Prasanna

A number composed of the sum of the place-values of its digits is called an expanded form of a number. Learn how a number can be expanded using the place values of its digits. We have given the process to expand a number and with different examples. All the examples are given along with the explanation. Practice all the problems available here and get a grip on the expanded form a number concept.

How to Expand a Number?

Check out the below steps to learn how to expand the number. Following these simple steps, you can expand a number easily. They are as follows

1. Firstly, take the unit place of a number and multiply it with 1.
2. Take the next number left to it and multiply it by 10.
3. Repeat the same process by increasing the 0 in the multiplicand.

Example: Write the expanded form for the number 2385?

Solution:
The given number is 2385.
Firstly, multiply the number 5 with 1. 5 × 1 = 5.
Multiply 8 with 10. 8 × 10 = 80.
Next, Multiply 3 with 100. 3 × 100 = 300.
Multiply 2 with 1,000. 2 × 1,000 = 2,000.
Add all the numbers to get the number.
2,000 + 300 + 80 + 5 = 2385.

Here, 2385 is the standard form and 2,000 + 300 + 80 + 5 is the expanded form.

Expanded Form of Numbers Examples

1. Write the expanded form for the number 70,34,51,395?

Solution:
The given number is 70,34,51,395.
Firstly, multiply the number 5 with 1. 5 × 1 = 5.
Multiply 9 with 10. 9 × 10 = 90.
Next, Multiply 3 with 100. 3 × 100 = 300.
Multiply 1 with 1000. 1 × 1000 = 1000.
Multiply 5 with 10000. 5 × 10000 = 50000.
Also, Multiply 4 with 100000. 4 × 100000 = 400000.
Multiply 3 with 1000000. 3 × 1000000 = 3000000.
Multiply 0 with 10000000. 0 × 10000000 = 0.
Next, Multiply 7 with 100000000. 7 × 100000000 = 700000000.
Add all the numbers to get the number.
700000000 + 0 + 3000000 + 400000 + 50000 + 1000 + 300 + 90 + 5 = 70,34,51,395.

Here, 70,34,51,395 is the standard form and 700000000 + 0 + 3000000 + 400000 + 50000 + 1000 + 300 + 90 + 5 is the expanded form.

2. Write the number in figures and then in words for the following expanded form
(i) 70000 + 6000 + 400 + 40 + 6

Solution:
Given expanded form is 70000 + 6000 + 400 + 40 + 6
Add the values to get a expanded form.
76446 (in figures)
Seventy-six thousand four hundred forty-six (in words)

(ii) 200000 + 40000 + 3000 + 20 + 7

Solution:
Given expanded form is 200000 + 40000 + 3000 + 20 + 7
Add the values to get an expanded form.
24327 (in figures)
Twenty-four thousand three hundred twenty-seven (in words)

3. Write in expanded form
(i) 39369

Solution:
The given number is 39369.
Firstly, multiply the number 9 with 1. 9 × 1 = 9.
Multiply 6 with 10. 6 × 10 = 60.
Next, Multiply 3 with 100. 3 × 100 = 300.
Multiply 9 with 1000. 9 × 1000 = 9000.
Multiply 3 with 10000. 3 × 10000 = 30000.
Add all the numbers to get the number.
30000 + 9000 + 300 + 60 + 9 = 39369.

Here, 39369 is the standard form and 30000 + 9000 + 300 + 60 + 9 is the expanded form.

(ii) 528,358

Solution:
The given number is 528,358.
Firstly, multiply the number 8 with 1. 8 × 1 = 8.
Multiply 5 with 10. 5 × 10 = 50.
Next, Multiply 3 with 100. 3 × 100 = 300.
Multiply 8 with 1000. 8 × 1000 = 8000.
Multiply 2 with 10000. 2 × 10000 = 20000.
Also, Multiply 5 with 100000. 5 × 100000 = 500000.
Add all the numbers to get the number.
500000 + 20000 + 8000 + 300 + 50 + 5 = 528,358.

Here, 528,358 is the standard form and 500000 + 20000 + 8000 + 300 + 50 + 5 is the expanded form.

Examples on Writing a Number in Expanded Form

I. Write the expanded form for the given numbers
(i) 6,01,217
(ii) 7,06,283
(iii) 5,33,786
(iv) 4,84,848
(v) 8,33,004

Solution

(i) The given number is 6,01,217.
Firstly, multiply the number 7 with 1. 7 × 1 = 7.
Multiply 1 with 10. 1 × 10 = 10.
Next, Multiply 2 with 100. 2 × 100 = 200.
Multiply 1 with 1000. 1 × 1000 = 1000.
Multiply 0 with 10000. 0 × 10000 = 0.
Also, Multiply 6 with 100000. 6 × 100000 = 600000.
Add all the numbers to get the number.
600000 + 0+ 1000 + 200 + 10 + 7 = 6,01,217.

Here, 6,01,217 is the standard form and 600000 + 0+ 1000 + 200 + 10 + 7 is the expanded form.

(ii) The given number is 7,06,283.
Firstly, multiply the number 3 with 1. 3 × 1 = 3.
Multiply 8 with 10. 8 × 10 = 80.
Next, Multiply 2 with 100. 2 × 100 = 200.
Multiply 6 with 1000. 6 × 1000 = 6000.
Multiply 0 with 10000. 0 × 10000 = 0.
Also, Multiply 7 with 100000. 7 × 100000 = 700000.
Add all the numbers to get the number.
700000 + 0 + 6000 + 200 + 80 + 3 = 7,06,283.

Here, 7,06,283 is the standard form and 700000 + 0 + 6000 + 200 + 80 + 3 is the expanded form.

(iii) The given number is 5,33,786.
Firstly, multiply the number 6 with 1. 6 × 1 = 6.
Multiply 8 with 10. 8 × 10 = 80.
Next, Multiply 7 with 100. 7 × 100 = 700.
Multiply 3 with 1000. 3 × 1000 = 3000.
Multiply 3 with 10000. 3 × 10000 = 30000.
Also, Multiply 5 with 100000. 5 × 100000 = 500000.
Add all the numbers to get the number.
500000 + 30000 + 3000 + 700 + 80 + 6 = 5,33,786.

Here, 5,33,786 is the standard form and 500000 + 30000 + 3000 + 700 + 80 + 6 is the expanded form.

(iv) The given number is 4,84,848.
Firstly, multiply the number 8 with 1. 8 × 1 = 8.
Multiply 4 with 10. 4 × 10 = 40.
Next, Multiply 8 with 100. 8 × 100 = 800.
Multiply 4 with 1000. 4 × 1000 = 4000.
Multiply 8 with 10000. 8 × 10000 = 80000.
Also, Multiply 4 with 100000. 4 × 100000 = 400000.
Add all the numbers to get the number.
400000 + 80000 + 4000 + 800 + 40 + 8 = 4,84,848.

Here, 4,84,848 is the standard form and 400000 + 80000 + 4000 + 800 + 40 + 8 is the expanded form.

(v) The given number is 8,33,004.
Firstly, multiply the number 4 with 1. 4 × 1 = 4.
Multiply 0 with 10. 0 × 10 = 0.
Next, Multiply 0 with 100. 0 × 100 = 0.
Multiply 3 with 1000. 3 × 1000 = 3000.
Multiply 3 with 10000. 3 × 10000 = 30000.
Also, Multiply 8 with 100000. 8 × 100000 = 800000.
Add all the numbers to get the number.
800000 + 30000 + 3000 + 0 + 0 + 4 = 8,33,004.

Here, 8,33,004 is the standard form and 800000 + 30000 + 3000 + 0 + 0 + 4 is the expanded form.

