Most Popular & Recommended Books for BANK EXAMS 2019 This blog is meant for all the job aspirants who are preparing for various competitive exams. Because, the pattern of the subjects are same for every competitive exam and comprises Quantitative Aptitude, Reasoning skills, General Awareness, Basic Computer skills. Here is the full details of best …

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]]>This blog is meant for all the job aspirants who are preparing for various competitive exams. Because, the pattern of the subjects are same for every competitive exam and comprises Quantitative Aptitude, Reasoning skills, General Awareness, Basic Computer skills. Here is the full details of best books for bank exams recommended by experts.

The bank job aspirants who are seriously preparing for bank exams like IBPS Bank exams, SBI Exams, RRBs and other private banks. To get success in the competitive exams, mainly the job aspirant must need to have a good determination and planning with right books/sources of materials is important than whether the candidate has been taken coaching or not. With our analysis, we can strongly say that some of the successful people may not have taken coaching but they succeeded in the competitive exams.

The selection of right sources Self-preparation with right books by more practice with proper planning makes them reach the target of winning the job. We suggest the below collection of books which are very helpful in their self-preparation and practice.

Here we are presenting some of the best books which are recommended by successful bankers. Which are very much useful to crack bank exams in their self-preparation. We are confident that this books can surely serve your requirements to practice more and more test papers using proper strategies and solving techniques.

These days’ competitive exams became a technique. If we follow proper strategies in the selection of materials, books and in their preparation for exams and also while attempting the exam. If the job aspirant who masters these techniques can succeed easily and can reach his goal of getting a job in a government sector.

Practicing more and more test papers is important to improve your speed and accuracy in the final exam.

For Full Length Bank Test Papers with complete Solutions to Practice: Kiran Series is the most popular and recommended book to practice by self. If you get any doubt in solving answers. The solutions are also provided by following easy to understand methods. The author Kiran Prakashan has published practice workbooks in different segments.

IBPS Exam Pattern 2017

IBPS Bank Exams consists of 200 questions under 4 sections

**Quantitative Aptitude****Reasoning****English****Banking GK and General Awareness/ Current Affairs****Computer Awareness**

In every competitive exam, quantitative aptitude is a compulsory paper. The candidates who are preparing for various competitive exams should have good knowledge of aptitude solving.

Once I suggest going through **NCERT Books/ State Govt. Mathematics Text Books** from Class VIII to Class X to recall basic knowledge of numerical ability.

Every Job aspirant must and should have this book with them to prepare for any type of competitive exam having quantitative aptitude is part of that exam. This is a well-known book and have special goodwill among students and job aspirants to appear for various competitive exams – Banks, Insurance, UPSC, SSC, Railways, Teaching jobs etc.,

Book Description: This book involves all the important topics of quantitative aptitude – numerical ability from basic level to high-level problems with complete solutions. This book is one of the most popular books and most recommended book for every competitive exam. It comprises the huge collection of questions and solutions (shortcut methods) on quantitative aptitude from easy to tricky. The new edition of Quantitative Aptitude by R.S. Agarwal have some positive added features than older editions. – Good in paper quality – Complete solutions from easy to tricky problems. Conclusion: Finally, I personally recommend this book for every job aspirant and also students who are preparing for entrance exams as well as any competitive exam.

**Book Details:**

**Author:** R.S.Agarwal

**Publisher: **S Chand publishing

**Publishing Date: 2018 (New Edition** *with good quality***)**

**No. of Pages: 960 pages**

**BOOK Cost: ****Rs.580**

**Quantitative Aptitude for Competitive Examinations by R.S. Agarwal (2019 Edition)**

The candidates who are preparing for competitive exams should know some tips and shortcuts to solve the paper in less time than usual methods. There is time constraint in every competitive exam so the candidates need to finish the exam in less time is also matters in one’s success. Quicker Math’s by M. Tyra is one of the topmost suggested book and magical book to learn speed maths tricks.

