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. $\newline$

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Grade 7

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Q. 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.

\newline

Define Variables: Let's call the number of students who failed only in English $E$ , only in Science $S$ , and only in Maths $M$ . According to the problem, $S = 2E$ and $M = 3E$ .
Set Up Equations: Now, let's denote the number of students who passed in all three subjects as $x$ . Then, the number of students who passed in exactly two subjects is $x + 5$ .
Substitute Variables: The number of students who passed in English and one other subject is also $x$ . This means the sum of students who passed in English and Maths, English and Science, and all three subjects is $2x$ .
Simplify Equation: We can set up an equation for the total number of students: $190$ (English) + $60$ (Maths) + $75$ (Science) - $2x$ (passed in two or all three) + $E$ + $S$ + $M$ + $x$ (all three) + $F$ (failed all) = $260$ .
Final Equation: Substitute $S$ and $M$ with $2E$ and $3E$ , respectively, and simplify the equation: $190 + 60 + 75 - 2x + E + 2E + 3E + x + F = 260$ .
Final Equation: Substitute $S$ and $M$ with $2E$ and $3E$ , respectively, and simplify the equation: $190 + 60 + 75 - 2x + E + 2E + 3E + x + F = 260$ .Combine like terms: $325 + 6E - x + F = 260$ .
Final Equation: Substitute $S$ and $M$ with $2E$ and $3E$ , respectively, and simplify the equation: $190 + 60 + 75 - 2x + E + 2E + 3E + x + F = 260$ .Combine like terms: $325 + 6E - x + F = 260$ .We know that $x = E + 5$ , so we can substitute $x$ with $E + 5$ in the equation: $325 + 6E - (E + 5) + F = 260$ .
Final Equation: Substitute $S$ and $M$ with $2E$ and $3E$ , respectively, and simplify the equation: $190 + 60 + 75 - 2x + E + 2E + 3E + x + F = 260$ .Combine like terms: $325 + 6E - x + F = 260$ .We know that $x = E + 5$ , so we can substitute $x$ with $E + 5$ in the equation: $325 + 6E - (E + 5) + F = 260$ .Simplify the equation: $M$ $0$ .
Final Equation: Substitute $S$ and $M$ with $2E$ and $3E$ , respectively, and simplify the equation: $190 + 60 + 75 - 2x + E + 2E + 3E + x + F = 260$ .Combine like terms: $325 + 6E - x + F = 260$ .We know that $x = E + 5$ , so we can substitute $x$ with $E + 5$ in the equation: $325 + 6E - (E + 5) + F = 260$ .Simplify the equation: $M$ $0$ .Further simplify to: $M$ $1$ .