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N S Balaji .

Chromepet,Chennai

₹ 2,000 per month

For the past 6 months ,iam taking quantitative aptitude for RRB

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Ankita S.

Sector 51,Gurgaon

₹ 5,000 per month

I have done M tech and Btech in electronics with 7 years of teaching experience. I teach B.tech electronics or electrical tuition's as that is my area of specialization. I also provide gate and net coaching. I like teaching as my one of favorite pass time apart from my well established career. I am having friendly attitude and and having passion in teaching. I am a private individual tutor to teach students at my home.

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Sanju Kumar

Uttam Nagar,Delhi

₹ 3,500 per month

Teaching mathematics with help of short short tricks and innovative techniques to trend students mind for the competitive exams. With these tricks students can find the answer within seconds and also check whether it is correct or not in the exam..!!! Maximum competitive exams have maths only till class 10th so it is important for student to learn basics very well so that he won't face any problem in future. My strengths includes teaching students short tricks along with proper method. Building confidence in students. Vedic maths along with self made tricks to ensure whether answer is correct or not.

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Ruchi A.

Shalimar Garden,Ghaziabad

₹ 1,200 per month

# I have taught so many students at my home of diifferent age groups. # I have cleared CRA in DMRC, # Cleared pre of ssc cgl 2017 with 170/200 marks and Mains with 320/400. # I am very much dedicated to my work especially teaching because it's my passion. i believe in learning the basics and concepts rather than mugging it up. # Maths is my favourite subject because of cleared basic concepts and most of my students also changed their mind about maths after my classes.

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Golconda Institute

Gaddiannaram,Hyderabad

₹ 6,000 per month

We provide excellent coaching for competitive exams like , IBPS ( Po's v , Clearks) , SSC ( CGL , CPO, CHSL & MTS) , Railways ( RPF& ASO ) , Sub- Inspector of Police & Constable for both Medium students. We provide Excellent Study material and conducting Mock Tests.

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Sanjay Singh Katheth

Rajinder Nagar,Dehradun

₹ 5,000 per month

I did b Tech in computer science and and currently I am taking tution classes for those aspirants who are preparing for the banking and SSC . I also have experience of teaching to class 6 to 10th currently I am applying for teacher for this post.

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Lesson Posted on 21/04/2018 Exam Coaching/Government Exams Coaching/Staff Selection Commission Exam Exam Coaching/Railway Exam Exam Coaching/Government Exams Coaching

How to prpeare for SSC CGL exam?

Siddhartha Singh Chandel

I did masters in Electronics in 2011 later I joined DRDO govt of India where I worked on artificial intelligence....

There are four sections in SSC CGL namely 1.General Intelligence and Reasoning2.English Language3.Quantitative Aptitude4.General AwarenessComing to clearing SSC CGL if you have prepared earlier for any other competitive exams you would have gone the syllabus once.But if that is not the case you have... read more

There are four sections in SSC CGL namely

1.General Intelligence and Reasoning
2.English Language
3.Quantitative Aptitude
4.General Awareness

Coming to clearing SSC CGL if you have prepared earlier for any other competitive exams you would have gone the syllabus once.But if that is not the case you have limited time here.

For the first part, i.e. Reasoning the best approach would be to go through the syllabus once get a rough idea of each type of problem and try to solve at least 2 or 3 problems of each type them.Because it is very necessary to learn how to tackle these type of problems and you should have some prior knowledge about each of them as every problem is a variation of its basic premise.

For the second part, i.e. English Language If you are an above average reader you would be able to handle it. It's not that tough and can be easily handled. If you struggle in English and do not have a good command on it the best way would be to read as much as you can till the exam as only reading will help you to get a sense of English.Try to increase your vocabulary by reading English newspaper and especially the editorial it helps in increasing your vocabulary and also helps about a how an idea is presented.

For the third part, i.e. Quantitative Aptitude this part is a tricky one. Firstly you have to cover all the formulas and understand their implementation in the questions then you will have to work yourself up to the level of question and here also try to every type of question as each type of question uses a different concept.

