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Bunuel
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If the Number of students learning exactly two subjects is maximum and 01 Nov 2019, 06:31
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The Number of students learning three subjects are 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


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B
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If the Number of students learning exactly two subjects is maximum and 08 Sep 2020, 10:29
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Bunuel
The Number of students learning three subjects are 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


Solution:

Let ME, MS, and ES be the number of students who study Math and English, Math and Science, and English and Science, respectively. So we have 40 - (ME + MS), 50 - (ME + ES), and 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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nick1816
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If the Number of students learning exactly two subjects is maximum and 01 Nov 2019, 13:11
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Assume number of students learning exactly two subjects is 'x', and number of students learning exactly one subject is 'y'


we have to minimize, y=40+50+35-2*x=125-2x
minimum value of y is 1 {odd-even= odd}, when x=62



Bunuel
The Number of students learning three subjects are 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


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If the Number of students learning exactly two subjects is maximum and 02 Nov 2019, 03:34
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total = M+E+S-2*all subjects
total = 40+50+35-2x
total = 125-2x
value will be minimized at value of x = 62 ; ie we then get = 125-124 ;1
IMO B

Bunuel
The Number of students learning three subjects are 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


Are You Up For the Challenge: 700 Level Questions
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If the Number of students learning exactly two subjects is maximum and 31 Dec 2019, 11:49
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nick1816
Assume number of students learning exactly two subjects is 'x', and number of students learning exactly one subject is 'y'
we have to minimize, y=40+50+35-2*x=125-2x
minimum value of y is 1 {odd-even= odd}, when x=62

Sorry where did you get this? Is is another formula?

The formula I am aware of is Total = A+B+C-(sum of exactly 2 overlap) -2(sum of all 3 overlap)+Neither

or, Total = 40+50+35-(sum of exactly 2 overlap) -2(0)+0
or total = 125 - (sum of exactly 2 overlap)
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If the Number of students learning exactly two subjects is maximum and 31 Dec 2019, 17:10
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N(maths)= a+b+d+e

N(English)= b+c+e+f

N(Science)= d+e+f+g

N(maths)+N(English)+N(Science)
= a+b+d+e+b+c+e+f+d+e+f+g
= a+c+g+2(b+d+f)+3e

Now we need to figure out students learning exactly one subject, that is a+c+g.

a+c+g= N(maths)+N(English)+N(Science)-2(b+d+f)-3e


It's always better to draw venn diagram and look for what you need to figure out. (Don't just cram the formulas)


AnirudhaS
nick1816
Assume number of students learning exactly two subjects is 'x', and number of students learning exactly one subject is 'y'
we have to minimize, y=40+50+35-2*x=125-2x
minimum value of y is 1 {odd-even= odd}, when x=62

Sorry where did you get this? Is is another formula?

The formula I am aware of is Total = A+B+C-(sum of exactly 2 overlap) -2(sum of all 3 overlap)+Neither

or, Total = 40+50+35-(sum of exactly 2 overlap) -2(0)+0
or total = 125 - (sum of exactly 2 overlap)

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If the Number of students learning exactly two subjects is maximum and 21 Apr 2020, 07:14
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Still not clear with the explanation.
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If the Number of students learning exactly two subjects is maximum and 11 Jun 2020, 12:18
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AnirudhaS
nick1816
Assume number of students learning exactly two subjects is 'x', and number of students learning exactly one subject is 'y'
we have to minimize, y=40+50+35-2*x=125-2x
minimum value of y is 1 {odd-even= odd}, when x=62

Sorry where did you get this? Is is another formula?

The formula I am aware of is Total = A+B+C-(sum of exactly 2 overlap) -2(sum of all 3 overlap)+Neither

or, Total = 40+50+35-(sum of exactly 2 overlap) -2(0)+0
or total = 125 - (sum of exactly 2 overlap)

I have the same question
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If the Number of students learning exactly two subjects is maximum and 03 Sep 2020, 12:54
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Maths = 40
English = 50
Science = 35

Total number of students = 125
maximum how many students can be assigned two subjects: 2 x (125/2) = 2 X(62.5) ~ 2 x (62) = 124.
so 1 student left when number of students learning exactly two subjects is maximum.
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If the Number of students learning exactly two subjects is maximum and 13 Sep 2020, 09:17
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bhupendersingh27
Maths = 40
English = 50
Science = 35

Total number of students = 125
maximum how many students can be assigned two subjects: 2 x (125/2) = 2 X(62.5) ~ 2 x (62) = 124.
so 1 student left when number of students learning exactly two subjects is maximum.

bhupendersingh27 Total number of students is not 125. Number of students in individual subjects contain the number the number of students common to other subjects.
Can you please explain the Formula used by you ? 2 x (125/2) = 2 X(62.5) ~ 2 x (62) = 124.
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If the Number of students learning exactly two subjects is maximum and 07 Jan 2022, 21:35
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nick1816

Could you draw a venn diagram specifying the distribution of the subject combinations showing how B works?
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Very simple way to solve the problem

Let a be sum of only one

b sum of only two

c sum of only three

a + 2b + 3c = 125

Since c = 0

a + 2b = 125

Since 2b is even, maximum number for 2b will be 124

So a is 1

Answer choice B
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If the Number of students learning exactly two subjects is maximum and 02 Mar 2025, 09:39
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Hi Nick, the formula you are referring to provides us with the total population value. However, considering we are focusing on "exactly one subject" we will subtract the remaining "sum of exactly 2 overlap" from the population. giving us A+B+C -2x
AnirudhaS
nick1816
Assume number of students learning exactly two subjects is 'x', and number of students learning exactly one subject is 'y'
we have to minimize, y=40+50+35-2*x=125-2x
minimum value of y is 1 {odd-even= odd}, when x=62

Sorry where did you get this? Is is another formula?

The formula I am aware of is Total = A+B+C-(sum of exactly 2 overlap) -2(sum of all 3 overlap)+Neither

or, Total = 40+50+35-(sum of exactly 2 overlap) -2(0)+0
or total = 125 - (sum of exactly 2 overlap)
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