Simplifications Problems:
We must remember all basic concepts and tricks used to solve these numerical on simplification.
1) Numbers
2) H.C.F. and L.C.M.
3) Decimal fractions
4) Square and cube roots
5) Surds and indices
Important points to Remember:
1) This chapter is very important because numerical on simplification is based on concepts related to decimal to fraction, square and cube roots, H.C.F. and L.C.M., numbers, etc.
2) How to find H.C.F. and L.C.M.
3) Tricks to solve decimal fractions
4) Number system, types of numbers
5) How to find square root and cube root
Basic Formulae: (Must Remember)
1) (a – b)2 = (a2 + b2 – 2ab)
2) (a + b)2 = (a2 + b2 + 2ab)
3) (a + b) (a – b) = (a2 – b2 )
4) (a3 + b3) = (a + b) (a2 – ab + b2)
5) (a3 – b3) = (a – b) (a2 – ab + b2)
6) (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7) (a3 + b3 + c3 – 3abc) = (a + b + c) (ar>2 + b2 + c2 – ab – bc – ac)
Quick Tips and Tricks:
1) Virnaculum : If a given expression contains a virnaculum (bar), then first simplify the given expression under virnaculum and then follow the rule of ‘BODMAS’.
Step1: When simplifying the given expressions, the first brackets are to be removed in the order: ‘––’, ‘( )’, ‘{ }’, ‘[ ]’
Step2: The operations are to be performed strictly in the order: Division, Multiplication, Addition, and Subtraction
2) ‘BODMAS Rule’: This rule is used to solve and find out the value of given expressions by performing the operations in a correct sequence.
BODMAS is the shortcut used to remember the procedure of simplification.
B: Bracket
O: Order (i.e. power, square root, etc.)
D: Division (Left to right)
M: Multiplication (Left to right)
A: Addition (Left to right)
S: Subtraction (Left to right)
Step1: When simplifying the given expressions, first brackets are to be removed in the order: ( ), { }, [ ]
Step2: The operations are to be performed strictly in the order: Division, Multiplication, Addition, and Subtraction
3) Modulus of a real number
The modulus of a real number or the absolute value is defined as follows:
The modulus of x is written as |x|
| |x| = | x, if x > 0 | |
| – x, if x < 0 |
Click here to download some example simplification problems: Simplification problems with an example question and answer pdf