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

If $$\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$$ then K is equal to -

A

2^{24}

B

2^{25}$$-$$ 1

C

2^{25}

D

(25)^{2}

$$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$

$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$

$$ \therefore $$ $$K = {2^{25}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$

$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$

$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$

$$ \therefore $$ $$K = {2^{25}}$$

2

The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3,
whenever k is a multiple of 4, is :

A

6^{5} $$ \times $$ (15)!

B

5^{6} $$ \times $$ 15

C

(15)! $$ \times $$ 6!

D

5! $$ \times $$ 6!

f(k) = 3m (3, 6, 9, 12, 15, 18)

for k = 4, 8, 12, 16, 20

6.5.4.3.2 ways

For rest numbers 15! ways

Total ways = 6! (15!)

for k = 4, 8, 12, 16, 20

6.5.4.3.2 ways

For rest numbers 15! ways

Total ways = 6! (15!)

3

Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n_{i}, the label of the ball drawn from the i^{th} box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n_{1} < n_{2} < n_{3} is :

A

164

B

240

C

82

D

120

Number of ways = ^{10}C_{3} = 120

4

There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :

A

12

B

9

C

7

D

11

Let m-men, 2-women

^{m}C_{2} $$ \times $$ 2 = ^{m}^{}C_{1} ^{2}C_{1} . 2 + 84

m^{2} $$-$$ 5m $$-$$ 84 = 0 $$ \Rightarrow $$ (m $$-$$ 12) (m + 7) = 0

m = 12

m

m = 12

Number in Brackets after Paper Name Indicates No of Questions

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