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How much marks are required to get Computer Science in DTU in Jee mains 2020 September because the competition is relatively high?


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techysparta 29th Jun, 2020
Answers (3)
Naveen Reddy 8th Aug, 2020

Hello'

DTU Delhi reports its cut off and merit list for different courses to get a confirmation in UG and PG programs each year. There are around 2 departments and colleges in DTU Delhi for which cutoff will be resolved dependent on the reservation categories and the seat accessibility. DTU Delhi profoundly and gives you the valid adjust shrewd cut information dependent on different sources. Here you will get DTU Delhi 2019 shorts, from first slice off to last cut off for every one of the 2 departments/universities and their courses.

For CSE cutoff rank is 11564 in the 2019 where as in 2018 cutoff rank is 9194.

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Sharon Rose Student Expert 8th Aug, 2020

Hello,

In 2019 the cut off was as follows for admission to CSE in DTU

Delhi region

  • General: 5896
  • EWS: 11798
  • OBC: 44546
  • SC: 77286
  • ST: 223920

Outside Delhi region

  • General: 1418
  • EWS: 3124
  • OBC: 5822
  • SC: 25375
  • ST: 49944

You have not mentioned your category and domicile. You need to score the minimum cut off rank to get admission in DTU and cut off changes every year depending on total number of candidates applied, difficulty of exam, number of seats available etc. We cannot predict the exact marks but a safe score in between 150-220 marks depending on your category. In January session 869010 candidates appeared and for September session 12,00,000 candidates are expected to appear in JEE Main so yes, there are chances of high competition.

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Vaishak V Prabhu 8th Aug, 2020
For a Delhite General Category student, the minimum target should be at least 195 for getting Computer Science in DTU. But it also depends on the difficulty level of the JEE Mains Paper of 2020. If it is easy then you set the target to be around 205.

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