Expected Percentile for MHT CET April 11, 2026 Shift 2

Q: Ques: What would be the predicted percentile for MHT CET April 11, 2026 Shift 2 with 130 marks?

Ans: The expected percentile for MHT CET 2026 with 130-150 marks would be 98 - 99.49 percentile, which will give candidates 7000 rank in MHT CET 2026.

Q: Ques: Is percentile the same as percentage of marks obtained in MHT CET 2026?

Ans: No, they differ. The percentile gives the number of students who obtained the same or less than you have obtained in the examination. The percentile calculation can be done using the State CET Cell s normalization process.

Q: Which colleges can I get with 130 Marks in MHT CET 2026 exam?

With 130 marks in MHT CET 2026, candidates can expect a within 1800 to 7000. At this level, you may get colleges like COEP Pune,VJTI Mumbai and top autonomous institutes, depending on branch choice.

Q: What will be my expected rank for 150 marks in MHT CET 2026 exam?

The expected rank for 150 marks in MHT 2026 exam would from 1800 to 7000.

Q: Ques: Is 130 marks sufficient for admission into a government engineering college in Maharashtra for the year 2026?

Ans: 130 marks are sufficient to apply for branches like mechanical, civil, or electrical engineering at decent government-aided colleges.

Q: Ques: Is it possible to enroll in Computer Engineering at a private autonomous college with 130-150 marks in MHT CET held on April 11, 2026, Shift 2?

Ans: Yes, 130-150 marks are adequate for admission into computer engineering in private autonomous colleges such as PICT Pune and VIT Pune.

Q: Ques: Does the percentile of 98.1-98.6 for 130-150 marks hold for all shifts in the MHT CET 2026 examination?

Ans: The percentile of 130-150 marks varies based on the difficulty level of MHT CET exam shifts.

The MHT CET Apr 11,2026 shift exam was concluded on 11th April,2026 and as per the student reactions, the shift 2 paper was moderate in difficulty level.

As the exam was moderate, the expected percentile for a score of 130-150 marks out of 200 marks in MHT CET April 11,2026 Shift 2 can be estimated to 98.00 to 99.49 percentile.

The PCM paper carries 150 questions for 200 marks to be attempted in 180 minutes.
If a candidate get marks above 160 then the percentile would be 99.5 which will get you expected in the top 2000 rank, while 120 marks get you to 96 percentile with 10000-12000 rank range.
A score of 130-150 marks in MHT CET 2026 Exam is likely to get you percentile 98.0-99.49, placing candidates in from 1800 to 7000 ranks.

Also Read:

Expected Percentile for MHT CET April 11, 2026 Shift 2 - Score 130-150 Marks

Key Summary

Candidates scoring 160+ marks are expected to 99.5 percentile, which puts them in a much stronger position for top colleges
In MHT CET 2026 exam the 130–150 marks range is highly competitive, where even a few marks can shift ranks by thousands
With 130 - 150 marks, students can expect decent colleges and branches, but getting top colleges or CSE/IT in top colleges will still be difficult

MHT CET April 11, 2026 Shift 2: Marks vs Percentile

Candidates with 160+ marks in MHT CET 2026 will get 99.5 percentile.
Candidates with 130-150 marks in MHT CET April 11,2026 are expected to get 98 - 99.49 percentile.
The percentile for 130-150 marks can vary depending on the difficulty level of the specific MHT CET exam 2026 exam shift.

Marks (Out of 200)	Expected Percentile
180 - 200	99.9 - 100
160 - 179	99.50 - 99.89
130 - 150	98.00 - 99.49
100 - 129	95.00 - 97.99
80 - 99	90.00 - 94.99
60 - 79	80.00 - 89.99
50 - 59	70.00 - 79.99
Below 50	Below 70.00

MHT CET April 11, 2026 Shift 2: Marks vs Percentile

Also Read:

130 Marks in MHT CET Percentile

Ques: What would be the predicted percentile for MHT CET April 11, 2026 Shift 2 with 130 marks?

Ans: The expected percentile for MHT CET 2026 with 130-150 marks would be 98 - 99.49 percentile, which will give candidates 7000 rank in MHT CET 2026.

Ques: Is percentile the same as percentage of marks obtained in MHT CET 2026?

Ans: No, they differ. The percentile gives the number of students who obtained the same or less than you have obtained in the examination. The percentile calculation can be done using the State CET Cell’s normalization process.

MHT CET 2026 Shift-Wise Marks vs Percentile Analysis

The marks corresponding to certain percentiles also change after normalization process in MHT CET 2026, since the difficulty level of each shift may vary.