Book Description: Quicker Maths is another important book to prepare for quantitative aptitude section in bank exams because time management is the basic problem to crack any competitive exam. Every exam has the time constraint and the aspirant needs to solve in time. Even though you are familiar with the questions you may not able to solve in time. Quicker Maths is an important book to learn speed maths tricks and maths formulas. By getting this book you can able to solve more problems in less time than before using speed maths tricks. First, you go through the R.S. Agarwal book, solve it and then take quicker maths book by M.tyra. Really a magical speed up calculations. This is one of the magical book on quantitative aptitude to solve problems in less time by using the speed math’s techniques explained by author M. Tyra. In every competitive exam time is the basic constraint so the aspirants have to learn techniques to finish the exam in less time. In Magical Book On Quicker Maths, M. Tyra easily explained the speed maths tricks in problem-solving. This book is recommended for all competitive exams conducted by different govt. organizations (Banks, UPSC, LIC, and other competitive exams). The concepts involved are explained simply and concisely so that students will find them easier to understand. Here, the author has written various short cut methods and speed math’s tricks. So we recommend this book is must read the book for everyone who is targeting various competitive exams.

Edition: 2019 (Revised Edition)

Once you get a proper grip on all topics of Quantitative Aptitude, Speed Maths, Shortcuts, and Formulas. Then you can go through this book for better practice. This book involves objective type questions and this will be helpful to solve mentally without using paper and pen practice just as in the main exam.

**Book Details:**

Author: Rajesh Verma

Total No. of pages: 400 Pages

Publisher: Arihant.

**Book Cost: Rs. 193**

This book involves the logical and analytical reasoning contains questions from high-level reasoning very much useful for SBI PO, IBPS PO, and Bank Specialist Officer Exams.

Analytical Reasoning book is a great source to learn core concepts of analytical reasoning.

**Book Details:**

Author: **M.K. Pandey**

Total Pages: 710.

**Book Cost: Rs. 220**

**For Probationary Officers (PO’s)/ Management Trainees(MT’s)**

**Book Title: IBPS PO/MTs Practice Work Book**

**Kiran Series Bank PO/MT/SO Online Exam CWE – VI Practice Work Book (With CD) **

Book Description: The main purpose of providing these sets is to make easy for the candidates to learn techniques on different topics to attain the required level of speed and accuracy. Every set is adequate, balanced and total in itself.

**Author:** Kiran Prakashan.

**Publisher:** Kiran Prakashan.

**Publishing Date:** 2016.

**No. of Pages:** 760

**BOOK Cost: ****Rs.380.**

The aspirants who are preparing for various bank exams can find this book as must need the book and is good and recommended in every aspect of standard and relevant and to the point for practicing opportunity. Totally, **32 sets** of full-length test papers are provided based on latest exam pattern

Book Description: This book is mainly targeted at the people who are preparing for bank exams at home. To help such candidates Kiran Prakashan has explained the topic wise guidance and full-length practice papers. Along with this book, CD consisting online test papers is given for practicing and to make the reader comfortable with online exams.The candidates who are preparing for various bank exams can find this book as the useful book.

English Preparation Made Easy with English Grammer by Wren & Martin.

**1. High School English Grammar & Composition Revised Edition (English) 1st Edition**

High School English Grammar & Composition

This book is very helpful to develop vocabulary.

Best Book for IBPS & SBI Bank PO Exam 2019 – 101 Speed Tests for Practice. This book covered all topics and each test is based on small topics which are most important for the Bank PO exams – SBI, IBPS.

