Confusing problem solving

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Confusing problem solving

by NeilWatson » Sat Aug 09, 2014 5:56 pm
At a certain committee meeting only associate professors and assistant professors are present. Each associate professor has brought 2 pencils and 1 chart to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a total of 10 pencils and 11 charts have been brought to the meeting, how many people are present?

a) 6
b) 7
c) 8
d) 9
e) 10

I have 3 weeks till my retake and I am trying to figure out how this problem might apply to others. I have read the explanation and I am not sure that if I saw something similar I would be able to answer quickly. Please help.
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by GMATGuruNY » Sat Aug 09, 2014 6:10 pm
NeilWatson wrote:At a certain committee meeting only associate professors and assistant professors are present. Each associate professor has brought 2 pencils and 1 chart to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a total of 10 pencils and 11 charts have been brought to the meeting, how many people are present?

a) 6
b) 7
c) 8
d) 9
e) 10
Approach 1:

This is a MIXTURE problem.

The MIXTURE:
A total of 10 pencils and 11 charts are brought to the meeting.
Thus, of the 22 items brought to the meeting, the fraction that are pencils = 10/21.

ASSOCIATE professors in the mixture:
Each associate professor brings 2 pencils and 1 chart.
Thus, of every 3 items brought by the associate professors, the fraction that are pencils = 2/3 = 14/21.

ASSISTANT professors in the mixture:
Each assistant professor brings 1 pencil and 2 charts.
Thus, of every 3 items brought by the assistant professors, the fraction that are pencils = 1/3 = 7/21.

To determine the ratio of associate professors to assistant professors in the "mixture", use ALLIGATION.

Step 1: Plot the 3 fractions on a number line, with the fractions for the two types of professors (14/21 and 7/21) on the ends and the fraction for the whole group (10/21) in the middle.
associate 14/21---------------------10/21----------------------7/21 assistant

Step 2: Calculate the distances between the fractions.
associate 14/21---------4/21--------10/21---------3/21---------7/21 assistant

Step 3: Determine the ratio in the mixture.
The ratio of associate professors to assistant professors is the RECIPROCAL of the distances in red.
associate : assistant = 3/21 : 4/21 = 3:4.

Since the sum of the parts of the ratio = 3+4 = 7, the total number of professors must be a multiple of 7.

The correct answer is B.

For another problem that I solved with alligation, check here:

https://www.beatthegmat.com/cars-n-trucks-t115617.html

Approach 2:

Let x = the total number of associate professors and y = the total number of assistant professors.

PENCILS:
Since each associate professor brought 2 pencils, the total number of pencils brought by the associated professors = 2x.
Since each assistant professor brought 1 pencil, the total number of pencils brought by the assistant professors = 1y = y.
Since a total of 10 pencils were brought:
2x+y = 10.

CHARTS:
Since each associate professor brought 1 chart, the total number of charts brought by the associated professors = 1x = x.
Since each assistant professor brought 2 charts, the total number of charts brought by the assistant professors = 2y.
Since a total of 11 charts were brought:
x+2y = 11.

Adding together 2x+y = 10 and x+2y = 11, we get:
(2x+y) + (x+2y) = 10+11
3x + 3y = 21
x+y = 7.

Thus, the total number of professors = 7.

The correct answer is B.
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by GMATGuruNY » Sat Aug 09, 2014 6:57 pm
At a certain committee meeting only associate professors and assistant professors are present. Each associate professor has brought 2 pencils and 1 chart to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a total of 10 pencils and 11 charts have been brought to the meeting, how many people are present?

a) 6
b) 7
c) 8
d) 9
e) 10
An alternate approach is PLUG IN THE ANSWERS, which represent the total number of people.

Since the assistant professors bring more charts than pencils, and the total number of charts (11) is slightly greater than the total number of pencils (10), there must be slightly more assistant professors than associate professors.

Answer choice D: 9 people
If there are 5 assistant professors and 4 associate professors, then:
The 5 assistant professors bring 5 pencils and 10 charts.
The 4 associate professors bring 8 pencils and 4 charts.
Total number of pencils = 5+8 = 13.
Total number of charts = 10+4 = 14.
Too many pencils and charts.
Eliminate D.

Answer choice B: 7 people
If there are 4 assistant professors and 3 associate professors, then:
The 4 assistant professors bring 4 pencils and 8 charts.
The 3 associate professors bring 6 pencils and 3 charts.
Total number of pencils = 4+6 = 10.
Total number of charts = 8+3 = 11.
Success!

The correct answer is B.
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by [email protected] » Sun Aug 10, 2014 12:21 am
Hi NeilWatson,

This question actually has a fantastic shortcut built into it that allows you to avoid almost all the long-winded math:

Associates bring 2 pencils and 1 chart; a total of 3 items
Assistants bring 1 pencil and 2 charts; a total of 3 items

There are 10 pencils and 11 charts; a total of 21 items

Since each person (regardless of his/her title) brings 3 items and there are 21 items total....

21/3 = 7

There must be 7 people in the meeting.

Final Answer: B

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by GMATinsight » Sun Aug 10, 2014 12:15 pm
NeilWatson wrote:At a certain committee meeting only associate professors and assistant professors are present. Each associate professor has brought 2 pencils and 1 chart to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a total of 10 pencils and 11 charts have been brought to the meeting, how many people are present?

a) 6
b) 7
c) 8
d) 9
e) 10

I have 3 weeks till my retake and I am trying to figure out how this problem might apply to others. I have read the explanation and I am not sure that if I saw something similar I would be able to answer quickly. Please help.
TRADITIONAL WAY:

associate professor has brought 2 pencils and 1 chart
Let, X Associate Prof
then Pencils = 2X
and Charts = X


assistant professor has brought 1 pencil and 2 charts
Let, Y AssistantProf
then Pencils = Y
and Charts = 2Y


Total Pencils = 2X + Y = 10
Total Charts = X + 2Y = 11


Solving the two equations, We get
X = 3 and Y = 4

Total people = 3 + 4 = 7

Answer: Option B
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