The NHB often announces job openings on its official website, offering positions such as General Manager, Deputy GM, Assistant GM, Manager, Assistant Manager, and various contractual roles like Chief Financial Officer, Economist, MIS, Chief Economist, Senior Application Developer, Application Developer, Senior Project Finance Officer, and Project Finance Officer. As we have mentioned earlier, the recruitment notification for NHB in 2024 has not been released. Exams were conducted in 2022 and the NHB Online Exam Result 2022 was announced on May 06, 2022 while the NHB Exam Results for 2023 were out on March 7, 2023. All the candidates who cleared the two phases of the exam and had their names in these result sections had automatically qualified for the NHB Interview in both 2022 and 2023

NHB Admit Card

The NHB Admit Card 2022 was released on February 24, 2022. The admit card is an important component of the exams for NHB. The card contains all the relevant details that you need to be aware of regarding the exams. So, for the next exam cycle, when the admit card is out online, don’t delay but rather download it and keep it safe immediately. Once the dates are announced, you can begin your preparation for the exam by checking out the NHB Exam Pattern and Syllabus. Don’t forget to use your time effectively for the exam preparation

NHB Exam Date

The NHB Exam Date 2022 was on March 6, 2022. The National Housing Bank (NHB) provides some of the best employment opportunities for candidates from any kind of background. Although the application period has already ended, if you are hopeful about this career path will be surely waiting for the NHB Assistant Manager Notification 2024. This is a career opportunity gives you one of the best future prospects for you. Working for the esteemed National Housing Bank is sure to be a source of great pride for fresh graduates. Once the NHB Assistant Manager exam dates for 2024 are announced, aspirants can begin their rigorous preparation for the exam. They need to visit the website constantly to look for updates in this regard..

National Housing Bank Notification

The NHB Notification for 2024 is not yet out. But those who are waiting for this important update will be those who are eager to appear for the exam and begin a prospective career with the organization. Once the notification is out, candidates can prepare for the exam by checking the eligibly details, exam pattern, salary structure and other important details about the recruitment. Don’t forget to constantly check the website for updates on the anticipated 2024 notification. The application window was open between December 1 and January 7, 2022. 

NHB Recruitment 2021 Overview

Earlier the last date for filling in the application was December 30, 2021. It was later extended to January 7, 2022.

NHB Recruitment Vacancy

You should be aware of the vacancy numbers for the previous years if you hope to prepare for the upcoming NHB recruitment. There were only 14 vacancies for Scale 1. This was later on increased for the subsequent cycle. Based on these trends, the number of vacancies is expected to be close to this number for the coming year. Despite speculation, all final numbers will only be out when the notification is out for 2024.

NHB Recruitment Eligibility

When the notification is out, the first thing to do is take a look at the NHB Eligibility Criteria to confirm that you are eligible for the exam. Without confirming your eligibility, you will be unable to know if you even qualify for the exam. Take a look at the age bracket and educational qualifications outlined for previous years eligible candidates in the table below.  

NHB Recruitment Exam Pattern

The exam pattern for all 3 scales differs slightly. You should take great precaution while preparing for the exam. So, read the details mentioned below carefully for the post for which you will apply. 

Selection Process for JMGS-I (AM)

• Phase I: Online Examination
• Phase II – Interview

Online Examination

Study carefully the details of the online examination provided below.

• The tests of Reasoning & Computer Aptitude, General Awareness, English Language and Quantitative Aptitude will be objective. 
• This online test will be followed by a descriptive test of English Language. 
• The Descriptive English test is qualifying in nature. 
• There is sectional allotment of time for all sections. 

Interview

The interview will be held for 50 marks. 

Selection Process for MMGS-II (DM)

• Phase I: Online Examination
• Phase II – Interview

Online Examination

Take a look at the details of the examination provided below.

• The General Aptitude test is qualifying in nature. 
• All sections will carry objective-type questions. 
• A composite time of 2 hours will be awarded. 

Interview

The interview will be held for 100 marks. 

Selection Process for SMGS-IV (RM)

The selection process for Regional Manager comprises 2 phases- Shortlisting and Interview. 

