Tricks To Solve JEE Main Maths MCQs Within Seconds

Tricks To Solve JEE Main Maths MCQs Within Seconds

Vishal kumarUpdated on 19 Apr 2024, 12:43 PM IST

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.

Tricks To Solve JEE Main Maths MCQs Within Seconds
Tricks to solve MCQs within seconds

Example 1

If A + B = 45°, then find the value of \frac{(1+tanA)}{(1+tanB)}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 \frac{(1+tanA)}{(1+tanB)}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 (\frac{(1+tanA)}{(1+tanB)}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°,

\frac{(1+tanA)}{(1+tanB)} = (1+tan0°)(1+tan45°) = (1 + 0)(1 + 1) = 1*2 = 2

So, option C is correct

Example 2

\text{Q: The value of} \ \sum_{r=1}^{n} \frac{r}{1+r^2+r^4} \text{ equals}

\\a.\ \frac{n^2+n}{1+n+n^2} \ \\b.\ \frac{n^2+n}{2(1+n+n^2)} \\c.\ \frac{1}{n+2} \ \\d. \ \frac{1}{n^2+n-1}

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 \frac{1}{1+1+1}= \frac{1}{3}

Checking what value option A for n = 1: \frac{1+1}{1+1+1}=\frac{2}{3}. 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: (\frac{1+1}{2(1+1+1)}=\frac{1}{3}. 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: \frac{1}{2(1+2)}=\frac{1}{3} 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: \frac{1}{1+1-1}=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

\frac{1}{1+1+1}+\frac{2}{1+4+16}=\frac{9}{21}=\frac{3}{7}

Checking what value option B for n = 2: \frac{4+2}{2(1+2+4)}=\frac{3}{7}. 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: \frac{1}{2+2}=\frac{1}{4}. 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

1645770629771

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

1645770629606

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.

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Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

Hi aspirant,

If you are willing to give jee and you are in 12 th then it is best to focus from now on as u have to give your boards this year as well . And according to the exam pattern you can prepare for your jee prelims with you boards at the same time it is hardly differ from each other . But if you have 0 preparation and even your basics are not clear then i suggest u to take a drop and prepare to the fullest for next year thankyou .

Hello Hari

You can find JEE Mains last 10 year Previous Year Questions (PYQs) with detailed solution at CAREERS360 website . CAREERS360 provide JEE Mains as well as JEE Advance questions with solutions in many languages like English and Hindi.

Here's the link: JEE Mains Last 10 Years PYQs by CAREERS360

Hope this link helps! Thank You!!!


Hello Aspirant,

If you already have a Class 12 from NIOS in April 2025 with 67%, you are technically considered a “pass”. But, now since you are reappearing for Class 12 through BOSSE (Sikkim) in October 2025, in order to increase your percentage to 75% (for eligibility) here is how it works:

In the JEE Main Application Form:

  • Choose "Appearing" for Class 12 (you will be giving the BOSSE oct 2025 exam).
  • Enter the passing year as 2025.

For JoSAA Counselling:

  • You must submit the latest valid marksheet, that states you have obtained more than 75%.
  • If your result from BOSSE (oct 2025) comes before counselling/document verification, then to submit BOSSE marksheet.
  • If you only submit NIOS marksheet (67%), you will not then meet the criteria for 75%, so you may be excluded from admission

Here’s a plan for JEE Mains 2026 in 4 months:

1. Divide time: 2 months for Class 12 syllabus, 1 month for Class 11, 1 month for full revision & mock tests.

2. Daily schedule: 6–7 hours study; 50% for theory & problem-solving, 50% for practice & revision.

3. Topic-wise focus: Prioritize high-weightage chapters and weak areas first.

4. Daily problem practice: Solve previous year questions and chapter-wise exercises.

5. Weekly tests: Take 1 full-length test weekly, analyze mistakes, and revise weak concepts.

6. Consistency: Avoid skipping days; maintain notes and formula sheets for quick revision.


If you want to crack JEE exam you read to dedicatedly prepared for that from the scratch to the advance focus on high weightage topic and prepare question in the time based and continuously practice the previous question this will help to know the pattern of JEE exam questions