shows how much a company pays out in dividends each year relative to its stock price. The dividend payout ratio is one way to assess the sustainability of a company’s dividends.

2. What is the difference between dividend rate and dividend yield?

The dividend rate is another way to say dividend, which is the dollar amount of the dividend paid on a dividend-paying stock. The dividend yield is the percentage relationship between the stock’s current price and the dividend currently paid.

3. What is the dividend rate per share?

Dividend per share is the sum of declared dividends issued by a company for every ordinary share outstanding. DPS is calculated by dividing the total dividends paid out by a business, including interim dividends, over a period of time, usually a year, by the number of outstanding ordinary shares issued.

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## Volume of Cuboid – Definition, Facts, Formula, Examples | How to Calculate Volume of Cuboid?

I think all the students of grade 9 are familiar with the solid figures. Here on this page, we will discuss in-depth the volume of cuboids like what the volume of cuboids exactly is, formulas, and how to calculate the volume of cuboids. Therefore the students who are looking forward to knowing about the volume of the cuboid can make use of our page and prepare well for the exams. In addition to this, you can find examples of the volume of the cuboid here.

## What is the Volume of Cuboid?

Volume is the quantity that is used to measure solid figures like cuboids. A cuboid is a three-dimensional geometric figure. Simply we can say that a cuboid is a rectangular three-dimensional figure. In a rectangular cuboid, all the angles are right angles. A cuboid has 12 edges, 6 faces, and 8 vertices. A cuboid has three dimensions such as length, breadth, and height.

So, we can measure the space occupied by the cuboid by using the volume formula. The units of volume of a cuboid are cubic units. The volume of a cuboid completely depends upon the length, breadth, and height of the object.

### Formula of Volume of Cuboid

The unit of volume of cuboids is cubic units like cu. meter, cu. cm, cu. inches etc.
The formula of volume of cuboid is l × b × h
where,
l = length
h = height

### Volume of Cuboid Prism | Volume of Rectangular Prism

A cuboid or rectangular prism is the same. A cuboid has six faces, eight vertices, and 12 edges. We can say that a cuboid is a solid rectangle. The top and bottom surfaces are the same in the cuboid. We can find the volume of the cuboid prism using the formula.
The volume of a rectangular prism or cuboid prism = length × breadth × height

### How to find Volume of Cuboid?

We know that the volume of the cuboid can be calculated using the dimensions of the given figure. So, to find the volume of the cuboid we have to follow some steps. The scenario to calculate the volume of the cuboid is shown below.
Step 1: Check the dimensions of the given cuboid.
Step 2: Check the length, breadth, and height and see whether all the units are the same.
Step 3: If the units are not the same then convert them and make them into the same units.
Step 4: And then apply the volume of the cuboid formula.
Step 5: At last write the obtained value and write it in cubic units.

Also, Check:

### Volume of Cuboid Questions

Check out the problems given below to know how to find the volume of a cuboid. If you learn these problems the students of 9th grade can solve any type of problem from the topic Volume of Cuboid.

Example 1.
Find the volume of a cuboid of dimensions 2 cm × 4 cm × 6 cm.
Solution:
Given that

Length = 2 cm
Height = 6 cm
We know that
The volume of cuboid = length × breadth × height.
Volume of cuboid = 2 cubic cm × 4 cubic cm × 6 cubic cm.
= 48 cubic cm.
Therefore, the volume of the cuboid = 48

Example 2.
A water tank is 10 cm long, 15 cm broad and 5 cm high. What is the volume of a water tank in cubic cm?
Solution:
Given that,

The length of the water tank = 10cm
The breadth of the water tank = 15 cm
The height of the water tank = 5 cm
We know that,
The volume of cuboid = length × breadth × height.
Volume of cuboid = 10 × 15 × 5 cuboid cm
Therefore, the volume of water tank = 750

Example 3.
Kiran made a bangle box for his sister with a length of 12 cm, breadth of 34 cm, and height of 18 cm. Find the volume of the bangle box.
Solution:
Given that,

Kiran made a bangle box for his sister of length = 12 cm
Kiran made a bangle box for his sister of breadth = 34 cm
Kiran made a bangle box for his sister of height = 18 cm
The volume of the bangle box = Length × breadth × height
= 12 × 34 × 18
= 7344cu cm.
Therefore, the volume of a bangle box = 7344 cu cm

Example 4.
Find the number of cubical boxes of 4 cm which can be accommodated in a carton of dimensions 24 cm × 6 cm × 8 cm.
Solution:
Given that,
Side of a box = 4
Volume of box = side × side × side.
= 4 × 4 × 4
= 64 cu. cm.
Volume of carton = length × breadth × height.

