Hii,

Hope you are doing good.

So, 55 Percentile is good. Most probably your rank would be between 47k - 53k.

Below link provides you with names you can get admission through mht-cet rank:

MHT CET Percentile Score = 100 * (Number of candidates in exam with normalized marks ≤ the candidate) ÷ Total number of candidates in the exam.

Some college names with respect to your rank:

1. Keystone school of engineering, Pune CSE--> 2020 cut off- 43007
2. KKEI's Trinity academy of engineering, Pune CSE-->2020 cut off- 46634
3. SKN Sinhgad Institute of Technology & Science, Kusgaon (BK), Pune. CSE-->2020 cut off- 56-59k

And many more. You can have a look bu entering your rank in the link provided above.

I hope this helps you.

Best wishes ahead.