II. Write the given number in standard form
(i) 40,000 + 3,000 + 600 + 3
(ii) 3,00,000 + 70,000 + 5,000 + 600 + 40
(iii) 7,00,000 + 50,000 + 3,000 + 200 + 80 + 4
(iv) 6,00,000 + 20,000 + 7,000 + 300 + 40 + 6
(v) 2,00,000 + 30,000 + 4,000 + 300 + 20 + 3
(vi) 8,00,000 + 3

Solution:
(i) Given expanded form is 40,000 + 3,000 + 600 + 3
Add the values to get an expanded form.
43603 (in figures)
Forty-three thousand six hundred three (in words)

(ii) Given expanded form is 3,00,000 + 70,000 + 5,000 + 600 + 40
Add the values to get a expanded form.
375640 (in figures)
Three hundred seventy-five thousand six hundred forty (in words)

(iii) Given expanded form is 7,00,000 + 50,000 + 3,000 + 200 + 80 + 4
Add the values to get an expanded form.
753284 (in figures)
Seven hundred fifty-three thousand two hundred eighty-four (in words)

(iv) Given expanded form is 6,00,000 + 20,000 + 7,000 + 300 + 40 + 6
Add the values to get a expanded form.
627346 (in figures)
Six hundred twenty-seven thousand three hundred forty-six (in words)

(iv) Given expanded form is 2,00,000 + 30,000 + 4,000 + 300 + 20 + 3
Add the values to get a expanded form.
234323 (in figures)
Two hundred thirty-four thousand three hundred twenty-three (in words)

(v) Given expanded form is 8,00,000 + 3
Add the values to get an expanded form.
8,00,003 (in figures)
Eight hundred thousand three (in words)

Do you want to easily learn the multiplication table of 15? Then, refer to the 15 Times Table Multiplication Chart and the explanation we have given below. By referring to the 15 multiplication table chart, students can easily gain confidence and also get better results in the exam. It is also easy to remember any number that is multiplied by 15. Also, the multiplication of any number with 15 is easy using a 15 Times Table Multiplication Chart. Check out the below explanation for a better understanding.

15 Times Table Multiplication Chart

See the below 15 Times Table which will helps you to learn all the multiplications with 15.

15 × 0 = 0
15 × 1 = 15
15 × 2 = 30
15 × 3 = 45
15 × 4 = 60
15 × 5 = 75
15 × 6 = 90
15 × 7 = 105
15 × 8 = 120
15 × 9 = 135
15 × 10 = 150
15 × 11 = 165
15 × 12 = 180

Tips to Learn 15 Times Table

It is really easy to remember 15 Times Table with the simple tricks. All you have to do is remember one’s place digits, ten’s place digits, odd numbers, and even numbers. Look at the below explanation to learn the trick to find the resultant values of the 15 Times Table.

The one’s place always follows the pattern of 5-0.

• 15 × 1 = 15
• 15 × 2 = 30

For ten’s place digit in resultants, follow two consecutive even numbers and next two consecutive odd numbers.

• 15 × 1 = 15 (odd number at ten’s place)
• 15 × 2 = 30 (odd number at ten’s place)
• 15 × 3 = 45 (even number at ten’s place)
• 15 × 4 = 60 (even number at ten’s place)

Solved Problems on Table of 15

1. Using 15 times table, Evaluate: a) 15 times 7, b) 15 times 6 minus 5, and c) 5 times 5 plus 15 times 6?

Solution:
a) First, we will write 15 times 7 mathematically.
Using 15 times table, we have 15 times 7 = 15 × 7 = 105.
Hence, 15 times 7 is 105.
b) First, we will write 15 times 6 minus 5 mathematically.
Using table of 15, we have 15 times 6 minus 5 = 15 × 6 – 5 = 90 – 5 = 85
Hence, 15 times 6 minus 5 is 95.
c) First, we will write 5 times 5 plus 15 times 6 mathematically.
Using table of 15, we have: 5 times 5 plus 15 times 6 = 5 × 5 + 15 × 6 = 25+ 90 = 115
Hence, 5 times 5 plus 15 times 6 is 115.

2. Using the 15 times table, find: a) What is the 18th multiple of 15? and b) What is the 16th multiple of 15?

Solution:
a) To find the 18th multiple of 15, we need to call out 15 times table till 18 or multiply 15 by 18. We get 15 × 18 = 270
Thus, the 18th multiple of 15 is 270.
b) Similarly, to find the 16th multiple of 15 we need to call out the table of 15 till 16 or multiply 16 by 15 i.e. 15 × 16 = 240
Thus, the 16th multiple of 15 is 240.

3. Sam wants to give 30 chocolates to each of her 15 friends. This means 1 friend will get 2 chocolates. Using the table of 15 find how many chocolates should Sam purchase?

Solution:
Total number of friends = 15 and 1 friend will get chocolate = 2
Therefore, the total number of chocolates = 15 × 2 = 30.

Everything cannot be explained with sentences. There are some rotations to represent some words. In maths, there are some notations given for sets of numbers. Learn the Different Notations in Sets and know how to use them while solving problems. We have given Notations of sets of numbers and also their explanation. Also, get a complete grip on notations of numbers. Refer to the entire article to improve your preparation level.

What are Different Notations in Sets?

Check out the different notations insets below.

∈ – Belongs to
∉ – Does not belong to
: or | – Such that
∅ – Null set or empty set
n(A) – Cardinal number of the set A
∪ – Union of two sets
∩ – The intersection of two sets
N – Set of natural numbers = {1, 2, 3, ……}
W – Set of whole numbers = {0, 1, 2, 3, ………}
I or Z – Set of integers = {………, -2, -1, 0, 1, 2, ………}
Z+ – Set of all positive integers
Q – Set of all rational numbers
Q+ – Set of all positive rational numbers
R – Set of all real numbers
R+ – Set of all positive real numbers
C – Set of all complex numbers

The above notations are the various types of problems on sets.

Note:
(i) The pair of curly braces { } denotes a set. All the elements of a set are placed inside the pair of curly braces. Those elements are separated using commas.
(ii) The set always represents a capital letter such as; A, B, C, ………
(iii) The elements of the sets are represented with small letters.
(iv) There is no particular order for elements of a set.
(v) Also, the elements of a set are not repeated.
(vi) The Greek letter Epsilon ‘∈’ is used for the words ‘belongs to’, ‘is an element of’, etc. Therefore, p ∈ S will be read as ‘p belongs to set S’ or ‘p is an element of the set S’.
(vii) The symbol ‘∉’ stands for ‘does not belong to’ also for ‘is not an element of’. Therefore, p ∉ S will read as ‘p does not belong to set S’ or ‘p is not an element of the set S’.

Read Related Articles:

• Intersection of Sets
• Subset

Union

Let A and B be two sets.
Now, we can define the following new set.
A ∪ B = {c | c ∈ A or c ∈ B}
(That is, c may be in A or in B or in both A and B)
A ∪ B is read as “A union B”
Now that A u B contains all the elements of A and all the elements of B and the figure given below illustrates this.
It is clear that A ⊆ A ∪ B and also B ⊆ A ∪ B.

Intersection

Let A and B be two sets.
Now, we can define the following new set.
A ∩ B = {c | c ∈ A and c ∈ B}
(That is c must be in both A and B)
A ∩ B is read as “A intersection B”
Now that A n B contains only those elements which belong to both A and B and the figure given below illustrates this.
It is trivial that A ∩ B ⊆ A and also A ∩ B ⊆ B.

Set Difference

Let A and B be two sets.
Now, we can define the following new set.
A \ B = {c | c ∈ A but c ∉ B}
(That is c must be in A and must not be in B)
A \ B is read as “A difference B”
Now that A \ B contains only elements of A that are not in B and the figure given below illustrates this. A \ B can also write as A – B

Symmetric Difference

Let A and B be two sets.
Now, we can define the following new set.
A Δ B = (A \ B) u (B \ A)
A Δ B is read as “A symmetric difference B”
Now that A Δ B contains all elements in A u B which are not in A n B and the figure given below illustrates this.

Complement

If A ⊆ U, where U is a universal set, then U \ A is called the complement of A with respect to U. If the underlying universal set is fixed, then we denote U \ A by A’ and it is called the complement of A.
A’ = U \ A
The difference set A \ B can also be viewed as the complement of B with respect to A.

Disjoint Sets

Two sets A and B are said to be disjoint if they do not have any common element. That is, A and B are disjoint if A n B = ᵩ
It is clear that n(A u B) = n(A) + n(B), if A and B are disjoint finite set.

Subset of a Set

A set M is a subset of set N if every element of M is also an element of N.
In symbol, we write m ⊆ n. We can read it as “M is a subset of N” or “M is contained in N”.
Read ⊈ as “M is a not subset of N” or “M is not contained in N”

Proper Subset

A set M is said to be a proper subset of set N if M ⊆ N and M ≠ N.
In symbol, we write M ⊂ N. We can read it as “M is a proper subset of N”

Power Set

The set of all subsets of X is said to be the power set of the set X.
The power set of X is denoted by P(X)

Superset

A set M is said to be a proper subset of set N if M ⊆ N and M ≠ N.
In symbol, we write M ⊂ N. We can read it as N is called superset of M.

