Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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 ]

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

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

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

2

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 $$1.$$

3

MCQ (Single Correct Answer)

A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after

A

19 months

B

20 months

C

21 months

D

18 months

Let required number of months $$=n$$

$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$

$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$

$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$

$$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$

$$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$

$$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$

$$ \Rightarrow {n^2} + 5n - 546 = 0$$

$$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$

$$\therefore$$ $$n = 21$$

$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$

$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$

$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$

$$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$

$$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$

$$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$

$$ \Rightarrow {n^2} + 5n - 546 = 0$$

$$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$

$$\therefore$$ $$n = 21$$

4

MCQ (Single Correct Answer)

A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1}$$ = $${a_2}$$ = ....= $${a_{10}}$$= 150 and $${a_{10}}$$, $${a_{11}}$$,.... are in an AP with common difference - 2, then the time taken by him to count all notes is

A

34 minutes

B

125 minutes

C

135 minutes

D

24 minutes

Till $$10$$^{th} minute number of counted notes $$ = 1500$$

$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$

$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$

$$ \Rightarrow n = 125,24$$

But $$n=125$$ is not possible

$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes.

$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$

$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$

$$ \Rightarrow n = 125,24$$

But $$n=125$$ is not possible

$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes.

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