Hello aspirant,

Hope you are doing great.

As per your question, basically if you see from the three consecutive years, almost 12 lakh students will apply for the CUET exam, and out of which at least 5 to 7 lakh might apply for the BA LLB admissions. However, remember that this will vary from one year to the another.

Hope this helps!

Hey,

MCA students in India can apply for internships based on their graduate degree and with the approval of their MCA college. Companies that provide internships for MCA students are IBM India, TATA Group, Deloitte India, Eli Lilly India, etc.

Hope this helps you.

Thank you.

Hey,

To get admission into the prescribed courses at the university is given based on the eligibility norms of the institute. Sciences, Medicine and Allied Sciences, are the streams in which AAU Jorhat admissions 2023 are offered. CETUGAAU23 scores are considered for AAU Jorhat UG admissions 2023. Valid GAT-B scores are accepted for AAU Jorhat M.Sc. admission and for AAU Jorhat M.V.Sc. admission. NTA scores obtained by the candidates in ICAR’s AIEEA (PG) or the (ET) entrance exam conducted by the university are accepted for other AAU Jorhat PG admissions.

For more information about admissions into MSc agriculture please visit : https://dqxeclau.top/university/assam-agricultural-university-jorhat/admission

hope this helps you.

Dear aspirant !!

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Hope it helps you !!

Thanking you !!

Hello,

firstly, lets denote:

F as the number of students who passed in finance,

M as the number of students who passed in marketing,

B as the number of students who passed in both subjects,

X as the total number of students in the class.

Given:

60% pass in finance----   F = 60X/100= 0.6X

Among those passed in finance, 33 1/3% passed in marketing:  M = 0.333F = 0.333(0.6X)

150 failed in both subjects: B = 150

1/3rd of students who passed in marketing failed in finance:  0.333M = 1/3* (0.6X - B)

To find the number of students in the class:

B = 0.333F + 0.667M - B

B = 0.333(0.6X) + 0.667(0.333(0.6X)) - 150

B = 0.2X + 0.133X - 150

B = 0.333X - 150

150 = 0.333X - B

150 = 0.333X - 150

300 = 0.333X

X= 300/0.333

X= 90(approx)

So, there are approximately 900 students in the class.

Now, to find the percentage of students who failed in finance only, we need to find the number of students who failed in finance but passed in marketing.

Failed in finance only=  0.6X - B

Failed in finance only= 0.6(900) - 150

Failed in finance only = 540 - 150

Failed in finance only= 390

To find the percentage, divide by the total number of students and multiply by 100:

Percentage of students who failed in finance only= (390/900)*100%

Percentage of students who failed in finance only= 43.33%

So, approximately 43.33% of students failed in finance only.

Hope this helps,

Thank you

Hi

hope you doing well

Based on previous years' cutoffs, to secure admission in NIT Trichy for the OBC category, Tamil medium state board students will need to aim for a competitive JEE Main percentile. Here's an overview:

General Trends:

The cutoff percentile for NIT Trichy varies each year depending on various factors, including the number of applicants and the difficulty level of the exam. However, based on past data:

• For OBC candidates , a percentile range of around 90-95 is generally required to secure a seat in NIT Trichy, especially for popular branches like Computer Science and Engineering (CSE) and Electronics and Communication Engineering (ECE).

Some Specific Branch Cutoff's:

• Computer Science and Engineering (CSE):
  Usually, the cutoff percentile is around 93-95+ for OBC candidates.

• Electronics and Communication Engineering (ECE):
  Generally falls in the range of 90-93+ .

• Other Engineering Branches:
  Cutoffs for branches like Civil, Mechanical, and Electrical Engineering might be slightly lower, around 85-90+ .

To increase your chances of getting into NIT Trichy under the OBC category , aim for a JEE Main percentile of at least 90-95.

Hope this info will help you

Best Wishes.