Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to :

A

100

B

110

C

$${{121} \over {10}}$$

D

$${{441} \over {100}}$$

Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$ \Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$ \Rightarrow k = 100$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$ \Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$ \Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$ \Rightarrow k = 100$$

2

MCQ (Single Correct Answer)

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :

A

$$2 - \sqrt 3 $$

B

$$2 + \sqrt 3 $$

C

$$\sqrt 2 + \sqrt 3 $$

D

$$3 + \sqrt 2 $$

Let $$a,ar,a{r^2}$$ are in $$G.P.$$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$

$$ \Rightarrow 4r = 1 + {r^2}$$

$$ \Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected

Hence, $$r = 2 + \sqrt 3 $$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$

$$ \Rightarrow 4r = 1 + {r^2}$$

$$ \Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected

Hence, $$r = 2 + \sqrt 3 $$

3

MCQ (Single Correct Answer)

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,........,is

A

$${7 \over {81}}\left( {179 - {{10}^{ - 20}}} \right)$$

B

$$\,{7 \over 9}\left( {99 - {{10}^{ - 20}}} \right)$$

C

$${7 \over {81}}\left( {179 + {{10}^{ - 20}}} \right)$$

D

$${7 \over 9}\left( {99 + {{10}^{ - 20}}} \right)$$

Given sequence can be written as

$${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... + $$ up to $$20$$ terms

$$ = 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$$ up to $$20$$ terms ]

Multiply and divide by $$9$$

$$ = {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,$$ $$+$$ up to $$20$$ terms ]

$$ = {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......$$ $$+$$ up to $$20$$ terms ]

$$ = {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]$$

$$ = {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]$$

$$ = {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]$$

$${7 \over {10}} + {{77} \over {100}} + {{777} \over {{{10}^3}}} + ..... + $$ up to $$20$$ terms

$$ = 7\left[ {{1 \over {10}} + {{11} \over {100}} + {{111} \over {{{10}^3}}} + ...... + } \right.\,\,$$ up to $$20$$ terms ]

Multiply and divide by $$9$$

$$ = {7 \over 9}\left[ {{9 \over {10}} + {{99} \over {100}} + {{999} \over {1000}} + ......} \right.\,\,$$ $$+$$ up to $$20$$ terms ]

$$ = {7 \over 9}\left[ {\left( {1 - {1 \over {10}}} \right)} \right. + \left( {1 - {1 \over {{{10}^2}}}} \right) + \left( {1 - {1 \over {{{10}^3}}}} \right) + ......$$ $$+$$ up to $$20$$ terms ]

$$ = {7 \over 9}\left[ {20 - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^{20}}} \right)} \over {1 - {1 \over {10}}}}} \right]$$

$$ = {7 \over 9}\left[ {{{179} \over 9} + {1 \over 9}{{\left( {{1 \over {10}}} \right)}^{20}}} \right]$$

$$ = {7 \over {81}}\left[ {179 + {{\left( {10} \right)}^{ - 20}}} \right]$$

4

MCQ (Single Correct Answer)

** Statement-1: ** The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000.

** Statement-2: ** $$\sum\limits_{k = 1}^n {\left( {{k^3} - {{(k - 1)}^3}} \right)} = {n^3}$$, for any natural number n.

A

Statement-1 is false, Statement-2 is true.

B

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

C

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

D

Statement-1 is true, Statement-2 is false.

$$n$$^{th } term of the given series

$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$

$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$

$$ = {n^3} - {\left( {n - 1} \right)^3}$$

$$ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} $$

$$ \Rightarrow 8000 = {n^3}$$

$$ \Rightarrow n = 20\,\,$$ which is a natural number.

Now, put $$n = 1,2,3,.....20$$

$${T_1} = {1^3} - {0^3}$$

$${T_2} = {2^3} - {1^3}$$

.

.

.

$${T_{20}} = {20^3} - {19^3}$$

Now, $${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$$

$$ \Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$$

Hence, both the given statements are true and statement $$2$$ supports statement $$1.$$

$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$

$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$

$$ = {n^3} - {\left( {n - 1} \right)^3}$$

$$ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} $$

$$ \Rightarrow 8000 = {n^3}$$

$$ \Rightarrow n = 20\,\,$$ which is a natural number.

Now, put $$n = 1,2,3,.....20$$

$${T_1} = {1^3} - {0^3}$$

$${T_2} = {2^3} - {1^3}$$

.

.

.

$${T_{20}} = {20^3} - {19^3}$$

Now, $${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$$

$$ \Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$$

Hence, both the given statements are true and statement $$2$$ supports statement $$1.$$

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