Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $${T_n}$$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $${T_{n + 1}} - {T_n}$$ = 10, then the value of n is :

A

7

B

5

C

10

D

8

Number of possible triangle using n vertices = ^{n}C_{3}

$$ \therefore $$ T_{n} = ^{n}C_{3}

then T_{n + 1} = ^{n + 1}C_{3}

Given, $${T_{n + 1}} - {T_n}$$ = 10

$$ \Rightarrow $$^{n + 1}C_{3} - ^{n}C_{3} = 10

$$ \Rightarrow $$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10

$$ \Rightarrow $$ 3n(n - 1) = 60

$$ \Rightarrow $$ n(n - 1) = 20

$$ \Rightarrow $$ n^{2} - n - 20 = 0

$$ \Rightarrow $$ (n - 5)(n + 4) = 0

$$ \therefore $$ n = 5

$$ \therefore $$ T

then T

Given, $${T_{n + 1}} - {T_n}$$ = 10

$$ \Rightarrow $$

$$ \Rightarrow $$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10

$$ \Rightarrow $$ 3n(n - 1) = 60

$$ \Rightarrow $$ n(n - 1) = 20

$$ \Rightarrow $$ n

$$ \Rightarrow $$ (n - 5)(n + 4) = 0

$$ \therefore $$ n = 5

2

MCQ (Single Correct Answer)

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:

A

880

B

629

C

630

D

879

For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n

Given 10 identical white balls, 9 identical green balls and 7 identical black balls.

To find number of ways for selecting atleast one ball.

Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]

Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]

Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]

Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)

Also, number of ways to choose a total of zero balls = 1

Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $$-$$ 1 = 879

3

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.

4

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 $${}^{6}{C_3}$$.

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

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

$$ = 120 - 20$$

$$ = 100$$

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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