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How many marks are required to get 100 percentile in Mathematics In jee mains


Prudhvi 16th Sep, 2020
Answer (1)
Esita Sau 16th Sep, 2020
Hello there,
Score Percentile
-75 - -20= 0.843517743614459 - 0.843517743614459
-19 - -10= 0.843517743614459 - 0.843517743614459
0 - 10= 0.843517743614459 - 9.69540662201048
11 - 20= 13.4958497103427 - 33.2291283360524
21 - 30= 37.6945295632834 - 56.5693109770195
31 - 40= 58.1514901857346 - 71.3020522957121
41 - 50= 73.2878087751462 - 80.9821538087469
51 - 60= 82.0160627661434 - 86.9079446541208
61 - 70= 87.5122250914779 - 90.7022005707394
71 - 80= 91.0721283110867 - 93.1529718505396
81 - 90= 93.4712312797351 - 94.7494792463808
91 - 100= 94.9985943180054 - 96.0648502433078
101- 110= 96.2045500677875 - 96.9782721725982
111 - 120 = 97.1429377776765 - 97.6856721385145
121 - 130 = 97.8112608696124 - 98.2541321080562
131 - 140= 98.3174149345299 - 98.6669358629096
141 - 150 =98.7323896268267 - 98.9902969950969
151 - 160= 99.0286140409721 - 99.2397377073381
161 - 170 =99.272084675244 - 99.4312143898418
171 - 180 =99.4569399985455 - 99.573193698637
181 - 190 =99.5973996511304 - 99.6885790237511
191 - 200= 99.7108311325455 - 99.7824720681761
201 - 210= 99.7950635053476 - 99.845212160289
211 - 220= 99.8516164257469 - 99.8937326121479
221 - 230= 99.9011137994553 - 99.9289017987302
231 - 240= 99.9349804235716 - 99.9563641573886
241 - 250= 99.9601632979145 - 99.9750342194015
250 - 262= 99.9772051568448 - 99.9888196721667
263 - 270 =99.9909906096101 - 99.9940299220308
271 - 280 =99.9946812032638 - 99.997394875068
281-300= 99.99989145
A minimum marks of 120 marks approximately is required for 100 percentile.
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https://engineering.careers360.com/articles/jee-main-marks-vs-percentile/amp
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https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer
Thank you.
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