Equipercentile Method explained by Patrick Dsouza

This is a post just to explain you the equipercentile method that is being used by DTE. I have already created a video on this topic. The link for the same is: “Equipercentile Method Explained: http://youtu.be/WeUVuFQ5-X4”. This is just to explain you the same method in form of a text. Equipercentile is used in IBPS and the formula used is borrowed from them.

The following is a table approximately depicting the top 10 CET rankers in the 4 slots. The number of students that has been taken in the example is just indicative and is not the actual numbers.

Slot 1		Slot 2		Slot 3		Slot 4
Stud	Mks	Stu	Mks	Stu	Mks	Stu	Mks
Tot Mks	200		200		200		200
Wrg Qs	19		18		11		11
A1	180	B1	176	C1	185	D1	174
A2	178	B2	174	C2	183	D2	172
A3	177	B3	173	C3	182	D3	172
A4	174	B4	173	C4	182	D4	172
A5	174	B5	172	C5	180	D5	170
A6	173	B6	170	C6	178	D6	169
A7	172	B7	170	C7	177	D7	169
A8	171	B8	169	C8	177	D8	169
A9	171	B9	168	C9	177	D9	168
A10	170	B10	167	C10	176	D10	167
Total Stud

Formula used to calculate Percentile:

My Percentile Rank = {[(Total no. of Students) – (My Rank)] / (Total no. of Students-1)} X 100

Lets start with the first slot. The total questions and marks is 200. There were 19 incorrect questions in slot 1 so the number of marks out of which you will get you score is 181. Now let us say that the person who has got the first rank has got 180. Now the second ranker has got a score of 178, two students have got 174 and they will get the same rank. We have assumed that there are 15340 students in the first slot. Using the formula you get the percentile of each student.

Slot 1
Stud	Mks	Rank	%ile
Tot Mks	200
Wrg Qs	19
A1	180	1	100
A2	178	2	99.99
A3	177	3	99.99
A4	174	4	99.98
A5	174	4	99.98
A6	173	6	99.97
A7	172	7	99.96
A8	171	8	99.95
A9	171	8	99.95
A10	170	10	99.94
Total Stud		15340

Similar calculations for other slots will give the following results:

Slot 2
Stu	Mks	Rank	%ile
Tot Mks	200
Wrg Qs	18
B1	176	1	100
B2	174	2	99.99
B3	173	3	99.99
B4	173	3	99.99
B5	172	5	99.97
B6	170	6	99.97
B7	170	6	99.97
B8	169	8	99.95
B9	168	9	99.95
B10	167	10	99.94
		14872

Slot 3
Stu	Mks	Rank	%ile
Tot Mks	200
Wrg Qs	11
C1	185	1	100
C2	183	2	99.99
C3	182	3	99.99
C4	182	3	99.99
C5	180	5	99.97
C6	178	6	99.96
C7	177	7	99.96
C8	177	7	99.96
C9	177	7	99.96
C10	176	10	99.93
		13847

Slot 4
Stu	Mks	Rank	%ile
Tot Mks	200
Wrg Qs	11
D1	174	1	100
D2	172	2	99.99
D3	172	2	99.99
D4	172	2	99.99
D5	170	5	99.97
D6	169	6	99.96
D7	169	6	99.96
D8	169	6	99.96
D9	168	9	99.94
D10	167	10	99.94
		14213

If you notice the calculations properly the 10^th rank in slot 3 has got 99.93%ile and the 10^th rank in the remaining slots have got 99.94%ile. The reason for this is that the number of students writing the test in slot 3 was lower. Also though the first ranker from Slot 4 has got less marks than 10^th ranker from slot 3 he has got a higher percentile. The low marks in slot 4 could mainly be due to the difficulty level of the paper as the assumption is that all the students in different slots are of equal capability. So the percentile is calculated purely based on your rank in your slot and the number of students who wrote the exam in your slot. You are not competing with students from other slots for the percentile. So even if the number of wrong questions are different in different slots, it should not matter as you are not competing with students for other slots for percentile calculations.

Combining the percentiles of all the four slots we get a consolidated list as follows.

%ile	Stud	Rank		%ile	Stud	Rank		%ile	Stud	Rank		%ile	Stud	Rank
100	A1	1		99.99	C3	5		99.97	B7	18		99.96	D8	24
100	B1	1		99.99	C4	5		99.97	C5	18		99.95	A8	32
100	C1	1		99.99	D2	5		99.97	D5	18		99.95	A9	32
100	D1	1		99.99	D3	5		99.96	A7	24		99.95	B8	32
99.99	A2	5		99.99	D4	5		99.96	C6	24		99.95	B9	32
99.99	A3	5		99.98	A4	16		99.96	C7	24		99.94	A10	36
99.99	B2	5		99.98	A5	16		99.96	C8	24		99.94	B10	36
99.99	B3	5		99.97	A6	18		99.96	C9	24		99.94	D9	36
99.99	B4	5		99.97	B5	18		99.96	D6	24		99.94	D10	36
99.99	C2	5		99.97	B6	18		99.96	D7	24		99.93	C10	40

If you notice there is no one slot which has an advantage over others. Students with same rank will then further be differentiated into different ranks based on other factors, but that would most likely be done at a later statge. As per DTE if students have got same percentiles then the merit ranks will be awarded based on:

In the case of candidates securing equal MAH-MBA/MMS-CET 2015 percentile, their relative merit will be determined on the basis of the following order of preference:

a) Higher percentage of marks scored in the aggregate at the HSC (Std. XII) or its equivalent examination will get preference.

b) Higher percentage of the marks scored in the aggregate at the SSC (Std. X) or its equivalent examination will get preference.

c) Birth date of the candidate (Elder candidates will get preference)

To sum it up - In an equipercentile method, you basically compete with students with only your slot for the percentiles and will purely depend upon your rank in your slot and the number of students who have written the test in your slot. Number of wrong questions does not matter as all the students in your slot would have the same number of wrong questions. Even different difficulty level in different slots should not make a difference as you are not competing with students of other slots for percentiles. 

Hope this gives you some idea about equipercentile method.

Video of the equipercentile method: