SSC CGL Quant โ€ข Official PYQs

Simplification, BODMAS & Surds

Practice 290+ SSC CGL Simplification & Surds PYQs online. Master VBODMAS rules, surd rationalization, continuous fractions & step-by-step solutions.

๐Ÿ“ 291 Solved PYQs
๐Ÿ“… Exam Years: 2019โ€“2024
๐Ÿ“Š Weightage: 8%โ€“10% (2 Questions in Tier 1)
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๐ŸŽฏ High-Yield Syllabus Breakdown

These high-frequency concepts account for over 90% of all questions asked in Simplification, BODMAS & Surds:

โšก Essential Formulas & Speed Shortcuts

Formulas and time-saving shortcuts to solve Simplification questions in under 45 seconds:

Speed Trick #1:

Square root of a ยฑ 2โˆšb: Find two numbers x and y such that x + y = a and xยทy = b; result is โˆšx ยฑ โˆšy.

Speed Trick #2:

Recurring Decimals: 0.abcฬ… = (abc - a) / 990.

Speed Trick #3:

Always evaluate "OF" before division in BODMAS.

๐Ÿ“– Sample Solved Exam Questions

Preview authentic questions from this chapter with full step-by-step solutions:

Question 1 of 3 SSC CGL Tier-II Exam. 2015
The value of 4 1 1 1 3 1 2 1 4 โˆ’ + + + is
(A) 1 8
(B) 1/64
(C) 1/16
(D) 1/32
๐Ÿ’ก View Step-by-Step Solution & Answer

Correct Answer: 1 8

  1. Step 1: Understand problem conditions and equations.
  2. Step 2: W e have, 4 5 1 1 3 1 2 1 4 4 5 1 1 3 1 9 4 4 5 1 1 3 4 9 4 5 1 9/31 4/31 5/40 160 โˆ’ + + + =โˆ’ + + =โˆ’ + + =โˆ’ + =โˆ’ ร— = โˆ’ โˆ’ ==155/40 5/40 1 8
  3. Step 3: Evaluating gives verified answer: 1 8.

โšก Speed Tip: Verify intermediate calculations and units.

Question 2 of 3 SSC CGL Tier-I Exam. 2015
The simplified value of 1 1/10 1/10 1 1/10 1/10 1 1/10 1/10 1 1/10 1 + + ( ( ) ) + + ( ( ) ) โˆ’ + ( ( ) ) โˆ’ + 1/10 1 1/10 1/10 1 1/10 1/10 ( ( ) ) โŽง โŽจ โŽช โŽชโŽช โŽฉ โŽช โŽช โŽช โŽซ โŽฌ โŽช โŽชโŽช โŽญ โŽช โŽช โŽช รท + + ( ( ) ) โˆ’ + ( ( ) ) โŽง โŽจโŽช โŽฉโŽช โŽซ โŽฌโŽช โŽญโŽช
(A) 20/101
(B) 100/101
(C) 2
(D) 90/101
๐Ÿ’ก View Step-by-Step Solution & Answer

Correct Answer: 20/101

  1. Step 1: Understand problem conditions and equations.
  2. Step 2: Assume that, a =+ + =+ =+ =1 1/10 1/10 1 1/101 10 1/10 101/111 101 and b =โˆ’ + =โˆ’ =โˆ’ =1 1/10 1/10 1 1/101 10 1/10 101/91 101 Now, Given expression = aa bb abร—โˆ’ ร—[] รท ab ab ab ab ab 22/111 101/91 101/111 101/91 101 1 โˆ’ = โˆ’() +() = โˆ’ + 1/11 101/91 101/20 101/202 101/111 101/91 101/20 202/111 91/4040 101 ร— = ร— ร— = ร— ร— = 0/01
  3. Step 3: Evaluating gives verified answer: 20/101.

โšก Speed Tip: Verify intermediate calculations and units.

Question 3 of 3 SSC CGL Tier-II Exam. 2014, 2015
On simplification, the expression 4 1 7 2 1 7/31 2/11 7 1 2 1 2 1 5 1 5 โˆ’ + รท + + โˆ’ is equal to
(A) 28/65
(B) 24/53
(C) 56/53
(D) 14/65
๐Ÿ’ก View Step-by-Step Solution & Answer

Correct Answer: 56/53

  1. Step 1: Understand problem conditions and equations.
  2. Step 2: W e have, 4 1 7 2 1 7/31 2/11 7 1 2 1 2 1 5 1 5/29 7/15 7 7 2 8 7 1 2 1 2 5/24 14 7/49 16 โˆ’ + รท + + โˆ’ = โˆ’ + รท + + = + 1/14 1 2/24 53 2/65 14/53 106/24 28/65 130/53 56/53 รท + =รท + =ร— =
  3. Step 3: Evaluating gives verified answer: 56/53.

โšก Speed Tip: Verify intermediate calculations and units.

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โ“ Frequently Asked Questions (FAQs)

How to simplify โˆš(7 + 2โˆš10)?

Find two numbers with sum 7 and product 10: they are 5 and 2. Therefore, โˆš(7 + 2โˆš10) = โˆš5 + โˆš2.

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