Candidates will be shortlisted based on their applications. Selected candidates will be invited for an interview. 

Probation Period

NHB Recruitment Syllabus

Take a look at the complete before you proceed. After all, you can’t expect to clear the exam without actually being aware and studying the topics well.  

NHB Recruitment Online Application

You should complete the NHB Online Recruitment application at the earliest if you are interested and eligible. Do ensure that you:

• Fill your application carefully. 
• Review your documents before making payment.
• Make online payment. 

NHB Recruitment Application Fee

Study Resources @ixamBee

At ixamBee, we have all the resources that can help you successfully and extensively study for this exam. Just follow the links given below to explore our resources further. 

1. National Housing Bank Online Course 
2. Free Mock Test for NHB Assistant Manager
3. National Housing Bank Assistant Manager Previous Year Papers
4. BeePedia for GA

If there’s anything else you’d like to know about the NHB Exam, do reach out to us in the comments below. 

We wish you the best! 

At ixamBee you can find free Online Test Series, GK updates in the form of BeePedia, as well as latest updates for Bank PO, Bank Clerk, SSC, RBI, NABARD and Other Government Jobs.

Also read

Everything to know about the IFSCA Grade A Officer Exam 2024 

Why is SEBI Called a Watchdog of the Indian Securities Market?

Some RBI Grade B Preparation Tips for Non-Banking Backgrounds

The post All About the NHB Exam 2022-2023 and its Results appeared first on ixambee.

Quadratic Equation Questions for RBI Grade B  Exam

The RBI Grade B exam has recently introduced a new pattern of Quadratic Equation questions. These questions require candidates to solve quadratic equations by using the given options, rather than by traditional methods of factoring or completing the square. This new pattern is aimed at testing a candidate’s ability to think logically, make accurate calculations, and utilize different problem-solving strategies. 

Practicing quadratic equation questions is essential for success in the RBI Grade B exam. Studying quants series to hone your skills and learn to solve using various methods. Take a look at some difficult questions below with their solutions.

Read the following statement and answer the following questions.

Equation 1: ax² + bx + c = 0

Equation 2: py² + qy + c = 0

Given that:

=> c is a single digit prime number greater than 2.

=> p and b are 2-digit prime number less than 20.

=> p is greater than 11.

=> b is greater than p.

=> Smallest roots of both the equations are same

=> No root is irrational.

=> 3c is greater than b.

=> q= b + 1

1. Find the value of p² + 2q.

1. 200
2. 219
3. 209
4. 229
5. None of these

2. Find the value of a.

1. 10
2. 11
3. 12.
4. 15
5. None of these

Read the following statement and answer the following questions.

HEquation 1: x² – 7x + t = 0

Equation 2: y² – 4y + (t – 9) =0

Roots of equation 1 are p and q and the roots of equation 2 are p and q – p.

3. Value of (t – 8)^q?

1. 256
2. 206
3. 236
4. 246
5. 276

4. Had equation 2 been multiplied with 2n and 1 been added to smallest root of the resultant quadratic equation newly formed then find the resultant number?

1. 1
2. 2
3. 0
4. 3
5. None of these

Answer the questions based on the information given below.

Given below are two equations i.e. ‘I’ and ‘II’

Note:

i.) ‘x’ and ‘y’ are roots of equation ‘II’, such that x < y.

ii.) One of the roots of the equation ‘I’ is (-25).

5. Which of the following is the sum of values obtained by putting values of ‘p’ as roots of equation ‘II’ in the expression (p² + 37p + 400)?

1. 1787/4
2. 1687/4
3. 1887/4
4. 1857/4
5. None of these

6. Find the difference between roots of equation I

1. 10
2. 12
3. 13
4. 14
5. None of these

7. Find the value of m^(3/2) + √(225)

1. 1015
2. 1005
3. 1115
4. 1215
5. None of these

Answer the questions based on the information given below.