**Book Cost: Rs.90/-**

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]]>Preparation Tips to crack SBI Clerk Exam 2018 How to Crack SBI Clerk Exam by Self-preparation? The candidate has to understand the questions well before going to solve it. For that, the candidate’s need to prepare well and proper care should be taken in their preparation just before one month of the exam. Though we are …

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__The candidate has to understand the questions well before going to solve it. For that, the candidate’s need to prepare well and proper care should be taken in their preparation just before one month of the exam. Though we are preparing for bank exams from 1 year onwards the last 1 or 2 months just after the notification is important and this is the precious time to speed up and concentrate more. Let’s have a look at some instructions on How to Crack SBI Clerk Exam by Self-preparation?__

Have knowledge on the complete syllabus of the exam in asking under different sections – Quantitative Aptitude, Reasoning, English, Banking Awareness, General Awareness, Computers, and Marketing & Daily current affairs.__Know the Syllabus of SBI Exam:__

– how many questions are asking under each section and marks allotted for each section and total duration of the exam.__Know the pattern of the exam__

For First 1 month, we have to concentrate more on gaining the knowledge on different topics which are frequently asked in the previous exams.__Understand and have knowledge on all the topics:__

They need to practice more and more of previous papers to improve their speed and accuracy of the exam.__Practice More & More Test papers:__

By practicing the full-length test papers we have to identify our weak sections and concentrate on those areas to speed up our performance.__Point Out Weak Sections:__

Allocate time to solve each section and stick at that time. The candidate need keep in mind about the sectional cutoff marks. Getting highest marks in one section and very fewer marks in other section won’t promote you to the next round of the selection process. So getting sectional wise cutoff marks and overall cutoff marks is important. We have to qualify in section as well as the overall cutoff.__Time Management To Get Qualifying Marks:__**Speed Maths Tricks**to do fast calculations in aptitude section.

7.** Planning Before Exam: **

You need to attempt, at least, minimum questions to qualify in each section of Quantitative Aptitude, English Language, Reasoning Ability, English, GA/Computer/Marketing section in fixed time as per your planning. After that allocate remaining time on your strong areas where you can get more score by analyzing the given questions.

** 8. Practice Previous Exam Papers:** Take mock tests as you are attempting the exam at exam center (that means one full-length test, one sitting online).

** 9. Practice…..Practice……Practice…. **

Practice more on mental calculations. Practicing more and more previous papers is the key point in our preparation. Practicing previous papers in one sitting one paper can improve the candidate’s ability to solve the paper with good accuracy and also the correct usage of formulas, shortcut methods tips and tricks wherever required to get quick solutions.

**How to attempt Exam:**

The preparation is not complete by only knowing the concepts and gaining knowledge and simply with practicing test papers. Preparation for the exam completes only when they attempt multiple numbers of previous exam papers with good score and accuracy. This makes them be more confident about the exam and enhances the speed and accuracy and time management. Always start with an easy section which takes less time to solve more questions.

** **There is a bit difference between preparing for an exam under guidance by staying in an environment where the people with the same goal around us and we ourselves preparing for an exam with self-motivation and self-learning. But it doesn’t mean that we can’t get success without taking coaching and it takes some extra time to do ourselves everything.

It’s proven that some people got success without taking coaching by preparing themselves with more dedication and self-motivation towards their goal. All that needs is more practice with a proper plan by knowing the shortcuts, tips, and tricks in solving the paper on different sections of the given exam syllabus with proper planning in the preparation.

In a coaching class, they have a mentor to discuss each topic and to guide in their preparation and also they can have the people with the same goal and so that they can get more information and they are required to dedicate few hours just for the preparation of SBI exams every day.

Apart from all these, All the candidates are required to solve more and more test papers of previous exams assuming the main exam environment and complete the whole test in one sitting and analyze the do’s and dont’s in next exam.

Gradually, you can improve the test performance in all aspects such as speed, accuracy, usage of formulas & shortcuts.

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]]>Simple Interest and Compound Interest Practice Problems with Easy Tricks: To solve SI and CI problems we have different methods as using formulas, shortcut methods and tricks but we have to apply appropriate method which is required to get answer in less time. At some places we get quick answer by applying direct formula and at …

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]]>To solve SI and CI problems we have different methods as using formulas, shortcut methods and tricks but we have to apply appropriate method which is required to get answer in less time. At some places we get quick answer by applying direct formula and at some places shortcut methods.

