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# At a certain committee meeting only associate professors and

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Re: At a certain committee meeting only associate professors and [#permalink]
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anon1 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
Attachment:

Capture.PNG [ 2.17 KiB | Viewed 11456 times ]

Now, Check -

2a + b = 10 ---------> ( I )
and
a + 2b = 11 ---------> ( II )

In ( I ) , 2a will always result in an even number , hence b must be Odd
In ( II ) , 2b will always result in an even number , hence a must be Odd

Plug in the values the correct result satisfying both the equations, a will be 3 and b will be 4

Hence Total number of people is 7 , correct answer will be (B)
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Re: At a certain committee meeting only associate professors and [#permalink]
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anon1 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

We can let the number of associate professors = a and the number of assistant professors = n, and create two equations:

Pencil equation:

2a + n = 10

Chart equation:

a + 2n = 11

Let’s add the two equations together:

3a + 3n = 21

a + n = 7

Thus, there were 7 people in the meeting.

Answer: B
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Re: At a certain committee meeting only associate professors and [#permalink]
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anon1 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

Notice that the question asks us to find that TOTAL number of people present.
In other words, it the question does NOT ask us to find the number of associate professors or the number of assistant professors present.

Since each person brings THREE items, and since a total of 21 items were brought (10 pencils and 11 charts), the number of people = 21/3 = 7

Answer: B

Cheers,
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Re: At a certain committee meeting only associate professors and [#permalink]
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Let the number of Associate professors be ‘x’ and the number of Assistant professors be ‘y’. A table can be drawn summarizing the information given, to look like the one below:

Attachment:

22nd June 2021 - Reply 5.JPG [ 43.04 KiB | Viewed 6084 times ]

From the table, it can be seen that the number of pencils brought by the Associates is 2x and those brought by the Assistants is y. Similarly, the number of charts brought by the Associates is x and those brought by the Assistants is 2y.

Therefore, 2x + y = 10 and x + 2y = 11.
Solving, we get y = 4 and x = 3. 7 people were present at the meeting.

The correct answer option is B.

Hope that helps!
Aravind B T
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Re: At a certain committee meeting only associate professors and [#permalink]
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anon1 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

A pencil-bringing rate for associate professors is combined with a pencil-bringing rate for assistant professors to yield a pencil-bringing rate for the MIXTURE of professors.
Since this is a MIXTURE problem, an alternate approach is to solve with ALLIGATION.

Let X = the number of associate professors and Y = the number of assistant professors.

Pencil-bringing rate for X $$= \frac{2-pencils}{3-items-brought} = \frac{2}{3} = \frac{14}{21}$$
Pencil-bringing rate for Y $$= \frac{1-pencil}{3-items-brought} = \frac{1}{3} = \frac{7}{21}$$
Pencil-bringing rate for the MIXTURE of X and Y $$= \frac{10-pencils}{21-items-brought} = \frac{10}{21}$$

Since the fractions have been put over the same denominator, the alligation can be performed with only the numerators.

Step 1: Plot the 3 numerators on a number line, with the numerators for X and Y on the ends and the numerator for the mixture in the middle.
X 14-----------10------------ 7 Y

Step 2: Calculate the distances between the numerators.
X 14-----4-----10-----3-----7 Y

Step 3: Determine the ratio in the mixture.
The ratio of X to Y is equal to the RECIPROCAL of the distances in red:
X:Y = 3:4

Since there are 3 X's for every 4 Y's -- and 3+4 = 7 -- the total number of people must be a multiple of 7.

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At a certain committee meeting only associate professors and [#permalink]
is this  considered algebra? ratio? number properties? Is there any other ways of solving this?­
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Re: At a certain committee meeting only associate professors and [#permalink]
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Sa800 wrote:
is this  considered algebra? ratio? number properties? Is there any other ways of solving this?­

­I'd categorize this question under Algebra and Word Problems. Different solution approaches have been presented in the discussion above. If anything remains unclear, please feel free to let me know.
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At a certain committee meeting only associate professors and [#permalink]
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­Keep it simple. System of equations:

­
At a certain committee meeting only associate professors and [#permalink]
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