Equation A: (a × a) – 3a – √(4a²) = – 6

Equation B: (b²) – √(81b²) = – 4 × (5)

Equation C: (c² √625c⁶) / 5c³ + (4 × 7) = 39c

Equation D: d² – (3 × 5)d = (7 × (-8) )

8. Find the difference between the larger and smaller roots of equation A.

1. 1
2. 2
3. -1
4. 3
5. None of these

9. Find the LCM of all the larger roots of equations A, B, C, D.

1. 840
2. 440
3. 740
4. 860
5. None of these

10. In which of the following equations difference between larger and smaller root is 1.

1. Only A and B
2. Only B
3. Only B and C
4. Only A, B and D
5. None of these

Solutions:

1. c.

From the given points:

c can be 3, 5 and 7

p can be 11, 13, 17 and 19

q can be 11, 13, 17 and 19

p > 11

b > p

3c > b => 3 x 7 > b => 21 > b, so we can say that value of c will be 7.

Now, we can form different cases from the above points

Given that, no roots are irrational, means that value of discriminant will be perfect square.

We can write discriminant as D = b² – 4ac

From Equation 2: py² + qy + c = 0

So, from Case I:

D = q² – 4pc

=> (18)² – 4 × 13 × 7 = 324 – 364 = -ve, so case I is not possible

From Case II:

D = q² – 4pc

=> (20)² – 4 × 13 × 7 = 400 – 364 = 36, this is perfect square, hence case II is possible.

From case III:

D = q² – 4pc

=> (20)² – 4 × 17 × 7 = 400 – 476 = -ve, so case III is also not possible.

so, p = 13, b = 19, q = 20 and c = 7.

=> 13² + 2 × 20 = 209

2. c.

From the given points:

c can be 3, 5 and 7

p can be 11, 13, 17 and 19

q can be 11, 13, 17 and 19

p > 11

b > p

3c > b => 3 x 7 > b => 21 > b, so we can say that value of c will be 7.

Now, we can form different cases from the above points

Given that, no roots is irrational, means that value of discriminant will be perfect square.

We can write discriminant as D = b² – 4ac

From Equation 2: py² + qy + c = 0

So, from Case I:

D = q² – 4pc

=> (18)² – 4 × 13 × 7 = 324 – 364 = -ve, so case I is not possible

From Case II:

D = q² – 4pc

=> (20)² – 4 × 13 × 7 = 400 – 364 = 36, this is perfect square, hence case II is possible.

From case III:

D = q² – 4pc

=> (20)² – 4 × 17 × 7 = 400 – 476 = -ve, so case III is also not possible.

Therefore, Equation 1: ax² + bx + c = 0 can be written as,

 ax² + 19x + 7 = 0

Therefore, Equation 2: py² + qy + c = 0 can be written as,

13y² + 20y + 7 = 0

=> 13y² + 13y + 7y + 7 = 0

=> 13y(y + 1) + 7(y + 1) = 0

=> (13y + 7)(y + 1) = 0

=> y = -7/13, -1

Smaller root = -1

Given that, smallest roots of both the equations are same, so putting x = -1 in eq 1,

ax² + bx + c = 0

=> a (-1)² + 19 × -1 + 7 = 0

=> a – 19 + 7 = 0

=> a = 12

3. a.

From equation 1:

p + q = 7 (sums of roots)

pq = t (product of roots)

Now, from equation 2

p + q – p = 4

so, q = 4

Now,

p + q = 7

p + 4 = 7

p = 3

so, p = 3 and q = 4

so, t = 12

so, equation 2 will be, y² – 4y + (t – 9) = 0

y² – 4y + 3 = 0

Required value = (12 – 8) ⁴ = 4 ⁴ = 256

4. b.

From equation 1:

p + q = 7 (sums of roots)

pq = t (product of roots)

Now, from equation 2

p + q – p = 4

so, q = 4

Now,

p + q = 7

p + 4 = 7

p = 3

so, p = 3 and q = 4

so, t = 12

so, equation 2 will be, y² – 4y + (t – 9) = 0

y² – 4y + 3 = 0

ATQ:

When 2ⁿ is multiplied by equation 2

 2ⁿ (y² – 4y + 3) = 0

2ⁿ (y² – 3y – y + 3) = 0

2ⁿ (y – 1)(y – 3) = 0

2ⁿ cannot be zero.