**SI & CI Formulas:**

Principal Amount – (P), Time – (T), Rate of Interest (R), Simple Interest (SI).

SI = (PTR) / 100.

CI = p[(1+R/100)^t – 1].

- Amount = P[1+(r/100)]
- Compound Interest (CI) = P [ (1+r/100)
^{t}-1 ]. - Simple Interest (SI) = (CI * rt) / 100 [(1+r/100)
^{t}-1]. - When the difference between SI and CI on the certain sum for 2 years at r% is d Rs.
**Sum = (d/r**^{2})*100^{2.} ^{ }When the difference between SI and CI on a certain sum for 3 years at r% is d Rs.**Sum = [d(100)**^{3 }/ r^{2}(300+r)].- If a sum A becomes B in t1 years at CI. Then After t2 years, Sum = (B)
^{t2 / t1 }/(A)(^{t2 / t1)-1}. - The rate of Interest = [ (Difference of amount after n years and (n+1) years )/ Amount after n years ]*100.

1. In how many years simple interest be equal to the principle amount at 25% p.a. interest rate?

SI of n years = P

=> P T R /100 = P

=> T = 100/R = 100/25 =** 4 years.**

2. The simple interest on certain principle amount P at 7% per anum for 4 years is 3584. What will be compound inerest on same principle for 2 years at 4% per anum?

Explaination: Here we have to use shortcut method than applying formula.

Shortcut method: if we have rate of interest (r%) and time period in years (1,2,3…n years)

SI for 2 years = (r% + r%) of P

CI for 2 years = [r% +r% + (r*r)/100] of P.

SI for 4 years = 1st year + 2nd year + 3rd year + 4th year.

= 7 + 7 + 7 + 7 = 28%.

Here, 28% = 3584

Then 100% = P

Therefore P = (100 * 3584)/28 = 100 * 512/4 = 100 * 128 = 12800.

CI of 2 years at 4% rate.

= 4 + 4 + 16/100 = 8+0.16 = 8.16%

we have from question that 28% = 3584.

Then 8.16% = ?

=> (8.16 * 3584)/28 = 128 * 8.16 =** 1****044.48.**

3. Akshaya received Rs.5175 as simple interest on the amount of Rs.15000 fixed for 2 years. Find out the rate of simple interest.

Explaination: Here we have to use the direct formula SI = PTR/100 i.e. R= (SI * 100 )/ PT. Answer is** 17.25%.**

4. The principle amount becomes 5 times of itself in 10 years. What should be the rate of interest per anum?

Assume p = 100 and it becomes 5 times means 500. Then the interest is 400 i.e. 4P.

PTR/100 = 4P

=> P*10*R/100 = 4P,

=> R = **40%.**

5. What would be the simple interest obtained on an amount of Rs.8930 at the rate of 8% per anum after 5 years?

Always principle amount (P) be 100%. SI for 5 years = 5 * 8 = 40%.

Here 100% = 8930.

=> 40% =?

=> (40 * 8930)/100 = **3572.**

6. What would be the compound interest (CI) obtained on an amount of Rs.4000 at the rate of 5% per anum after 3 years?

P= 100%, CI for 3 years = 1st calculate for 2years

for 2 years = 5 + 5 + 25/100 = 10.25

then for 3rd year = 10.25 + 5 + 0.52 = 15.77

100% = 4000

15.77% = ? => 15.77 * 4000/100 =** 631.**

7. What will be the simple interest accrued on a sum of Rs.5224 at a rate of 5% per anum in 6 years?

Always Principle amount (P) = 100% and SI for 6 years = 6*5 =30%.

100% = 5224

30% = ?

