__VBODMAS RULE__

We’ve been cramming **BODMAS **since school, but its actually **VBODMAS**. A yes!! That’s why the rule here is **VBODMAS RULE.**

**V**, here is Vinculum which is actually called bar.

We can easily learn anything if it’s in most simpler form. Let me make it simpler for you by making it in tabular form.

Sr.No. | ALPHABET | FULL NAME | SYMBOL |

1. | V | Vineculum or bar | (—) |

2. | B | Brackets | () , { }, [ ] |

3. | O | Of | Of repalced by × |

4. | D | Division | ÷ or / |

5. | M | Multiplication | × or dot(.) |

6. | A | Addition | + |

7. | S | Subtraction | – |

Nothing is more clear if you don’t solve an example. Right??

So here is an example

**E.g.,** {2+9×2 of 4} + 12 – 9 ÷ 3

**Step 1:**– ** V** i.e. bar 12 – 9 = 3

**Step 2:-** ** B** , here we have {}

But wait!!! inside the bracket we have **of**,** + **and **× **so we solve here for these using VBODMAS again….

Lets first solve inside the bracket

We have, 2+9×2 of 4

According to BODMAS, {Ahan!! its VBODMAS but never mind for simplicity you can.}

We get, 2+9×8 =2+72=74

**Step3:-** Now we have,

74+3÷3= 74+1= 75.

It was easy. Isn’t it?

Look at the question asked below:-

e.g., 5÷5 of 5×2+2÷2 of 2×5-(5-2)÷6×2 **( SSC-CGL)**

Using **VBODMAS**,

__Step1__**:- Bracket**

So, 5-2=3

__Step2__**:- Of**

5÷(5×5)×2+2÷(2×2)×5-3÷6×2 {I have used( ) to just have clarity }

=5÷25×2+2÷4×5-3÷6×2

__STEP3__**:- Division**

(5÷25)×2+(2÷4)×5-(3÷6)×2

=1/5 × 2+ 1/2 × 5 – 1/2 × 2

__STEP4__**:- Multiplication**

=2/5 + 5/2 – 1

__STEP4__**:- Addition**

**=**25/10 -1

And finally,

__Step5__**:- Subtraction**

**=** 29 – 10 / 10=19/10.

So, here we have our answer.

Thus, you can see that VBODMAS makes even tedious calculation easy.

As now we are friendly with VBODMAS , let’s look into fractions which is a common topic we function with even in our day to day lives.

You must have heard quite commonly, one- fourth of a percent, 3/4^{th} part of a cake, and of course we say quite often that half (1/2) of my syllabus is incomplete etc

__FRACTIONS__

We are already introduced to fractions. We shall now have a quick glance at the types of fractions.

__TYPES OF FRACTIONS__** :-**

**1. **Proper fraction:

Fractions whose numerators are less than the denominators are called proper fractions. (Numerator < denominator)

e.g., 2/3, 3/4, 4/5, 5/6, 6/7, 7/8 2/7, 1/5, etc are proper fractions.

**2.** Improper fraction:

Fractions with the numerator either equal to or greater than the denominator are called improper fraction. (Numerator = denominator or, Numerator > denominator)

Fractions like 6/4, 23/15, 56/29 etc. are improper fractions.

**REMARKS:-**

- Every natural number can be written as a fraction in which 1 is it’s denominator. For example, 3 can be written as 3/1, 23 =23/1 , etc.

So every natural number is an improper fraction.

- The value of an improper fraction is always equal to or greater than 1.

**3.** Mixed fraction:

A combination of a proper fraction and a whole number is called a mixed fraction*.*

are examples of mixed fraction.

Let’s look at some properties of mixed fractions

__Property 1:__

A mixed fraction may always be converted into an improper fraction.

This can be done as –

Multiply the number by the denominator and add to the numerator. This new numerator over the denominator is the required fraction.

= 9×1+4= 13.

Was it simple, right?

__Property 2:__

An improper fraction can be always be converted into a mixed fraction.**Example:** Like you have .

What you can simply do is – simple divide as,

Dividing 14 by 9, we get quotient = 1 and remainder = 5.

Therefore, 14/9 = 1^{5}/_{9}

__Note:__

- Proper fraction is between 0 to 1.
- Improper fraction is 1 or greater than 1.
- Mixed fraction is greater than 1.

This is something you should generally know, alright then.

As you are now quite familiar with fraction and its types so without further ado let’s look some of the questions asked in the QUANT SECTION OF SSC CGL**.**

We begin by converting these mixed fractions into improper fractions.

By solving these we get,

As, we have already learnt about VBODMAS it won’t be difficult to solve the given equation using it.

__FRACTIONS AND DECIMALS__

We generally talk about big fractions in the form of decimal. Like, you call half as 0.5 or one- fourth as 0.25.

There is nothing new to it. We just divide the numerator with the denominator and insert a decimal when we see that the numerator is smaller than the denominator and add 0 on the right hand side of the numerator. For example, look below –:

If I want to divide 1 by 2 and convert it into decimal form then,

2)10(0.5

This was quite basic, no? But converting decimals into fractions is what is more important as per exam perspective.

There are two types of decimals-

**Recurring (or non-terminating)-**A decimal in which a digit or a set of digit is repeated continuously.

e.g. 1/3 = 0.3333…..

8/3 = 2.6666…

They are written as –

**Non- recurring (or terminating)-**we know these types of decimals.

**Like, ** 5/2 = 2.5 , 34/17 =2 , etc.

__CONVERTING RECURRING DECIMALS INTO FRACTIONS__

This can be better understood with the help of an example.

Express the below one in the form of fractions.

Here is the way to solve this

Let x= 0.66666 ——(1)

Multiply eq. (1) by 10 ,we get

10x= 6.6666 ——(2)

Subtract (2) by (1)

- 9x= 6
- x= 6/9 = 2/3

A recurring decimal is equivalent to a fraction which is formed by the recurring digits for its numerator and for its denominator the number which has for its digits as many nines as there are digits in the period.

EXAMPLE- 0.225 = 225/999

As exam point of view following examples have been asked in the form of decimals in SSC CGL QUANT.

**QUES:- **(5.6 × 0.36 + 0.42 × 3.2)/ 0.8 × 2.1 =? **(SSC CGL, TIER-II)**

**Solution:- ** We can easily solve this by applying BODMAS bt to make it simpler we can separate the two addition terms with same base,as-

5.6 × 0.36/ 0.8 × 2.1 + 0.42 × 3.2/ 0.8 × 2.1 = 1.2+0.8 = 2 Ans.

**QUES:- ** If A= 0.216 + 0.008/ 0.36 + 0.04 – 0.12 and B = 0.729 – 0.027 / 0.81 + 0.09 + 0.27

then,find the value of (A^{2} + B^{2})^{2 }.

**Solution:-** Given,

A = 0.216 + 0.008/ 0.36 + 0.04 – 0.12

= (0.6)^{3} + (0.2)^{3 }/(0.6)^{2}+(0.2)^{2}– 0.6×0.2 [ since, a^{3}+b^{3}= a^{2}+b^{2}-a.b]

= 1

Similarly,

B = 0.729 – 0.027 / 0.81 + 0.09 + 0.27

=(0.9)^{3} + (0.3)^{3} /(0.9)^{2}+(0.3)^{2}– 0.9×0.3[ since, a^{3}+b^{3}= a^{2}+b^{2}+a.b]

= 1

**Therefore,** (A^{2} + B^{2})^{2 }= (1+1)^{2}= 4.