# Simplification Problems

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# Simplification Problems

### Introduction

Simplification Problems is based on BODMAS rule, where

B → Brackets,
O → Of,
D → Division,
M → Multiplication,
S → Subtraction.

### Methods

BODMAS Rule:
BODMAS is about simplifying an expression by firstly removing the brackets in the order i.e. (), {}, []. Removal of brackets is followed by addition, subtraction, multiplication, division, square roots, cube roots, powers, cancellation of numerator/ denominator and so on.
Example 1: Simplify using BODMAS rule: 25 - 48 ÷ 6 + 12 × 2
Solution:
25 - 48 ÷ 6 + 12 × 2
= 25 - 8 + 12 × 2, (Simplifying ‘division’ 48 ÷ 6 = 8)
= 25 - 8 + 24, (Simplifying ‘multiplication’ 12 × 2 = 24)
= 17 + 24, (Simplifying ‘subtraction’ 25 - 8 = 17)
= 41, (Simplifying ‘addition’ 17 + 24 = 41)

Example 2: Simplify using BODMAS rule: 78 - [5 + 3 of (25 - 2 × 10)]
Solution:
78 - [5 + 3 of (25 - 2 × 10)]
= 78 - [5 + 3 of (25 - 20)], (Simplifying ‘multiplication’ 2 × 10 = 20)
= 78 - [5 + 3 of 5], (Simplifying ‘subtraction’ 25 - 20 = 5)
= 78 - [5 + 3 × 5], (Simplifying ‘of’)
= 78 - [5 + 15], (Simplifying ‘multiplication’ 3 × 5 = 15)
= 78 - 20, (Simplifying ‘addition’ 5 + 15 = 20)
= 58, (Simplifying ‘subtraction’ 78 - 20 = 58)

Example 3: Simplify using BODMAS rule: 52 - 4 of (17 - 12) + 4 × 7
Solution:
52 - 4 of (17 - 12) + 4 × 7
= 52 - 4 of 5 + 4 × 7, (Simplifying ‘parenthesis’ 17 - 12 = 5)
= 52 - 4 × 5 + 4 × 7, (Simplifying ‘of’)
= 52 - 20 + 4 × 7, (Simplifying ‘multiplication’ 4 × 5 = 20)
= 52 - 20 + 28, (Simplifying ‘multiplication’ 4 × 7 = 28)
= 32 + 28, (Simplifying ‘subtraction’ 52 - 20 = 32)
= 60, (Simplifying ‘addition’ 32 + 28 = 60)

Modulus of a real number:
In many engineering calculations you will come across the symbol "| |". This is known as the modulus.
The modulus of a number is its absolute size. That is, we disregard any sign it might have.
Examples:

• The modulus of −8 is simply 8.

• The modulus of −$\frac{1}{2}$ is $\frac{1}{2}$.

• The modulus of 17 is simply 17.

• The modulus of 0 is 0.

So, the modulus of a positive number is simply the number.
The modulus of a negative number is found by ignoring the minus sign.
The modulus of a number is denoted by writing vertical lines around the number.
This observation allows us to define the modulus of a number quite concisely in the following way. $| a | = \begin{cases} a & \quad \text{if } a > 0\\ -a & \quad \text{if } a < 0 \end{cases}$
Examples:

1. | 9 | = 9
2. | − 11 | = 11
3. | 0.25 | = 0.25
4. | − 3.7 | = 3.7

Virnaculum (or Bar):
When an expression contains Virnaculum, before applying the 'BODMAS'rule, we simplify the expression under the Virnaculum.
Example 1: Simplify: $78 \,- [24 \,- {16 \,- (5 \,– \overline{4 \,- 1})}]$
Solution:

$78 \,- [24 \,- {16 \,- (5 \,– \overline{4 \,- 1})}]$
= 78 - [24 - {16 - (5 - 3)}] (Removing vinculum)
= 78 -[24 - {16 - 2}] (Removing parentheses)
= 78 - [24 – 14] (Removing braces)
= 78 - 10
= 68.