= 24 × 6 × 8
=1152 cu. cm.
A number of boxes = Volume of carton/Volume of each box.
= 1152/64
Therefore, the number of cubical boxes = 18.

Example 5.
How many bricks each 15 cm long, 3 cm wide, and 7cm thick will be required for a wall 2 m long, 22 m high, and 6 m thick? If bricks sell at $600 per thousand what will it cost to build the wall? Solution: Given that Length of the brick = 15 cm Width of the brick = 3 cm Thick of the brick = 7 cm Volume of the wall = 15 m × 3 m × 7 m = 2 × 100 cm × 22 × 100 cm × 6 × 100 cm = 264000000 Volume of brick = 15 cm × 3 cm × 7 cm = 315 Number of bricks = Volume of the wall/Volume of the brick = 200 × 2200 × 600/15 × 3 × 7 = 264000000/315 The number of bricks = 838095 The cost of 1 thousand bricks =$ 600
The cost of building the wall = $600 × 838095 =$ 502857000

### FAQs on Volume of Cuboid

1. How do we define the volume of a cuboid?

A volume of cuboids is defined as the amount of space occupied by the cuboid surface in a three-dimensional figure.

2. If the units of dimensions of a cuboid are different, then how to find the volume?

If the units of length, width, and height are different then, we need to convert them into the same unit first and then find the volume.

3. What is the formula for the volume of cube and cuboid?

The volume of the cuboid is the product of length, width, and height.
Volume of cuboid = lbh
Volume of cube = s × s × s

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## Mean of Ungrouped Data – Definition, Formula, Explanation, Examples | How to Calculate Arithmetic Mean of Raw Data

Mean of Ungrouped Data: Ungrouped data is the type of distribution where individual data is presented in a raw form. The mean of data shows how the data are scattered throughout the central part of the distribution. Hence, the arithmetic numbers are called the measures of central tendencies.

Here, the mean is also known as the arithmetic mean or average of all the observations in the data. In this article, we will be explaining what is the mean of ungrouped data, the formula to find the mean for ungrouped data, steps to calculate mean deviation for raw data, some practice Examples on Mean of Arrayed Data.

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## Mean of Raw Data or Arrayed Data or Ungrouped Data

In Statistics, Mean is nothing but the measurement of average and it defines the central tendency of a given set of data.

In short, the mean of the given data set is estimated by adding all the observations and then dividing by the total number of observations.

The mean of ungrouped data is denoted by the mathematical symbol or notation ie, $$\overline{x}$$

The mean of the ungrouped data or arrayed data when it is raw can be measured by utilizing the following formula:

The mean of n observations (variables) x1, x2, x3, x4, ….., xn is given by the formula:

Mean = (x1+ x2 + x3 + x4 +…..+ xn ) / n = ∑xi / n

where ∑xi = x1+ x2 + x3 + x4 +…..+ xn

For instance, let’s take the scores of 10 students are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.

Hence, the mean scores of 10 students = ∑xi / n

= 5+10+15+20+25+30+35+40+45+50 / 10 = 27.5(approx).

### How to Find Assumed Mean of Ungrouped Data?

Want to know more about the assumed mean of ungrouped data? Please have a look at the below stuff and understand the concept of it clearly.

In the method of assumed mean, the values that are taken from the data or not can be used as assumed mean. Yet, it must be centrally positioned in the data so that to determine the mean of the given data via easy calculations.

The formula for the Assumed mean of ungrouped data is A+ sd/N

Where, A is the assumed mean,
sd is the summation of X-A for all figures, and
N is the frequency or the number of elements in the given data.

Apply the formula directly and calculate the assumed mean of the given data with ease and confidence.

### Steps to Determine the Mean Deviation for Ungrouped data

The following steps are mainly helpful for all students to calculate the mean for ungrouped data. Simply have a look at them and solve the arithmetic mean of raw data easily.

Let x1, x2, x3, x4, ….., xn observations consist in the given set of data.

Step 1: In the first step, we have to find out the mean deviation of the measure of central tendency. Assume that the measure is a.
Step 2: Find the absolute deviation of each variable from the measure of central tendency which is obtained in step 1 ie.,
|x1 – a|, |x2 – a|, |x3 – a|, …., |xn – a|
Step 3: Estimate the mean of all absolute deviations. At last, it provides the mean absolute deviation (M.A.D) about a for ungrouped data ie.,

If the central tendency measure is mean then the resulted equation can be rewritten as:
Where, $$\overline{x}$$ is the mean.

Let’s understand these calculating steps very clearly by practicing with the solved mean of ungrouped data questions with answers.

