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Entrance Exam Tips: Well-Kept Secrets to Crack IIT JEE Lakhs of students prepare for IIT JEE every year and only a few thousand get selected. The competition is very high. You get only three chances to clear JEE Main (for admission to NITs, IIITs and other Centrally Funded Technical Institutions) and only two to clear JEE Advanced (for admission to most coveted IITs. As IIT JEE toppers will tell you, engineering entrance exam preparation has two major parts: studying for the exam, and learning tips and tricks to solve the MCQs in the exam quickly and accurately. Here, we bring you 15 tips to cheat at IIT JEE MCQs – straight from the toppers’ diaries: Don’t miss the Minutiae in Questions IIT JEE preparation is the time to train your eyes and mind to be scientifically and technically correct. Do not read the question at just a glance and start solving it. Instead pay attention to each and every letter carefully. Here are examples where missing crucial information can lead you to wrong assumptions: Example1: Determine the points of maximum and minimum of the function f(x) = (1/8) log_{e} x – bx + x^{2}, x > 0, when b is any positive constant. Hence, find such a value of b such that there is no maximum or minimum of the function. Most students miss out on “x > 0” and are not able to solve this question correctly. Example 2: A job can be done by 3 persons in 4 days. How many persons will be required to complete the job in a day and a half? (A) 8 (B) 12 (C) 9 (D) 18 This is an easy question for a cautious candidate. However, the examiner has set a trap for careless students here. If you do not read that you need to calculate the number of persons needed to complete the work in day ‘and a half’, you might calculate the answer 3X4=12. The correct answer is (a) 8 though. Read The Question Carefully To make sure you do not miss out on anything in a quesion, underline all the key conditions. Try to peg the key concept of the question and what is being asked by the examiner. Sometimes, by just reading the question carefully, you will be able to derive at the correct option directly. Start Solving the Paper It is no use wasting time to read the entire paper at once and fish out easy questions first. Start solving the paper. If you find a particular question difficult or time-taking, mark it and move on to the next one. Time management can be done on the go. Assigning Values to Variables To solve a complex equation, put values instead of variables. It will help you quickly reach the correct option. Example: If a+b = pi/2 and b+c=a, then tan a equals (IIT 2001) (A) 2(tan b + tan c) (B) tan b + tan c (C) tan b + 2 tan c (D) 2 tan b+ tan c To solve this question, if you put a=b=pi/4, then c=0 and tan a = 1. Hence, (b) or (c) can be the correct answers. Now, let us suppose that a = pi/8, then b = 3pi/8. Then c = - pi/4. Then ‘tan b + tan c’ will not fit. Hence, (c) is the correct answer. Simplify Geometrical Shape questions In geometry, if you get general form questions, try using an easy value for its variables to simplify them. Example: In any triangle ABC (with usual notations of the sides a,b,c) ( IIT 2005) (A) b-c/2 sin A = a cos A (B) (b-c) cos A/2 = a sin B-C/2 (C) b+c/2 cos A/2 = a sin B-C/2 (D) (b+c) cos A/2 = 2a sin B+C/2 Here, make it simple by assuming that the triangle is an isosceles one. Hence, b=c, i.e. B = C. Only answer (b) satisfies the condition (0=0). Hence, it is the correct alternative. Dimensional Analysis Method Dimensional Analysis can save us a lot of time in questions like the one given below: Example: A simple pendulum of mass m and length l carries a charge q. Find its time period when it is suspended in a uniform electric field region as shown in figure The time period of the pendulum will have the dimension of length in the ‘numerator’ and dimension of acceleration in the ‘denominator’. Options ‘c’ and ‘d’ do not satisfy this condition and hence, we will ignore them. Approximation Method In the exponential series questions, approximating values work quite well. Example: Find the value of 1/2!+1/4!+1/6!+....... (AIEEE 2004) A.(e2-1) / 2 B.(e2-2) / e C.(e2-1) / 2e D.(e-1)2 / 2e This is an irritating problem and would require a sound knowledge of the exponential series. Normally, it would be tough, almost impossible, to solve this question in less than a minute. However, if you examine the series carefully you may write it as, 1/2!+1/4!+1/6!+....... = 1/2 + 1/4.3.2.1 +1/6.5.4.3.2.1 +.......... = 0.5 + 1/24 + 1/120 +.......... = 0.5 + 0.05 (approx) + 0.01 (approx)+ .... Only first two terms are worth approximation as the succeeding terms become negligible. Thus, 1/2!+1/4!+1/6!+....... = 0.5 + 0.05 (approximately) It means our answer should be closer to 0.55. Now, let’s analyze the given options one by one: A .(e2-1) / 2 =(2.712 -1) / 2 = 3.5 take 2.712 ~= 8 (i.e. a little less than 3^2). We know that this can’t be the correct answer. Similarly, B .(e2-2) / e = (8-2)/2.71 ~= 2 We know that this does not fit our answer too. C.(e2-1) / 2e =(8-1)/(2.71*2) > 1 D.