# Simplification tricks, rule, methods with example BODMAS Rule (Mathematics)- Simplification BODMAS rule, rule, methods, solution with the help of various examples and short methods to solve simplification questions in exam

## Rule for Simplification BODMAS  rule, tricks, methods, solution with the help of various examples and short methods to solve simplification questions in exam

Name of Rule:   BODMAS Rule

B:                            Bracket

O:                           of

D:                            Division

M:                          Multiplication

S:                            Subtraction

Thus, simplifying an expression, first of all brackets must be removed, strictly in the Order:

1. ()
2. {}
3. []

After removing the bracket, we must use the following operations strictly in order

1. Of
2. Division
3. Multiplication
5. Subtraction

Question 1)        5005 – 5000 ÷ 10

Steps 1) Apply BODMAS

Step 2) Change Sign of Divide into Multiply i.e ( ÷ 10 ) convert with ( ×  )

Solution:

= 5005 – 5000 ×

= 5005 – 500

Question 2)    Question regarding – Simplification tricks, rule, methods, solution

4   +  3   +  ?  +  2   =  13

Solution)

4   +  3   +  ?  +  2   =  13

Step 1) Convert mixed fraction into simple fraction

i.e. mixed fraction (4   ) convert into (   ) simple fraction

—————————

Method of Conversion mixed fraction into Simple fraction

= 4

= 4 +

=

——————————–

+      +  x  +    =

Firstly find the value of x

Shift these values to the right hand side ( Note: while shifting, if the values are positive before shifting, then the values become negative after shifting )

x =   – (    +  +   )

x =   – (   )

x =   –

x =   – 10

x =

x =   or It can also be written as  3

Question 3 :

1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 ÷ 2 )}]

Solution:

Apply BODMAS Rule

Step 1) Change Sign of Divide into Multiply i.e ( ÷ 2 ) convert with ( ×  )

= 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 + 1 ×   )}]

Step 2) Start solving from the inner most bracket

= 1 ÷ [1 + 1 ÷ { 1 + 1 ÷ ( 1 +   )}]

= 1 ÷ [1 + 1 ÷ { 1 + 1 ÷    }]

= 1 ÷ [1 + 1 ÷ { 1 + 1 ÷    }]

= 1 ÷ [1 + 1 ÷ { 1 + 1 ×    }]

= 1 ÷ [1 + 1 ÷ { 1 +    }]

= 1 ÷ [1 + 1 ÷    ]

= 1 ÷ [1 + 1 ×    ]

= 1 ÷ [1 +    ]

= 1 ÷

= 1 ×

=

Question 4) Question regarding –  Simplification tricks, rule, methods, solution for the same.

A man divides Rs. 8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew and each son receives five times as much as each nephew, how much does each daughter receive?

Solution)

In the above statement the man is sharing money between Sons, Daughters and nephew and everybody is receiving more money that nephew.

So, Let’s take the share for nephew = x

Share for each daughter = Rs. 4x (i.e. 4 times more that nephew)

Share for each son = Rs. 5x (i.e. 5 times more that nephew)

Therefore, Total amount 8600 = 5 Sons (5x) + 4 Daughters (4x) +2 nephew (x)

So,

8600 = 5 (5x) + 4 (4x) + 2x

8600 = 25x + 16x + 2x

8600 = 43x

= x

200 = x

Share of Daughter’s = 4x = 4 × 200 = 800

Check / Verify:

Total amount Rs. 8600 = 5 Sons × (5 × 200) + 4 Daughters × (4× 200) +2 nephew × 200

= 5000 + 3200 + 400

= 8600