Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If
one of the sides passing through the origin makes an angle 30^{o} with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :

A

$$2\sqrt 3 - 1$$

B

$$2\sqrt 3 - 2$$

C

$$\sqrt 3 - 2$$

D

$$\sqrt 3 - 1$$

Let, coordinate of point A = (x, y).

$$\therefore\,\,\,$$ For point A,

$${x \over {\cos {{30}^ \circ }}}$$ = $${y \over {\sin {{30}^ \circ }}}$$ = 2

$$ \Rightarrow $$ x = $$\sqrt 3 $$

and y = 1

Similarly, For point B,

$${x \over {\cos {{75}^ \circ }}}$$ = $${y \over {\sin {{75}^ \circ }}}$$ = 2$$\sqrt 2 $$

$$\therefore\,\,\,$$ x = $$\sqrt 3 - 1$$

y = $$\sqrt 3 + 1$$

For point C,

$${x \over {cos{{120}^ \circ }}}$$ = $${y \over {sin{{120}^ \circ }}}$$ = 2

$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$1

y = $$\sqrt 3 $$

$$\therefore\,\,\,$$ Sum of the x - coordinate of the vertices

= 0 + $$\sqrt 3 $$ + $$\sqrt 3 $$ $$-$$ 1 + ($$-$$ 1) = 2$$\sqrt 3 $$ $$-$$ 2

2

A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is
the origin and the rectangle OPRQ is completed, then the locus of R is :

A

3x + 2y = 6xy

B

3x + 2y = 6

C

2x + 3y = xy

D

3x + 2y = xy

Let coordinate of point R = (h, k).

Equation of line PQ,

(y $$-$$ 3) = m (x $$-$$ 2).

Put y = 0 to get coordinate of point p,

0 $$-$$ 3 = (x $$-$$ 2)

$$ \Rightarrow $$ x = 2 $$-$$ $${3 \over m}$$

$$\therefore\,\,\,$$ p = (2 $$-$$ $${3 \over m}$$, 0)

As p = (h, 0) then

h = 2 $$-$$ $${3 \over m}$$

$$ \Rightarrow $$ $${3 \over m}$$ = 2 $$-$$ h

$$ \Rightarrow $$ m = $${3 \over {2 - h}}$$ . . . . . . (1)

Put x = 0 to get coordinate of point Q,

y $$-$$ 3 $$=$$ m (0 $$-$$ 2)

$$ \Rightarrow $$ y = 3 $$-$$ 2m

$$\therefore\,\,\,$$ point Q = (0, 3 $$-$$ 2m)

And From the graph you can see Q = (0, k).

$$\therefore\,\,\,$$ k = 3 $$-$$ 2m

$$ \Rightarrow $$ m = $${{3 - k} \over 2}$$ . . . . (2)

By comparing (1) and (2) get

$${3 \over {2 - h}} = {{3 - k} \over 2}$$

$$ \Rightarrow $$ (2 $$-$$ h)(3 $$-$$ k) = 6

$$ \Rightarrow $$ 6 $$-$$ 3h $$-$$ 2K + hk = 6

$$ \Rightarrow $$ 3 h + 2K = hk

$$\therefore\,\,\,$$ locus of point R is 3x + 2y= xy

3

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of $$\Delta $$ ABC (in sq. units) is :

A

12

B

4

C

5

D

9

Median through C is x = 4

So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and

C(4, y) is D which lies on the median through B.

$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$

Now, $${{1 + 4 + 2 + y} \over 2}$$ = 5 $$ \Rightarrow $$ y = 3.

So, C $$ \equiv $$ (4, 3).

The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).

The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC

= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.

So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and

C(4, y) is D which lies on the median through B.

$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$

Now, $${{1 + 4 + 2 + y} \over 2}$$ = 5 $$ \Rightarrow $$ y = 3.

So, C $$ \equiv $$ (4, 3).

The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).

The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC

= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.

4

The sides of a rhombus ABCD are parallel to the lines, x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :

A

$${5 \over 2}$$

B

$${7 \over 4}$$

C

2

D

$${7 \over 2}$$

Let the coordinate A be (0, c)

Equations of the given lines are

x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0

We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

$$\therefore\,\,\,$$ equation of angle bisectors is given as :

$${{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}$$

5x $$-$$ 5y + 10 = $$ \pm $$ (7x $$-$$ y + 3)

$$\therefore\,\,\,$$ Parallel equations of the diagonals are 2x + 4y $$-$$ 7 = 0

and 12x $$-$$ 6y + 13 = 0

$$\therefore\,\,\,$$ slopes of diagonals are $${{ - 1} \over 2}$$ and 2.

Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $$-$$ c)

$$\therefore\,\,\,$$ 2 $$-$$ c = 2 $$ \Rightarrow $$ c = 0 (not possible)

$$ \therefore $$$$\,\,\,$$ 2 $$-$$ c = $${{ - 1} \over 2}$$ $$ \Rightarrow $$ c = $${5 \over 2}$$

$$\therefore\,\,\,$$ Coordinate of A is $${5 \over 2}$$.

Equations of the given lines are

x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0

We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

$$\therefore\,\,\,$$ equation of angle bisectors is given as :

$${{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}$$

5x $$-$$ 5y + 10 = $$ \pm $$ (7x $$-$$ y + 3)

$$\therefore\,\,\,$$ Parallel equations of the diagonals are 2x + 4y $$-$$ 7 = 0

and 12x $$-$$ 6y + 13 = 0

$$\therefore\,\,\,$$ slopes of diagonals are $${{ - 1} \over 2}$$ and 2.

Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $$-$$ c)

$$\therefore\,\,\,$$ 2 $$-$$ c = 2 $$ \Rightarrow $$ c = 0 (not possible)

$$ \therefore $$$$\,\,\,$$ 2 $$-$$ c = $${{ - 1} \over 2}$$ $$ \Rightarrow $$ c = $${5 \over 2}$$

$$\therefore\,\,\,$$ Coordinate of A is $${5 \over 2}$$.

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (4) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *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*

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*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*