(e-1) 2/ 2e =(2.71-1)^2/(2*2.71) < 1. Notice that option ‘D’ is closest to our answer. Hence, ‘D’ is the correct option. Playing with Initial Values Just put the initial values in the series in variables and check the options. Example: The sum of the first n terms of the series 12+2.22 + 32 +2.42 +52+2.62 + ............. is n (n+1) 2/2 when n is even, when n is odd the sum is (AIEEE 2004) A. 3n(n+1)/2 B.[n(n+1)/2]^{2} C. n(n+1)^{ 2}/4 D.n^{2}*(n+1)/2 Luckily for us, the first number in series is ‘1’ which is odd (remember, we need to calculate the sum only if n is an odd number). So, let’s assume n to be 1. Now, the required sum = 1^{2 }(the first term of the series only) = 1. Let’s verify the options: A. 3n(n+1)/2 =3*1(1+1)/2 =3 We know that this can’t be the correct answer. B. [n(n+1)/2] 2 =[1(2)/2] 2 = 1 C. n(n+1) 2/4 =1 D.n2*(n+1)/2=1 Here, we see that B,C and D – all options are giving us the correct answer. So, let’s put another odd value from the series as ‘n’. Put n=3. The required sum now is 1+8+9=18. Check the remaining options: B.[n(n+1)/2]^{ 2 }=[3(4)/2]^{ 2 }= 36 Not the correct answer. C. n(n+1)^2/4 =36 Not the correct answer. D.n^2*(n+1)/2=18 This is the correct answer. Smarter ones would observe that ‘n=1’ will give same answer for three alternatives, and straightaway substitute ‘n=3’ instead to save time. Digit-by-Digit Calculations Example: Find the value of 36C3 A. 7282 B.7140 C. 6954 D.8326 Most students will be able to calculate it correctly. The answer is 7140. But the trick here will be able to save you about 45 seconds. You probably solved the question in the following manner: 36C3= (36*35*34)/(3*2*1) = ( 6*35*34) = 7140 Did you spend the last minute to actually calculate the value of 6*35*34? You didn’t need to. You could have just checked what should be the last digit of the final product (by multiplying 6*5*4) and it would be sufficient to reveal the correct answer - as only ‘B’ has ‘0’ as the last digit. With this trick, you should take less than 15 seconds to solve the question. More Marks per Minute It might be more satisfying to get to the answer of a challenging question but this strategy can be your undoing in the JEE exam or any other engineering entrance exam you want to appear for. Opt for more marks per minute. If you can solve 2 questions in a minute correctly, go for them first. Do not dwell too much on one question. Try Extreme Values first Example: In complex domain questions, don’t solve the function for each given value. Just put in extreme values first to check if they satisfy the function or not. (A) [0,5) U (6,7) (B) (0,5) U (6,7] (C) [0,5] U [6,7] (D) (0,5) U (6,7) Put extreme values ( 5,6 and 7) and see whether these values satisfy the function or not. From this you can guess the answer. Trace Back the Steps Do you remember how we used to look back at the answer to a Maths question first and then, try to solve it later to reach it. In MCQ-based exam pattern, you have correct answer right in front of you. So, you can try tracing back the steps and see if a particular option fits the question or not. Precaution for Pen-and-Paper based Exams In pen-and-paper based entrance exams, you have to be very careful while marking the answers in the ORS sheet. If you choose wrong option for the wrong answer, you will only earn penalty marks. Questions with No Negative Marking Know the exam pattern well and read the instructions carefully. If there is no negative marking for certain questions, make sure to choose any option for them. Since, there is no harm if you mark the incorrect answer, take your guess even for the questions you do not know at all. Do not get swayed by Emotions If you don’t know an answer for sure and employed all the tips to make an educated guess, leave it at that. Do not get swayed by your emotions and nervousness and start second-guessing them. This tendency often leads you in the wrong direction. For more preparation tips for JEE Main and JEE Advanced, Please fill the form below:
Lakhs of students prepare for IIT JEE every year and only a few thousand get selected. The competition is very high. You get only three chances to clear JEE Main (for admission to NITs, IIITs and other Centrally Funded Technical Institutions) and only two to clear JEE Advanced (for admission to most coveted IITs. As IIT JEE toppers will tell you, engineering entrance exam preparation has two major parts: studying for the exam, and learning tips and tricks to solve the MCQs in the exam quickly and accurately. Here, we bring you 15 tips to cheat at IIT JEE MCQs – straight from the toppers’ diaries:
IIT JEE preparation is the time to train your eyes and mind to be scientifically and technically correct. Do not read the question at just a glance and start solving it. Instead pay attention to each and every letter carefully. Here are examples where missing crucial information can lead you to wrong assumptions:
Example1: Determine the points of maximum and minimum of the function f(x) = (1/8) log_{e} x – bx + x^{2}, x > 0, when b is any positive constant. Hence, find such a value of b such that there is no maximum or minimum of the function.