Formula to Find Number of Subsets

If M is denoted as the given set and it has n number of elements, then we can find the formula of subsets as
Number of subsets = 2ⁿ
Formula to find the number of proper subsets
Number of proper subsets = 2^n-1

Null set or Empty set

A set that contains no element is called a null set or an empty set. And it is denoted by { } or ∅. Hence, the cardinal number of a null set is zero.

Multiplication of Fractions steps and methods are here. Check rules, tricks, and tips to solve fraction multiplication problems. Refer to the important formulae and also types involved in it. Know the fraction multiplication solved examples, types, and parts of fractions, variables, etc. Go through the below sections to get the complete details regarding multiplication methods, formulae, rules, etc.

Multiplication of Fractions – Introduction

Fractions multiplication starts with numerators multiplication followed by denominators multiplication. The resultant fraction of the multiplication fraction can be simplified further and can be reduced to its lowest terms. Fraction Multiplication is not the same as adding or subtracting the fraction values.

Any two or more fractions with different denominators can easily be multiplied. The main thing to be considered is the fractions should not be mixed fractions, they should be either proper or improper fractions. There are various steps involved in multiplying the fractions. They are:

1. In the fractions multiplication, we multiply the numerator with the numerator term to get the desired result of the numerator.
2.  In the fractions multiplication, we multiply the denominator with the denominator term to get the desired result of the denominator.
3. After finding the resultant numerator and denominator values, check for simplification if possible
4. Once the simplification is done, we get the final resultant value.

How to Multiply Mixed Fractions?

Consider a mixed fraction which is the form of a \(\frac { b }{ c } \).

In the above-mixed fraction, convert the fraction value into an improper fraction. After converting it into an improper fraction, apply all the above steps we do in the multiplication of fractions. To convert mixed fractions into improper fractions, we apply the following steps:

• Multiply the whole number (a) with the denominator (c). We get the result value (a * c)
• To the above result value (a * c), add the numerator value (b). After the addition of the numerator, we find the numerator of the improper fraction.
• The denominator value of the improper fraction will be the denominator of the same mixed fraction.
• Generally, we can write it as a \(\frac { b }{ c } \) = \(\frac { c*a + b}{ c } \)

Also Read:

• Addition and Subtraction of Fractions

Multiplication of Proper Fractions

Multiplication of proper fractions is the easiest form of all the fractions multiplication.

Example:

Solve the equation \(\frac { 2 }{ 3 } \) × \(\frac { 4 }{ 6 } \)?

Solution:

As given in the question,

The equation is \(\frac { 2 }{ 3 } \) × \(\frac { 4 }{ 6 } \)

Here, \(\frac { 2 }{ 3 } \), \(\frac { 4 }{ 6 } \) are the proper fractions. To multiply the proper fractions, we have to follow the steps.

Step 1: First of all, multiply the numerators together i.e., 2 and 4. The solution is 2 * 4 = 8

Step 2: Next, multiply the denominators together ie., 3 and 6. The solution is 3 * 6 = 18. The fraction value can be written as \(\frac { 2*4 }{ 3*6 } \) = \(\frac { 8 }{ 18 } \)

Step 3: Check if you can simplify the resultant equation. On simplification, we can write it as \(\frac { 4 }{ 9 } \)

Multiplication of Improper Fractions

An improper fraction is the one that has a greater denominator than the numerator. If we multiply an improper fraction, we result in an improper fraction.

Example: 

Solve the equation \(\frac { 3 }{ 2 } \) × \(\frac { 7 }{ 5 } \) of the improper fractions?

Solution:

As given in the question,

The equation is \(\frac { 3 }{ 2 } \) × \(\frac { 7 }{ 5 } \)

Here, \(\frac { 3 }{ 2 } \), \(\frac { 7 }{ 5 } \) are improper fractions. To multiply the improper fractions, we have to follow the steps.

Step 1: First of all, multiply the numerators together i.e., 3 and 7. The solution is 3 * 7 = 21

Step 2: Next, multiply the denominators together i.e., 2 and 5. The solution is 2 * 5 = 10. The fraction value can be written as \(\frac { 3*7 }{ 2*5 } \) = \(\frac { 21 }{ 10 } \)

Step 3: Check if you can simplify the resultant equation. In the above equation, simplification is not possible.

Step 4: Now, convert the fraction into an improper fraction. Hence, the result is 2[/latex] = \(\frac { 1 }{ 10 } \)

Multiplication of Mixed Fractions

Mixed Fractions are those fractions which have a whole number and a fraction like 2 [/latex] = \(\frac { 1 }{ 2 } \). When multiplying both the mixed fractions, we have to convert the mixed fractions into improper fractions.

Example:

Multiply the fractions 2\(\frac { 2 }{ 3 } \) and 3\(\frac { 1 }{ 4 } \)

Solution:

The given equation is 2\(\frac { 2 }{ 3 } \) x 3\(\frac { 1 }{ 4 } \)

Here, 2\(\frac { 2 }{ 3 } \) and 3\(\frac { 1 }{ 4 } \) are mixed fractions. To multiply the mixed fractions, we have to follow the following steps.

Step 1: First of all, convert the mixed fractions to improper fractions. To convert the mixed fraction of 2\(\frac { 2 }{ 3 } \), we write it as \(\frac { (3×2+2) }{ 3 } \). The result is 8/3. To convert the mixed fraction of 3\(\frac { 1 }{ 4 } \), we write it as \(\frac { (4×3+1) }{ 4 } \). The result is 13/4

Step 2: Multiply the numerators together i.e., 8 and 13. The solution is 8 * 13 = 104

Step 3: Next, multiply the denominators together i.e., 3 and 4. The solution is 3 * 4 = 12. The fraction value can be written as \(\frac { 8*13 }{ 3*4 } \) = \(\frac { 104 }{ 12 } \)

Step 4: Check if you can simplify the resultant equation. In the above equation, the simplification can be done as \(\frac { 26 }{3 } \).

Step 5: Now, convert the fraction into an improper fraction. Hence the result is 8\(\frac { 2 }{ 3 } \)

Multiplying Fractions Examples

Problem 1: 

A recipe calls for \(\frac { 3 }{ 4 } \) cups of sugar. Amari is tripling the recipe. How much amount of sugar will be needed?

Solution:

As given in the question,

Amount of sugar for recipe = \(\frac { 3 }{ 4 } \)

No of times Amari multiplied the recipe = 3

Therefore, to find the amount of sugar we apply the multiplication of fractions

Hence, \(\frac { 3 }{ 4 } \) * \(\frac { 3 }{ 1 } \) = \(\frac { 9 }{ 4 } \)

Now, convert the proper fraction into an improper fraction i.e., 2\(\frac { 1 }{ 4 } \)

Thus, the final solution is 2\(\frac { 1 }{ 4 } \)

Problem 2:

\(\frac { 4 }{ 5 } \) of all students at Riverwood High School are involved in an extracurricular activities. Of those students, \(\frac { 2 }{ 3 } \) are involved in a fall activity. What fraction of students at Riverwood are involved in a fall activity?

Solution:

As given in the question,

No of students involved in extracurricular activities = \(\frac { 4 }{ 5 } \)

No of students involved in fall activity = \(\frac { 2 }{ 3 } \)

To find the fraction of students involved in a fall activity, we have to apply the multiplication of fractions

Hence \(\frac { 2 }{ 3 } \) x \(\frac { 4 }{ 5 } \) = \(\frac { 8 }{ 15 } \)

Thus, \(\frac { 8 }{ 15 } \) fraction of students are involved in a fall activity.

Therefore, the final solution is \(\frac { 8 }{ 15 } \)

Problem 3:

Jimmy has a collection of 18 video games. Of the 18 video games. \(\frac { 1 }{ 3 } \) are sports games. How many of his games are sports games?

Solution:

As given in the question,

No of video games = 18

Part of sports games = \(\frac { 1 }{ 3 } \)

To find the no of sports games, we apply multiplication of fractions

Hence \(\frac { 1 }{ 3 } \) x \(\frac { 18 }{ 1 } \)

=\(\frac { 18 }{ 3 } \)

On further simplications, we get the result as 6.

Therefore, Jimmy has a collection of 6 sports games.

In practice tests on square and square roots, we are providing different methods of problems with solutions. We are providing four options for every question. Practice the test paper on square roots and find the correct answer for the given questions. Improve your preparation level for every problem. Different methods included for every problem. Therefore, choose the best and easy method and learn the problems.

