
Numerical and Quantitive Aptitude
 Alligation or Mixture
 Area
 Average
 HCF and LCM
 Indices and Surds
 Number System
 Orders of Magnitude
 Partnership
 Percentage
 Pipes and Cisterns
 Probability
 Problem on Ages – MCQ
 Problems on Trains
 Profit and Loss
 Ratio and Proportion
 Ratio and Proportion
 Simple & Compound Interest
 Simplification
 Square Root & Cube Root
 Time and Distance
 Time and Work
 Unitary Method
 Decimal Fraction

Data Interpretation

English

Reasoning
 AlphaNumeric Sequence Test
 Analogy
 Arithmetical Reasoning
 Blood Relation
 Cause and Effect
 Classification
 Direction Sense
 Input Output
 Logical Sequence of Words
 Logical Venn Diagram
 Mathematical Operations
 Number, Ranking and Time Sequence Test
 Series Completion
 Sitting Arrangements
 Meaning of Analogy
 Syllogism
 Statement & Assumptions

Computer Knowledge

Marketing Aptitude

General Awareness/ Current Affairs

IBPS PO  Mock Test
 IBPS PO preliminary test series1
 IBPS PO preliminary test series3
 IBPS PO mains test series1
 IBPS PO preliminary test series4
 IBPS PO preliminary test series5
 IBPS PO preliminary test series2
 IBPS PO mains test series3
 IBPS PO mains test series2
 IBPS RRB PO mains test series1
 IBPS RRB PO mains test series2
 IBPS RRB PO mains test series4
 IBPS RRB PO mains test series5
 IBPS RRB PO mains test series6
 IBPS RRB PO Prelims test series10
 IBPS RRB PO prelims test series7
 IBPS RRB PO Prelims test series8
 IBPS RRB PO prelims test series9
Mathematical Operations
Mathematical Operations
The basic approach for the problems of this type is more or less similar to that of coding and decoding. One has to study the symbols or the geometrical figures and their meanings carefully. Then, the meanings are to be used in place of those symbols in answering the questions.
The questions can be categorised into two types Symbols and Notations
Symbols and Notations
Symbols for these types of questions stand for mathematical operations like +, –, ×, ÷, > , <, ³, £, = and #. So, the studentsmust replace the symbols by mathematical operations and apply the BODMAS’ rule to find the value of the given expression. Other symbols which can be used are DÑ, *, @, $, etc, with proper definitions. Some examples are given below
Example 1: If ‘+’ means ‘–’, ‘–’ means ‘×’, ‘×’ means and ‘means’ ‘÷’, and ‘÷’ mean ‘+’ then 15 × 3 ÷ 15 + 5 – 2 = ?
(a) 0
(b) 10
(c) 20
(d) 6
Solution. (b): 15 × 3 = 15 + 5 – 2 after changing the signs = 15 = 3 + 15 – 5 × 2 = 5 + 15 – 10 =10
Example 2: If 2 * 3 = 12, 3 * 4 = 20 and 4 * 5 = 30, then 2 * 6 is
(a) 18
(b) 12
(c) 21
(d) None of these
Solution. (c) The numbers on both sides of * are increased by one and then multiplied to get the answer. 2*6 = 3 × 7 = 21
Example 3: If x $ y = (x + y + xy –1) (x + y + xy + 1), then the value of (4 $ 10) is.
(a) 2915
(b) 2195
(c) 2951
(d) 2955
Solution. (a) As per the definition of $, (4$10) would be (4 + 10 + 4 × 10 – 1) (4 + 10 + 4 × 10 + 1) = 2915. So, answer is (a)
Example 4: If * means “is greater than”, @ means `is less then’; and $ means “is equal to” and if a $ b and b @ c, then
(a) c * b
(b) b * c
(c) c * a
(d) Both (a) and (c)
Solution. (d) Replace the symbols with the meaning given against them. If a $ b and b @ c would become a = b and b < c, then c > b or c > a ie, c * b, c * a are true from the given options.
