Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

A

480

B

240

C

360

D

120

In the word ''GARDEN'', there are two vowels A and E present, and A should come always before E.

$$\therefore\,\,\,$$ Total no of ways = $${{6!} \over {2!}}$$ = 360

Here A and E has fixed order that is why we divide by 2!.

$$\therefore\,\,\,$$ Total no of ways = $${{6!} \over {2!}}$$ = 360

Here A and E has fixed order that is why we divide by 2!.

2

MCQ (Single Correct Answer)

If $${}^n{C_r}$$ denotes the number of combination of n things taken r at a time, then the expression $$\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$$ equals

A

$$\,{}^{n + 1}{C_{r + 1}}$$

B

$${}^{n + 2}{C_r}$$

C

$${}^{n + 2}{C_{r + 1}}$$

D

$$\,{}^{n + 1}{C_r}$$

Arrange it this way,

$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$

$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$

$$\left[ \, \right.$$ Now use the rule,

$$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$

$$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$

$$ = {}^{n + 2}{C_{r + 1}}$$

$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$

$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$

$$\left[ \, \right.$$ Now use the rule,

$$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$

$$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$

$$ = {}^{n + 2}{C_{r + 1}}$$

3

MCQ (Single Correct Answer)

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A

346

B

140

C

196

D

280

No of ways student can answer 10 questions = $${}^5{C_4} \times {}^8{C_6}$$ = 140

No of ways student can answer 10 questions = $${}^5{C_5} \times {}^8{C_5}$$ = 56

$$\therefore$$ Total ways = 140 + 56 = 196

4

MCQ (Single Correct Answer)

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

A

$$7!\, \times 5!\,\,$$

B

$$6!\, \times 5!$$

C

$$30!$$

D

$$5!\, \times 4!$$

6 men can sit at the round table = $$\left( {6 - 1} \right)! = 5!$$ ways

Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.

So total no of ways 6 men and 5 women can dine at the round table

= $$5!\, \times {}^6{C_5} \times 5!$$

= $$5!\, \times 6 \times 5!$$

= $$5!\, \times 6!$$

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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