Best reasoning books for all competitive Examination S. No. Book name Author/Publisher About the book 1. Verbal and Non-Verbal Reasoning Dr. R.S. Aggarwal This book cover UPSC, RBI, SBI, IBPS, PO, SSC, Management, and all other Entrance Recruitment and Aptitude Test 2. A New Approach to Reasoning B.S. Sijwali and S. Sijwali This helpful for all competitive and Entrance Examination. 3. Analytic Reasoning M.K. Pandey This book guides you to UPSC, RBI, IBPS, PO, SCC, Management, and all other Entrance Examination and Aptitude test. 4. Multi-Dimensional Reasoning Mishra and Kumar Dr. Lal This book covers all type of competitive examination and all other Entrance Examination. 5 Reasoning Book for Competitive Examinations Pearson This helps in Bank PO/Cle...

Simplification  Simplification mean find the most simplest form of given question with the help of different operations of maths in sequence . 1) Rule : VBODMAS                                                V→ Bar (Virnaculum)                     B→ Bracket                                     [{( )}] O→ Of                                             × D→ Division                                   ÷ M→ Multiplication                         × A→ Additi...

Arithmetic Progression  (AP) An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference (d) of AP, it can be positive and negative First term of an AP is (a) 1) General Form of an AP is     a, a + d, a + 2d, a + 3d,..... 2) nth Term of an AP     a n  =  a + ( n - 1 ) d Example 1) How many two- digit numbers are divisible by 3 ? Solution : The list of two-digit numbers divisible by 3 is : 12, 15, 18, .........99 . a n  =  a +  ( n - 1 ) d a = 12, d = 3,  a n  = 99 99 = 12 + ( n - 1 ) 3 99 - 12 = ( n - 1 ) 3 87 = n - 1  3 29 = n - 1 29 + 1 = n n = 30 So, there are 30, two-digit numbers divisible by 3. 2) Find the 10th term of the AP : 2, 7, 12,....... Solution : a n  =  a +  ( n - 1 ) d Here a = 2, d = 7 - 2 = 5, and n = 10 a 10  = 2 + ( 10 - 1 ) 5     ...