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Simplification Questions for Bank Exams: The simplification topic covers a big portion in the Upcoming government exams of the Banking sector. If you are preparing to appear in the IBPS, SBI or any other banking exams, you should practice the simplification question well. It will boost your confidence and give you some tricks/hacks to solve the questions. Without delaying anymore, let’s have a look on Simplification Questions for Bank exams with answers.
Table of Contents
Tips to Solve Simplification Questions for Bank Exams
Simplification questions are an important part of bank exams like IBPS, SBI, RBI, and other clerical and PO exams. These questions test your basic arithmetic skills and speed. Here are some tips and strategies to help you solve simplification problems efficiently:
- Follow BODMAS Rule: Solve expressions in the order of Brackets, Orders, Division, Multiplication, Addition, and Subtraction.
- Learn Tables & Squares: Memorize multiplication tables (up to 20), squares (up to 30), and cubes (up to 20).
- Use Estimation: Round off numbers for quick calculations if exact answers are not required.
- Master Fractions & Percentages: Know common fractions and their decimal forms (e.g., 1/2=0.5, 1/4=0.25).
- Practice Mental Math: Regularly practice mental arithmetic to speed up your calculations.
- Break Down Complex Problems: Split large numbers into smaller, manageable parts (e.g., 234×5=(230+4)×5).
- Skip Lengthy Steps: Identify shortcuts and avoid unnecessary calculations (e.g., cancel out common factors).
20+ Simplification Questions for Bank Exam
Following are some questions that you can practice on Simplification for the banking exams.
Solution:
We can simplify by first evaluating the expressions inside the parentheses: 12/3 + 5(2 – 1) = 4 + 5(1) Then, we can multiply and add: 5(1) = 5 4 + 5 = 9 Therefore, 12/3 + 5(2 – 1) simplifies to 9.
Solution:
We can simplify by first evaluating the expressions inside the parentheses: 2(4 + 6) – 3(2 + 5) = 2(10) – 3(7) Then, we can multiply and subtract: 2(10) = 20 and 3(7) = 21 20 – 21 = -1 Therefore, 2(4 + 6) – 3(2 + 5) simplifies to -1.
Solution:
Combine like terms: (3x – 7) + (2x + 5) = (3x + 2x) + (-7 + 5) = 5x – 2 Therefore, the simplified expression is 5x – 2.
Solution:
Distribute the coefficients: 2(3x + 4) = 6x + 8 2(2x – 3) = 4x – 6 Combining like terms: 6x + 8 – 5x + 4x – 6 = (6x – 5x + 4x) + (8 – 6) = 5x + 2 Therefore, the simplified expression is 5x + 2.
Solution:
Express the decimal as a fraction: 0.875 = 875/1000. Find the greatest common factor of 875 and 1000: 875 = 5 x 5 x 7 x 2 x 5 1000 = 2 x 2 x 5 x 5 x 5 The greatest common factor is 125. Divide both numerator and denominator by 125: 875/1000 = (875 ÷ 125)/(1000 ÷ 125) = 7/8. Therefore, the simplified decimal is 7/8.
Solution:
Invert the denominator and multiply: (2/3)/(4/5) = (2/3) x (5/4) = 10/12. Reduce the fraction: 10/12 = (10 ÷ 2)/(12 ÷ 2) = 5/6. Therefore, the simplified fraction is 5/6.
Solution:
Using the difference of squares formula: (a^2 – b^2) = (a + b)(a – b) (2x + 3y)^2 – (2x – 3y)^2 = [(2x + 3y) + (2x – 3y)][(2x + 3y) – (2x – 3y)] = (4x)(6y) = 24xy Therefore, the simplified expression is 24xy.
Solution:
Distribute the coefficients and combine like terms: = 10x + 15 – 6x + 2 + 4x = 8x + 1.
Solution:
First, square the terms inside the parentheses: = (9a² – 12ab + 4b²) – 4(a² + 2ab + b²) Distribute the negative sign: = 9a² – 12ab + 4b² – 4a² – 8ab – 4b² Combine like terms: = 5a² – 20ab
Solution:
Find a common denominator: 3/4 + 1/6 = (3 x 3)/(4 x 3) + (1 x 2)/(6 x 2) = 9/12 + 2/12 = 11/12. Therefore, the simplified fraction is 11/12.
Also Read: How to Prepare for Bank Exams
Directions (11-20): What value should come in the place of (?) in the following questions.
A.78
B.86
C.68
D.96
E.103
Solution: D
√729 * 8 + ? = 8(2/3) * √1296
216 + ? = 26/3 * 36
? = 96
A.4500
B.4200
C.4800
D.5100
E.5300
Solution: A
(155 ÷ 1665) * (74 ÷ 3100) * ? = 10
? = 10 * 3100 * 1665/155 * 74
? = 4500
A.35
B.45
C.39
D.42
E.25
Solution: A
(155 ÷ 1665) * (74 ÷ 3100) * ? = 10
? = 10 * 3100 * 1665/155 * 74
? = 4500
A.32
B.37
C.41
D.29
E.39
Solution: A
80% of 750 + 40% of 1060 = ?2
600 + 424 = ?2
? = 32
A.180
B.190
C.200
D.210
E.220
Solution: C
1680 ÷ 8 + 1360 ÷ 16 = ? + 19 * 5
210 + 85 = ? + 95
? = 200
A.312
B.298
C.308
D.317
E.327
Solution: D
48 % of 325 + 115% of 140 = ?
156 + 161 = ?
317 = ?
A.-100
B.-120
C.-140
D.-160
E.-180
Solution: A
? = 23 * √169 – √361 * 21
? = 299 – 399
? = -100
A.140
B.120
C.160
D.180
E.200
Solution: C
(66 ÷ 225) * (45 ÷ 528) * ? = 4
? = 160
A.-14
B.-8
C.–15
D.–10
E.0
Solution: D
(133 ÷ 7) – (145 ÷ 5) = ?
19 – 29 = ?
? = -10
A.205
B.210
C.220
D.225
E.240
Solution: A
40 % of 300 + 15 * 24 = ? + 275
120 + 360 = ? + 275
? = 205
Importance of Simplification Questions
Check the given points of importance of the simplification questions:
- The questions are to test the ability of the candidates to solve complex questions.
- It arrives in most banking exams like SBI PO, RBI Grade B, NABARD GRADE A, and more.
- These questions are easy to solve and carry most of the marks.
FAQs
You can use the BODMAS rule to solve these types of questions.
Simplification questions will be asked in two ways, so you all have to know both types,
Yes, it is easy to crack but requires the right tips and tricks to be implemented.
1. Memorize table up to 20
2. Memorize squares and cubes up to 30.
3. Learn basic mental calculations.
4. Practice regularly
These are the basic tips and tricks to solve the simplification questions for banking exams.
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