Simplification questions are the most frequently asked questions in quantitative aptitude section for various Government exams. Candidates those who are preparing for various competitive exams must have to know the simple tricks to solve the simplification questions.

What is simplification

A combination of numbers connected by one or more of the symbols +, -, x,÷ and of is called a numerical expression . Thus (i) 15 +8÷ 4-6 (ii) 16-4 $\times$ 3+8÷ 2 are examples of numerical expression. Brackets are used to separate various parts of an expression. There are four kinds of brackets:

(i) Bar or Vinculum $\overline{}$

$\overline{ }$ (

)

(ii) Round brackets or small brackets ( )

(iii) Curly brackets or braces { }

(iv) Square brackets or Big brackets [ ]

The order of working with brackets is given below.

(i)Bar (ii) Small brackets (iii) Curly brackets (iv) Big brackets

The rule to simplify an expression is ‘BODMAS‘ means

B -brackets O- Order(powers/indices or roots) D-Division M= Multiplication A-Addition

S-Subtraction

BODMAS

Note : In the absence of any sign before a bracket , we take the sign as multiplication.

Simplification tips and tricks

No matter which exam you are preparing for, simplification is one topic that can really help you to improve your marks. Given below are some of the tips and tricks by which you can solve Simplification problems easily.

Learn the concept of BODMAS and practice .
Memorize tables upto 30 and learn the different divisibility rules as it will help you to increase your speed.
Memorize cubes and squares of numbers up to 30.
Learn some basic algebraic identities. some of them are given below .

Learn the concept of percentages (conversion of fractions to percentage & percentage to fractions) .

Simplification question (Solved examples)

BODMAS Simplification question

(i) Simplify 36 − [18 − {14 − (15 − 4 ÷ 2 × 2)}]

Sol. Given Expression

36 − [18 − {14 − (15 − 4 ÷ 2 × 2)}]

= 36-[18-{14-(15-2 x 2 )}] (solving division)

= 36-[18-{14-(15-4)}} (performing multiplication)

=36-[18-{14-11}} ( removing small brackets)

=36-[18-3] (removing curly braces)

=36-15 (removing square brackets)
= 21

(ii) Solve 27-[18-{16-(5- $\overline{4-1}$ )}]

Given Expression 27-[18-{16-(5- $\overline{4-1}$ )}]

=27-[18-{16-(5-3)}] (removing bar)

=27-[18-{16-2}] (removing small brackets)

=27-[18-14] (removing curly braces)

=27-4 ((removing square brackets)

=23

Fraction simplification

(i) Solve

Sol.

= = $\large \frac{172-123+14}{42}=\frac{63}{42}=\frac{3}{2}$

(ii) $\large \mathbf{999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}=?}$

Sol. $999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}$ $=999+999+999+999+999+999+\left ( \frac{1}{7}+\frac{2}{7}+\frac{3}{7}+\frac{4}{7}+\frac{5}{7}+\frac{6}{7}\right )$

= $(6\times 999)+ \frac{\left ( 1+2+3+4+5+6 \right )}{7}=5994+\frac{21}{7}=5994+3$

=5997

(iii) If $\mathbf{x+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}=2}$ then $\mathbf{x=?}$

Sol. $x+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}=2\Rightarrow x+\frac{1}{1+\frac{1}{\frac{13}{4}}}=2$

$\Rightarrow x+\frac{1}{\frac{17}{13}}=2$

$\Rightarrow x+\frac{13}{17}=2\Rightarrow x=(2-\frac{13}{17})$ . Thus $x=\frac{21}{17}$

Decimal Simplification

(i) Solve $\mathbf{\frac{0.07\times 0.07\times 0.07-0.05\times 0.05\times 0.05}{0.07\times 0.07+ 0.07\times 0.05 +0.05\times 0.05}}$

Sol. $\frac{0.07\times 0.07\times 0.07-0.05\times 0.05\times 0.05}{0.07\times 0.07+ 0.07\times 0.05 +0.05\times 0.05}=\frac{ (0.07)^{3}-(0.05)^{3}}{(0.07)^{2}+0.07\times 0.05+(0.05)^{2}}$

= $\frac{(0.07-0.05)((0.07)^{2}+0.07\times 0.05+(0.05)^{2})}{(0.07)^{2}+0.07\times 0.05+(0.05)^{2}}$

(using $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ )

$=0.07-0.05= 0.02$

Simplification based on algebraic identities

(i) Solve $\mathbf{\frac{47\times 47\times 47+35\times 35\times 35+26\times 26\times 26-47\times 35\times 78}{47\times 47+35\times 35+26\times 26-47\times 35-35\times 26-47\times 26}}$

Sol. $\frac{47\times 47\times 47+35\times 35\times 35+26\times 26\times 26-47\times 35\times 78}{47\times 47+35\times 35+26\times 26-47\times 35-35\times 26-47\times 26}$

= $\frac{(47)^{3}+(35)^{3}+(26)^{3}-3\times 47\times 35\times 26}{(47)^{2}+(35)^{2}+(26)^{2}-47\times 35-35\times 26-47\times 26}$

= $\frac{(47+35+26)((47)^{2}+(35)^{2}+(26)^{2}-47\times 35-35\times 26-47\times 26)}{(47)^{2}+(35)^{2}+(26)^{2}-47\times 35-35\times 26-47\times 26}$

Now using $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

=47+35+26= 108

(ii) $\mathbf{\frac{(941+149)^{2}+(941-149)^{2}}{941 \times 941+149\times 149}}$

Sol. $\frac{(941+149)^{2}+(941-149)^{2}}{941 \times 941+149\times 149}$ = $\frac{941^{2}+149^{2}+2 \times 941 \times 149+941^{2}+149^{2}-2 \times 941 \times 149}{941^{2}+149^{2}}$

( $\because (a+b)^{2} =a^{2}+b^{2}+2 ab$ and $(a-b)^{2} =a^{2}+b^{2}-2 ab$ )

= $\frac{2(941^{2}+149^{2})}{(941^{2}+149^{2})}$

(iii) $\mathbf{\frac{888\times 888\times 888-222\times 222\times 222}{888\times 888+888\times 222+222\times 222}}$

Sol. $\frac{888\times 888\times 888-222\times 222\times 222}{888\times 888+888\times 222+222\times 222}=\frac{888^{3}-222^{3}}{888^{2}+888\times 222+222^{2}}$