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 $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the
equation :

A

n^{2} + 3n − 108 = 0

B

n^{2} + 5n − 84 = 0

C

n^{2} + 2n − 80 = 0

D

n^{2} + n − 110 = 0

$${{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11$$

$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$

$$ \Rightarrow $$ (n + 2)! = 11.6! (n $$-$$ 2)!

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11.6!

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 6 . 5 . 4 . 3 . 2 . 1

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 10 . 9 . 8

$$ \therefore $$ n = 9

This value of n satisfy the equation,

n^{2} + 3n $$-$$ 108 = 0

$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$

$$ \Rightarrow $$ (n + 2)! = 11.6! (n $$-$$ 2)!

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11.6!

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 6 . 5 . 4 . 3 . 2 . 1

$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 10 . 9 . 8

$$ \therefore $$ n = 9

This value of n satisfy the equation,

n

2

The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :

A

(11)!

B

10 $$ \times $$ (11!)

C

101 $$ \times $$ (10!)

D

11 $$ \times $$ (11!)

$$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!$$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$

$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$

$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$

$$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$$=$$ 11.11! $$-$$ 1.1! $$-$$ $$\sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!$$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ [(2! $$-$$ 1!) + (3! $$-$$ 2!) + . . . .+ (11! $$-$$ 10!)]

$$=$$ (11.11! $$-$$ 1) $$-$$ (11! $$-$$ 1)

$$=$$ 11.11! $$-$$ 11!

$$=$$ 11! (11 $$-$$ 1)

$$=$$ $$10\,.\,\left( {11!} \right)$$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$

$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$

$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$

$$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$$=$$ 11.11! $$-$$ 1.1! $$-$$ $$\sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!$$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]} $$

$$=$$ (11.11! $$-$$ 1) $$-$$ [(2! $$-$$ 1!) + (3! $$-$$ 2!) + . . . .+ (11! $$-$$ 10!)]

$$=$$ (11.11! $$-$$ 1) $$-$$ (11! $$-$$ 1)

$$=$$ 11.11! $$-$$ 11!

$$=$$ 11! (11 $$-$$ 1)

$$=$$ $$10\,.\,\left( {11!} \right)$$

3

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are
ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X
and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in
this party, is:

A

468

B

469

C

484

D

485

X(7 Friends) | Y(7 Friends) | |||
---|---|---|---|---|

4 Ladies | 3 Men | 3 Ladies | 4 Men | |

Case 1 | 3 | 0 | 0 | 3 |

Case 2 | 0 | 3 | 3 | 0 |

Case 3 | 2 | 1 | 1 | 2 |

Case 4 | 1 | 2 | 2 | 1 |

In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.

$$\therefore$$ No of ways 6 friends can be invited =

$$({}^4{C_3} \times {}^3{C_0} \times {}^3{C_0} \times {}^4{C_3})$$ + $$({}^4{C_0} \times {}^3{C_3} \times {}^3{C_3} \times {}^4{C_0})$$ + $$\left( {{}^4{C_2} \times {}^3{C_1} \times {}^3{C_1} \times {}^4{C_2}} \right)$$ + $$\left( {{}^4{C_1} \times {}^3{C_2} \times {}^3{C_2} \times {}^4{C_1}} \right)$$

= 16 + 1 + 324 + 144 = 485

4

If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :

A

44^{th}

B

45^{th}

C

46^{th}

D

47^{th}

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