Joint Entrance Examination

Graduate Aptitude Test in Engineering

NEW

New Website Launch

Experience the best way to solve previous year questions with **mock
tests** (very detailed analysis), **bookmark your favourite questions**, **practice** etc...

1

MCQ (Single Correct Answer)

A

Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.

B

Statement - 1 is true, Statement - 2 is false.

C

Statement - 1 is false, Statement - 2 is true.

D

Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1.

Let XA_{}, X_{B}, X_{C} and X_{D} represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

X_{A} $$\ge$$ 1, X_{B} $$\ge$$ 1, X_{C} $$\ge$$ 1 and X_{D} $$\ge$$ 1

$$\Rightarrow$$ X_{A} $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X_{B} $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X_{C} $$-$$ 1 $$\ge$$ 0 and $$\Rightarrow$$ X_{D} $$-$$ 1 $$\ge$$ 0

t_{A} $$\ge$$ 0, t_{B} $$\ge$$ 0, t_{C} $$\ge$$ 0 and t_{D} $$\ge$$ 0

According to the question,

X_{A} + X_{B} + X_{C} + X_{D} = 10

$$\Rightarrow$$ (X_{A} $$-$$ 1) + (X_{B} $$-$$ 1) + (X_{C} $$-$$ 1) + (X_{D} $$-$$ 1) = 6

$$\Rightarrow$$ t_{A} + t_{B} + t_{C} + t_{D} = 6

Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6

From formula we know, n things can be distributed among r people in $${}^{n + r - 1}{C_{r - 1}}$$ ways where each people can have either 0 or more things.

$$\therefore$$ 6 balls can be distributed among 4 boxes in $${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$$ ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$ ways. But Statement - 2 is not the correct explanation of Statement - 1.

2

MCQ (Single Correct Answer)

These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:

A

$$N \le 100$$

B

$$100 < N \le 140$$

C

$$140 < N \le 190\,$$

D

$$N > 190$$

We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$

Here 6 points are on the same line so we can't make any triangle with those 6 points.

So subtract $${}^{6}{C_3}$$.

$$\therefore$$ $$N = {}^{10}{C_3} - {}^6{C_3}$$

$$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$$

$$ = 120 - 20$$

$$ = 100$$

3

MCQ (Single Correct Answer)

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

A

36

B

66

C

108

D

3

Thus number of ways $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$

4

MCQ (Single Correct Answer)

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is :

A

at least 500 but less than 750

B

at least 750 but less than 1000

C

at least 1000

D

less than 500

From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways

And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways

Let 4 novels are N_{1}, N_{2}, N_{3}, N_{4} and 1 dictionary is D_{1}.

Dictionary should be in the middle. So the arrangement will be like this

_ _ D_{1} _ _

On those 4 blank places 4 novels N_{1}, N_{2}, N_{3}, N_{4} can be placed. And 4 novels can be arrange $$4!$$ ways.

$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080

And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways

Let 4 novels are N

Dictionary should be in the middle. So the arrangement will be like this

_ _ D

On those 4 blank places 4 novels N

$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Main 2021 (Online) 1st September Evening Shift (1)

JEE Main 2021 (Online) 25th July Evening Shift (1)

JEE Main 2021 (Online) 18th March Morning Shift (1)

JEE Main 2021 (Online) 17th March Evening Shift (1)

JEE Main 2021 (Online) 17th March Morning Shift (1)

JEE Main 2021 (Online) 16th March Evening Shift (1)

JEE Main 2021 (Online) 26th February Evening Shift (1)

JEE Main 2021 (Online) 26th February Morning Shift (1)

JEE Main 2021 (Online) 25th February Morning Shift (1)

JEE Main 2021 (Online) 24th February Morning Shift (1)

JEE Main 2020 (Online) 6th September Morning Slot (1)

JEE Main 2020 (Online) 5th September Evening Slot (1)

JEE Main 2020 (Online) 3rd September Morning Slot (1)

JEE Main 2020 (Online) 2nd September Evening Slot (1)

JEE Main 2020 (Online) 9th January Morning Slot (1)

JEE Main 2020 (Online) 8th January Morning Slot (1)

JEE Main 2020 (Online) 7th January Evening Slot (1)

JEE Main 2020 (Online) 7th January Morning Slot (1)

JEE Main 2019 (Online) 12th April Evening Slot (1)

JEE Main 2019 (Online) 12th April Morning Slot (1)

JEE Main 2019 (Online) 10th April Evening Slot (1)

JEE Main 2019 (Online) 10th April Morning Slot (1)

JEE Main 2019 (Online) 9th April Morning Slot (1)

JEE Main 2019 (Online) 8th April Evening Slot (1)

JEE Main 2019 (Online) 8th April Morning Slot (1)

JEE Main 2019 (Online) 12th January Evening Slot (1)

JEE Main 2019 (Online) 12th January Morning Slot (1)

JEE Main 2019 (Online) 11th January Evening Slot (1)

JEE Main 2019 (Online) 10th January Evening Slot (1)

JEE Main 2019 (Online) 9th January Evening Slot (2)

JEE Main 2019 (Online) 9th January Morning Slot (1)

JEE Main 2018 (Online) 16th April Morning Slot (1)

JEE Main 2018 (Offline) (1)

JEE Main 2018 (Online) 15th April Evening Slot (1)

JEE Main 2018 (Online) 15th April Morning Slot (1)

JEE Main 2017 (Online) 9th April Morning Slot (1)

JEE Main 2017 (Online) 8th April Morning Slot (1)

JEE Main 2017 (Offline) (1)

JEE Main 2016 (Online) 10th April Morning Slot (2)

JEE Main 2016 (Online) 9th April Morning Slot (2)

JEE Main 2016 (Offline) (1)

JEE Main 2015 (Offline) (1)

JEE Main 2013 (Offline) (2)

AIEEE 2012 (1)

AIEEE 2011 (2)

AIEEE 2010 (1)

AIEEE 2009 (1)

AIEEE 2008 (2)

AIEEE 2007 (1)

AIEEE 2006 (1)

AIEEE 2005 (1)

AIEEE 2004 (2)

AIEEE 2003 (3)

AIEEE 2002 (4)

Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations