Simplifications and Approximations forms one of the most common topics in the section on quantitative aptitude. In the section on simplification, we will learn tips and tricks that have been asked in various banking exams and other graduate level courses. We will try and form memory maps, state and develop formulas, and see how to apply them. Let us begin!

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## Approximations and Simplifications

Let us start by stating important rules and formulae that we shall be using in the below space to solve problems. Let us start with the infamous “BODMAS” or the “PEMDAS” rule.

### BODMAS Rule

BODMAS rule: This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of a given expression. Here ‘B’ stands for ‘Brackets’, ‘O’ for ‘of’, ‘D’ for ‘Division’, ‘M’ for ‘Multiplication’, ‘A’ for ‘Addition’ and ‘S’ for Subtraction. These steps have to be followed in that order to get a correct answer from all arithmetic question.

Thus, in simplification, an expression, first of all, the brackets must be removed (by solving the expressions inside), strictly in the order (), {} and [].

After removing the brackets, we must use the following operations strictly in the order: (i) of – also as Order (Power/ Exponents) (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.

“PEMDAS” is the same rule. Here ‘P’ stands for Parenthesis, ‘E’ for exponents i.e. Powers, ‘M’ for Multiplication, ‘D’ for Division, ‘A’ for addition, and ‘S’ for subtraction. So basically both are the same.

### Modulus Of A Real Number

The modulus or the magnitude of a real number ‘a’, written as |a|, is defined as:

|a| = a, if a > 0 and |a| = -a, if a < 0. For example the |5| = 5 and |-5| = -(-5) = 5. The modulus is defined in a way such that it always returns a positive value.

Let us solve some examples to get more aware of the concept and also some practice.

## Solved Examples For You

### Part I

Example 1: What value will replace the question mark in the following equation?

4(1/2) + 3(1/6) + ? + 2(1/3) = 13(2/5).

A) 3(2/5) B) 3(3/5) C) 2(1/2) D) Data not sufficient

Answer: Let ‘x’ be the missing fraction. We will have: 9/2 + 19/6 + x + 7/3 = 67/5.

Then, x = 67 – [(9/2) + (19/6) + (7/3)]. Using the BODMAS or the PEMDAS rule, we can write that: x = 67/5 – [(27+ 19 = 14)/6] = [(67/5) – (60/6)].

In other words we can write: x = [67/5 – 10] = 17/5 = 3(2/5). Hence the missing fraction is = 3(2/5) and the correct option is A) 3(2/5).

Example 2: Simplify the following: b – [b – (a + b) – {b – (b – (a -b))} + 2a]?

A) 4a B) 6a C) 8a D) 10a

Answer: Let us rewrite the expression as b -[b – (a+b) – {b – (b – a + b)} + 2a].

Which gives = b – [b – a – b – {b – 2b + a} + 2a]

We can write it as: b – [- a a{b – 2b + a + 2a}] or b – [ -a -{-b + 3a}] = b – [-a + b – 3a]

In other words we have: b – [-4a + b] = b + 4a – b = 4a. Thus the correct answer is A) 4a.

Example 3: 4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number? [SBI – PO 2009]

A) 112 B) 216 C) 315 D) 412

Answer: Let the number be x. Then, 4/15 of 5/7 of x – 4/9 of 2/5 of x = 8. This implies that (4/21)x – (8/45)x = 8. In other words, we have [4/21 – 8/45]x = 8.

Thus we have [(60 – 56)/315]x = 8 or 4x/315 = 8. This gives the value of ‘x’ as = [(8×315)/4] = 630 or (1/2)x = 315. Hence the required number is 315 and the correct option is C) 315.

### Part II

Example 4: If 2x + 3y + z = 55, x + z – y = 4 and y – x + z = 12, then what are the values of x, y and z? [Bank P.O. 2003]

A) 7, 11, 8 respectively B) 12, 4, 2 respectively C) 11, 19, 15 respectively D) 13, 14, 15 respectively

Answer: The given equations are:

2x + 3y + z = 55 ……….(i); x + z – y = 4 ………(ii); y – x + z = 12 …….(iii)

Subtracting (ii) from (i), we get: x + 4y = 51 ……(iv)

Subtracting (iii) from (i), we get: 3x = 2y = 43 …..(v)

Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x = 35 or x = 7. Putting x = 7 in (iv), we get: 4y = 44 or y = 11. Putting x = 7, y = 11 in (i), we get z = 8. Hence the correct option is A) 7, 11, 8 respectively.

Example 5: A man divides Rs. 8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much ass each nephew, and each son receives five times as much as each nephew, how much does each daughter receive? [SSC 2000]

A) Rs. 800 B) Rs. 600 C) Rs. 1600 D) Rs. 1800

Answer: Let the share of each nephew be Rs. x. Then, share of each daughter = Rs. (4x); share of each son = Rs. (5x). So, 5(5x) + 4(4x) + 2x = 8600. This implies that 25x + 16x + 2x = 8600.

In other words, we have 43x = 8600 and thus x = 200.

Therefore the share of each daughter = Rs. (4×200) = Rs. 800. Thus the correct option is A) Rs. 800

## Practice Questions

Q 1: A man spends 2/5 of his salary on his salary on house rent, 3/10 of his salary on food and 1/8 of his salary on conveyance. if he has Rs. 1400 left with him, find his expenditure on food and conveyance.

A) Rs 200 B) Rs 1400 C) Rs 1000 D) Rs 2100

Ans: C) Rs 1000

Q 2: In a certain office, 1/3 of the workers are women, 1/2 of the women are married and 1/3 of the married women have children. if 3/4 of the men are married and 2/3 of the married men have children, what part of workers are without children?

A) 1/2 of all workers B) 11/12 of all workers C) 11/16 of all workers D) 11/18 of all workers

Ans: D) 11/18 of all workers