Careers360 Logo
ask-icon
share
    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.

    Articles
    |
    Upcoming Engineering Exams
    Ongoing Dates
    IEMJEE Exam Date

    5 Jan'26 - 30 May'26 (Online)

    Ongoing Dates
    SNUSAT Application Date

    1 Apr'26 - 31 May'26 (Online)

    Certifications By Top Providers
    Online Certificate Course on Cyber Laws
    Via Indian Law Institute, New Delhi
    Basic Programming using Python
    Via Indian Institute of Technology Bombay
    Software Engineering Essentials
    Via Technical University of Munich, Munich
    Material Science
    Via Indira Gandhi National Open University, New Delhi
    B.Tech Engineering Technology
    Via Birla Institute of Technology and Science, Pilani
    Principles of Digital Communications
    Via Indian Institute of Technology Delhi
    Udemy
     1525 courses
    Swayam
     817 courses
    NPTEL
     773 courses
    Coursera
     697 courses
    Edx
     608 courses
    Explore Top Universities Across Globe

    Questions related to JEE Main

    On Question asked by student community

    Have a question related to JEE Main ?

    Hi,

    Yes, with 74% in Class 12 you are eligible for admission in MITS. For B.Tech admission, MITS mainly considers JEE Main score along with eligibility criteria. Usually, minimum eligibility is around 45% in PCM subjects, so your 74% is good enough.

    Admission depends more on your JEE Main rank

    Hi,

    With 70 percentile in JEE Main and 564 rank in JKCET, you may have a good chance for CSE in many private/state engineering colleges, especially through JKCET counselling. Eligibility for CSE depends on the specific college cutoff and counselling round.

    Your JKCET rank of 564 is quite decent and

    Hi,

    With 989 in IPE, you have a very strong board score. Since you applied through Stream 1 in SASTRA, your JEE score will also be considered along with IPE marks. Based on previous trends, you have a fair chance of getting CSE specialization branches like CSE with AI/DS or

    Hello Dear Student,
    With a JEE Mains rank of 1,55,800, getting into top-tier NITs, IIITs, or GFTIs for CSE or ECE will be difficult through JoSAA/CSAB counselling.

    You can check, find and access more information here: https://engineering.careers360.com/jee-main-college-predictor
    Hope it helps!

    Hello Dear Student,

    With around 145 marks in JEE Advanced , your expected OBC-NCL rank may roughly fall around 1.5k–3k , depending on paper difficulty and normalization.

    This is a strong score and you can have good chances in many IITs for branches like:

    • Mechanical

    • Civil

    • Chemical

    • Electrical in some