Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

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Reinforced Cement Concrete

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General Aptitude

1

If the fractional part of the number $$\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$$, then k is equal to :

A

8

B

14

C

6

D

1

$${{{2^{403}}} \over {15}}$$

$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$

$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$

$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$

$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$

$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$

$$ = {8 \over {15}} + 8$$ (integer)

$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$

According to the question,

$${k \over {15}} = {8 \over {15}}$$

$$ \Rightarrow $$ K $$=$$ 8

$$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$$

$$ = {8 \over {15}}{\left( {16} \right)^{100}}$$

$$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$$

$$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$$

$$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$$

$$ = {8 \over {15}} + 8$$ (integer)

$$ \therefore $$ Fractional part $$ = {8 \over {15}}$$

According to the question,

$${k \over {15}} = {8 \over {15}}$$

$$ \Rightarrow $$ K $$=$$ 8

2

The coefficient of t^{4} in the expansion of $${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$ is :

A

14

B

15

C

10

D

12

$${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$$

= (1 $$-$$ t^{6})^{3} (1 $$-$$ t)^{$$-$$3}

= (1 $$-$$^{3}C_{1}t^{6} + ^{3}C_{2}t^{12} $$-$$ ^{3}C_{3}t^{18}) $$ \times $$ (1 $$-$$ t)^{$$-$$3}

coefficient of t^{4} is 1 $$ \times $$ coefficient of t^{4} in (1 $$-$$ t)^{$$-$$3}

= 1 $$ \times $$^{3+4$$-$$1}C_{4} (By multinomial theorem)

=^{6}C_{4} = 15

= (1 $$-$$ t

= (1 $$-$$

coefficient of t

= 1 $$ \times $$

=

3

If $${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$ then k is equal to

A

100

B

200

C

50

D

400

$${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$

$$ \Rightarrow 100 = k$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$$

$$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$$

$$ \Rightarrow 100 = k$$

4

If the third term in the binomial expansion

of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -

of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -

A

$$2\sqrt 2 $$

B

$$4\sqrt 2 $$

C

$${1 \over 8}$$

D

$${1 \over 4}$$

$${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$

$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$

$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$

$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$

$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$

$$ \Rightarrow 2{({\log _2}x)^2} = 8$$

$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$

$$ \Rightarrow \,\,{\log _2}x = 2$$ or $$-$$ 2

$$x = 4$$ or $${1 \over 4}$$

$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$

$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$

$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$

$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$

$$ \Rightarrow 2{({\log _2}x)^2} = 8$$

$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$

$$ \Rightarrow \,\,{\log _2}x = 2$$ or $$-$$ 2

$$x = 4$$ or $${1 \over 4}$$

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