Get clarity on every concept of Square root by the provided links on our website.

Objective Questions on Square and Square Roots

1 . Which of the following numbers is a perfect square?
(a) 131
(b) 144
(c) 165
(d) 251

2. A perfect square number can never have the digit …… at the unit’s place.
(a) 6
(b) 4
(c) 8
(d) 5

3. Which of the following is a Pythagorean triplet?
(a) (2, 4, 6)
(b) (5, 8, 2)
(c) (2, 9, 5)
(d) (3, 4, 5)

4. Sum of the first n natural numbers is
(a) n^2
(b) n(n + 1) / 2
(c) n² + 1
(d) (2n² – 1) / 2

5. √625 = ?
(a) 45
(b) 35
(c) 25
(d) 55

6. What least number must be added to 720 to make it a perfect square?
(a) 5
(b) 2
(c) 7
(d) 9

7. √0.81 = ?
(a) 0.9
(b) 0.09
(c) 0.99
(d) 0.96

8. √0.00065536 is equal to
(a) 0.256
(b) 0.0256
(c) 0.00256
(d) 0.000256

9. √1.0219 = ?
(a) 1.0109
(b) 1.03
(c) 0.505
(d) 1.056

10. √0.25 × √1.6 = ?
(a) 0.15
(b) 0.2
(c) 0.25
(d) 20

11. √289/√256 = ?
(a) 3 / 25
(b) 30 √32
(c) 17 / 16
(d) 1.25

12. √6084/144 = ?
(a) 22 / 17
(b) 25 / 12
(c) 32 / 4
(d) 39 / 6

13. Find the square root of 3844
(a) 56
(b) 52
(c) 62
(d) 65

14. Evaluate √4562496
(a) 2256
(b) 2136
(c) 2563
(d) 2222

15. Find the value of 49^2 X 25^2 = ?
(a) 1501621
(b) 1500625
(c ) 1502624
(d)1500725

16. 15^2 + √0.81 + 25 = ?
(a) 250.9
(b) 252.9
(c) 251.9
(d) 255.9

17. 25% of 125 + ? = 250
(a) 218
(b) 220.75
(c) 225.75
(d) 218.75

It is really easy to find Square root of numbers that are perfect squares when it comes to Square Root of Numbers that are Not Perfect Squares, it is difficult to find them. If you need to find the square root of a number m places of decimals, then the number of digits in the decimal part must be 2m. If the decimals are less than 2m, then add the required zeros at the right to the decimal part. Check out the complete concepts of Square Root on our website.

Square Root of Decimal Numbers using Long Division Method

Have a look at the procedure on how to solve square roots that aren’t perfect squares. They are as follows

• If we have to find a square root of number correct up to ‘n’ places of decimal, then we need to find the square root of number up to (n + 1) places of decimal.
• Also, if the digit place is (n + 1) decimal place is equal to 5 or greater than 5, then the digit at ‘n’ place increases by 1.
• Furthermore, if the digit place is (n + 1) decimal place is less than 5, then the digit at ‘n’ place remains the same and deletes the digit at (n + 1) place.

Square Root of Numbers that are not Perfect Squares Examples

1. Evaluate √2 correct up to two places of decimal?

Solution:
The given number is √2.
The square root of 1 is 1. Also, the square root of 2.25 is 1.5.
The 2 is in between 1 and 2.25.
Therefore, to find the square root of a 2, add 6 zeros after the decimal point.
√2.000000 = √(1.414 × 1.414)
Grouping the factors into the pairs of equal factors.
(1.414 × 1.414)
1.414 is the square root of 2.

The two places of decimal after the number is 1.41.

2. Evaluate √3 correct up to 3 places of decimal?

Solution:
The given number is √3.
The square root of 2.25 is 1.5. Also, the square root of 3.0625 is 1.75.
The 3 is in between 2.25 and 3.0625.
Therefore, to find the square root of a 3, add 10 zeros after the decimal point.
√3.0000000000 = √(1.7324 × 1.7324)
Grouping the factors into the pairs of equal factors.
(1.7324 × 1.7324)
1.7324 is the square root of 3.

The three places of decimal after the number is 1.732.

3. Evaluate √0.8 correct up to two places of decimal?

Solution:
The given number is √0.8.
The square root of 0.49 is 0.7. Also, the square root of 0.81 is 0.9.
The 0.8 is in between 0.7 and 0.9.
Therefore, to find the square root of 0.8, add 5zeros after the decimal point.
√0.800000 = √(0.894 × 0.894)
Grouping the factors into the pairs of equal factors.
(0.894 × 0.894)
0.894 is the square root of 0.8.

The two places of decimal after the number is 0.89.

Square Root of Number in the Fraction Form is explained clearly in this article. Finding natural numbers is easy but when it comes to fractions it is a little bit difficult to find the Square Root of Numbers. So, to make the process easy, we have given the step by step procedure on How do you find the Square Root of a Fraction. Check all the sample problems on finding the square root of numbers in fraction form and get to know the concept involved.

How to Find the Square Root of a Fraction?

If m and n are squares of some numbers then, √(m ÷ n) = √m ÷ √n. If you find any problem in mixed form, then convert it into fractions.

1. Find the square root of √(144 ÷ 81)?

Solution:
Given the number is √(144 ÷ 81).
Firstly, separate the numerator and denominator.
√(144) ÷ √(81)
Find the separate Square Roots of the numerator and denominator.
Find the prime factors of √(144)
√(144) = 12 × 12
Grouping the factors into the pairs of equal factors.
(12 × 12)
Take one number from each group and multiply them to find the number whose square is 144.
12
The square root of 144 is 12.
Find the prime factors of √(81)
√(81) = 9 × 9
Grouping the factors into the pairs of equal factors.
(9 × 9)
Take one number from each group and multiply them to find the number whose square is 81.
9.
The square root of 81 is 9.
The final answer is 12 ÷ 9 = 4/3

Therefore, the square root of √(144 ÷ 81) = 4/3.

2. Evaluate √(256 ÷ 64)?

Solution:
Given the number is √(256 ÷ 64).
Firstly, separate the numerator and denominator.
√(256) ÷ √(64)
Find the separate Square Roots of the numerator and denominator.
Find the prime factors of √(256)
√(256) = 16 × 16
Grouping the factors into the pairs of equal factors.
(16 × 16)
Take one number from each group and multiply them to find the number whose square is 256.
16
The square root of 256 is 16.
Find the prime factors of √(64)
√(64) = 8 × 8
Grouping the factors into the pairs of equal factors.
(8 × 8)
Take one number from each group and multiply them to find the number whose square is 64.
8.
The square root of 64 is 8.
The final answer is 256 ÷ 64 = 16/8

Therefore, the square root of √(256 ÷ 64) = 16/8.

3. Evaluate √(49 ÷ 121)?

Solution:
Given the number is √(49 ÷ 121).
Firstly, separate the numerator and denominator.
√(49) ÷ √(121)
Find the separate Square Roots of the numerator and denominator.
Find the prime factors of √(49)
√(49) = 7 × 7
Grouping the factors into the pairs of equal factors.
(7 × 7)
Take one number from each group and multiply them to find the number whose square is 49.
7
The square root of 49 is 7.
Find the prime factors of √(121)
√(121) = 11 × 11
Grouping the factors into the pairs of equal factors.
(11 × 11)
Take one number from each group and multiply them to find the number whose square is 121.
11.
The square root of 121 is 11.
The final answer is 49 ÷ 121 = 7/11

Therefore, the square root of √(49 ÷ 121) = 7/11.

4. Evaluate √(25 ÷ 169)?

Solution:
Given the number is √(25 ÷ 169).
Firstly, separate the numerator and denominator.
√(25) ÷ √(169)
Find the separate Square Roots of the numerator and denominator.
Find the prime factors of √(25)
√(25) = 5 × 5
Grouping the factors into the pairs of equal factors.
(5 × 5)
Take one number from each group and multiply them to find the number whose square is 25.
5
The square root of 25 is 5.
Find the prime factors of √(169)
√(169) = 13 × 13
Grouping the factors into the pairs of equal factors.
(13 × 13)
Take one number from each group and multiply them to find the number whose square is 169.
13.
The square root of 169 is 13.
The final answer is 25 ÷ 169 = 5/13

Therefore, the square root of √(25 ÷ 169) = 5/13.