=> (30 * 5224)/100 = **1306.**

**(If we learn this percentage based solving trick we can get answer in less than 10 seconds.)**

8. What will be the amount of compound interest (CI) accrued on an amount of Rs.16,400 at the rate of 8% in two years?

P = Rs.16,400. CI rate for 2 years = 8 + 8 + (8*8)/100 = 16 + 0.64 = 16.64%.

100% = 16,400

16.64% = ?

=> (16400 * 16.64)/100 = 164 * 16.64 = **2729.**

9. The compound interest on a certain amount for 2 years at the rate of 8% is Rs.312. What will be the simple interest on the same amount and at the same interest rate and same time?

Here P= 100% = Rs.x and CI for 2 years = 8 + 8 + (8*8)/100 = 16.64% = Rs.312(given)

Then, 16.64% = 312

100% = ? (Find P)

=> P= 312 * 100 /16.64 = 1875.

We have to find SI for 2 years at 8% Rate on 1875.

=> 100% = 1875

16% = ?

=> SI for 2 years = (16* 1875)/100 = **300.**

10. What will be the total amount will ramya get in 2 years if she invests Rs.5000 to obtain compound interest at 5% per anum rate?

Here Principle P = 100%, CI for 2 years = 5 + 5 + 25/100 = 10.25% Then Total Amount = Principle +Interest = 100% + 10.25% = 110.25%

We have 100% = 5000

110.25% = ? (Total amount)

Total amount = (110.25 * 5000)/100 = **Rs.5512.50**

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]]>Addition Tricks for Bank Exams – SBI Clerks and PO’s 2017 To crack SBI clerk’s exam we need to be smart in doing calculations faster and accurate. So we have to use some maths tricks while solving aptitude section that traditional method of solving. Here we are providing some simple addition tricks with examples which …

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To crack SBI clerk’s exam we need to be smart in doing calculations faster and accurate. So we have to use some maths tricks while solving aptitude section that traditional method of solving. Here we are providing some simple addition tricks with examples which are very much useful to crack upcoming SBI exams 2016.

Some Tricky Additions – Simplifications

If you came across these type of additions we can follow this easy trick to solve in less time.

** 1) 6 + 66 + 666+ 6666+ 66666+ 666666**

= 6 × (123456) -> * [Explanation: *In above problem the number 6 repeated orderly 1 time, 2 times …up to 6 times).]

**2) 4 + 44 + 444 + 4444 + 44444 + 444444**

= 4 × (123456)

**= 493824.**

**3) 7 + 77 + 777 + 7777 + 77777**

= 7 × (1 2 3 4 5)

**= 86415**

**1) 6.6 + 66.66 + 666.666 + 6666.6666 + 66666.66666**

= 6 × (12345) + [6 × (54321)]-> [**Explanation:** In above problem the number 6 repeated orderly 1 time, 2 times …up to 6 times).]
= 74070 + 3.25926 ( 5 numbers after decimal point)

**= 74073.25926**

**2) 4.4 + 44.44 + 444.444 + 4444.4444**

= [4× ( 1 2 3 4)] + [ 4×(4 3 2 1)]

= 4936 + 1.7284 ( 4 times repeatations after decimal point so keep decimal point upto 4 digits )

**= 4937.7284.**

**Step 1:**

Before decimal point 9 repeated up to 4 times orderly from 1 time, 2 times …4 times. so multiply 9 with 1234.

[9 ×( 1234) ] = **11106**

**Step 2:**

After decimal point 4 repeated up to 4 times orderly from 1 to 4 numbers. So multiply 4 with (4321) and put the decimal point after 4 digits by counting from right to left.

[4 × (4321) ] = **1. 7284**

**Step 3:**

Add step 1 + step2

11106 + 1.7284 = **11107.7284**

**What is the sum of all even natural numbers below 100**

**Solution:**

2 + 4+ 6 + 8 + 10 +……+98

Formula: Sum =n/2 (f + l) [n = total numbers, f = first number, l= last number].

