It is currently 19 Apr 2018, 08:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Several friends in a dinner group decide to contribute equally to the

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44566
Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

17 Nov 2014, 12:34
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

65% (01:07) correct 35% (00:52) wrong based on 127 sessions

### HideShow timer Statistics

Tough and Tricky questions: Word Problems.

Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group? (1) The number of people in the group is equal to the size of each person’s contribution, in dollars. (2) If three more people joined the group, each person’s individual contribution would fall by$2.

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________
Manager
Joined: 12 May 2013
Posts: 75
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

18 Nov 2014, 07:24
1
KUDOS
Hey !

According to me the answer should be D

The only solution possible for statement 1 is 6*6 = 36

from statement 2 we can make an equation.
Let the number of contributes be =N
now,(N+3)((36/N)-2)= 36
solving the above equation we get N= -9 and N= 6
The value cannot be in negative hence N=6.

so i go with D.
Math Expert
Joined: 02 Sep 2009
Posts: 44566
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

18 Nov 2014, 07:57
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:

Tough and Tricky questions: Word Problems.

Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group? (1) The number of people in the group is equal to the size of each person’s contribution, in dollars. (2) If three more people joined the group, each person’s individual contribution would fall by$2.

Kudos for a correct solution.

Official Solution:

Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group? Let’s call the number of people in the group $$n$$, and let’s call each contribution$$$x$$. Then we know from the stem that $$36 = nx$$. We are asked for $$n$$, which is equivalent to asking for $$x$$ (because of the equation we are given).

(1) SUFFICIENT. This statement tells us that $$n = x$$. We can substitute into the given equation:

$$36 = n^2$$

Since $$n$$ represents a number of people, only the positive root makes sense, and $$n$$ must be equal to 6.

(2) SUFFICIENT. This statement requires more work. We are told that if the number of people increased by 3, the contribution would fall by $2. In other words, the new number of people is $$n + 3$$, and the new contribution is $$x - 2$$. The product will still be$36.

Thus, we know that $$36 = (n + 3)(x - 2)$$. We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$. Let's expand the new equation and swap out $$x$$.
$$36 = (n + 3)(x - 2) = nx + 3x - 2n - 6$$

Since nx equals 36, we can substitute in 36 for nx as follows:
$$36 = 36 + 3x - 2n - 6$$
$$6 = 3x - 2n$$

Now substitute in $$\frac{36}{n}$$ for $$x$$:
$$6 = 3 * \frac{36}{n} - 2n$$
$$6 = \frac{108}{n} - 2n$$
$$2n^2 + 6n - 108 = 0$$

Before factoring this quadratic, we should divide the entire equation by 2. Every term in the equation is even, so we will still have integers, and it is much easier to factor equations in which the $$x^2$$ term has a coefficient of 1.

$$n^2 + 3n - 54 = 0$$

Since the 54 in the distributed quadratic equation has a minus sign in front of it, we know that one of the missing numbers is negative and that the other one is positive. This means that one of the solutions for $$n$$ will be positive, while the other one will be negative. We could stop here, since only one positive solution exists.

If we wanted to keep going with the factoring, we could observe that we need a pair of factors relatively close in value to each other, since their difference (after multiplying one of the factors by 2) is only 3. The pair of factors that works is {6, 9}, as we can see by trial and error.
$$(n - 6)(n + 9) = 0$$

$$n = 6$$ or $$n = -9$$

The negative solution is impossible, so we know that $$n$$ is 6.