Geometrical Figures
These figures will be divided and subdivided into a number of parts, each part is filled with a number or a letter except one part. The numbers or the letters in the figures follow certain pattern. The objective is to identify the pattern and find the missing number or letter. The problems based on geometrical figures are of various types.
Type 1
Based on circles: In these type of, three circles with numbers outside the circle will be given. In the firsttwo circles, the number inside the circle is written according to a particular pattern. The objective of the student is to
find the missing number of third circle.
Example 5:
(a) 7.16
(b) 9.25
(c) 6.23
(d) 8.33
Type 2
Circle divided into parts: In such questions, circle is divided into three parts. An arithmetic operation on the numbers gives the missing number
Type 3
Problems based on triangles: In such questions, three triangles are given with numbers inside and outside. The number inside the triangle is obtained by operating some arithmetic operators on the numbers outside the triangle.
Type 4
Problems based on squares: Such questions consist of three squares with five numbers inside the square of the 5 numbers, 4 numbers are at the four corners of the square and one middle number follow some arithmetic
operation.
Type 5
Problems based on matrix: In such questions, a square divided into nine parts, three rows and three columns, out of the nine parts, eight parts are filled with one part left vacant. The students need to apply some
arithmetic operation to find out the missing number.
MCQ
Directions (Q. No. 15: In each of the following questions, the two expessions on either side of the sign (=) will have the same value if two terms on either side on the same side are interchanged. The correct terms to be interchanged have been given as one of the five altenatives under the expressions. Find the corect altenative in each case.
1. 5 + 3 × 6 – 4 ÷ 2 = 4 × 3 – 10 ÷ 2 + 7
(1) 4, 7
(2) 5, 7
(3) 6, 4
(4) 6, 10
(5) 3, 7
2. 7 × 2 – 3 + 8 ¸ 4 = 5 + 6 × 2 – 24 ¸ 3
(1) 2, 6
(2) 6, 5
(3) 3, 24
(4) 7, 6
(5) 6, 4
3. 15 + 3 × 4 – 8 ¸ 2 = 8 × 5 + 16 ¸ 2 –1
(1) 3, 5
(2) 15, 5
(3) 15, 16
(4) 3, 1
(5) 20, 5
4. 6 × 3 + 8 ¸ 2 – 1 = 9 – 8 ¸ 4 + 5 × 2
(1) 3, 4
(2) 3, 5
(3) 6, 9
(4) 9, 5
(5) 15, 16
5. 8 ¸ 2 × – 11 + 9 = 6 × 2 – 5 + 4 ¸ 2
(1) 5, 9
(2) 8, 5
(3) 9, 6
(4) 11, 5
(5) 9, 5
Directions (Q. No 610): In each of the following questions, which one of the fou interchanges in signs and numbes would make the given equation correct?
6. 6 × 4 + 2 = 16
(1) + and ×, 2 and 4
(2) + and ×, 2 and 6
(3) + and ×, 4 and 6
(4) × and –, 6 and 2
(5) None of these
7. (3 ¸ 4) + 2 = 2
(1) + and ×, 2 and 3
(2) + and ¸, 2 and 4
(3) + and ¸, 3 and 4
(4) No intechange, 3 and 4
(5) + and ×, 4 and 3
8. 4 × 6 – 2 = 14
(1) × to ¸, 2 and 4
(2) – ¸, 2 and 6
(3) – to +, 2 and 6
(4) × to +, 4 and 6
(5) – to, 4 and 6
9. (6 ¸ 2) × 3) = 0
(1) ¸ and ×, 2 and 3
(2) × to –, 2 and 6
(3) ¸ and ×, 2 and 6
(4) × to –, 2 and 3
(5) ¸ to –, 2 and 6
10. Select the correct set of symbols which will fit in the given equation 5 0 3 5 = 20.
(1) ×, ×, ×
(2) –, +, ×
(3) ×, +, ×
(4) +, –, ×
(5) +, ×, +