### Mean of Ungrouped Data Example Problems

Example 1:
In the competition of banana eating, the number of bananas consumed by 7 contestants in an hour is as follows: 10, 13, 16, 19, 22, 25, 30. Find the mean deviation from the mean of the given raw data.
Solution:
Given the number of bananas eaten by 7 contestants are 10, 13, 16, 19, 22, 25, 30
Let’s apply the above steps for finding the M.A.D about the mean.
Step 1: The mean of the following data can be given by,
$$\overline{x}$$ = $$\frac { 10+13+16+19+22+25+30 }{ 7 }$$
= $$\frac { 135 }{ 7 }$$ = 19.2(appox)
Step 2: Now find the absolute deviation around each observation,
|x1 – $$\overline{x}$$| = |10-19| = 9
|x2 – $$\overline{x}$$| = |13-19| = 6
|x3 – $$\overline{x}$$| = |16-19| = 3
|x4 – $$\overline{x}$$| = |19-19| = 0
|x5 – $$\overline{x}$$| = |22-19| = 3
|x6 – $$\overline{x}$$| = |25-19| = 6
|x7 – $$\overline{x}$$| = |30-19| = 11
Step 3: Finally, calculate the mean deviation for ungrouped data by using the following formula:
M.A.D(x) = ∑ni=1|xi−a| / n
= $$\frac { 9+6+3+0+3+6+11 }{ 7 }$$
= $$\frac { 38 }{ 7 }$$ = 5.428

Example 2:
The mean length of ropes in 20 coils is 12 m. A new coil is added in which the length of the rope is 16 m. What is the mean length of the ropes now?
Solution:
Given that, the mean length of ropes in 20 coils is 12 m. Let’s find the sum of length for each rope using mean formula:
Mean(length) A = x1+ x2 + x3 + x4 +…..+ x20 / 20
⟹ 12 = x1+ x2 + x3 + x4 +…..+ x20 / 20
⟹ x1+ x2 + x3 + x4 +…..+ x = 240 …….(i)
Now, add one coil and find the mean of new coils of rope,
A = x1+ x2 + x3 + x4 +…..+ x20 + x21 / 21
Here, length of new rope is 16m and use equation (i)
= x1+ x2 + x3 + x4 +…..+ x20 + x21 / 21
= $$\frac { 240 + 16}{ 21 }$$
= $$\frac { 256 }{ 21 }$$
= 12.19 (Appox)
Hence, the required new mean length is 12.19 m approximately.

Example 3:
The ages in years of 6 teachers of a school are 32, 28, 54, 40, 65, 20. What is the mean age of these teachers?
Solution:
Mean age of the teachers = $$\frac { Sum of the age of teachers}{ Number of teachers }$$
= $$\frac { (32+28+54+40+65+20) }{ 6 }$$
= $$\frac { 239 }{ 6 }$$
= 39.8 (approx) years.

## BODMAS/PEMDAS Rules – Involving Decimals | Order of Operations with Decimals Questions and Answers

We can easily simplify the arithmetic expression which involves decimals by using the BODMAS Rules or PEMDAS Rules. Some children will get confused while simplifying an arithmetic expression but by using the BODMAS rule, they can simply solve the expressions. Check out the Order of Operations and solved examples of BODMAS Rules Involving Decimals in this article. We have clearly given questions and answers along with the explanations for your best practice.

Also, Check:

## Order of Operations Involving Decimals Questions and Answers

Example 1.

First priority for Bracket terms: Solve inside the Brackets/parenthesis before Of, Multiply, Divide, Add or Subtract.
For example 1.2 + (1.5 – 2.3).

Solution:

The given expression is 1.2 + (1.5 – 2.3).
1.2 + (1.5 – 2.3) = 1.2 + (- 0.8) (subtraction of bracket term).
= 1.2 – 0.8.
= 0.4.
Finally, 1.2 + (1. 5 – 2. 3) is equal to 0.4.

Example 2.

Order terms are second priority: Then, solve Of part (Exponent, Powers, Roots, etc.,) before Multiply, Divide, Add or Subtract.
For example 0.2 X (1.2 + 1.4) + (0.2)^2.
Solution:

The given expression is 0.2 X (1.2 + 1.4) + (0.2)^2.
0.2 X (1.2 + 1.4) + (0.2)^2 = 0.2 X 2.6 + (0.2)^2 ( bracket terms addition first).
= 0.2 X 2.6 + 0.04 (order term (0.2)^2 = 0.04).
= 0.52 + 0.04 (multiplication 0.2 X 2.6 = 0.52).
= 0.56 (addition 0.52 + 0.04 = 0.56).
Therefore, 0.2 X (1.2 + 1.4) + (0.2)^2 is equal to 0.56.

Example 3.

