JEE

Suppose it is 187 and 61 percentile in JEE Mains 2025 January Exam and it seems that your rank would be between 173239 to 326517. With such strength of merit, it would be extremely difficult to secure an admission in the highly ranked government colleges such as NITS or ILCA IIITs. But you may get a

Of course, it is extremely possible to learn AI without JEE or NEET. You may also consider B.Sc. or BCA courses with Artificial Intelligence for specialization, which are usually taken on the basis of 12th marks or your result in the exam, to be conducted by some universities. Moreover, very many on

No, it is unlikely to get top branches of VNIT  like computer Science or electrical branches you may get metallurgy branch maybe from the score of 95.90 in JEE Mains for general unreserved or EWS female student. It doesn't depend on your home state.

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Yes a student can get Nit Suruthkal's EEE branch with a score of 80 percentile in JEE Mains 2025 exam because the students are from home state and their category is St the excepted rank for admission in EEE branch in this college is 30000 to 50000 and if you are St category caste and you are from th

To avail affordable B.Tech CSE in Andhra Pradesh and Telangana, some of the popular government colleges are Andhra University College of Engineering (Visakhapatnam), JNTUK University College of Engineering (Kakinada), Sri Venkateswara University College of Engineering (Tirupati) and University Colle

A 96 percentile in JEE Mains will put you in the general percentage of score between 33000 and 48000. The score is not enough to get an admission in best IITs or most popular faculty members (such as CSE ) in best NITs. But you can easily get into mid-tier NITs (e.g. NIT Agartala, NIT Raipur, NIT SI

1. Amity University
2. Bharati Vidyapeeth's College of Engineering
3. Maharaja Agrasen Institute of Technology
4. Maharaja Surajmal Institute of Technology

sin (A + B) = sinA cosB + cosA sinB

sin (A - B) = sinA cosB - cosA sinB

cos (A + B) = cosA cosB - sinA sinB

cos (A - B) = cosA cosB + sinA sinB

tan (A + B) = (tanA + tanB) / (1 - tanA tanB)

tan (A - B) = (tanA - tanB) / (1 + tanA tanB) 

Double Angle Formulas:

sin 2? = 2 sin? cos?

cos 2? = cos²? - sin²

For JEE Main April 2nd shift 2025, a score of 151 marks is expected to fall within the 96.5 to 98 percentile range. The exact percentile can vary slightly based on the difficulty level of the specific paper and the normalization process used. If the paper is considered moderate to slightly difficult

To get into SPA Vijayawada for B.Arch through JEE Main Paper 2, a good percentile to aim for is generally above 95, especially for reserved categories like OBC-NCL. A percentile around 94.93 has a good chance for OBC-NCL candidates with an estimated rank between 7,000 to 10,000. However, it's best t