For the fourth part, i.e. General Awareness the GA part has a question from everything which you have studied till now from your childhood. It has questions from history, geography, economics, biology, physics, chemistry, general studies and current affairs.The current affairs part can be learned from any site which provides daily updates on ga.
Finally, some points to give you a scenario on how to get things done

  • Give a mock test to know your position what you know and what you need to improve(Mahendra's test series is a good option)
  • Work your way to remove the problems which you have dealing with specific sections.
  • Try to cover every type of question
  • Slowly try to increase you're no. of attempts
  • Only mark those question which you have got the exact answers for in QA and Reasoning.
  • DO NOT LOSE HOPE because“Everything is going to be fine in the end.
  • If it's not fine, it's not the end.”
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Lesson Posted on 18/04/2018 Exam Coaching/Bank Clerical Exam Coaching SBI PO/SBI PO Mains/General Awareness/Banking & Financial Awareness IBPS CLERICAL +8 IBPS PO/IBPS PO Prelims IBPS CLERICAL/QUANTITATIVE APTITUDE IBPS PO/IBPS PO Prelims/Reasoning Ability IBPS PO/IBPS PO Prelims/Reasoning Ability/Series Exam Coaching/Government Exams Coaching/Staff Selection Commission Exam Exam Coaching/Railway Exam Campus Placement Training Exam Coaching/IBPS Exam Coaching/Interview prep less

Series

Resoluter Learning Solutions

Guidance for SSC, Banking, Railways, Management Entrances & other Competitive Examination: IBPS PO /...

INTRODUCTION Questions on series are now extremely common in all the Prelims and Mains exams. They are generally of three types – number, letter and alphanumeric – but the most commonly asked questions are on number series. NUMBER SERIES You are given a sequence of numbers that... read more

INTRODUCTION
    


Questions on series are now extremely common in all the Prelims and Mains exams. They are generally of three types – number, letter and alphanumeric – but the most commonly asked questions are on number series.


NUMBER SERIES
    


You are given a sequence of numbers that follow a common pattern. You need to understand this pattern and either identify the next term (more common) or find a missing term. There are infinite number of ways to create a number series by using one or more patterns. However, with practice you can start identifying the more common patterns such as difference, product, ratio, squares, cubes, powers factorials etc. Some basic series are:


Even Numbers: 2, 4, 6, 8…

Odd Numbers: 1, 3, 5, 7, 9 …

Prime Numbers: 2, 3, 5, 7, 11, 13, 17 …

Composite Numbers: 4, 6, 8, 9, 10, 12, 14, 15 …

Squares: 1, 4, 9, 16, 25, 36, 49, ….

Cubes: 1, 8, 27, 64, 125, 216, …

Factorials: 1, 2, 6, 24, 120, 720, ….


Note that ‘1’ is neither prime nor composite. Depending on the numbers used, a series can be increasing, decreasing or alternate. The common patterns used to create number series are:


DIFFERENCE BASED
    


This is the most basic and common form of series. If possible, this is the first thing you should check when you try to find a pattern in the series. You should take the difference between consecutive terms of a series. This difference can be constant (e.g. +5, −3 etc) or have some pattern of its own. This can be one of the patterns explained above e.g. if the series is: 32 33 41 68 132, take the difference to get 1 8 27 64. Since these numbers are clearly a series of perfect cubes, the next number in this sequence is 125. Hence, the next number in the original series is 132 + 125 = 257. Occasionally, you might not get a pattern after a first level of difference. Ten, you should take a difference of these differences and check.

 

Directions for examples 1 to 5: Find the missing term to complete the given series.


Example 1:

58, 70, 84, 100, ___


(1) 112(2) 119(3) 116(4) 118(5) 122


Solution:

Consider the difference between consecutive terms: 12, 14, 16, …

Observe that the difference is a series of consecutive even numbers.

Hence, next difference = 18 and next number = 100 + 18 = 118

Hence, option 4.


Example 2:

    

3, 9, 18, 30, ___


(1) 33(2) 36(3) 45(4) 39(5) 42


Solution:

This is an example of multiple patterns applying to the same set of numbers. In each case, the answer turns out to be the same.


Logic 1: Take the difference of consecutive terms i.e. 6, 9, 12, ….

The difference is a series of consecutive multiples of 3.

Hence, next difference = 15 and next term = 30 + 15 = 45

Hence, option 3.


Logic 2: Each term can be written as:

3 = 3 × 1; 9 = 3 × 3; 18 = 3 × 6; 30 = 3 × 10

Observe that the multiples of 3 in the above series themselves form a series 1, 3, 6, 10. The difference between consecutive terms of this new series is 2, 3, 4 and so on.

 

 

Hence, the next difference should be 10 + 5 = 15. Thus, the required multiple of 3 is 15.

Required term = 3 × 15 = 45

Hence, option 3.


Logic 3:

3 = 3 × 1

9 = 3 + 3 × 2

18 = 9 + 3 × 3

30 = 18 + 3 × 4

The pattern is (Previous term) + (3 × Position of term in the sequence)

Therefore, the next term should be 30 + 3 × 5 = 45

Hence, option 3.


Note:

Though the answer can be obtained through various patterns, you should always select the pattern that you first think of in the exam. Sometimes, the different patterns possible give different answers. In that case, you should look at the options and select the pattern whose answer is given in the options.


