# Simplification Problems

#### Chapter 4

5 Steps - 3 Clicks

# Simplification Problems

### Introduction

Simplification Problems is based on BODMASÂ rule, where

B â†’ Brackets,

O â†’ Of,

D â†’ Division,

M â†’ Multiplication,

S â†’ Subtraction.

### Methods

BODMAS Rule:

BODMAS is about simplifying an expression by firstly removing the brackets in the order i.e. (), {}, []. Removal ofÂ brackets isÂ followed byÂ addition, subtraction, multiplication, division,Â square roots, cube roots, powers, cancellation of numerator/ denominator and so on.

Example 1:
Simplify using BODMAS rule: 25 – 48 Ã· 6 + 12 Ã— 2

Solution:

25 – 48 Ã· 6 + 12 Ã— 2

= 25 – 8 + 12 Ã— 2, (Simplifying â€˜divisionâ€™ 48 Ã· 6 = 8)

= 25 – 8 + 24, (Simplifying â€˜multiplicationâ€™ 12 Ã— 2 = 24)

= 17 + 24, (Simplifying â€˜subtractionâ€™ 25 – 8 = 17)

= 41, (Simplifying â€˜additionâ€™ 17 + 24 = 41)

Example 2:
Simplify using BODMAS rule: 78 – [5 + 3 of (25 – 2 Ã— 10)]

Solution:

78 – [5 + 3 of (25 – 2 Ã— 10)]

= 78 – [5 + 3 of (25 – 20)], (Simplifying â€˜multiplicationâ€™ 2 Ã— 10 = 20)

= 78 – [5 + 3 of 5], (Simplifying â€˜subtractionâ€™ 25 – 20 = 5)

= 78 – [5 + 3 Ã— 5], (Simplifying â€˜ofâ€™)

= 78 – [5 + 15], (Simplifying â€˜multiplicationâ€™ 3 Ã— 5 = 15)

= 78 – 20, (Simplifying â€˜additionâ€™ 5 + 15 = 20)

= 58, (Simplifying â€˜subtractionâ€™ 78 – 20 = 58)

Example 3:
Simplify using BODMAS rule: 52 – 4 of (17 – 12) + 4 Ã— 7

Solution:

52 – 4 of (17 – 12) + 4 Ã— 7

= 52 – 4 of 5 + 4 Ã— 7, (Simplifying â€˜parenthesisâ€™ 17 – 12 = 5)

= 52 – 4 Ã— 5 + 4 Ã— 7, (Simplifying â€˜ofâ€™)

= 52 – 20 + 4 Ã— 7, (Simplifying â€˜multiplicationâ€™ 4 Ã— 5 = 20)

= 52 – 20 + 28, (Simplifying â€˜multiplicationâ€™ 4 Ã— 7 = 28)

= 32 + 28, (Simplifying â€˜subtractionâ€™ 52 – 20 = 32)

= 60, (Simplifying â€˜additionâ€™ 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 âˆ’8 is simply 8.

• The modulus of âˆ’$$\frac{1}{2}$$ 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. | âˆ’ 11 | = 11

3. | 0.25 | = 0.25

4. | âˆ’ 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: $$78 \,- [24 \,- {16 \,- (5 \,â€“ \overline{4 \,- 1})}]$$

Solution:

$$78 \,- [24 \,- {16 \,- (5 \,â€“ \overline{4 \,- 1})}]$$

= 78 – [24 – {16 – (5 – 3)}] (Removing vinculum)

= 78 -[24 – {16 – 2}] (Removing parentheses)

= 78 – [24 â€“ 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 â€“ [1/9 {42 + (56 â€“ 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: $$95 \,- [144 \,Ã· (12 \,\times 12) \,- (-4) \,- \{3 \,â€“ \overline{17 \,- 10}\}]$$

Solution:

$$95 \,- [144 \,Ã· (12 \,\times 12) \,- (-4) \,- \{3 \,â€“ \overline{17 \,- 10}\}]$$

= 95 – [144 Ã· (12 x 12) – (-4) – {3 -7}]

= 95 – [144 Ã· 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]

â‡’a-[a-(a+b)-{a-a+b-a}+2b]

â‡’a-[a-(a+b)-{b-a}+2b]

â‡’a-[a-(a+b)-b+a+2b]

â‡’a-[a-a-b-b+a+2b]

â‡’a-a = 0

Therefore a-[a-(a+b)-{a-(a-b+a)}+2b] = 0

2. Find the value of $$x$$ if $$(\frac{12.24 Ã· x}{3.2 Ã— 0.2})$$ = 2

Solution:

Given expression is $$(\frac{12.24 Ã· x}{3.2 Ã— 0.2})$$ = 2

â‡’$$\frac{12.24}{x}$$ = 2 x 3.2 x 0.2

â‡’$$x$$ = $$\frac{12.24}{1.28}$$

â‡’$$x$$ = 0.011

Hence, the value of $$x$$ = 0.011

3. Find the value of $$\sqrt{30} Ã— \sqrt{10} = ?$$

Solution:

Given that $$\sqrt{30} Ã— \sqrt{10} = ?$$

Consider $$\sqrt{30} Ã— \sqrt{10}$$

â‡’$$\sqrt{{30} Ã— {10}}$$

â‡’$$\sqrt{6 Ã— 5 Ã— 5 Ã—2}$$

â‡’5$$\sqrt{6 Ã— 2}$$

â‡’5$$\sqrt{12}$$

â‡’5$$\sqrt{4 Ã— 3}$$

â‡’5 Ã— 2$$\sqrt{3}$$

â‡’10$$\sqrt{3}$$

Therefore the value of $$\sqrt{30} Ã— \sqrt{10}$$ = 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$$\frac{3}{11}$$

â‡’$$\frac{56}{9}$$ + $$\frac{25}{9}$$ – $$\frac{47}{11}$$ + $$\frac{14}{11}$$

Now L.C.M. of 9, 9, 11, 11 is 99

â‡’$$\frac{616 + 275 – 423 + 126}{99}$$

â‡’$$\frac{1017 – 423}{99}$$

â‡’$$\frac{594}{99}$$

â‡’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,

â‡’ a = 4$$b^2$$

Now consider $$\frac{b^2}{a}$$

â‡’$$\frac{1}{4}$$ or 0.25

âˆ´ The value of $$\frac{b^2}{a}$$ = $$\frac{1}{4}$$ or 0.25

6. Find the unknown value from 25% of 180 = ? Ã· 0.25

Solution:

Assume the unknown value as $$x$$

Given 25% of 180 = ? Ã· 0.25

Substitute $$x$$ i.e.

$$\frac{25}{100}$$ Ã— 180 = $$x$$ Ã— $$\frac{1}{0.25}$$

â‡’$$x$$ = $$\frac{25 Ã— 180 Ã— 0.25}{100}$$

â‡’$$x$$ = $$\frac{1125}{100}$$

â‡’$$x$$ = 11.25

Therefore, the unknown value $$x$$ = 11.25