Example 2: Simplify: $197 \,- [1/9 \{42 \,+ (56 \,- \overline{8 \,+ 9})\} \,+ 108]$
Solution:
$197 \,- [1/9 \{42 \,+ (56 \,- \overline{8 \,+ 9})\} \,+ 108]$
= 197 – [1/9 {42 + (56 – 17)} + 108] (Removing vinculum)
= 197 - [1/9 {42 + 39} + 108] (Removing parentheses)
= 197 - [(81/9) + 108] (Removing braces)
= 197 - [9 + 108]
= 197 - 117
= 80

Example 3: Simplify: $95 \,- [144 \,÷ (12 \,\times 12) \,- (-4) \,- \{3 \,– \overline{17 \,- 10}\}]$
Solution:
$95 \,- [144 \,÷ (12 \,\times 12) \,- (-4) \,- \{3 \,– \overline{17 \,- 10}\}]$
= 95 - [144 ÷ (12 x 12) - (-4) - {3 -7}]
= 95 - [144 ÷ 144 - (-4) - {3-7)]
= 95 - [1 - (-4) - (-4)] [Performing division]
= 95 - [1 + 4 + 4]
= 95 - 9
= 86

### Samples

1. Simplify a-[a-(a+b)-{a-(a-b+a)}+2b]?
Solution:
Given that a-[a-(a+b)-{a-(a-b+a)}+2b]
⇒a-[a-(a+b)-{a-a+b-a}+2b]
⇒a-[a-(a+b)-{b-a}+2b]
⇒a-[a-(a+b)-b+a+2b]
⇒a-[a-a-b-b+a+2b]
⇒a-a = 0
Therefore a-[a-(a+b)-{a-(a-b+a)}+2b] = 0

2. Find the value of $x$ if $(\frac{12.24 ÷ x}{3.2 × 0.2})$ = 2
Solution:
Given expression is $(\frac{12.24 ÷ x}{3.2 × 0.2})$ = 2
⇒$\frac{12.24}{x}$ = 2 x 3.2 x 0.2
⇒$x$ = $\frac{12.24}{1.28}$
⇒$x$ = 0.011
Hence, the value of $x$ = 0.011

3. Find the value of $\sqrt{30} × \sqrt{10} = ?$
Solution:
Given that $\sqrt{30} × \sqrt{10} = ?$
Consider $\sqrt{30} × \sqrt{10}$
⇒$\sqrt{{30} × {10}}$
⇒$\sqrt{6 × 5 × 5 ×2}$
⇒5$\sqrt{6 × 2}$
⇒5$\sqrt{12}$
⇒5$\sqrt{4 × 3}$
⇒5 × 2$\sqrt{3}$
⇒10$\sqrt{3}$
Therefore the value of $\sqrt{30} × \sqrt{10}$ = 10$\sqrt{3}$.

4. Simplify 6$\frac{2}{9}$ + 2$\frac{7}{9}$ - 4$\frac{3}{11}$ + 1$\frac{3}{11}$?
Solution:
Given that
6$\frac{2}{9}$ + 2$\frac{7}{9}$ - 4$\frac{3}{11}$ + 1$\frac{3}{11}$
⇒$\frac{56}{9}$ + $\frac{25}{9}$ - $\frac{47}{11}$ + $\frac{14}{11}$
Now L.C.M. of 9, 9, 11, 11 is 99
⇒$\frac{616 + 275 - 423 + 126}{99}$
⇒$\frac{1017 - 423}{99}$
⇒$\frac{594}{99}$
⇒6
Therefore 6$\frac{2}{9}$ + 2$\frac{7}{9}$ - 4$\frac{3}{11}$ + 1$\frac{3}{11}$ = 6.

5. If $\sqrt{a}$ = 2b then find the value of $\frac{b^2}{a}$?
Solution:
Given
$\sqrt{a}$ = 2b
By taking Square root to other side,
⇒ a = 4$b^2$
Now consider $\frac{b^2}{a}$
⇒$\frac{1}{4}$ or 0.25
∴ The value of $\frac{b^2}{a}$ = $\frac{1}{4}$ or 0.25

6. Find the unknown value from 25% of 180 = ? ÷ 0.25
Solution:
Assume the unknown value as $x$
Given 25% of 180 = ? ÷ 0.25
Substitute $x$ i.e.
$\frac{25}{100}$ × 180 = $x$ × $\frac{1}{0.25}$
⇒$x$ = $\frac{25 × 180 × 0.25}{100}$
⇒$x$ = $\frac{1125}{100}$
⇒$x$ = 11.25
Therefore, the unknown value $x$ = 11.25