Division and Multiplication: Then, calculate Multiply or Divide before Add or Subtract start from left to right.
For example 1.4 + 1.6 ÷ 0.2 X 0.2.
Solution:

The given expression is1.4 + 1.6 ÷ 0.2 X 0.2.
1.4 + 1.6 ÷ 0.2 X 0.2 = 1.4 + 8 X 0.2 (division 1.6 ÷ 0.2 = 8).
= 1.4 + 0.4 (multiplication 8 X 0.2 = 0.4).
= 1.8 (addition 1.4 + 0.4 = 1.8).
Therefore, 1.4 + 1.6 ÷ 0.2 X 0.2 is equal to 1.8.

Example 4.

Addition and Subtraction: At last Add or Subtract start from left to right.
For example 1.5 + (3.2 – 1.6) + 0.1.
Solution:

The given expression is 1.5 + (3.2 – 1.6) + 0.1.
1.5 + (3.2 – 1.6) + 0.1 = 1.5 + 1.6 + 0.1 (subtraction of bracket terms 3.2 – 1.6 = 1.6).
= 3.2 (addition 1.5 + 1.6 + 0.1 = 3.2).
Therefore, 1.5 + (3.2 – 1.6) + 0.1 is equal to 3.2.

### BODMAS Rules Involving Decimals Questions and Answers

Problem 1.

Simplify the below expressions by using the BODMAS rules.
(i) 0.4 + 0.8 ÷ 0.2 + 0.2 X 0.4.
(ii) 0.2 X 0.5(1.2 + 2.1) + 2.5.
(iii) 0.6 – 0.2 X 1.2 + 1.6.
(iv) 1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2.
(v) 0.8 + 1.2 X 0.5 + 1.2.

Solution:

(i) 0.4 + 0.8 ÷ 0.2 + 0.2 X 0.4.
Solution:
The given expression is 0.4 + 0.8 ÷ 0.2 + 0.2 X 0.4.
0.4 + 0.8 ÷ 0.2 + 0.2 X 0.4 = 0.4 + 4 + 0.2 X 0.4 (division 0.8 ÷ 0.2 = 4).
= 0.4 + 4 + 0.08 (multiplication 0.2 X 0.4 = 0.08).
= 4.48.
By using the BODMAS Rule, 0.4 + 0.8 ÷ 0.2 + 0.2 X 0.4 is equal to 4.48.

(ii) 0.2 X 0.5(1.2 + 2.1) + 2.5.
Solution:
The given expression is 0.2 X 0.5(1.2 + 2.1) + 2.5.
0.2 X 0.5(1.2 + 2.1) + 2.5 = 0.2 X 0.5(3.3) + 2.5 (addition 1.2 + 2.1 = 3.3).
= 0.2 X 0.5 X 3.3 + 2.5
= 0.33 + 2.5 (multiplication 0.2 X 0.5 X 3.3 = 0.33).
= 2.83 (addition 0.33 + 2.5 = 4.83).
By using the BODMAS rule, the given expression 0.2 X 0.5(1.2 + 2.1) + 2.5 is simplified as 2.83.

(iii) 0.6 – 0.2 X 1.2 + 1.6.
Solution:
The given expression is0.6 – 0.2 X 1.2 + 1.6.
0.6 – 0.2 X 1.2 + 1.6 = 0.6 – 0.24 + 1.6 (multiplication 0.2 X 1.2 = 0.24).
= 2.2 – 0.24 (addition 0.6 + 1.6 = 2.2).
= 1.96 (subtraction 2.2 – 0.24 = 1.96).
Finally, by using the BODMAS rule, the given expression 0.6 – 0.2 X 1.2 + 1.6 is simplified as 1.96.

(iv) 1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2.
Solution:
The given expression is 1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2.
1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2 = 1.2 + [2.5 + (1.0) – 2.4)] – 3.2 (subtraction in brackets 2.6 – 1.6 = 1.0).
1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2 = 1.2 + [3.5 – 2.4] – 3.2 (addition 2.5 + 1.0 = 3.5).
1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2 = 1.2 + 1.1 – 3.2 (subtraction 3.5 – 2.4 = 1.1).
1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2 = 2.3 – 3.2 (addition 1.2 + 1.1 = 2.3).
= – 0.9 (subtraction 2.3 – 3.2 = – 0.9).
By using the BODMAS rule, the given expression 1.2 + [2.5 + (2.6 – 1.6) – 2.4)] – 3.2 is simplified as – 0.9.

(v) 0.8 + 1.2 X 0.5 + 1.2.
Solution:
The given expression is 0.8 + 1.2 X 0.5 + 1.2.
0.8 + 1.2 X 0.5 + 1.2 = 0.8 + 0.6 + 1.2 (multiplication first 1.2 X 0.5 = 0.6).
= 2.6 (addition 0.8 + 0.6 + 1.2 = 1.4 + 1.2 = 2.6).
By using the BODMAS rule, 0.8 + 1.2 X 0.5 + 1.2 is simplified as 2.6.