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

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

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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

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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 18 Nov 2014, 07:24
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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.:)
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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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.

The new equation now reads:
\(n^2 + 3n - 54 = 0\)

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.

Answer: D.
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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?
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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 \(36 = nx\).
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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

Answer: D

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.

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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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.

The answer is D.
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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?
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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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New post 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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Several friends in a dinner group decide to contribute equally to the [#permalink]

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Re: Several friends in a dinner group decide to contribute equally to the   [#permalink] 30 Mar 2018, 22:19
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