So, y = 1 and y = 3

y = 1 + 1 = 2 

5. c.

We have; 2q² + 9q = 18

=> 2q² + 9q – 18 = 0

=> 2q² + 12q – 3q – 18 = 0

=> 2q(q + 6) – 3(q + 6) = 0

=> (q + 6)(2q – 3) = 0

=> q = -6, q = (3/2)

Since, x < y, x = -6 and y = (3/2)

Since, -25 is a root of the give equation, the value of equation will be ‘0’, when we put ‘x’ = 25.

So, (-25)² + (37 x -25) + (400 – m) = 0

=> 625 – 925 + 400 – m = 0

=> m = 100

Now, equation I become: p² + 37p + 400 = 0

Roots of equation II = (-6) and (3/2)

So, value of equation (I), when x = -6 = (-6)² + 37 x (-6) + 300 = 114

And, value of equation (I), when x = (3/2) = (3/2)² + (37 x 3/2) + 300 = (1431/4)

So, required sum = 114 + (1431/4) = (1887/4)

6. c.

We have; 2q² + 9q = 18

=> 2q² + 9q – 18 = 0

=> 2q² + 12q – 3q – 18 = 0

=> 2q(q + 6) – 3(q + 6) = 0

=> (q + 6)(2q – 3) = 0

=> q = -6, q = (3/2)

Since, x < y, x = -6 and y = (3/2)

Since, -25 is a root of the give equation, the value of equation will be ‘0’, when we put ‘x’ = 25.

So, (-25)² + (37 x -25) + (400 – m) = 0

=> 625 – 925 + 400 – m = 0

=> m = 100

Now, equation I become: p² + 37p + 400 = 0

Equation I = p² + 37p + 300 = 0

=> p² + 25p + 12p + 300 = 0

=> p(p + 25) + 12(p + 25) = 0

=> (p + 25)(p + 12) = 0

So, p = -25 and p = -12

So, required value = (-12) – (-25) = 13

7. a.

Common sol:

We have; 2q² + 9q = 18

=> 2q² + 9q – 18 = 0

=> 2q² + 12q – 3q – 18 = 0

=> 2q(q + 6) – 3(q + 6) = 0

=> (q + 6)(2q – 3) = 0

=> q = -6, q = (3/2)

Since, x < y, x = -6 and y = (3/2)

Since, -25 is a root of the give equation, the value of equation will be ‘0’, when we put ‘x’ = 25.

So, (-25)² + (37 x -25) + (400 – m) = 0

=> 625 – 925 + 400 – m = 0

=> m = 100

Now, equation I become: p² + 37p + 400 = 0

Required value = (100)^3/2 + √ (225) = (10²)^3/2 + 15 = 1000 + 15 = 1015

8. a.

First, we will simplify the equation and find the roots of these quadratic equation:

A: (a × a) – 3a – √(4a²) = – 6

=> a² – 3a – 2a + 6 = 0

=> a² – 5a + 6 = 0

=> a² – 3a – 2a + 6 = 0

=> a = 3, 2

B: (b²) – √(81b²) = – 4 × (5)

=> b² – 9b + 20 = 0

=> b² – 5b – 4b + 20 = 0

=> b = 5, 4

C: (c² √625c⁶) / 5c³ + (4 × 7) = 39c

=> {c² × 25c³}/5c³ – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 35c – 4c + 28

=> c = 7, 0.8

D: d² – (3 × 5)d = (7 × (-8))

=> d² – 15d + 56 = 0

=> d² – 8d – 7d + 56 = 0

=> d = 8, 7

Now,

Required difference = 3 – 2 = 1

9. a.

