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# A certain office has 6 employees. If R employees

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A certain office has 6 employees. If R employees  [#permalink]

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24 Oct 2016, 08:36
2
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5
00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:06) correct 60% (01:07) wrong based on 274 sessions

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A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

NOTE: Kudos for all correct solutions

Our video solution to this question can be found here: https://www.gmatprepnow.com/module/gmat ... /video/799

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Re: A certain office has 6 employees. If R employees  [#permalink]

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24 Oct 2016, 08:42
1
GMATPrepNow wrote:
A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

NOTE: Kudos for all correct solutions

Our video solution to this question can be found here: https://www.gmatprepnow.com/module/gmat ... /video/799

S1 - Sufficient (R can not be negative & you get R = 4 on solving the equation)
S2 - Insufficient. Cannot get any value of R on solving the equation - 6Cr = 15

Hence A
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Re: A certain office has 6 employees. If R employees  [#permalink]

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24 Oct 2016, 09:14
2
GMATPrepNow wrote:
A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

NOTE: Kudos for all correct solutions

Our video solution to this question can be found here: https://www.gmatprepnow.com/module/gmat ... /video/799

S1: By solving, (R-2)*(R-4)=0;
R can't be negative so, R=4 >>> SUFFICIENT
S2: There are two possible values i.e. R=2 or R=4 >>> INSUFFICIENT

Hence, A
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Re: A certain office has 6 employees. If R employees  [#permalink]

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24 Oct 2016, 09:21
1
St 1:

R^2 + 2R -8=0

(R+4)(R-2)=0

R=-4 or 2

R can't be negative

So R = 2

Sufficient

Statement 2:

6C4 = 6C2 = 15

So R can be 2 or 4

Insufficient

A

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A certain office has 6 employees. If R employees  [#permalink]

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05 Nov 2016, 17:54
mankodim wrote:
GMATPrepNow wrote:
A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

NOTE: Kudos for all correct solutions

Our video solution to this question can be found here: https://www.gmatprepnow.com/module/gmat ... /video/799

S1 - Sufficient (R can not be negative & you get R = 4 on solving the equation)
S2 - Insufficient. Cannot get any value of R on solving the equation - 6Cr = 15

Hence A

mankodim please note that your reasoning for Stmt 2 is not correct. The statement is Insufficient because you can get 15 for two values of R i.e. 2 and 4 and not because you can not solve for R

Others have provided more detailed solution that you can refer above.
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Posts: 533
Location: India
GMAT 1: 780 Q51 V46
Re: A certain office has 6 employees. If R employees  [#permalink]

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05 Nov 2016, 20:49
2
Top Contributor
Hi,

My two cents to this question

nCr means selecting “r” things out of “n”.

Remember this, if in a question total number of ways is given (the value of nCr) and how many are selected is also given (the value of r), then “n” will always a unique value.

For an example, nC2 = 10, then “n” has to be 5.

But if in a question total number of ways is given (the value of nCr) and total number of things is given (the value of n), then “r” may have more than one value.

For an example, 5Cr = 10 means “r” could be 2 or 3.

Just to add to that, remember nCr = nC(n-r).

So “r” could have more than one value.

So here in this question, n is given,

So we are finding “R”.

Statement I is sufficient:

No need to factorize this,

Just remember in a quadratic equation of the form Ax^2+Bx+ C = 0,

If B is positive and C is negative, then one root positive and one root will be negative.

So here also one value of R will be positive and one value of R will be negative.

Here negative value doesn’t make sense because we are selecting things.

So it is sufficient.

So answer has to be either A or D.

Statement II is insufficient:

It’s the first case which mentioned above,

So it could be 6C2 or 6C4

So not sufficient.

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Re: A certain office has 6 employees. If R employees  [#permalink]

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06 Nov 2016, 04:37
1
GMATPrepNow wrote:
A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

FROM STATEMENT - I (POSSIBLE)

$$r^2 + 2r - 8 = 0$$

$$r^2 + 4r - 2r - 8 = 0$$

$$r ( r + 4 ) - 2 ( r + 4 ) = 0$$

$$r = 2 , - 4$$

Since, r can not be -ve value of r must be 2

FROM STATEMENT - I (NOT POSSIBLE)

$$6c_r$$ = 15

So, r can be 2 or 4

We can not have a unique solution for r from this statement ...

Hence, Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, answer will be (A)

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A certain office has 6 employees. If R employees  [#permalink]

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11 Jul 2017, 20:06
1
GMATPrepNow wrote:
A certain office has 6 employees. If R employees are chosen to be on the party planning committee, what is the value of R?

(1) R² + 2R - 8 = 0

(2) There are 15 different ways to choose R employees to be on the party planning committee

NOTE: Kudos for all correct solutions

Our video solution to this question can be found here: https://www.gmatprepnow.com/module/gmat ... /video/799

What is tested in statement 2 is whether one can use the fact nCr = nC(n-r) and therefore see that two values of R are possible and in statement 1 of course, that R cannot be negative.
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Re: A certain office has 6 employees. If R employees  [#permalink]

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26 Nov 2018, 03:56
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Re: A certain office has 6 employees. If R employees &nbs [#permalink] 26 Nov 2018, 03:56
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