Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are

A

312

B

3125

C

120

D

216

Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.

So total no of arrangement = 1$$ \times $$2$$ \times $$3$$ \times $$4$$ \times $$5 = 5!

So total no of arrangement = 4$$ \times $$4$$ \times $$3$$ \times $$2$$ \times $$1 = 4.4!

$$\therefore$$ Total arrangement = 5! + 4.4! = 216

2

MCQ (Single Correct Answer)

Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is

A

125

B

105

C

374

D

625

There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed

**Case 1 :** First digit is 1 = 1 _ _ _

Possible numbers starting with 1 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.

**Case 2 :** First digit is 2 = 2 _ _ _

Possible numbers starting with 2 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

**Case 3 :** First digit is 3 = 3 _ _ _

Possible numbers starting with 3 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

Total possible numbers = 124 + 125 + 125 = 374

Possible numbers starting with 1 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.

Possible numbers starting with 2 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

Possible numbers starting with 3 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125

Total possible numbers = 124 + 125 + 125 = 374

3

MCQ (Single Correct Answer)

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are

A

216

B

375

C

400

D

720

$$\therefore$$ Total no of ways = 5$$ \times $$6$$ \times $$6$$ \times $$$${}^4{C_1}$$ = 720

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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