Candidates who get more than 180 marks in the difficult shift will be able to get a 99.9 percentile while for the easy shift the percentile will be 99.5 only because there will be more competition.
For 140-130 marks range corresponds to 97.5 percentile in difficult shift and 96 percentile in easy shift after normalization in MHT CET 2026 exam.
For 150 to 140 marks you can get percentile of 98.5 for difficult shift whereas 98 and 97 for medium and easy shift.
130 marks falls under another range which is 130 - 120 marks where for difficult shift the percentile would be 97 whereas in easy shift the percentile would be 95.

Marks Range	Easy Shift	Medium Shift	Difficult Shift
180+	99.5+	99.8+	99.9+
180 - 160	98.5+	99+	99.5+
150 - 140	97+	98+	98.5+
140 - 130	96+	97+	97.5+
130 - 120	95+	96+	97+
120 - 110	94+	95+	96+
110 - 100	93+	94+	95+
100 - 95	91+	92+	93+
95 - 90	90+	91+	92+
90 - 85	86+	87+	89+
85 - 75	70+	72+	73+
75 - 60	52+	55+	60+
60 - 40	45+	50+	55+
Below 40	<45	<50	<55

MHT CET 2026 Shift-Wise Marks vs Percentile Analysis

Also Read:

MHT CET April 11, 2026 Shift 2: Marks vs Rank

Candidates with 180+ marks in MHT CET 2026 will be in the top 100 ranks in MHT CET 2026 exam.
With 130 - 150 marks in MHT CET 2026 exam candidates will get expected rank between 1800 to 7000.
Candidates who have scored between 100-119 marks fall into a very competitive category where even one mark can cause a thousand ranks jump.

Marks (Out of 200)	Expected Rank
180 - 200	1 - 100
160 - 179	100 - 1800
130 - 150	1800 - 7000
100 - 129	7000 - 20000
80 - 99	20000 - 40000
60 - 79	40000 - 55000
50 - 59	55000 - 70000
Below 50	Above 70000

MHT CET April 11, 2026 Shift 2: Marks vs Rank

Also Read:

Ques. Which colleges can I get with 130 Marks in MHT CET 2026 exam?

Ans. With 130 marks in MHT CET 2026, candidates can expect a within 1800 to 7000. At this level, you may get colleges like COEP Pune,VJTI Mumbai and top autonomous institutes, depending on branch choice.

Ques. What will be my expected rank for 150 marks in MHT CET 2026 exam?

Ans. The expected rank for 150 marks in MHT 2026 exam would from 1800 to 7000.

130-150 Marks in MHT CET 2026 Exam: What Colleges Can You Get?

Scoring 130-150 marks in MHT CET 2026 get you rank range from 1800 to 7,000, which is decent but not strong enough for the very top colleges
Candidates can get admission in colleges like COEP Pune, VJTI Mumbai, and WCE Sangli for branches Civil, mechanical with above 99 percentile.
Even top private colleges like PICT Pune and VIT Pune you can get admission with percentile 98 and above.
Candidates can get into government colleges with 130 marks in MHT CET 2026 exam such as SGGS Nanded and Shivaji University, Kolhapur.
Candidate with 130-150 marks in MHT CET 2026, it is difficult to get a seat in CSE or IT in top colleges but candidate can get seat in branches like mechanical, civil or electrical.

College Name	Branch	Type	Percentile Cutoff
COEP Pune	Mechanical, Civil, Electrical	Government	99.4
VJTI Mumbai	Mechanical, Civil, Electrical	Government	98.2
PICT Pune	CSE, IT, AI-DS, ECE	Private	99.1
MIT Pune	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Private	96.18
VIT Pune	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Private	98.9
PCCOE Pune	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Private	98.6
DY Patil Pune	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Private	98.1
WCE Sangli	CSE, IT, ECE, Mechanical, Civil, Electrical	Government	99.3
SGGS Nanded	CSE, IT, ECE, Mechanical, Civil, Electrical	Government	96.67
SPCE Mumbai	Mechanical, Civil, Electrical	Government	99.05
JSPM Pune	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Private	89.68
Shivaji University, Kolhapur	CSE, IT, AI-DS, ECE, Mechanical, Civil, Electrical	Government	93.09

Also Read:

Ques: Is 130 marks sufficient for admission into a government engineering college in Maharashtra for the year 2026?

Ans: 130 marks are sufficient to apply for branches like mechanical, civil, or electrical engineering at decent government-aided colleges.

Ques: Is it possible to enroll in Computer Engineering at a private autonomous college with 130-150 marks in MHT CET held on April 11, 2026, Shift 2?

Ans: Yes, 130-150 marks are adequate for admission into computer engineering in private autonomous colleges such as PICT Pune and VIT Pune.

