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# Some good questions..Require solution

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Intern
Joined: 07 Sep 2009
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08 Nov 2009, 12:18
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1.
A certain company assigns employees to offices in such a way that some of the offices
can be empty and more than one employee can be assigned to an office. In how many
ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

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VP
Joined: 05 Mar 2008
Posts: 1467

Kudos [?]: 307 [0], given: 31

Re: Some good questions..Require solution [#permalink]

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08 Nov 2009, 12:51
alok2171 wrote:
1.
A certain company assigns employees to offices in such a way that some of the offices
can be empty and more than one employee can be assigned to an office. In how many
ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

8 ways

Assuming 1 office
# of ways with 3 employees = 1
# of ways with 2 employees = 3
total = 4
two office so answer = 8

Kudos [?]: 307 [0], given: 31

VP
Joined: 05 Mar 2008
Posts: 1467

Kudos [?]: 307 [0], given: 31

Re: Some good questions..Require solution [#permalink]

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08 Nov 2009, 13:02
alok2171 wrote:

2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

a1 = 2
a2 = -3
a3 = 5
a4 = -1
a5=a5-4= a1
a6=a2 and so on and so forth

In other words, the 4 numbers given are the only ones in the sequence
We need 97 terms total. Therefore, each term will be seen at least 24 times and 4 times 24 = 96 and the first number will be seen 5 times because we need 97 total:

2 x 25 = 50
-3 x 24 = -72
5 x 24 = 120
-1 x 24 = -24

multiply and add and you get 74

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132646 [0], given: 12326

Re: Some good questions..Require solution [#permalink]

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08 Nov 2009, 13:04
alok2171 wrote:
1.
A certain company assigns employees to offices in such a way that some of the offices
can be empty and more than one employee can be assigned to an office. In how many
ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

Each of 3 employees has 2 offices to be assigned so the total # of ways is 2^3=8

alok2171 wrote:
2.The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

I guess that an=an-4? If so than:

a1=a5, a2=6, a3=a7 etc. So we'll have the sequence 2,-3,5,-1, 2,-3,5,-1, 2,-3,5,-1, ... 97 terms.

The sum of group of four terms 2,-3,5,-1 is 3. We'll have 97/4=24 full groups of four, plus 97th term 2.

The sum would be 24*3+2=74

_________________

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Intern
Joined: 26 Jan 2010
Posts: 7

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02 May 2010, 09:18
The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72
B. 74
C. 75
D. 78
E. 80

Kudos [?]: 2 [0], given: 0

Senior Manager
Joined: 25 Jun 2009
Posts: 298

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02 May 2010, 11:20
chaitu1315 wrote:
The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72
B. 74
C. 75
D. 78
E. 80

Is An = A (n-4) ?

If yes then the answer should be B

A5= A (5-4) = A 1 = 2
similarly A 6 = -3 and this sequence will repeat after every 4 terms

A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 3

now Sum of first 96 terms = 3 *24
Sum of 97 terms = 72 + 2 = 74

Last edited by cipher on 02 May 2010, 11:54, edited 1 time in total.

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Intern
Joined: 26 Jan 2010
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02 May 2010, 11:50
nitishmahajan wrote:
chaitu1315 wrote:
The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72
B. 74
C. 75
D. 78
E. 80

Is An = A (n-4) ?

If yes then the answer should be B

A5= A (5-4) = A 1 = 2
similarly A 6 = -3 and this sequence will repeat after every 4 terms

A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 2

now Sum of first 96 terms = 2 *24
Sum of 97 terms = 72 + 2 = 74

I understood until the point that as after 4 we have repeat of the sequence, you can say 96/4= 24...as sum of 4 terms is 2, so 2*24= 48...from where did 72 come from...can you explain it further....

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Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 151 [0], given: 6

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02 May 2010, 11:54
chaitu1315 wrote:
nitishmahajan wrote:
chaitu1315 wrote:
The infinite sequence a1, a2,..., an,... is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an =
an-4 for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72
B. 74
C. 75
D. 78
E. 80

Is An = A (n-4) ?

If yes then the answer should be B

A5= A (5-4) = A 1 = 2
similarly A 6 = -3 and this sequence will repeat after every 4 terms

A1 + A 2+ A3 + A4 = 2- 3+ 5-1 = 2

now Sum of first 96 terms = 2 *24
Sum of 97 terms = 72 + 2 = 74

I understood until the point that as after 4 we have repeat of the sequence, you can say 96/4= 24...as sum of 4 terms is 2, so 2*24= 48...from where did 72 come from...can you explain it further....

my bad again the was 3 and not 3 edited the post ..!

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Re: Sequence   [#permalink] 02 May 2010, 11:54
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