Example 3:

    

336, 305, 268, 227, 184, ___


(1) 137 (2) 163(3) 146(4) 133 (5) 129


Solution:

Consider the difference between consecutive terms: −31, −37, −41, −43.

If you ignore the minus sign, the differences form a series of consecutive primes.

Since the next prime is 47, the required difference = −47.

Hence, required term = 184 – 47 = 137

Hence, option 1.

 


Example 4:

    


1, 1, 2, 3, 5, 8, ___


(1) 9(2) 10(3) 12(4) 13(5) 15


Solution:

Observe that, starting from the third term, each term is the sum of the 2 terms immediately preceding it.

Hence, the required term = 5 + 8 = 13.

Hence, option 4.

 

Note:
The series given in this example is a special series known as the Fibonacci Series in which the sum of two successive terms is the next term.

 

PRODUCT BASED
    


Just like difference based series, product based series either have a common ratio between terms or the ratio also shows some pattern. Common patterns are powers, factorials and multiples. A common feature of such series is that the value of consecutive terms increases/decreases quite sharply. However, first level subtraction often helps in identifying the underlying pattern.


PRODUCT

The series may be based on simple application of factors or multiples.


Example 6:

    

Find the missing term. 2, 12, 30, 56, ___


(1) 77(2) 90(3) 79(4) 72(5) 92


Solution:

Each number can be expressed as a product of consecutive numbers:

1 × 2 = 2; 3 × 4 = 12; 5 × 6 = 30; 7 × 8 = 56

Hence, required term = 9 × 10 = 90

Hence, option 2.


Alternatively,

Consider the difference between consecutive terms: 10, 18, 26.

These differences form a series increasing by 8. Hence, next difference =

26 + 8 = 34.

Hence, the required term = 56 + 34 = 90.

Hence, option 2.


Example 7:

    

Find the missing term. 24, 12, 12, 18, 36, ___


(1) 90(2) 92(3) 78(4) 67(5) 77


Solution:

Consider the ratio between consecutive terms: 0.5, 1, 1.5, 2.

Since the ratio keeps increasing by 0.5, the next ratio is 2.5.

Hence, required term = 36 × 2.5 = 90

Hence, option 1.


POWER

These include questions where the pattern is related to squares, cubes or higher powers. Here, the value of the term increases even more sharply.


Example 8:

    

Find the missing term. 2, 6, 30, 260, ___


(1) 420(2) 500(3) 3140(4) 610(5) 3130


Solution:

The sharp increase in value indicates that the pattern may be based on powers.

The given series can be expressed as (11 + 1), (22 + 2), (33 + 3), (44 + 4)


Therefore, the required term = 55 + 5 = 3130

Hence, option 5.


Example 9:

    

Find the missing term. 11, 24, 39, 416, ___


(1) 626(2) 525(3) 552(4) 523(5) 5025


Solution:

Here, the numbers do not show an obvious pattern either using differences or simple products. If we split each number in 2 parts, we see that one part is the square of the other part.

Each term is of the form nn2.


Hence, the 5th term should be 552 i.e. 525.

Hence, option 2.


FACTORIAL

This involves the fastest growth in values since the factorials of the first 7 natural numbers are 1, 2, 6, 24, 120, 720 and 5040.


Example 10:

    

Find the missing term. 1, 1, 4, 36, ___


(1) 36(2) 51(3) 576(4) 81(5) 225


Solution:

The pattern seen in the above series is (n!)2.

Each term can be expressed as: (0!)2, (1!)2, (2!)2, (3!)2 and so on.

Therefore the required term should be (4!)2 = (24)2 = 576.

Hence, option 3.


Consider ratio between consecutive terms: 1, 4, 9. These are squares of consecutive natural numbers.
Hence, next ratio = 16 and required term = 36 × 16 = 576

Hence, option 3.


ALTERNATING SERIES
    


An alternating series is a combination of two or more series. Each series can have different patterns applied to it and then combined to form a series. In a combination of 2 series, alternate terms follow the same pattern. An alternating series can be a combination of more than 2 series as well. If you are asked to find two or more values, it is very likely to be an alternating series.


Example 11:

    

Find the missing term. 0, 3, 3, 4, 6, 5, 9, 6, ___


(1) 9(2) 5(3) 4(4) 10(5) 12


Solution:

Since the values increase and decrease, you should check for alternate series.

Taking alternate terms together, we get (0, 3, 6, 9, ?) and (3, 4, 5, 6). One is a series of consecutive multiples of 3 while the other is a series of consecutive integers.

Hence, the required number is the next multiple of 3 i.e. 12.

