# In a class of 65, there are twice as many maths students as biology students.

I have a task:

In a class of 65, there are twice as many maths students as biology students. If 12 biology students do not take maths and 15 students take neither of these subjects, how many students take maths but not biology?

A. 7
B. 19
C. 26
D. 31
E. This is impossible

I found out that the result is 31 (Answer D) but only by using the answer as help. Is there another mathematical way? I tried out a lot but I couldn't find out.

Thanks!

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Effectively there are only $50$ students to worry about.

There are $50-12=38$ students who take math, and therefore $19$ who take biology. Thus $19-12$, that is, $7$ take both. That leaves $38-7$ students who take math but not biology.

Remark: A picture (Venn diagram) may be useful in keeping track of things.

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Nice. Definitely the algebraically-simplest answer. – alexqwx Jul 16 '14 at 15:28

Let the number of students who take strictly Biology be $b$, the number who take strictly maths be $m$, the number who take both be $d$, and the number who take neither be $n$. Then $$b + m + d + n = 65$$ We are also given that $$m + d = 2(b + d)$$ and that $b = 12$ and $n = 15$. Then $$m + d = 38$$ and $$m - d = 2b = 24$$ So $$2m = 24 + 38 = 62$$ So $m = 31$.

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This could be a bit more explicit about where each equation comes from, but this is definitely a good example of how to approach word problems. Turn everything into variables, describe the relationships with equations, and then go wild with your math skills until you get an answer (or get stuck and need to make some logical leap). Then come back and interpret what the answer means. – jpmc26 Jul 16 '14 at 4:48

Let $B$ be the set of biology students and $M$ be the set of Math students and $U$ be the set of all the students. Then we are given that $$|U|=65, \quad |M|=2|B|, \quad |B-M|=12, \quad |U-(M \cup B)|=15.$$ What you want is $|M-B|$. Let $y=|M-B|$ and $x=|M \cap B|$, then $|M|=x+y$, so we have $$(x+y)=2(12+x).$$ and $$y+x+12+15=65.$$ Now solve for $y$ to get your answer. If you are familiar with Venn diagrams then you will be able to follow the argument I have stated fairly easily.

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Although this is a good answer, it seems a bit too advanced for the asker. – Milo Jul 16 '14 at 10:02

Let t be total no. of students in the class, m be no. of students having maths only, b be students having biology only, n be students having neither and x be students having both Maths and Biology.

We are given that

t= 65

b=12

n=15

(m+x)=2*(b+x)

Also

(m+x) + b + n = t

Replacing it with with values.

2*(b+x) + 12 + 15 = 65

(b+x) = 38/2 = 19, No. of biology students

There for No. of Maths students = t - n - (b+x) = 65 - 15 - 19 = 31

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Total number of students, T = 65

Number of biology students, B

Number of maths students, M = 2B

Number of other students, O = 15

Number of students studying either maths or biology = T - O = 65 - 15 = 50

Number of students studying just biology = 12 (given)

Number of maths students( just Maths + both maths and biology), M = 50 - 12 = 38

As, M = 2B <=> B = 38/2 = 19

Number of students who do both maths and biology = 19 - 12 = 7

Therefore, Number of maths students, M = 38 - 7 = 31 (Option D)

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Making a picture often helps. For sets such pictures are called Venn diagrams.

Consider for instance the Venn diagram below:

We know that there are twice as many math students as there are biology students, so it must hold that $|A_1| + |A_3| = 2 (|A_1| + |A_2)$ ($|S|$ denotes the number of elements in the set $|S|$), or equivalently that $|A_3| \overset{|A_2|=12}{=} |A_1| + 2 |A_2| = 24 + |A_1|$. Also, we know that $|A_1| + |A_2| + |A_3| = 50$, or equivalently, that $|A_3| \overset{|A_2|=12}{=} 38 - |A_1|$.

So now we have two equations:

1. $|A_3| = 24 + |A_1|$
2. $|A_3| = 38 - |A_1|$

From 2 it follows that $|A_1| = 38 - |A_3|$,

so that from 1 it follows that $|A_3| = 24 + |A_1| = 24 + 38 - |A_3|$, or equivalently

$2|A_3| = 62 \rightarrow |A_3| = 31$.

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