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The Number of students learning Three subjects are as

Expert replies
by M7MBA » Wed May 16, 2018 1:27 am
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The Number of students learning Three subjects is as follows:

- Maths = 40
- English = 50
- Science = 35

If the Number of students learning exactly two subjects is maximum and no student learns all the three subjects, then what is the minimum number of students learning exactly one subject?

A. 0
B. 1
C. 25
D. 35
E. 45

The OA is B.

Can someone give me some help? I am confused and I need a clarification. Thanks in advance.
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Source: — Problem Solving |

by [email protected] » Wed May 16, 2018 7:57 pm
Hi M7MBA,

We're told that the number of students learning three subjects is as follows: Maths = 40, English = 50, Science = 35, the number of students learning EXACTLY two subjects is MAXIMIZED and NO student learns all the three subjects. We're asked for the minimum number of students learning EXACTLY ONE subject. While this question might look complex, the answer choices are 'spread out' enough that you can answer this question without too much math (just a little basic Arithmetic).

We know that many of the students in the three listings have been 'counted' TWICE (for example, a student who took Math and English was 'counted' in BOTH lists). The sum of those 3 numbers is 40+50+35 = 125. Since 125 is an ODD number, then there's no way that every student took 2 classes (in that situation, the total would have to be an EVEN number). Thus, there's at least 1 student who didn't take two classes.

From this point, you could 'play around' with the given numbers and forms 'groups of 2':

IF...
25 students took English and Math
25 students took English and Science
10 students took Math and Science
That would leave 5 students to just take Math. However, since 5 is a possible outcome - but answers C, D and E are all much bigger than 5 - none of those three possibilities could be the answer. There's only one answer that fits.

Final Answer: B

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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Math, Science, and English

by GMATGuruNY » Fri May 18, 2018 1:44 am
M7MBA wrote:The Number of students learning Three subjects is as follows:

- Maths = 40
- English = 50
- Science = 35

If the Number of students learning exactly two subjects is maximum and no student learns all the three subjects, then what is the minimum number of students learning exactly one subject?

A. 0
B. 1
C. 25
D. 35
E. 45
Image

The values in blue represent the number of students studying exactly 1 subject.
Sum of the values in blue:
125 - 2(a+b+c).
To MINIMIZE the value of this expression, we must MAXIMIZE the value of a+b+c.
Since the expression must be nonnegative, a+b+c ≤ 62.

Goal:
Choose values so that a+b+c = 62.
Since a+c belongs to the most studied subject -- English -- try to maximize a+c.
Since there are 50 English students, the greatest possible option for a+c=50, with the result that b = 62-50 = 12.
In the Venn Diagram, insert b=12 and maximize the values of b and c so that b+c = 50:
Image
In the resulting Venn Diagram, one student studies only science, implying that the minimum number of students studying exactly one subject = 1.

The correct answer is B.
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by Scott@TargetTestPrep » Tue May 07, 2019 5:37 pm
M7MBA wrote:The Number of students learning Three subjects is as follows:

- Maths = 40
- English = 50
- Science = 35

If the Number of students learning exactly two subjects is maximum and no student learns all the three subjects, then what is the minimum number of students learning exactly one subject?

A. 0
B. 1
C. 25
D. 35
E. 45

The OA is B.

Let ME, MS and ES be the number students who study Math and English, Math and Science, and English and Science, respectively. So we have 40 - (ME + MS), 50 - (ME + ES), 35 - (MS + ES) as the number of students who study only Math, only English and only Science, respectively. Therefore, the total number of students who study only one subject is:

40 - (ME + MS) + 50 - (ME + ES) + 35 - (MS + ES)

125 - 2(ME + MS + ES)

Since we want to minimize this value and we see that if ME + MS + ES = 62, the value of 125 - 2(ME + MS + ES) will be minimized and that value is 125 - 2(62) = 1.

Answer: B

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