Most students miss out on “x > 0” and are not able to solve this question correctly.
Example 2: A job can be done by 3 persons in 4 days. How many persons will be required to complete the job in a day and a half?
(A) 8 (B) 12 (C) 9 (D) 18
This is an easy question for a cautious candidate. However, the examiner has set a trap for careless students here. If you do not read that you need to calculate the number of persons needed to complete the work in day ‘and a half’, you might calculate the answer 3X4=12. The correct answer is (a) 8 though.
To make sure you do not miss out on anything in a quesion, underline all the key conditions. Try to peg the key concept of the question and what is being asked by the examiner. Sometimes, by just reading the question carefully, you will be able to derive at the correct option directly.
It is no use wasting time to read the entire paper at once and fish out easy questions first. Start solving the paper. If you find a particular question difficult or time-taking, mark it and move on to the next one. Time management can be done on the go.
To solve a complex equation, put values instead of variables. It will help you quickly reach the correct option.
Example: If a+b = pi/2 and b+c=a, then tan a equals (IIT 2001)
(A) 2(tan b + tan c) (B) tan b + tan c
(C) tan b + 2 tan c (D) 2 tan b+ tan c
To solve this question, if you put a=b=pi/4, then c=0 and tan a = 1.
Hence, (b) or (c) can be the correct answers.
Now, let us suppose that a = pi/8, then b = 3pi/8. Then c = - pi/4.
Then ‘tan b + tan c’ will not fit. Hence, (c) is the correct answer.
In geometry, if you get general form questions, try using an easy value for its variables to simplify them.
Example: In any triangle ABC (with usual notations of the sides a,b,c) ( IIT 2005)
(A) b-c/2 sin A = a cos A (B) (b-c) cos A/2 = a sin B-C/2
(C) b+c/2 cos A/2 = a sin B-C/2 (D) (b+c) cos A/2 = 2a sin B+C/2
Here, make it simple by assuming that the triangle is an isosceles one. Hence, b=c, i.e. B = C.
Only answer (b) satisfies the condition (0=0). Hence, it is the correct alternative.
Dimensional Analysis can save us a lot of time in questions like the one given below:
Example: A simple pendulum of mass m and length l carries a charge q. Find its time period when it is suspended in a uniform electric field region as shown in figure
The time period of the pendulum will have the dimension of length in the ‘numerator’ and dimension of acceleration in the ‘denominator’. Options ‘c’ and ‘d’ do not satisfy this condition and hence, we will ignore them.
In the exponential series questions, approximating values work quite well.
Example: Find the value of 1/2!+1/4!+1/6!+....... (AIEEE 2004)
A.(e2-1) / 2 B.(e2-2) / e C.(e2-1) / 2e D.(e-1)2 / 2e
This is an irritating problem and would require a sound knowledge of the exponential series. Normally, it would be tough, almost impossible, to solve this question in less than a minute. However, if you examine the series carefully you may write it as,
1/2!+1/4!+1/6!+....... = 1/2 + 1/4.3.2.1 +1/6.5.4.3.2.1 +.......... = 0.5 + 1/24 + 1/120 +..........
= 0.5 + 0.05 (approx) + 0.01 (approx)+ .... Only first two terms are worth approximation as the succeeding terms become negligible.
