Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

The number of four letter words that can be formed using the letters of the word **BARRACK** is :

A

120

B

144

C

264

D

270

When all the four letters different then no of words

=

When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words

=

When out of four letters two letters are A and other two different letters are chosen from B, R, C, K then the no of words

=

When word is formed using two R and two A then number of words

= $${{4!} \over {2!2!}}$$ = 6

So, total number of 4 letters words possible = 120 + 72 + 72 + 6 = 270

2

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :

A

24

B

30

C

36

D

48

Here number should be divisible by 3, that means sum of numbers should be divisible by 3.

Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are

(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)

(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)

__Case 1__ :

When 4 digits are (0, 2, 3, 4) then

$$\therefore\,\,\,\,$$ Total possible numbers = $$^3{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

= 3 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 18

__Case 2__ :

When 4 digits are (0, 1, 2, 3) then,

$$\therefore\,\,\,\,$$ Total possible number in this case = $$^2{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

= 2 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 12

$$\therefore\,\,\,\,$$ Total possible numbers will be = 18 + 12 = 30

Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are

(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)

(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)

When 4 digits are (0, 2, 3, 4) then

$$\therefore\,\,\,\,$$ Total possible numbers = $$^3{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

= 3 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 18

When 4 digits are (0, 1, 2, 3) then,

$$\therefore\,\,\,\,$$ Total possible number in this case = $$^2{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$

= 2 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 12

$$\therefore\,\,\,\,$$ Total possible numbers will be = 18 + 12 = 30

3

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same
team, is :

A

500

B

350

C

200

D

300

From 5 girls 2 girls can be selected

=^{5}C_{2} ways

From 7 boys 3 boys can be selected

=^{7}C_{3} way

$$ \therefore $$ Total number of ways we can select 2 girls and 3 boys

=^{5}C_{2} $$ \times $$ ^{7}C_{3} ways

When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.

So, no of ways 3 boys can be chosen when A and B should must be chosen =^{5}C_{1} ways

$$ \therefore $$ Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team =^{5}^{}C_{1} $$ \times $$ ^{5}C_{2} ways

$$ \therefore $$ Required number of ways

= Total number of ways $$-$$ when A and B are always included.

=^{5}C_{2} $$ \times $$ ^{7}C_{3} $$-$$ ^{5}C_{1} $$ \times $$ ^{5}C_{2}

= 300

=

From 7 boys 3 boys can be selected

=

$$ \therefore $$ Total number of ways we can select 2 girls and 3 boys

=

When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.

So, no of ways 3 boys can be chosen when A and B should must be chosen =

$$ \therefore $$ Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team =

$$ \therefore $$ Required number of ways

= Total number of ways $$-$$ when A and B are always included.

=

= 300

4

Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is :

A

9

B

18

C

36

D

32

Area = $${1 \over 2}$$ h. k = 50

h. k = 100

h. k = 2

Total divisors

= (2 + 1) (2 + 1) = 9

if h > 0, k > 0

But $${\matrix{ {h > 0,} & {k < 0} \cr {h < 0,} & {k > 0} \cr {h < 0,} & {k < 0} \cr } }$$

all are possible so that total no. of positive case

9 + 9 + 9 + 9 = 36

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (4) *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*