Hence, option 5.


Example 12:

    

Find the missing term. 1, 2, 7, 12, 21, 70, 43, _____


(1) 124(2) 224(3) 184(4) 150(5) 212


Solution:

Since the values increase and decrease, you should check for alternate series.

Taking alternate terms together, we get (1, 7, 21, 43) and (2, 12, 70, ?).

The first series is of the form [n + (n – 1)2]; where n is the series of consecutive odd numbers starting from 1.

The second series is of the form [n + (n – 2)3]; where n is the series of even numbers starting from 2.

The required number is part of the second series.

Hence, required number = 8 + (8 – 2)3 = 8 + 63 = 8 + 216 = 224

Hence, option 2.


MISCELLANEOUS
    


These can involve a combination of patterns or series and cannot be directly classified.


Example 13:

    

Find the missing term. 10, 103, 18, 187, ___


(1) 979(2) 26(3) 9(4) 251(5) 34


Solution:

The difference of consecutive terms does not form a logical series.

Here, look at the actual digits of each term. The sum of digits of each term is:

10: 1 + 0 = 1

103: 1 + 0 + 3 = 4

18: 1 + 8 = 9

187: 1 + 8 + 7 = 16

Thus, the sum of digits for each term is the square of consecutive natural numbers.

Hence, the next term should have sum of digits = 52 = 25.

Only 979 satisfies this condition.

Hence, option 1.


Example 14:

    

Find the missing term. 0, 1, 0, 8, 2, 7, 6, ___


(1) 2(2) 3(3) 4(4) 5(5) 8


Solution:

It is a series of cubes of natural numbers but each digit of the number is written as a separate term. Here, each cube has been expressed as a two digit number split into two parts i.e. 1 as 01, 8 as 08 and so on.

The next cube is 43 = 64. Since ‘6’ is already present, the next term = 4.

Hence, option 3.

 

 

 

Tips: 

  • For solving problems on numbers series, first observe the difference between the numbers. The difference may be constant or may form a pattern.
  • If the first level of subtraction does not show a pattern, subtract the differences from each other again. This may show a pattern. Continue till a consistent pattern is found.
  • If the difference between consecutive terms is very large and there is no constant pattern in the difference, it may be a product series. In a product series, the terms increase/decrease at a greater rate compared to a difference based series. Multiplicative series are generally based on pure multiplication, powers or factorials.
  • First check if the terms show some relationship with the multiple or factor of some number or group of numbers (such as even multiples of prime numbers).
  • If no factor/multiple based relationship is found, check if the numbers lie close to squares or cubes or higher powers of any number.
  • Remember the factorials of the first 7-8 numbers and see if the terms lie close to the factorials.
  • If the terms increase and decrease alternately, it may be an alternate series with two different patterns. If a question has more than 1 blank, in most cases it implies a combination of 2 or more series. If nothing works out, check the relationship between the digits of the terms, especially if some terms are very small in value and some are very large or if the terms seem to be random.
  • If the middle term is missing, check the pattern between the 1st and 2nd term, as well as the 4th and 5th term to see if there is an alternating pattern. Then use the answer option to see which alternative suits the best.
  • In letter series, always write down the position of the letters in the alphabet and then find the relationship.
  • In alphanumeric as well as letter series, focus on one element at a time e.g. first letter, first number. Based on its pattern, you may be able to eliminate some options.
  • If you are unable to get the pattern for a series based question in approximately one minute, leave it for the time being. Do not spend too much time on it.
  • If two patterns are visible for a given series, check if the answer by both is the same or different. If it is different, check which answer is given in the options. If very rare cases where both are given, you can either of the two as the answer.

 

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Lesson Posted on 23/10/2017 Exam Coaching/Bank Clerical Exam Coaching Exam Coaching/Railway Exam Tuition/Class IX-X Tuition

Average Speed

Vijaya Kumar Jayanthi

I am M.Sc and M.B.A and having more than 10 years of experience in handling CBSE, ICSE as well as state...

Average speed is total distance travelled in total time. We can calculate average only when we are given any of these two. In case, we are given only speed, then this formula is helpful. V travelled half of his journey at X kmph and the remaining half at Y kmph. What is his average speed? Average... read more

Average speed is total distance travelled in total time. We can calculate average only when we are given any of these two. 

In case, we are given only speed, then this formula is helpful.

V travelled half of his journey at X kmph and the remaining half at Y kmph. What is his average speed?

Average speed of V is = 2XY / X + Y.

Example: A travelled half of a distance is with 30Kmph and remaining half distance at 10Kmph. Then his average speed is:

2 × 30 × 10/30 + 10

= 600/40.

= 15Kmph.

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