We all are familiar with Multiple Choice Questions (MCQs). These are the questions which give us some options, usually 4, out of which one, or sometimes more than one, is correct. The Joint Entrance Examination Main, or JEE Main, has 30 Questions per subject out of which 20 are MCQs with only one correct option. Generally in mathematics for some questions, you will find the options in integer values like chapters trigonometry, sequence and series and others for which you can easily get the answer by using the values given in options directly to solve questions and verify it in less time. In this article, we are going to explain some ticks and trips with the help of examples which you can implement in various questions. Let us understand the trick with the help of some examples.
Example 1
If A + B = 45°, then find the value of
equals
a)1
b)0
c)2
d)1
Let us first try to understand the question. It means that if A + B = 45°, then the value of
is always constant, which is equal to one of the options. So, for all combinations of angles A and B whose sum is 45°, the value of (
will always be the same, which is equal to one of the four options.
So, whether (A = 0° and B = 45°) or (A = 1 ° and B = 44°) or (A = 10° and B = 35°), they all will give the same value of (1+tanA)(1+tanB). So, we can easily replace A and B with some convenient values of angles and get the required value of (1+tanA) * (1+tanB). Let us put A = 0° and B = 45°,
= (1+tan0°)(1+tan45°) = (1 + 0)(1 + 1) = 1*2 = 2
So, option C is correct
Example 2


Now we have to select the option that gives the correct sum for all natural number values of n. So, the correct option has to be true for n = 1, 2, 3, …and all other natural number values of n. So, if any option is giving the wrong sum for n = 1, we can be sure that that option is incorrect.
Also Read,
Let us try to find the options that are incorrect.
For n = 1, the actual sum of the given series is
= 
Checking what value option A for n = 1:
. So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Checking what value option B for n = 1: (
. So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option C for n = 1:
So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option D for n = 1:
So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Now both options A and D are eliminated, and one of option B or C is correct. Let us now see which of these two is giving the correct answer for n = 2.
For n = 2, the actual sum of the given series is

Checking what value option B for n = 2:
. So, it gives the correct answer for n = 2 as well. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option C for n = 2:
. So, it gives the wrong answer for n = 2, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Now we know that 3 out of 4 options are wrong, and only option B is left. So, we can safely mark option B as correct.
Example 3

If the value of the determinant given equals ka3b3c3, then the value of k is
a)1
b)0
c)-1
d)2
This question means that the value of the determinant always equals ka3b3c3 for all sets of values of a, b and c. In such questions, we can substitute some values of a, b and c and check to see the values of the determinant and the value of ka3b3c3. Also try to keep values of a, b and c such that ka3b3c3 does not become 0. So we will keep non-zero values of a, b and c. Let us put a = b = c = 1.
The value of the determinant is

which equals 0 (As two columns are the same) and the value of ka3b3c3 is k. So comparing these, we get k = 0.
Know when to use this trick and when not to. Don't use it for questions with more than one right answer or for multiple-choice questions where 'None of these' or 'All of these' is the only correct option.
On Question asked by student community
The marks needed for a 99+ percentile in the JEE Main January attempt depend on the difficulty of the paper and the total number of candidates. Generally, you need roughly *180–200* marks out of 300 to hit the 99+ percentile. The exact cutoff varies each session, so checking the official NTA percentile score calculator or previous year cutoffs gives a more precise idea.
Hello aspirant
JEE Main accepts NIOS, so you can appear if you meet the basic eligibility.
BITS does not accept marks from two different boards, so this option won’t work for BITS.
VIT and SRM generally accept NIOS, but having two separate mark sheets can be an issue. You should check their official eligibility rules before applying.
Thankyou I hope this help
Hi
The NCHM JEE 2026 registration is expected to start from the second or third week of December 2025. The official notification for the registration will be released in mid-December 2025. The last date of application may be in February or March, and the correction window will be open in the third week of March 2026. The exam date will take place in the second week of April 2026.
Thank you.
Hello,
Here are some Government colleges that generally do not require 75% CBSE board criteria for admission through JEE mains based or university counselling.
I hope it will help you. Kindly check the latest eligibility rules for the specific year.
Thank you.
Hello
If you want to get admission in IIT Delhi, then you will score well in JEE Advanced. Only the JEE Mains score is not sufficient to get admission in the IIT Delhi. You need to score 250+ marks for the general category in CSE. For the reservation category, you also need to score much higher marks and give your best performance. If you score quite well in JEE Advanced, then you will have a chance to get admission through JOSAA counselling.
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