Thus, 1/2!+1/4!+1/6!+....... = 0.5 + 0.05 (approximately)
It means our answer should be closer to 0.55. Now, let’s analyze the given options one by one: A .(e2-1) / 2 =(2.712 -1) / 2 = 3.5 take 2.712 ~= 8 (i.e. a little less than 3^2).
We know that this can’t be the correct answer. Similarly,
B .(e2-2) / e = (8-2)/2.71 ~= 2 We know that this does not fit our answer too.
C.(e2-1) / 2e =(8-1)/(2.71*2) > 1
D.(e-1) 2/ 2e =(2.71-1)^2/(2*2.71) < 1. Notice that option ‘D’ is closest to our answer.
Hence, ‘D’ is the correct option.
Just put the initial values in the series in variables and check the options.
Example: The sum of the first n terms of the series 12+2.22 + 32 +2.42 +52+2.62 + ............. is n (n+1) 2/2
when n is even, when n is odd the sum is (AIEEE 2004)
A. 3n(n+1)/2 B.[n(n+1)/2]^{2} C. n(n+1)^{ 2}/4 D.n^{2}*(n+1)/2
Luckily for us, the first number in series is ‘1’ which is odd (remember,
we need to calculate the sum only if n is an odd number). So, let’s assume n to be 1.
Now, the required sum = 1^{2 }(the first term of the series only) = 1. Let’s verify the options:
A. 3n(n+1)/2 =3*1(1+1)/2 =3 We know that this can’t be the correct answer. B. [n(n+1)/2] 2 =[1(2)/2] 2 = 1 C. n(n+1) 2/4 =1 D.n2*(n+1)/2=1
Here, we see that B,C and D – all options are giving us the correct answer.
So, let’s put another odd value from the series as ‘n’.
Put n=3. The required sum now is 1+8+9=18. Check the remaining options:
B.[n(n+1)/2]^{ 2 }=[3(4)/2]^{ 2 }= 36 Not the correct answer. C. n(n+1)^2/4 =36 Not the correct answer.
D.n^2*(n+1)/2=18 This is the correct answer.
Smarter ones would observe that ‘n=1’ will give same answer for three
alternatives, and straightaway substitute ‘n=3’ instead to save time.
Example: Find the value of 36C3
A. 7282 B.7140 C. 6954 D.8326
Most students will be able to calculate it correctly. The answer is 7140.
But the trick here will be able to save you about 45 seconds.
You probably solved the question in the following manner:
36C3= (36*35*34)/(3*2*1) = ( 6*35*34) = 7140
Did you spend the last minute to actually calculate the value of 6*35*34? You didn’t need to.
You could have just checked what should be the last digit of the final product (by multiplying 6*5*4) and it would be sufficient to reveal the correct answer - as only ‘B’ has ‘0’ as the last digit. With this trick, you should take less than 15 seconds to solve the question.
It might be more satisfying to get to the answer of a challenging question but this strategy can be your undoing in the JEE exam or any other engineering entrance exam you want to appear for. Opt for more marks per minute. If you can solve 2 questions in a minute correctly, go for them first. Do not dwell too much on one question.
Example: In complex domain questions, don’t solve the function for each given value. Just put in extreme values first to check if they satisfy the function or not.
(A) [0,5) U (6,7) (B) (0,5) U (6,7]
(C) [0,5] U [6,7] (D) (0,5) U (6,7)
Put extreme values ( 5,6 and 7) and see whether these values satisfy the function or not. From this you can guess the answer.
Do you remember how we used to look back at the answer to a Maths question first and then, try to solve it later to reach it. In MCQ-based exam pattern, you have correct answer right in front of you. So, you can try tracing back the steps and see if a particular option fits the question or not.
In pen-and-paper based entrance exams, you have to be very careful while marking the answers in the ORS sheet. If you choose wrong option for the wrong answer, you will only earn penalty marks.
Know the exam pattern well and read the instructions carefully. If there is no negative marking for certain questions, make sure to choose any option for them. Since, there is no harm if you mark the incorrect answer, take your guess even for the questions you do not know at all.
If you don’t know an answer for sure and employed all the tips to make an educated guess, leave it at that. Do not get swayed by your emotions and nervousness and start second-guessing them. This tendency often leads you in the wrong direction.
For more preparation tips for JEE Main and JEE Advanced, Please fill the form below:
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