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17 Nov 2014, 12:34
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:37) correct
33%
(01:48)
wrong
based on 449
sessions
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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.
(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.
18 Nov 2014, 07:57
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Bunuel
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.
General Discussion
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.
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.