First, we will simplify the equation and find the roots of these quadratic equation:

A: (a × a) – 3a – √(4a²) = – 6

=> a² – 3a – 2a + 6 = 0

=> a² – 5a + 6 = 0

=> a² – 3a – 2a + 6 = 0

=> a = 3, 2

B: (b²) – √(81b²) = – 4 × (5)

=> b² – 9b + 20 = 0

=> b² – 5b – 4b + 20 = 0

=> b = 5, 4

C: (c² √625c⁶) / 5c³ + (4 × 7) = 39c

=> {c² × 25c³}/5c³ – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 35c – 4c + 28

=> c = 7, 0.8

D: d² – (3 × 5)d = (7 × (-8))

=> d² – 15d + 56 = 0

=> d² – 8d – 7d + 56 = 0

=> d = 8, 7

Now,

LCM of all the larger roots of equations A, B, C, D = 3 × 5 × 7 × 8 = 840

10. d.

First, we will simplify the equation and find the roots of these quadratic equation:

A: (a × a) – 3a – √(4a²) = – 6

=> a² – 3a – 2a + 6 = 0

=> a² – 5a + 6 = 0

=> a² – 3a – 2a + 6 = 0

=> a = 3, 2

B: (b²) – √(81b²) = – 4 × (5)

=> b² – 9b + 20 = 0

=> b² – 5b – 4b + 20 = 0

=> b = 5, 4

C: (c² √625c⁶) / 5c³ + (4 × 7) = 39c

=> {c² × 25c³}/5c³ – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 39c + 28 = 0

=> 5c² – 35c – 4c + 28

=> c = 7, 0.8

D: d² – (3 × 5)d = (7 × (-8))

=> d² – 15d + 56 = 0

=> d² – 8d – 7d + 56 = 0

=> d = 8, 7

Now,

In A, B and D equations difference between larger and smaller root is 1.

To help you prepare 50% faster for competitive exams, ixamBee provides free Mock Test Series and all the Current Affairs in English and Current Affairs in Hindi in the BeePedia capsules for GA Preparation. You can also get the latest updates for Bank PO, Bank Clerk, SSC, RBI  NABARD and Other Government Jobs.

Conclusion

If you have set your sights on clearing the RBI Grade B exam in 2023, then you must prioritize your preparation for quadratic equations with ixambee RBI Grade B Syllabus. It is vital to comprehend the question pattern on this topic thoroughly if you wish to score well in the exam. In order to achieve a high score and excel in the test, it is crucial to gain a strong grasp of the quadratic equations. With the right study plan and consistent practice offered by ixamBee, you can enhance your likelihood of success and secure your dream job.

Read Also

New pattern Question on Series For RBI Grade B exam 

From Preparation to Performance: Roshan’s Tips for Acing the RBI Grade B Exam

The post RBI Grade B Exam: New Pattern Quadratic Equation Questions appeared first on ixambee.

Nainital Bank Result 2022

The Nainital Bank Result 2022 was announced on March 28, 2022. Candidates can check their result by logging in with their registration number and D.O.B credentials.

Team ixamBee wishes all successful candidates the very best for the Interview.

Nainital Bank Admit Card 2022

The Nainital Bank Admit Card 2022 was released on March 8, 2022. You can download your Nainital Bank Call Letter 2022 by logging in with your Registration No/Roll No and Password.

Study with ixamBee

Now that you are so close to the Nainital Bank PO Exam 2022, we encourage you to take the Nainital Bank Mock Tests as giving mock tests will assist you in identifying your weak areas and judge your preparation before the exam. These mock tests are totally free.

Nainital Bank PO Free Mock Test

Free Mock Test for Nainital Bank Clerk

Nainital Bank 2022 Notification

The Nainital Bank Recruitment 2022 Notification was released on February 1, 2022. A total of 100 vacancies for Clerk & Management Trainee (MT/PO) posts were announced.

Nainital Bank Recruitment 2022: Important Update

The Nainital Bank application date was extended to February 25, 2022. Initially the last date to apply was February 15, 2022.

Also, the application fee was reduced by ₹500. The application fee is now ₹1000. Candidates who have already paid ₹1500 were refunded ₹500.

Nainital Bank Recruitment 2022 Overview

Let us briefly make an overview of the Nainital Bank Clerk and Management Trainee exam 2022.

Nainital Bank Recruitment 2022 Exam Calendar

Take a look at the important dates for Nainital Bank recruitment 2022 cycle.

