Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

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Engineering Mathematics

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General Aptitude

1

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is -

A

1356

B

1256

C

1365

D

1465

$$\sum\limits_{r = 2}^{13} {(7r + 2) = 7.{{2 + 13} \over 2}} \times 6 + 2 \times 12$$

= 7 $$ \times $$90 + 24 = 654

$$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$$

Total = 654 + 702 = 1356

= 7 $$ \times $$90 + 24 = 654

$$\sum\limits_{r = 1}^{13} {(7r + 5) = 7\left( {{{1 + 13} \over 2}} \right)} \times 13 + 5 \times 13 = 702$$

Total = 654 + 702 = 1356

2

Let a_{1}, a_{2}, a_{3}, ..... a_{10} be in G.P. with a_{i} > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $$ \in $$ N (the set of natural numbers) for which

$$\left| {\matrix{ {{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \cr {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \cr {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \cr } } \right|$$ $$=$$ 0.

Then the number of elements in S, is -

$$\left| {\matrix{ {{{\log }_e}\,{a_1}^r{a_2}^k} & {{{\log }_e}\,{a_2}^r{a_3}^k} & {{{\log }_e}\,{a_3}^r{a_4}^k} \cr {{{\log }_e}\,{a_4}^r{a_5}^k} & {{{\log }_e}\,{a_5}^r{a_6}^k} & {{{\log }_e}\,{a_6}^r{a_7}^k} \cr {{{\log }_e}\,{a_7}^r{a_8}^k} & {{{\log }_e}\,{a_8}^r{a_9}^k} & {{{\log }_e}\,{a_9}^r{a_{10}}^k} \cr } } \right|$$ $$=$$ 0.

Then the number of elements in S, is -

A

10

B

4

C

2

D

infinitely many

Apply

C_{3} $$ \to $$ C_{3}_{} $$-$$ C_{2}

C_{2}_{} $$ \to $$ C_{2} $$-$$ C_{1}

We get D = 0

C

C

We get D = 0

3

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $${{27} \over {19}}$$.Then the common ratio of this series is :

A

$${4 \over 9}$$

B

$${1 \over 3}$$

C

$${2 \over 3}$$

D

$${2 \over 9}$$

$${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$$

$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$

$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$

$$ \Rightarrow r = {2 \over 3}\,\,$$

as $$\left| r \right| < 1$$

$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$

$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$

$$ \Rightarrow r = {2 \over 3}\,\,$$

as $$\left| r \right| < 1$$

4

Let a_{1}, a_{2}, . . . . . ., a_{10} be a G.P. If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals

A

5^{3}

B

2(5^{2})

C

4(5^{2})

D

5^{4}

a_{1}, a_{2}, . . . . ., a_{10} are in G.P.,

Let the common ratio be r

$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$

$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$

Let the common ratio be r

$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$

$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$

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