= 49/2 ( 2 + 98) = 49 × 50 =** 2450**

**(a) 6/7 **

**(b)** 7/6

**(c)** 8/7

**(d)** 5/9

Answer: **4/13.**

**1. 2546 + 5480 + 3210**

Here add all first digits in step 1, and then add all second digits and then 3^{rd} digits and so on as shown below.

2546 + 5480 + 3210

10000 + 1100 + 130 + 6 = ** 11, 236.**

**2. 256 + 516 + 310.**

1000 + 70 + 12 = 1082.

**3. 5467 + 6752 + 8971 +8605.**

**4. 679 + 760 + 896**

Add all first digits in step 1, and then add all second digits and then 3^{rd} digits and so on as shown below.

2100 + 220 + 15.

2335.

**5.6587 + 6540 + 5555 + 2345.**

19000 + 1800 + 210 + 17

add from right to left,

21, 027

**(a**= (a-b) (a^{3 }– b^{3})^{2}+ ab + b^{2})**(a**= (a+b) (a^{3 }+ b^{3})^{2}– ab + b^{2})**(a + b)**= 2( a^{2}+ (a – b)^{2}^{2}+ b^{2})**a**= (a –b) (a+b) (a^{4}– b^{4}^{2}– b^{2})**(a**=(a-b) [a^{5}-b^{5})^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4}].

**Sum of first n odd numbers**

= 1 + 3 + 5 + ………… + (2n-1)= **n ^{2}**

**Sum of first n even numbers**

= 2 + 4 + 6 + …………+ 2n =** n (n+1).**

**Sum of first n natural numbers**

= 1+2+3 +4 + ……..+n = **n(n+1)/2.**

**Sum of squares of first n natural numbers**

=1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + …………+n^{2}.

= **[n (n+1)(2n+1)]/6.**

**Sum of cubes of first n natural numbers**

= 1^{3} + 2^{3}+3^{3} + 4^{3} + …… + n^{3}

= **[n(n+1)/2] ^{2}**

**1 + 2/3 + (2/3) ^{2} + (2/3)^{3} + ……(2/3)^{n}**=

**1/1.3 + 1/3.5 + 1/5.7 +.……. + 1/[(2n+1) (2n-1)]** =?

**Shortcut method:**

1/gap [1/first – 1/(2n-1)]= n/(2n+1) last number = 2n-1 =7

= 2n=8

=n=4.

= 4/(2*4 + 1) = 4/9

a + ar^{2} + ar^{3} + ar^{4 }+ …………… + ar^{n}

^{ }= Total sum = a (1-r^{n})/ 1-r. [ a = first number]

Example: 2 + 2×3^{2} + 2×3^{3} + 2×3^{4} + 2×3^{5}

= 2( 1- 3^{5})/1-3

= 2( 1- 243)/(-2)

=242

The sum of a, (a+d), (a+2d), (a+3d),……… (a+(n-1)d)

Total sum = n/2 [2a + (n-1) d]

= **n/2 (first +last).**

2 + 4+ 6 + 8 + 10 +……+98

**Formula:**

Sum =n/2 (f + l) [n = total numbers, f = first number, l= last number].

= 49/2 ( 2 + 98) = 49 × 50 = **2450**

Here, we can split 12 as 4×3

Ans. is 4 .

Here, we can split 12 as 4 × 3.

we take small number 3 as answer.

Formula:

A^{(2 ^n -1)/2^n)}

3^{(2 ^4 -1)/2^4)}

= 3^{(15/16)}

= 6

sqrt| 4 + 5/16 + 6/25 + 7/36 ( solve from right to left)

Explaination: ( sqrt (49) = 7, then 7+ 29 =36, sqrt(36) =6, 6+19 =25, sqrt (25) =5…..)

sqrt(4 +5) = sqrt(9) = 3

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