The Nainital Bank application date was extended to February 25, 2022. Initially, the last date to apply was February 15, 2022.

Nainital Bank Recruitment 2022 Vacancy

The Nainital Bank notification 2022 released 100 Clerk & PO/MT vacancies.

Eligibility Criteria

The Nainital Bank recruitment 2022 eligibility criteria for PO and Clerk have been provided below. The details of these criteria are as per the Nainital Bank MT & Clerk official notification. 

Age

Education Qualification

Nainital Bank Recruitment 2022 Exam Pattern

The Nainital Bank 2022 selection procedure for PO and Clerk will comprise 2 phases. They are:

1. Online examination
2. Interview

Let us take a look at the details of both phases of the Nainital Bank recruitment process.

Online Written Exam

The details of the Nainital Bank Online written exam have been provided in the table below. The exam pattern for both PO and Clerk is the same.

Please note

• The online written exam will be conducted in English.
• All questions will be MCQ-type questions with 5 options.
• There will be a penalty of 0.25 marks for each wrong answer.

Interview

Candidates who clear the online written exam will be invited for a personal interview.

Nainital Bank Recruitment 2022 Syllabus

Start preparing for the Nainital Bank recruitment exam 2022 by reviewing the syllabus. Follow the links to read the detailed syllabus for Nainital Bank PO & Clerk.

Nainital Bank syllabus 2022 for PO/MT

Nainital Bank Clerk Syllabus 2022.

Nainital Bank Recruitment 2022 Pay Scale

Given below are the details of Nainital Bank PO salary and the salary of Nainital Bank Clerks. Please take a look.

Indemnity Bond

The selected candidates will be required to execute an Indemnity Bond for the specified amount with two sureties for serving the Bank for a specified period. If the selected candidate  resigns/leaves/abandons the service before completion of the specified period, he/she will indemnify the Bank for the amount of the Bond for all costs/charges/expenses incurred by the bank.

The selected candidates have to execute the Indemnity Bond before joining/absorption in the Bank’s service. The details regarding the amount of the Bond is given in the table below-

Nainital Bank Recruitment 2022 Application Process

The application process is threefold-

• Application Registration
• Payment of Fees
• Document Scan and Upload

Application Registration

• Visit the Nainital Bank website- nainitalbank.co.in
• To register your application, click on “Click here for New Registration” and enter the necessary details.
• Fill the application form slowly and carefully to ensure that you do not miss out any important information.
• Validate and save your details by clicking on the “Validate your details” and “Save & Next” button.
• Proceed to upload your photo, signature and other details. 
• Preview your application form and verify the whole application before completing your registration.
• Click on “Final Submit” after verifying/modifying the details in the form.
• Click on “Payment Tab” and proceed for payment.
• Click on the “Submit” button. 

Payment of Fee

The application fees for the post of Management Trainees and Clerks is as follows-

Earlier, the application was fee was ₹1500. Now, the fee has been reduced by ₹500. Applicants who have already paid ₹1500 will be refunded.

Document Scan and Upload

• Photograph- It must be a recent passport size photo against a light coloured background (preferably white).
• Signature- The signature must be on a white paper with black ink pen. It must not be in capital letters.
• Left Thumb Impression- The left thumb impression must be on a white paper with black/blue ink.
• Handwritten Declaration- It must not be written in capital letters.
• There will be separate links for uploading the photograph, signature, left thumb impression and hand-written declaration. Click on the respective link and upload the documents.

Please Note- The Online Application will not be registered unless you upload your photo, signature, left thumb impression and handwritten declaration.

Similarly, you can also find all the resources for Nainital Bank Clerk exam 2022 on our website as well.

We encourage you to apply at the earliest.

For any other queries, please also take a look at this detailed notification video below.

We wish you the best.

At ixamBee, you can find free Online Test Series, GK updates in the form of BeePedia, as well as latest updates for SBI PO, Bank Clerk, SSC, RBI, NABARD, and Other Government Jobs.

Also Read 

How to Prepare for RBI Grade B 2022?

NRA CET 2022: Exam Date, Syllabus & Exam Pattern

SSC CGL Syllabus 2022 in Detail

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