Find total number of students who failed the exam?

Question

In an examination of three subjects, out of 260 students, 190 passed in English, 60 in Maths and 75 in Science. For every person who failed only in English, there were 2 person who failed only in science and 3 who failed in Maths alone. The number of students who passed in exactly two subjects was 5 more than the number of students who passes in all three. Also, those who passed in English along with only one other subject were equal in number to those who passed all three subjects. Find the number of students who failed in all the three subjects.

I made the following Venn diagram. But I can not solve the equation very easily am I interpreting this question wrong.

Answer 1

I often find that when you don't know where to go, it helps to just write down what you know.

The students who failed only in English also passed math and science. That's represented in your diagram by $f$. Similarly, the students who failed only in math also passed English and science, so they are represented in your diagram by $d$, and the students who failed only in science also passed English and math so they are represented in your diagram by $b$.

Therefore, there are $f$ students who failed only in English, so $b=2f, d=3f$ and there are $6f$ students who passed exactly $2$ classes, which means $e=6f-5$ students passed all three classes. We also know that $b+d=5f=6f-5$, so $f=5, b=10, d=15, e=25$.

Thus, $a=190-50=140, c=60-40=20, g=75-45=30$.

Adding everything up, we find that $245$ students passed at least one class, so there are $15$ students who didn't pass anything.

Answer 2

Alternative (Truth Table) Approach

$\underline{\text{Equations In Words}}$

EQ1 : out of 260 students
EQ2 : 190 passed in English, 
EQ3 : 60 in Maths 
EQ4 : 75 in Science.
EQ5 : For every person who failed only in English, 
      there were 2 person who failed only in science
EQ6 : For every person who failed only in English,
      and 3 who failed in Maths alone.
EQ7 : The number of students who passed in exactly two
      subjects was 5 more than the number of students
      who passes in all three.
EQ8 : those who passed in English along with only one
      other subject were equal in number to those who
      passed all three subjects.
??? : Find the number of students who failed in all the
      three subjects.

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$\underline{\text{Variables}}$

T = Passed, F = Failed.

\begin{array}{| l | l | l | l| } \hline \text{English} & \text{Math} & \text{Science} & \text{Variable} \\ \hline \text{T} & \text{T} & \text{T} & x_1 \\ \hline \text{T} & \text{T} & \text{F} & x_2 \\ \hline \text{T} & \text{F} & \text{T} & x_3 \\ \hline \text{T} & \text{F} & \text{F} & x_4 \\ \hline \text{F} & \text{T} & \text{T} & x_5 \\ \hline \text{F} & \text{T} & \text{F} & x_6 \\ \hline \text{F} & \text{F} & \text{T} & x_7 \\ \hline \text{F} & \text{F} & \text{F} & \color{red}{x_8 = ~?} \\ \hline \end{array}

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$\underline{\text{Equations In Algebra}}$

Var :   x1   x2   x3   x4   x5   x6   x7   x8 
EQ1 :    1    1    1    1    1    1    1    1  = 260
EQ2 :    1    1    1    1                      = 190
EQ3 :    1    1              1    1            =  60 
EQ4 :    1         1         1         1       =  75
EQ5 :         1             -2                 =   0
EQ6 :              1        -3                 =   0
EQ7 :   -1    1    1         1                 =   5
EQ8 :   -1    1    1                           =   0

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$\underline{\text{Analysis}}$

From EQ7 - EQ8, $x_5 = 5.$
Then, EQ5 and EQ6 give you $~x_2 = 10, ~x_3 = 15.$

Known Values :  x1  x2  x3  x4  x5  x6  x7  x8   
                    10  15       5

Now, EQ7 gives you $~x_1 = 25.$

Known Values :  x1  x2  x3  x4  x5  x6  x7  x8   
                25  10  15       5

Now, EQ2 gives you $~x_4 = 140.$

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                25  10  15  140   5

Now, EQ3 gives you $~x_6 = 60 - (25 + 10 + 5) = 20.$

Similarly, EQ4 gives you $~x_7 = 75 - (25 + 15 + 5) = 30.$

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                25  10  15  140   5  20  30    

Finally, EQ1 gives you:
$x_8 = 260 - (x_1 + x_2 + \cdots + x_7) = \color{red}{15.}$

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                25  10  15  140   5  20  30  15