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

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