5. Find the value of √(196 ÷ 9)?

Solution:
Given the number is √(196 ÷ 9).
Firstly, separate the numerator and denominator.
√(196) ÷ √(9)
Find the separate Square Roots of the numerator and denominator.
Find the prime factors of √(196)
√(196) = 13 × 13
Grouping the factors into the pairs of equal factors.
(13 × 13)
Take one number from each group and multiply them to find the number whose square is 196.
13
The square root of 196 is 13.
Find the prime factors of √(9)
√(9) = 3 × 3
Grouping the factors into the pairs of equal factors.
(3 × 3)
Take one number from each group and multiply them to find the number whose square is 9.
3.
The square root of 9 is 3.
The final answer is 196 ÷ 9 = 13/3

Therefore, the square root of √(196 ÷ 9) = 13/3.

6. Find out the value of √25 × √16?

Solution:
Given that √25 × √16.
Find the separate Square Roots of the two numbers.
Find the prime factors of √(25)
√(25) = 5 × 5
Grouping the factors into the pairs of equal factors.
(5  × 5)
Take one number from each group and multiply them to find the number whose square is 25.
5
The square root of 25 is 5.
Find the prime factors of √(16)
√(16) = 4 × 4
Grouping the factors into the pairs of equal factors.
(4 × 4)
Take one number from each group and multiply them to find the number whose square is 16.
4.
The square root of 16 is 4.
The final answer is √25 × √16 = 5 × 4 = 20

Therefore, √25 × √16 = 20.

Square Root of Numbers in the Decimal Form is finding the Square Root of given Decimal numbers. We can use the prime factorization method or long division method to find out the Square Root of a Decimal. Choose the best method and know the procedure, then find the Square Root of a Decimal number. We have given all concepts of Square root on our website along with examples.

How to find the Square Root of a Decimal?

Let us have a look at the procedure to find out the Square Root of a Decimal. Following the simple steps listed below, you can find the square root of a decimal easily. They are along the lines

1. Make the number of decimal places even by placing a zero on the extreme right of the decimal part.
2. Coming to an integral part, keep the periods as done while finding the square root of a perfect square of some natural number.
3. In the given decimal part, keep the periods on every pair of digits those starts with the first decimal place.
4. Now, find the square root of a given number.
5. Put the decimal point in the square root as soon as the integral part is exhausted.

Finding Square Root of a Decimal Number Examples

1. Evaluate √42.25?

Solution:

Solution:
Given the number is √42.25,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 42.25.
42.25 = 6.5 × 6.5
Grouping the factors into the pairs of equal factors.
(6.5 × 6.5)
Take one number from each group and multiply them to find the number whose square is 42.25.
6.5.

The square root of 42.25 is 6.5.

2. Evaluate √1.96?

Solution:
Given the number is √1.96,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 1.96.
1.96 = 1.4 × 1.4
Grouping the factors into the pairs of equal factors.
(1.4 × 1.4)
Take one number from each group and multiply them to find the number whose square is 1.96.
1.4.

The square root of 1.96 is 1.4.

3. Evaluate √39.69?

Solution:
Given the number is √39.69,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 39.69.
39.69 = 6.3 × 6.3
Grouping the factors into the pairs of equal factors.
(6.3 × 6.3)
Take one number from each group and multiply them to find the number whose square is 39.69.
6.3.

The square root of 39.69 is 6.3.

4. Evaluate √547.56?

Solution:
Given the number is √547.56,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 547.56.
547.56 = 23.4 × 23.4
Grouping the factors into the pairs of equal factors.
(23.4 × 23.4)
Take one number from each group and multiply them to find the number whose square is 547.56.
23.4.

The square root of 547.56 is 23.4.

5. Evaluate √0.064516?

Solution:
Given the number is √0.064516,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 0.064516.
0.064516 = 0.254 × 0.254
Grouping the factors into the pairs of equal factors.
(0.254 × 0.254)
Take one number from each group and multiply them to find the number whose square is 0.064516.
0.254.

The square root of 0.064516 is 0.254.

Are you searching to learn Square Root of a Perfect Square by Using the Long Division Method? Then, you can get the process to find a Square Root of a Perfect Square with the help of a Long Division Method here. We have included different types of problems, solutions, and also their explanation. Therefore, without any delay, you can begin your practice and get a grip on the complete concept.

We have given complete concepts on Square and Square Root on our website. Have a look at them and immediately start your practice.

How to find Square Root of a Perfect Square Number using Long Division Method?

Check the below steps to learn the procedure for finding the Square Root of a Number using the Long Division method.

• Group the digits in pairs using a period that starts with the digit in the units place. Each pair and the remaining digit (if any).
• Set the largest number whose square is equal to or just less than the first period. Then, you have to take this number as the divisor and also as the quotient.
• Do subtract the product of the divisor and the quotient from the first period.
• Then, write the next period to the right of the remainder and consider it as a new dividend.
• Divide the new dividend until the digit is equal to or just less than the new dividend.
• Repeat the above steps till all the periods have been taken up. Now, the quotient so obtained is the required square root of the given number.

Square Root of a Perfect Square Number using Long Division with Examples

1. Find the square root of 784 by the long-division method?

Solution:
The given number is 784.
To find the square root of a perfect square by the long division method, we have to follow the below procedure.
784 = 2 x 2 x 2 x 2 x 7 x 7
Grouping the factors into the pairs of equal factors.
(2 × 2) × (2 × 2) × (7 × 7)
Take one number from each group and multiply them to find the number whose square is 784.
2 × 2 × 7 = 28.

The square root of 784 is 28.

2. Evaluate √5329 using the long-division method?

Solution:
The given number is √5329.
To find the square root of a perfect square by the long division method, we have to follow the below procedure.
√5329 = 73 x 73
Grouping the factors into the pairs of equal factors.
(73 × 73)
Take one number from each group and multiply them to find the number whose square is 5329.
73.

The square root of 5329 is 73.

3. Evaluate √16384?

Solution:
The given number is √16384.
To find the square root of a perfect square by the long division method, we have to follow the below procedure.
√16384 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Grouping the factors into the pairs of equal factors.
(2 × 2) x (2 × 2) x (2 × 2) x (2 × 2) x (2 × 2) x (2 × 2) x (2 × 2)
Take one number from each group and multiply them to find the number whose square is 16384.
2 x 2 × 2 x 2 × 2 x 2 × 2 = 128.

The square root of 16384 is 128.

4. Evaluate √10609?

Solution:
The given number is √10609.
To find the square root of a perfect square by the long division method, we have to follow the below procedure.
√10609 = 103 x 103
Grouping the factors into the pairs of equal factors.
(103 x 103)
Take one number from each group and multiply them to find the number whose square is 10609.
103

The square root of 10609 is 103.

5. Evaluate √66049?

Solution:
The given number is √66049.
To find the square root of a perfect square by the long division method, we have to follow the below procedure.
√66049 = 257 x 257
Grouping the factors into the pairs of equal factors.
(257 x 257)
Take one number from each group and multiply them to find the number whose square is 66049.
257

The square root of 66049 is 257.

6. Find the cost of erecting a fence around a square field whose area is 4 hectares if fencing costs $ 3.50 per meter?

Solution:
Area of the square field = (4 × 1 0000) m² = 40000 m²
Length of each side of the field = √40000 m = 200 m.
Perimeter of the field = (4 × 200) m = 800 m.
Cost of fencing = $(800 × 7/2) = $2800.

7. Find the least number that must be added to 6412 to make it a perfect square?

Solution:
Given the number is 6412,
To find the square root of a perfect square by the long division method, we have to find the factors of 6412.
6412 = (80 × 80)
Grouping the factors into the pairs of equal factors.
(80 × 80)
We observe here that (80)² < 6412 < (81)²
The required number to be added = (81)² – 6412
= 6561 – 6412
= 149

8. What least number must be subtracted from 7250 to get a perfect square? Also, find the square root of this perfect square?

Solution:
Given the number is 7250,
To find the square root of a perfect square by the long division method, we have to find the factors of 7250.
7250 = (85 × 85)
Grouping the factors into the pairs of equal factors.
(85 × 85)
So, the least number to be subtracted from 7250 is 25.
Required perfect square number = (7250 – 25) = 7225
And, √7225 = 85.

9. Find the greatest number of four digits which is a perfect square?

Solution:
The greatest number of four digits is 9999.
The square root of 9999 is
To find the square root of a perfect square by the long division method, we have to find the factors of 9999.
9999 = (99 × 99)
Grouping the factors into the pairs of equal factors.
(99 × 99)
This shows that (99)² is less than 9999 by 198.
So, the least number to be subtracted is 198.
Hence, the required number is (9999 – 198) = 9801.

