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 tower T_{1} of height 60 m is located exactly opposite to a tower T_{2} of height 80 m on a straight road. Fromthe top of T_{1}, if the angle of depression of the foot of T_{2} is twice the angle of elevation of the top of T_{2}, then the width (in m) of the road between the feetof the towers T_{1} and T_{2} is :

A

$$10\sqrt 2 $$

B

$$10\sqrt 3 $$

C

$$20\sqrt 3 $$

D

$$20\sqrt 2 $$

Let the distance between T_{1} and T_{2} be x

From the figure

EA = 60 m (T_{1}) and DB = 80 m (T_{2})

$$\angle DEC = \theta $$ and $$\angle BEC = 2\theta $$

Now in $$\angle DEC$$,

$$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$$

and in $$\Delta BEC$$,

$$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$$

We know that

$$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$$

$$ \Rightarrow $$ $${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$$

$$ \Rightarrow $$ $${x^2} = 1200$$ $$ \Rightarrow $$ $$x = 20\sqrt 3 $$

From the figure

EA = 60 m (T

$$\angle DEC = \theta $$ and $$\angle BEC = 2\theta $$

Now in $$\angle DEC$$,

$$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$$

and in $$\Delta BEC$$,

$$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$$

We know that

$$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$$

$$ \Rightarrow $$ $${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$$

$$ \Rightarrow $$ $${x^2} = 1200$$ $$ \Rightarrow $$ $$x = 20\sqrt 3 $$

2

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30^{o} to 45^{o} ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

A

$$9\left( {1 + \sqrt 3 } \right)$$

B

$$18\left( {1 + \sqrt 3 } \right)$$

C

$$18\left( {\sqrt 3 - 1} \right)$$

D

$${9 \over 2}\left( {\sqrt 3 - 1} \right)$$

Assume height of tower = AB = h

From $$\Delta $$ABD,

tan45

$$ \Rightarrow $$$$\,\,\,$$ $${{AB} \over {AD}} = 1$$ [ as tan45

$$ \Rightarrow $$ $$\,\,\,$$ AB = AD

$$\therefore\,\,\,$$ AD = h

From $$\Delta $$BAC,

tan30

$$ \Rightarrow $$ $$\,\,\,$$ $${h \over {AC}} = {1 \over {\sqrt 3 }}$$

$$ \Rightarrow $$ $$\,\,\,$$ AC = h $$\sqrt 3 $$

As, AC = AD + DC

$$ \Rightarrow $$$$\,\,\,$$ DC = AC $$-$$ AD = $$\sqrt 3 h$$ $$-$$ h

Given that, time taken to reach from point C to D = 18 min.

$$\therefore\,\,\,$$ Car speed = $${{distance} \over {time}}$$ = $${{CD} \over {18}}$$ = $${{(\sqrt 3 - 1)h} \over {18}}$$

$$\therefore\,\,\,$$ Time taken to move from D to A

= $${{Distan ce\,\,of\,\,DA} \over {speed}}$$

= $${h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}$$

= $${{18} \over {\left( {\sqrt 3 - 1} \right)}}$$

= $${{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}$$

= 9 $$\left( {\sqrt 3 + 1} \right)$$ min.

3

Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30^{o} at B. The height (in m) of the lamp-post is -

A

$$2\sqrt {21} $$

B

$${3 \over 2}\sqrt {21} $$

C

$$7\sqrt {3} $$

D

$${2 \over 3}\sqrt {21} $$

BD = hcot30

So, 7

$$ \Rightarrow $$ 37 = 3h

$$ \Rightarrow $$ 3h

$$ \Rightarrow $$ h = $$\sqrt {{{28} \over 3}} = {2 \over 3}\sqrt {21} $$

4

With the usual notation, in $$\Delta $$ABC, if $$\angle A + \angle B$$ = 120^{o}, a = $$\sqrt 3 $$ $$+$$ 1, b = $$\sqrt 3 $$ $$-$$ 1 then the ratio $$\angle A:\angle B,$$ is -

A

9 : 7

B

7 : 1

C

5 : 3

D

3 : 1

A + B = 120^{o}

$$\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)$$

$$ = {{\sqrt 3 + 1 - \sqrt 3 + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3 = 1$$

$${{A - B} \over 2} = {45^o}$$

$$ \Rightarrow A - B = {90^o}$$

$$ \ A + B = {120^o}$$

$$2A = {210^o}$$

$$A = {105^o}$$

$$B = {15^o}$$

$$ \therefore $$ $$\angle A:\angle B,$$ = 7 : 1

$$\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)$$

$$ = {{\sqrt 3 + 1 - \sqrt 3 + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3 = 1$$

$${{A - B} \over 2} = {45^o}$$

$$ \Rightarrow A - B = {90^o}$$

$$ \ A + B = {120^o}$$

$$2A = {210^o}$$

$$A = {105^o}$$

$$B = {15^o}$$

$$ \therefore $$ $$\angle A:\angle B,$$ = 7 : 1

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

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Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

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Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

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