**In this post, we will discuss few smart methods to solve Simplification Questions that are asked in IBPS PO Exam**.

Simplification questions are often the simplest but the most time-consuming questions in IBPS PO Exam. We always tend to avoid Simplification Questions as they are time-consuming. But if you understand the concept of BODMAS it becomes a cakewalk for you to solve these simplification questions.

This post is the second part of the series on simplification. In the first post, Simplification I- Simplification Tricks to solve simplification Problems, we have discussed the concept of BODMAS and solved three basic examples of simplification questions using Simplification Tricks. In this post, we will solve few more examples of simplification questions using the smart method which will help you to save time in IBPS PO, SSC CGL and SBI PO Exam.

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**Example 4: Simplification Questions for IBPS PO and SSC CGL Exam**

**Question:**

1) 2/5 2) 3/5 3) 5/2 4) 5/3 5) 4/7

**Solution:**

**Step 1:**

First, convert the mixed fraction into improper fraction

55/17 ÷ 75/34 - 34/25 = (?)²

**Step 2:**

According to the BODMAS rule, we first need to perform division operation.

Instead of division we can reciprocate and multiply the numbers-

(55/17 x 34/75) - 34/25 = (?)²

**Step 3:**

Then performing the multiplication operation, we get

114/75 - 34/25 = (?)²

**Step 4:**

Now as we need to subtract the numbers, we take the LCM of the denominator and subtract the digits.

(114/75) - (34/25) = (?)² (when 34/25 is multiplied by the digit 3)

(114 - 102)/75 = (?)²

12/75 = (?)² (When simplified)

√4 / √25 = ?

? = 2/5

Therefore, the answer is option 1: 2/5

**Example 5: Simplification Questions for IBPS PO and SSC CGL Exam**

**Question:**

1) 23/13 2)13/23 3)25/13 4)25/26 5) None of these

**Solution:**

**Step 1:**

According to BODMAS rule, we first need to solve the equation present inside the brackets,

(1 ÷ 28/9 ) = 1 ÷ 26/9 = 1 x 9/26 = 9/26

**Step 2:**

9/2 + 9/26 - 40/13

Take the LCM of the denominators and perform the operations of addition and subtraction-

9/2 + 9/26 - 40/13 = (117 + 9 - 80) / 26 = 46/26 =23/13

Therefore, the answer is option 1: 23/13

**Example 6: Simplification Questions for IBPS PO and SSC CGL Exam**

**Question:**

1) 18 2)17 3)28 4)19 5)None of these

**Solution:**

**Step 1:**

Though this problem appears to be complex, it is one of the simplest simplification questions if you solve it in the correct order. First, simplify the numerator.

6 x 136 ÷ 8 + 132 = ?

As per BODMAS first, perform the division operation

136/8 = 17

Now perform the multiplication operation

6 x 17 = 102

At last perform the addition operation

102 + 132 = 234

Now you get the numerator as 234

**Step 2:**

Now take the denominator and start to simplify it using BODMAS method.

628 ÷ 16 - 26.25

As per BODMAS first perform division operation-

628 ÷ 16 = 39.25

Now, perform the subtraction operation-

39.25 - 26.25 = 13

**Step 3:**

Finally, combine the numerator and denominator together-

234/13 = 18

Therefore, the answer is option 1: 18.

**Example 7: Simplification Questions for IBPS PO and SSC CGL Exam**

**Question:**

{(441)1/2 x 207 x (343)1/3} ÷ {(14)²x (529)

1) 5.75 2)6.75 3)2.75 4)6.25 5)6.50^{1/2}} = ?**Solution:**

**Step 1:**

According to the BODMAS rule, first, we need to solve the numbers which are present inside the brackets. We are supposed to move inside out. So anything that is inside the bracket should be solved first and then we need to move outside the bracket.

(441) = 21

(343)

^{1/3}= 7
(14)² = 196

(529)

^{1/2}= 23**Step 2:**

Simplify the numerator and denominator separately

(21 x 207 x 7)/(196 x 23) = 6.75

Therefore, the answer is option 2: 6.75

Do write down in the comment section how this blog made it easy for you to solve simplification Questions for IBPS PO Exam

Stay tuned for more Simplification Questions.

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