The square root of a number can found by the prime factorization method easily. We have explained the step by step procedure to find the square root of a number. Check out the examples to find the square root of a number for better understanding. Also, we have given a clear explanation for the examples below. All square and square root concepts are given on our website for free of cost.

How to Find Square Root of a Perfect Square Number using Prime Factorization?

Let’s have a look at the below steps to find the square root of a perfect square by using the prime factorization method.

• Firstly, we will resolve the number inside the square root into prime factors.
• Then, inside the square root, for every two same numbers multiplied and one number can be taken out of the square root.
• In the next step, we will combine the like square root terms using mathematical operation.

Here in this finding square root of a Perfect Square by prime factorization method, we will solve various types of questions to make you familiar with the concept.

Square Root of a Perfect Square using Prime Factorization Examples

1. Find the square root of 256 by the prime factorization method?

Solution:
Given the number is 256,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 256.
256 = 4 × 4 × 4 × 4
Grouping the factors into the pairs of equal factors.
(4 × 4) ×(4 × 4)
Take one number from each group and multiply them to find the number whose square is 256.
4 × 4 = 16.

The square root of 256 is 16.

2. Find the square root of 5184?

Solution:
Given the number is 5184,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 5184.
5184 = 8 × 8 × 9 × 9
Grouping the factors into the pairs of equal factors.
(8 × 8) × (9 × 9)
Take one number from each group and multiply them to find the number whose square is 5184.
8 × 9 = 72.

The square root of 5184 is 72.

3. Find out the square root of 576?

Solution:
Given the number is 576,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 576.
576 = 2 × 2 × 3 × 3 × 4 × 4
Grouping the factors into the pairs of equal factors.
(2 × 2) × (3 × 3) × (4 × 4)
Take one number from each group and multiply them to find the number whose square is 576.
2 × 3 × 4 = 24.

The square root of 576 is 24.

4. Evaluate √3600?

Solution:
Given the number is √3600,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 3600.
√3600 = √(2 × 2 × 3 × 3 × 10 × 10)
Grouping the factors into the pairs of equal factors.
√[(2 × 2) × (3 × 3) × (10 × 10)]
Take one number from each group and multiply them to find the number whose square is 3600.
2 × 3 × 10 = 60.

The square root of 3600 is 60.

5. Evaluate √112896?

Solution:
Given the number is √112896,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 112896.
√3600 = √(2 × 2 × 4 × 4 × 6 × 6 × 7 × 7)
Grouping the factors into the pairs of equal factors.
√[(2 × 2) × (4 × 4) × (6 × 6) × (7 × 7)]
Take one number from each group and multiply them to find the number whose square is 112896.
2 × 4 × 6 × 7 = 336.

The square root of 112896 is 336.

6. In an auditorium, the number of rows is equal to the number of tables in each row. If the capacity of the auditorium is 484, find the number of tables in each row?

Solution:
Given that in an auditorium, the number of rows is equal to the number of tables in each row.
Let the number of chairs in each row be x.
Then, the number of rows = x.
Total number of chairs in the auditorium = (x × x) = x²
But, the capacity of the auditorium = 484(given).
Therefore, x² = 484.
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 484.
√484 = √(2 × 2 × 11 × 11)
Grouping the factors into the pairs of equal factors.
√[(2 × 2) × (11 × 11)]
Take one number from each group and multiply them to find the number whose square is 484.
2 × 11 = 22.

Hence, the number of tables in each row = 336.

7. Find the smallest number by which 1452 must be multiplied so that the product becomes a perfect square?

Solution:
Given the number is 1452,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 1452.
1452= √(2 × 2 × 11 × 11 × 3)
Grouping the factors into the pairs of equal factors.
[(2 × 2) × (11 × 11) × (3)]
Multiply 3 to make it a perfect square.
[(2 × 2) × (11 × 11) × (3 × 3)]
Take one number from each group and multiply them to find the number whose square is 1452.
2 × 11 × 3 = 66.
The square root of 1452 is 66.

Therefore, the smallest number by which 1452 must be multiplied so that the product becomes a perfect square is 3.

8. Find the smallest number by which 180 must be divided so that the quotient is a perfect square?

Solution:
Given the number is 180,
To find the square root of a perfect square by the prime factorization method, we have to find the factors of 180.
180 = √(2 × 2 × 3 × 3 × 5)
Grouping the factors into the pairs of equal factors.
[(2 × 2) × (3 × 3) × (5)]
Divide 5 to make it a perfect square.
[(2 × 2) × (3 × 3)]
Take one number from each group and multiply them to find the number whose square is 180.
2 × 3 = 6.
The square root of 36 is 6.

Therefore, the smallest number by which 180 must be divided so that the quotient is a perfect square is 5.

9. Find the least square number divisible by each one of 4, 9, and 12?

Solution:
The least number divisible by each one of 4, 9, 12 is their LCM.
Now, LCM of 4, 9, 12 = (2 × 2 × 9 × 6) = 216.
By prime factorization, we get
216 = 2 x 2 x 2 x 3 × 3 x 3
To make it a perfect square it must be multiplied by (2 × 3) i.e., 6.

Hence, the required number = (216 × 6) = 1296.

1. Note down the numerator and denominator values as a product of primes.
2. Now that you know the common prime factor, divide the numerator and denominator by each of the common prime factors. Division of fractions is indicated by the slanted line through each factor. This complete process is called canceling common factors.
3. The product value of the remaining factors in the numerator and also the product value of the remaining factors in the denominator are prime numbers and the fraction value is reduced to the lowest terms.

Example:

Suppose that 6/18 is the fraction number. On dividing the fractions we can write it as 2*3/2*3*3 = 1/3 is the final result.

In the above equation, 1 and 3 are relative primes.

Method 2: Dividing Out Common Factors

1. First, divide the denominator and numerator by a number of factors which is common to both. Note down the quotient which is above the original number.
2. Continue the process till the denominator and numerator are relatively prime numbers.

Example:

Suppose 25/30 is the fraction number. Now, consider the common whole number which divides both the numbers. Therefore, 5 is the common number that divides both numerator and denominator. After dividing the numbers by 5, the result will be 5/6. Here 5 and 6 are prime relatively.

Method 3: HCF Method

To reduce the fractions, find the HCF of the denominator and numerator of the given fraction. To reduce the fraction to its lowest terms, we divide the denominator and numerator by their HCF numbers.

Example:

Reduce the fraction 21/56 to its reduced form?

First of all, divide the number 21 by the number 56

The final quotient when dividing the values is 7.

Therefore HCF of 21 and 56 is 7.

Now, divide the denominator and numerator in the fraction by HCF value 7.

Then the final result value will be 3/8

Method 4: Prime Factorization Method

First, find the common factors of both numerators and denominators. Then express both the denominator and numerator of the fraction as the product values of the prime factors. Once you find the equation, then cancel the common factors from them to get the final result value.

Example:

Reduce 120/360 to the lowest term

First of all, find the factors of both numerators and denominators.

The factors of 120 are 2,2,2,3,5

The factors are 360 are 2,2,2,3,3,5

On further simplification, we get the final result as 1/3.

Example Problems on Reducing Fractions to Lowest Terms

Problem 1:

Mr. Lee is planting a vegetable garden. The garden will have no more than 16 equal sections. 3/4 of the garden will have tomatoes. What other fractions could represent the part of the garden that will have tomatoes?

Solution:

As given in the question,

No of equal sections the garden will have = 16

Part of the garden will have tomatoes = 3/4

Now, we have to find fractions that are equivalent to 3/4.

We can use multiplication to find equivalent fractions.

3*2/4*2 = 6/8

3*3/4*3 = 9/12

3*4/4*4 = 12/16

Few parts = Larger Parts = Smaller Parts

Each numerator represents the parts of Mr. Lee’s garden that have tomatoes

Each denominator represents how many parts there are in all the gardens.

As the denominator becomes the greater number, the size of the parts becomes smaller.

Therefore the final result is 1/16

Thus, 1/16th part of the garden has tomatoes.

Problem 2:

Sophia is making bracelets with beads. Each bracelet has 4 beads and 3/4 of the beads are red. If Sophia makes 5 bracelets, how many red beads does she need?