_________________
Manager
Joined: 21 Jul 2014
Posts: 126
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

18 Nov 2014, 08:25
Quote:
We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$. Let's expand the new equation and swap out $$x$$.
$$36 = (n + 3)(x - 2) = nx + 3x - 2n - 6$$

Since nx equals 36, we can substitute in 36 for nx as follows:
$$36 = 36 + 3x - 2n - 6$$
$$6 = 3x - 2n$$

Question about the part quoted above from the official answer: How can we know that nx = 36 if we are evaluating only statement (2)? Shouldn't we start by evaluating statement two independently of statement 1?
Math Expert
Joined: 02 Sep 2009
Posts: 44566
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

18 Nov 2014, 09:59
LighthousePrep wrote:
Quote:
We also still know that $$36 = nx$$, or $$\frac{36}{n} = x$$. Let's expand the new equation and swap out $$x$$.
$$36 = (n + 3)(x - 2) = nx + 3x - 2n - 6$$

Since nx equals 36, we can substitute in 36 for nx as follows:
$$36 = 36 + 3x - 2n - 6$$
$$6 = 3x - 2n$$

Question about the part quoted above from the official answer: How can we know that nx = 36 if we are evaluating only statement (2)? Shouldn't we start by evaluating statement two independently of statement 1?

We know that from the stem itself:
Let’s call the number of people in the group $$n$$, and let’s call each contribution $$$x$$. Then we know from the stem that $$36 = nx$$. _________________ Manager Status: Please do not forget to give kudos if you like my post Joined: 19 Sep 2008 Posts: 110 Location: United States (CA) Re: Several friends in a dinner group decide to contribute equally to the [#permalink] ### Show Tags 27 Nov 2014, 03:03 Good question! x = number of friends in group. #n = each persons contribution. #n =$36/x

1. only thing that work is 6x6 = 36. SURF
2. taking hint from 1, if x=6 then x+3=9 which bring #n to 4 which satisfies the statement. so lets find is that the only option.
lets say x = 3, so x+3=6 which bring #n to 6 diff is +3 does not satisfy.
lets say x = 9, so x+3 = 12 which brings #n to 3, -6 does not satisfy.
this tells us that X should be in between. so only number that works is X = 6. other number will not work.
So Suff

Bunuel wrote:

Tough and Tricky questions: Word Problems.

Several friends in a dinner group decide to contribute equally to the purchase of a $36 gift. How many people are in the group? (1) The number of people in the group is equal to the size of each person’s contribution, in dollars. (2) If three more people joined the group, each person’s individual contribution would fall by$2.

Kudos for a correct solution.

_________________

[Reveal] Spoiler:

Director
Joined: 23 Jan 2013
Posts: 591
Schools: Cambridge'16
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

27 Nov 2014, 03:55
St1. Of course it is 6*6=36 (eliminate B,C,E)

some my test for possible D answer in value questions after you know that St1 is sufficient as in this question

St2. Consider 6 as correct and test St2, so 6+3=9 and to be 36 contribution should be 4, it is 2$fewer. The same answer, i.e. 6 people, means that both statement alone sufficient D Manager Joined: 12 Sep 2014 Posts: 162 Concentration: Strategy, Leadership GMAT 1: 740 Q49 V41 GPA: 3.94 Re: Several friends in a dinner group decide to contribute equally to the [#permalink] ### Show Tags 28 Nov 2014, 14:52 Statement 1 tells us that 36 is a square of summering. So the answer must be 6 people each posting$6.
Statement two gives us a system of linear simultaneous equations that we can since to get back to 6 people paying $6. Whereas with 9 people, each pays$4.

Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 605
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

30 Nov 2014, 19:26
Hi Bunuel,

The question says that "Several friends in a dinner group "
So, isn't it that "not all: of them are contributing. Several out of All are contributing? Going by that logic I thought the answer would be E?
_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

Math Expert
Joined: 02 Sep 2009
Posts: 44566
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

01 Dec 2014, 03:50
nktdotgupta wrote:
Hi Bunuel,

The question says that "Several friends in a dinner group "
So, isn't it that "not all: of them are contributing. Several out of All are contributing? Going by that logic I thought the answer would be E?

I think the question means that there are several friends and all of them contributed some amount.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 6641
Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

### Show Tags

30 Mar 2018, 22:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Several friends in a dinner group decide to contribute equally to the   [#permalink] 30 Mar 2018, 22:19
Display posts from previous: Sort by