Ques: Does the percentile of 98.1-98.6 for 130-150 marks hold for all shifts in the MHT CET 2026 examination?

Ans: The percentile of 130-150 marks varies based on the difficulty level of MHT CET exam shifts.

Expected Percentile for MHT CET April 11, 2026 Shift 2 - Score 130-150 Marks

Key Summary

MHT CET April 11, 2026 Shift 2: Marks vs Percentile

MHT CET 2026 Shift-Wise Marks vs Percentile Analysis

MHT CET April 11, 2026 Shift 2: Marks vs Rank

130-150 Marks in MHT CET 2026 Exam: What Colleges Can You Get?

MHT CET Questions

1.
A particle starts oscillating simple harmonically from its mean position with time period \(T\). At time \(t = \frac{T}{6}\), find the ratio of potential energy to kinetic energy of the particle.

2.
Evaluate the integral: \(\int \frac{2x}{x^2 - 5x + 4} \, dx\)

3.
If \(\tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\), find \(\alpha\).

4.
Find the heat energy that must be supplied to 14 g of nitrogen at room temperature to raise its temperature by \(48^\circ\text{C}\) at constant pressure. (Molecular weight of nitrogen = 28; \(R\) is the gas constant; \(C_p = \frac{7}{2}R\) for a diatomic gas.)

5.
If \(A, B, C\) are vertices of a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then find the position vector of the point \(D\) where the angle bisector from vertex \(A\) meets \(BC\).

6.
Two progressive waves \(Y_1 = \sin 2\pi \left(\frac{t}{0.4} - \frac{x}{4}\right)\) and \(Y_2 = \sin 2\pi \left(\frac{t}{0.4} + \frac{x}{4}\right)\) superpose to form a standing wave (\(x\) and \(y\) in SI units). Find the amplitude of the particle at \(x = 0.5\) m.

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Key Summary

MHT CET April 11, 2026 Shift 2: Marks vs Percentile

MHT CET 2026 Shift-Wise Marks vs Percentile Analysis

MHT CET April 11, 2026 Shift 2: Marks vs Rank

130-150 Marks in MHT CET 2026 Exam: What Colleges Can You Get?

MHT CET Questions

1.A particle starts oscillating simple harmonically from its mean position with time period \(T\). At time \(t = \frac{T}{6}\), find the ratio of potential energy to kinetic energy of the particle.

2.Evaluate the integral: \(\int \frac{2x}{x^2 - 5x + 4} \, dx\)

3.If \(\tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\), find \(\alpha\).

4.Find the heat energy that must be supplied to 14 g of nitrogen at room temperature to raise its temperature by \(48^\circ\text{C}\) at constant pressure. (Molecular weight of nitrogen = 28; \(R\) is the gas constant; \(C_p = \frac{7}{2}R\) for a diatomic gas.)

5.If \(A, B, C\) are vertices of a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then find the position vector of the point \(D\) where the angle bisector from vertex \(A\) meets \(BC\).

6.Two progressive waves \(Y_1 = \sin 2\pi \left(\frac{t}{0.4} - \frac{x}{4}\right)\) and \(Y_2 = \sin 2\pi \left(\frac{t}{0.4} + \frac{x}{4}\right)\) superpose to form a standing wave (\(x\) and \(y\) in SI units). Find the amplitude of the particle at \(x = 0.5\) m.

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1.
A particle starts oscillating simple harmonically from its mean position with time period \(T\). At time \(t = \frac{T}{6}\), find the ratio of potential energy to kinetic energy of the particle.

2.
Evaluate the integral: \(\int \frac{2x}{x^2 - 5x + 4} \, dx\)

3.
If \(\tan^{-1}(-1) + \tan^{-1}(5) + \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{4}\right) = \pi + \tan^{-1}\left(\frac{\alpha}{2}\right)\), find \(\alpha\).

4.
Find the heat energy that must be supplied to 14 g of nitrogen at room temperature to raise its temperature by \(48^\circ\text{C}\) at constant pressure. (Molecular weight of nitrogen = 28; \(R\) is the gas constant; \(C_p = \frac{7}{2}R\) for a diatomic gas.)

5.
If \(A, B, C\) are vertices of a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then find the position vector of the point \(D\) where the angle bisector from vertex \(A\) meets \(BC\).

6.
Two progressive waves \(Y_1 = \sin 2\pi \left(\frac{t}{0.4} - \frac{x}{4}\right)\) and \(Y_2 = \sin 2\pi \left(\frac{t}{0.4} + \frac{x}{4}\right)\) superpose to form a standing wave (\(x\) and \(y\) in SI units). Find the amplitude of the particle at \(x = 0.5\) m.