Solution:

As given in the problem,

No of beads each bracelet has = 4

No of red beads = 3/4

No of bracelets = 5

For 1 bracelet, the number of red beads = 3

For 5 bracelets, the number of red beads = 15

Therefore, the total no of beads = 20

Hence, Sophia will need 15 red beads to make 5 bracelets.

Problem 3:

Tala cut a pizza in half. She cuts each half into 2 pieces and cut each piece into 2 slices. What fraction of the pizza did Tala eat?

Solution:

As given in the question,

No of halves she cut = 1

No of pieces she cut half piece = 2

Therefore, Tala ate = 2/8 or 1/4

Problem 4:

2/5 of the students voted for Tala to be the class president, 1/3 voted for Emma, and 4/15 voted for Tim. Which candidate got the most votes?

Solution: 

As given in the question,

We need to make equivalent fractions to solve this

Multiples of 5: 5,10,15…

Multiples of 3: 3,6,9,12,15….

Tals’s voting = 2/5*3/3 = 6/15

Emma’s voting = 1/3*5/5 = 5/15

Hence, Tala got the most votes

The square root of a number is the value that multiplied by itself gives the original number. If m is the square root of n, then it is represented as m = √n. Also, we can write this expression as m² = n. √ is the root of numbers. The square root of the square of a positive number gives the original number. For example, the square of 4 is 16, 4² = 16, and the square root of 16, √16 = 4. Let us say n is a positive integer, such that √(n . n) = √(n²) = n.

The square root of a negative number represents a complex number. √-m = mi, where i is the imaginary number.

Quick Links of Square Root Concepts

Below is the list of several concepts available in the Square Roots Chapter. You can get a grip on them by simply tapping on the direct links available. You just need to tap on them and learn the concept individually. Practice the Problems on finding Square Root of Numbers in Decimal Form, Fraction Form, etc.

• Square Root of a Perfect Square by using the Prime Factorization Method
• Square Root of a Perfect Square by Using the Long Division Method
• Square Root of Numbers in the Decimal Form
• Square Root of Number in the Fraction Form
• Square Root of Numbers that are Not Perfect Squares
• Table of Square Roots
• Practice Test on Square and Square Roots

How to Find the Square Root?

The square root of any number can be easily found for a given number using the given method. If the given number is a perfect square, then we can find the factors by the prime factorization method. If the number is an imperfect square, then we can use the long division method to find the root.

Example: Square of 8 = 8 x 8 = 8² = 64
The square root of 64, √64 = 8.

Square Root Examples

(i) 1² = 1
Therefore, the square root of 1 is 1. Also, it can write as √1 = 1.
(ii) 2² = 4
Therefore, the square root of 4 is 2. Also, it can write as √4 = 2.
(iii) 3² = 9
Therefore, the square root of 9 is 3. Also, it can write as √9 = 3.
(iv) 5² = 25
Therefore, the square root of 25 is 5. Also, it can write as √25 = 5.
(v) 4/3 is the square root of 16/9. Since 4²/3² = 4/3
Or we can write it as √(16/9) = 4/3 (Square root of 16/9 is 4/3)
(vi) 0.3 is the square root of 0.09. Since 0.3² = 0.09
Or we can write it as √0.09 = 0.3 (Square root of 0.09 is 0.3)

In general; if n = m², then m is the square root of n, i.e., m = √n

How to Solve the Square Root Equation?

The square root equations are solved using the below steps. Square both the sides of the given equation then simplify it to find the answer.

Example: Solve √(9a + 4) – 5 = 0
Solution: Given, √(9a + 4) – 5 = 0
Isolate the square root term first. Then the equation becomes, √(9a + 4) = 5
Now on squaring both the sides, we get; 36a + 16 = 25
36a + 16 = 25
36a = 25 – 16
a = 9/36
a = 1/4

A cube of a binomial can be defined by multiplying itself three times. You can also check both the sum of cubes and the difference of cubes formulas in this article. Find out different problems on a cube of a binomial, procedure to find a cube of a binomial along with detailed steps. Learn How to Expand a Cubic Binomial in Algebraic Expressions and solve the related problems easily. Check out the solved examples on How to Cube Binomials and get to know the concept involved behind them.

Sum of Cubes Formula

The sum of a cubes of two binomials is represented as (a + b)³ = a³ + 3a²b + 3ab² + b³. Add the cube of the first term, three times the square of the first term by the second term, three times the first term by the square of the second term, and also the cube of the second term. By adding these terms you can get the Sum of cubes.

(a + b)³ = a³ + 3a²b + 3ab² + b³
= a³ + 3ab (a + b) + b³

Difference of Cubes Formula

The difference of cubes of two binomials is represented as (a – b)³ = a³ – 3a²b + 3ab² – b³. Subtract the cube of the first term and three times the square of the first term by the second term. Then, add three times the first term by the square of the second term, then subtract the cube of the second term from it.

(a – b)³ = a³ – 3a²b + 3ab² – b³
= a³ – 3ab (a – b) – b³

Solved Problems on Expansion of Cubic Binomial

Simplify the following by cubing
1. (a + 5b)³ + (a – 5b)³

Solution:
Given expression is (a + 5b)³ + (a – 5b)³
In the given expression, the first term is in the form of (a + b)³  and the second term is in the form of (a – b)³ 
Compare the first term (a + 5b)³ with (a + b)³  and expand it.
We know that (a + b)³ = a³ + 3a²b + 3ab² + b³ where a = a, b = 5b
(a + 5b)³ = a³ + 3 . a² . 5b + 3 . a . (5b)² + (5b)³
Compare the second term (a – 5b)³ with (a – b)³  and expand it.
We know that (a – b)³ = a³ – 3a²b + 3ab² – b³ where a = a, b = 5b
(a – 5b)³ = a³ – 3 . a² . 5b + 3 . a . (5b)² – (5b)³
Now, write (a + 5b)³ + (a – 5b)³ = a³ + 3 . a² . 5b + 3 . a . (5b)² + (5b)³ + a³ – 3 . a² . 5b + 3 . a . (5b)² – (5b)³
= a³ + 15a²b + 75ab² + 125 b³ + a³ – 15a²b + 75ab² – 125 b³
= 2a³ + 150ab²
The final answer is 2a³ + 150ab²

Therefore, (a + 5b)³ + (a – 5b)³ = 2a³ + 150ab²

2. (2x + 3y)³ + (2x – 3y)³

Solution:
Given expression is (2x + 3y)³ + (2x – 3y)³
In the given expression, the first term is in the form of (a + b)³  and the second term is in the form of (a – b)³ 
Compare the first term (2x + 3y)³ with (a + b)³  and expand it.
We know that (a + b)³ = a³ + 3a²b + 3ab² + b³ where a = 2x, b = 3y
(2x + 3y)³ = (2x)³ + 3 . (2x)² . (3y) + 3 . (2x) . (3y)² + (3y)³
Compare the second term (2x – 3y)³ with (a – b)³  and expand it.
We know that (a – b)³ = a³ – 3a²b + 3ab² – b³ where a = 2x, b = 3y
(2x – 3y)³ = (2x)³ – 3 . (2x)² . (3y) + 3 . (2x) . (3y)² – (3y)³
Now, write (2x + 3y)³ + (2x – 3y)³ = (2x)³ + 3 . (2x)² . (3y) + 3 . (2x) . (3y)² + (3y)³ + (2x)³ – 3 . (2x)² . (3y) + 3 . (2x) . (3y)² – (3y)³
= 8x³ + 36x²y + 54xy² + 27y³ + 8x³ – 36x²y + 54xy² – 27y³
= 16x³ + 108xy²
The final answer is 16x³ + 108xy²

Therefore, (2x + 3y)³ + (2x – 3y)³ = 16x³ + 108xy²

3. (2 – 3a)³ – (5 + 3a)³

Solution:
Given expression is (2 – 3a)³ – (5 + 3a)³
In the given expression, the first term is in the form of (a – b)³ and the second term is in the form of (a + b)³
Compare the first term (2 – 3a)³ with (a – b)³  and expand it.
We know that (a – b)³ = a³ – 3a²b + 3ab² – b³ where a = 2, b = 3a
(2 – 3a)³ = (2)³ – 3 . (2)² . (3a) + 3 . (2) . (3a)² – (3a)³
Compare the second term (5 + 3a)³ with (a + b)³  and expand it.
We know that (a + b)³ = a³ + 3a²b + 3ab² + b³ where a = 5, b = 3a
(5 + 3a)³ = (5)³ + 3 . (5)² . (3a) + 3 . (5) . (3a)² + (3a)³
Now, write (2 – 3a)³ – (5 + 3a)³ = (2)³ – 3 . (2)² . (3a) + 3 . (2) . (3a)² – (3a)³ + (5)³ + 3 . (5)² . (3a) + 3 . (5) . (3a)² + (3a)³
= {8 – 36a + 54 a² – 27 a³} – {125 + 225a + 135a² + 27 a³}
=  8 – 36a + 54 a² – 27 a³ – 125 – 225a – 135a² – 27 a³
= 8 – 125 – 36a – 225a + 54 a² – 135a² – 27 a³ – 27 a³
= -117 – 261a – 81 a² – 54 a³
The final answer is -117 – 261a – 81 a² – 54 a³

