Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

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Reinforced Cement Concrete

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Fluid Mechanics and Hydraulic Machines

Hydrology

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Geomatics Engineering Or Surveying

Environmental Engineering

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Engineering Mathematics

General Aptitude

1

If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

A

2

B

4$${^{{1 \over 3}}}$$

C

4$${^{{2 \over 3}}}$$

D

4

a, b and c are in AP.

$$ \therefore $$ a + c = 2b

As, abc = 8

$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8

$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4

$$ \therefore $$ ac = 4 and a + c = 4

Then,

b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2

$$ \therefore $$ a + c = 2b

As, abc = 8

$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8

$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4

$$ \therefore $$ ac = 4 and a + c = 4

Then,

b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2

2

Let

S_{n} = $${1 \over {{1^3}}}$$$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$$

If 100 S_{n} = n, then n is equal to :

S

If 100 S

A

199

B

99

C

200

D

19

n^{th} term, T_{n} = $${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$$

T_{n} = $${{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}$$

$$ \Rightarrow $$ T_{n} = $${2 \over {n\left( {n + 1} \right)}}$$ = $$2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$$

$$ \therefore $$ S_{n} = $$\sum {{T_n}} $$

= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$

= $$2\left( {1 - {1 \over n}} \right)$$

= $${{{2n} \over {n + 1}}}$$

Given that,

100 S_{n} = n

$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n

$$ \Rightarrow $$ n + 1 = 200

$$ \Rightarrow $$ n = 199

T

$$ \Rightarrow $$ T

$$ \therefore $$ S

= $$2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} $$

= $$2\left( {1 - {1 \over n}} \right)$$

= $${{{2n} \over {n + 1}}}$$

Given that,

100 S

$$ \Rightarrow $$ 100 $$ \times $$ $${{{2n} \over {n + 1}}}$$ = n

$$ \Rightarrow $$ n + 1 = 200

$$ \Rightarrow $$ n = 199

3

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} ...........

If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to

1

If B - 2A = 100$$\lambda $$, then $$\lambda $$ is equal to

A

496

B

232

C

248

D

464

Sum of square of first n odd terms

1

Given,

1

A = Sum of first 20 terms

$$\therefore\,\,\,$$A = 1

Arrange those terms this way,

A = [1

A = [ 1

A = $$ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$$

A = $$ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$$

A =70 $$ \times $$ 19 + 70 $$ \times $$ 44

A = 70 $$ \times $$ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [1

B = [1

B = $${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$$

B = 260 $$ \times $$ 41 + 560 $$ \times $$ 41

B = 41 $$ \times \,\,\,820$$

$$\therefore\,\,\,$$ B $$-$$ 2A = 41 $$ \times \,$$ 820 $$-$$ 2 $$ \times \,$$ 70 $$ \times \,$$ 63 = 24800

Given that B $$-$$ 2A = 100 $$\lambda $$

$$\therefore\,\,\,$$ 100 $$\lambda $$ = 24800

$$ \Rightarrow \,\,\,\lambda $$ = 248

4

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to

A

33

B

66

C

68

D

34

a_{1}, a_{2}, a_{3} . . . a_{43} are in AP

So, a_{2} = a_{1} + d

a_{3} = a_{1} + 2d

.

.

.

a_{49} =a_{1} + 48d

Now given, $${a_9} + {a_{43}} = 66$$

$$ \Rightarrow \,\,\,\,$$ a_{1} + 8d + a_{1} + 42d = 66

$$ \Rightarrow \,\,\,\,$$ 2a_{1} + 50d = 66

$$ \Rightarrow \,\,\,\,$$ a_{1} + 25d = 33 . . . . . (1)

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + a_{5} + a_{9} + a_{13} +. . . . . 13 items = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + a_{1} + 4d + a_{1} + 8d + . . . . a_{1} + 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 4d +8d + 12d + . . . . . 48d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 4 (1+ 2 + 3 + . . . + 12) d = 416

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a_{1} + 24 $$ \times$$ 13d = 416

$$ \Rightarrow \,\,\,\,$$ a_{1} + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and $${a_1} = 8$$

$$\therefore\,\,\,$$ a_{1} = 8

a_{2} = 8 + 1 = 9

a_{3} = 8 + 2 = 10

.

.

.

a_{17} = 8 + 16 = 24

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (1^{2} +2^{2} + . . . . +24^{2}) $$-$$ (1^{2} + 2^{2} + . . . . .+ 7^{2}) = 140 m

$$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$$

$$ \Rightarrow \,\,\,\,\,$$ 490 $$-$$ 140 = 140 m

$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m

$$ \Rightarrow \,\,\,\,\,$$ m = 34

So, a

a

.

.

.

a

Now given, $${a_9} + {a_{43}} = 66$$

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ 2a

$$ \Rightarrow \,\,\,\,$$ a

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} $$ = 416

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ a

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times $$d = 416

$$ \Rightarrow \,\,\,\,$$ 13a

$$ \Rightarrow \,\,\,\,$$ a

Solving (1) and (2) we get,

d = 1

and $${a_1} = 8$$

$$\therefore\,\,\,$$ a

a

a

.

.

.

a

Now, $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$$

$$ \Rightarrow \,\,\,\,$$ $$a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$$

$$ \Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$$

We can write above series like this,

$$ \Rightarrow \,\,\,\,\,$$ (1

$$ \Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$$

$$ \Rightarrow \,\,\,\,\,$$ 490 $$-$$ 140 = 140 m

$$ \Rightarrow \,\,\,\,\,$$4760 = 140 m

$$ \Rightarrow \,\,\,\,\,$$ m = 34

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