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 value of $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)$$ is equal to :

A

560

B

680

C

1240

D

1085

We know,

$${{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}$$

$$ \therefore $$ $${{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}$$

$$ \therefore $$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)$$

$$=$$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)$$

$$ = \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)} $$

$$ = 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} } $$

$$ = 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}$$

$$=$$ 120 $$ \times $$ 16 $$-$$ 40 $$ \times $$ 31

$$=$$ 680

$${{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}$$

$$ \therefore $$ $${{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}$$

$$ \therefore $$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)$$

$$=$$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)$$

$$ = \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)} $$

$$ = 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} } $$

$$ = 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}$$

$$=$$ 120 $$ \times $$ 16 $$-$$ 40 $$ \times $$ 31

$$=$$ 680

2

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

3

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)

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

4

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

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*

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Complex Numbers *keyboard_arrow_right*

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Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

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Application of Derivatives *keyboard_arrow_right*

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Definite Integrals and Applications of Integrals *keyboard_arrow_right*

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