Therefore, (2 – 3a)³ – (5 + 3a)³ = -117 – 261a – 81 a² – 54 a³

4. (5x + 2y)³ – (5x – 2y)³

Solution:
Given expression is (5x + 2y)³ – (5x – 2y)³
In the given expression, the first term is in the form of (a + b)³  and the second term is in the form of (a – b)³ 
Compare the first term (5x + 2y)³ with (a + b)³  and expand it.
We know that (a + b)³ = a³ + 3a²b + 3ab² + b³ where a = 5x, b = 2y
(5x + 2y)³ = (5x)³ + 3 . (5x)² . (2y) + 3 . (5x) . (2y)² + (2y)³
Compare the second term (5x – 2y)³ with (a – b)³  and expand it.
We know that (a – b)³ = a³ – 3a²b + 3ab² – b³ where a = 2x, b = 3y
(5x – 2y)³ = (5x)³ – 3 . (5x)² . (2y) + 3 . (5x) . (2y)² – (2y)³
Now, write (5x + 2y)³ – (5x – 2y)³ = (5x)³ + 3 . (5x)² . (2y) + 3 . (5x) . (2y)² + (2y)³ + (5x)³ – 3 . (5x)² . (2y) + 3 . (5x) . (2y)² – (2y)³
= {125 x³ + 150 x² y + 60 xy² + 8 y³} – {125 x³ – 150 x² y + 60 xy² – 8 y³}
= 125 x³ + 150 x² y + 60 xy² + 8 y³ – 125 x³ + 150 x² y – 60 xy² + 8 y³
= 125 x³ – 125 x³ + 150 x² y + 150 x² y + 60 xy² – 60 xy² + 8 y³ + 8 y³
= 300 x² y + 16 y³
The final answer is 300 x² y + 16 y³

Therefore, (5x + 2y)³ – (5x – 2y)³ = 300 x² y + 16 y³

In the previous articles, we have various details regarding all the fractions such as like and unlike fractions, equivalent fractions, etc. Check the conversion of fractions here along with the formulae and methods. Refer to all the steps to convert decimal to fractions as well as fractions to decimals. The detailed process for the conversion of fractions is given here. Go through the below sections to know the rules, methods, and formulae, etc.

Conversion of Fractions – Introduction

Conversion of the fractions involves various steps. Each step is explained in detail here. We are providing the complete guide to converting fractions to decimals and also decimals to fractions.

How to Convert Decimals to Fractions?

To convert the decimals to fractions, we have a process to follow. Below are the steps to convert decimals to fractions.

1. Note the decimal fraction number as a fraction of the digits number to the right of the numerator i.e., decimal period and denominator to the power of 10.
2. Find the GCD (greatest common divisor) of the denominator and the numerator.
3. Now, reduce the fraction value by dividing the denominator and numerator with the greatest common divisor.

Converting a Terminal Decimal to a Fraction

Terminating decimals are any decimal numbers which has finite digits. In other words, terminating decimal has an end.

Examples: 

0.5,0.234,0.864721 etc.

These decimals are the common decimals you will see and they are the easiest way to convert to fractions.

Step 1:

Write the decimal in the form divided by one.

For suppose, you have the decimal number 0.5. Your first step is to write out the decimal. Hence, it looks like 0.5/1

Step 2:

In the next step, you have to multiply both the bottom and top of the new fraction by 10 to all the digits that are at the left of the decimal point.

For suppose, 0.55 is the decimal fraction. In the given fraction, 0.55 has two digits after the decimal point. Therefore, we have to multiply the entire equation by 10 * 10, or 100. If we multiply the fraction with 100/100 gives the result 55/100

Step 3:

The last step is to reduce the fraction to its simplest form.

Suppose that 0.5 is the number, then to denote the fraction value we write it as 5/10 = 1/2. For the number 0.15, we write it as 15/100. Then the result will be 3/20.

Note:

Use the last digits place value to define the fraction with a denominator of 10,100,1000 etc. Then try for further simplification to get the exact result value.

Converting a Fraction into a Decimal

To convert the fraction values into a decimal value, there are certain steps to be followed.

Step 1:

First of all, change the given fraction value into equivalent fraction values with denominators 10,100,1000, etc.

Suppose that 3/4 is a fraction value. Now, divide the denominator value i.e., 3/4 by 25, so that the value becomes 75/100. We have to concentrate on changing the decimal value to an equivalent fraction.

Step 2:

Multiply both numerator and denominator by that number.

Suppose that 3/4 is a fraction value. Now, divide the denominator value i.e., 3/4 by 25, so that the value becomes 75/100. We have to concentrate on changing the decimal value to an equivalent fraction.

Step 3:

Take the count of no of zeros in the denominator after the 1st digit. Now, put the decimal point in the numerator, starting from the extreme of right, and then move that decimal point to the left and equal the number of zeros.

Converting Fractions to Recurring Decimals

In some of the cases, conversion of fraction value leads to the result of repeating decimal value i.e., the same decimal value recurs forever throughout the similar number pattern.

Example:

Consider 2/3 as the fraction value, to convert the fraction value into the decimal value. First, divide the numerator value 2 by the denominator value 3. To continue the process of division, we have to add trailing zeros to the number 2.

Therefore, we can notice that the division continues, no matter the number of trailing zeros you add to the number 2. In this case, 2/3 = 0.6666….

In the above example, the bar is placed above the repeating integers to define that the number recurs forever.

There is also another case where more than 1 integer recurs in the decimal number either by alternating or consecutively.

Example:

Suppose that you have to convert the fraction value 5/11 to the decimal fraction. First, divide the numerator value 5 by the denominator value 11. To continue the process of division, we have to add trailing zeros to the number 5.

We can notice that the pattern of division repeats every integer 4 and 5. If we add trailing zeros to the original decimal value, then the string out the pattern indefinitely. The final result can be written as 0.45454545…..

Conversion of Fraction Value to Decimal Value when the denominator is a Multiple of 10

Whenever the denominator value of a fraction is the multiple of 10 i.e., 10,100,1000,10000, etc. Then conversion of a fraction value to a decimal value is a straightforward process.

First, the total number of zeros present in the denominator are counted, and then the numerator is written down

When the denominator of a fraction value is a multiple of 10, 100, 1000, 10000, etc. Then conversion of the fraction value to a decimal value is a straightforward process. The numerator is written by placing the decimal point before the number where a number of zeros are present from right to left.

Example:

Suppose 25/100 is a fraction value.

To convert the value 25/100 into a decimal value

We have to check for the number of zeros in the denominator, Hence the denominator value is 100, there are 2 zeros present in the denominator. Therefore, we have to shift the decimal point by 2 points in the numerator. Then the final result is 0.25

Example 2:

Suppose 276/1000 is a fraction value.

To convert the value 276/1000 into a decimal value.

We have to check for the number of zeros in the denominator, Hence the denominator value is 1000, there are 3 zeros present in the denominator. Therefore, we have to shift the decimal point by 3 points in the numerator. Then the final result is 0.276

Conversion by long division method

Conversion of the fraction value to decimal value by using the long division method involves various steps. The steps to be followed are:

Step 1:

In the first step, we convert the dividend value to the most suitable equivalent decimal value.

Whenever the digit to the right of the decimal value is bought down, we have to insert the decimal point in the quotient.

Example: Convert 3/4 into decimals.
Solution:

As already given 3/4 is the fractional value. Here 3 has to be divided by 4. As 3 is less than 4, it cannot be f=divided by 4

Therefore, we can write the value as 3.00 which can be divided by 4.

30 can be divided by 4 for 7 times. Therefore we write the quotient as 0.7, on further division, we get the final result as 0.75