There's probably an algebraic way to solve this, but I doubt most people could get it in two minutes. I would backsolve this problem by testing answer choice C, and assigning 4 associate professors and 4 assistant professors. In that case we would have:
Associate professors: 2 pencils * 4 = 8 pencils
1 chart * 4 = 4 charts
Assistant professors: 1 pencil * 4 = 4 pencils
2 charts * 4 = 8 charts
This would give a total of 12 pencils and 12 charts. If we just remove 1 associate professor, we would have 10 pencils and 11 charts. Thus the correct combination is 3 associate professors and 4 assistant professors, for a total of 7 people.
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Source: Beat The GMAT — Problem Solving |
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TheGmatTutor
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Lets see...
A = Associate professors
B = Assistant professors
Total no of pencils and charts = (2p+c)A + (p+2c)B
10p + 11c = p(2A+B) +c(A+2b)
Compare the coefficients of P and C... and you'll get two equations:
2A + B = 10 and A + 2B = 11
Solve them to get the answer..
A = Associate professors
B = Assistant professors
Total no of pencils and charts = (2p+c)A + (p+2c)B
10p + 11c = p(2A+B) +c(A+2b)
Compare the coefficients of P and C... and you'll get two equations:
2A + B = 10 and A + 2B = 11
Solve them to get the answer..
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GMATGuruNY
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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
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
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Heres's the official solution, and I solved the problem in the same way. But I just think the equations do not make sense. Please help.
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GMATGuruNY
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Let x = the total number of associate professors and y = the total number of assistant professors.George7 wrote:Heres's the official solution, and I solved the problem in the same way. But I just think the equations do not make sense. Please help.
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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Thank youGMATGuruNY wrote:Let x = the total number of associate professors and y = the total number of assistant professors.George7 wrote:Heres's the official solution, and I solved the problem in the same way. But I just think the equations do not make sense. Please help.
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.
