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S97-05

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S97-05  [#permalink]

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New post 16 Sep 2014, 01:51
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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.

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New post 16 Sep 2014, 01:51
Official Solution:


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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S97-05  [#permalink]

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New post 20 Feb 2016, 18:37
Bunuel wrote:
Official Solution:


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



Is it safe to not conitue on with solving for Statement 2 all the way through since we know we have one unknown and one equation ... stop at" \(36 = (n + 3)(x - 2)\). We also still know that \(36 = nx\), or \(\frac{36}{n} = x\)"
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Re: S97-05  [#permalink]

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New post 21 Jul 2016, 03:20
We need to go till the quadratic equation forming step, otherwise we cannot be sure that this equation will have only 1 answer. Like for the above question, if the quadratic equation would have resulted in two positive values then Statement 2 would be NOT SUFFICIENT to answer the question, and the answer would have been Option A.
Hope it clears your doubt.
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New post 31 Dec 2017, 10:23
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!
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Re: S97-05  [#permalink]

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New post 31 Dec 2017, 11:41
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New post 31 Dec 2017, 13:13
Bunuel wrote:
rnz wrote:
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!


Can you please point out which step is unclear by quoting or highlighting it in the solution? I could not find 6= 36/n -2n in the solution at all.

Meanwhile you can check other solutions here: https://gmatclub.com/forum/several-frie ... 88676.html


Sorry, it's this part of the equation after substituting x for 36/n.......6=3∗36n−2n. Where did that 3 come from? Thanks!
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New post 31 Dec 2017, 13:19
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rnz wrote:
Bunuel wrote:
rnz wrote:
I'm stuck at 6= 36/n -2n. Why did you multiply the right side by 3? Where did it come from? Thanks!


Can you please point out which step is unclear by quoting or highlighting it in the solution? I could not find 6= 36/n -2n in the solution at all.

Meanwhile you can check other solutions here: https://gmatclub.com/forum/several-frie ... 88676.html


Sorry, it's this part of the equation after substituting x for 36/n.......6=3∗36n−2n. Where did that 3 come from? Thanks!


We have \(6 = 3x - 2n\) and \(\frac{36}{n} = x\). Substitute: x:

\(6 = 3*\frac{36}{n} - 2n\)
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New post 31 Dec 2017, 13:29
We have \(6 = 3x - 2n\) and \(\frac{36}{n} = x\). Substitute: x:

\(6 = 3*\frac{36}{n} - 2n\)[/quote]

oh, duh!! wow thanks. Don't know how I missed that
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Re: S97-05   [#permalink] 31 Dec 2017, 13:29
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