Quantitative Aptitude - SPLessons

Simplification Problems

Chapter 4

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Simplification Problems

shape Introduction

Simplification Problems is based on BODMAS rule, where


    B → Brackets,

    O → Of,

    D → Division,

    M → Multiplication,

    A → Addition, and

    S → Subtraction.


shape 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


shape 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