Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

A

5040

B

6210

C

385

D

1110

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$

So, total no of ways he can give votes

= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$

= 385

**Note :** Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule.

Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$

Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$

Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$

Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$

So, total no of ways he can give votes

= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$

= 385

2

MCQ (Single Correct Answer)

If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number

A

601

B

600

C

603

D

602

3

MCQ (Single Correct Answer)

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

A

$${}^8{C_3}$$

B

21

C

$${3^8}$$

D

5

To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects

= $${}^{n - 1}{C_{p - 1}}$$

For this question, n = 8 and p = 3

$$ \therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21

= $${}^{n - 1}{C_{p - 1}}$$

For this question, n = 8 and p = 3

$$ \therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21

4

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!.

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

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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