-1
$\begingroup$

In a survey of a group; the number of students who passed in Maths only, science only and Nepali only are respectively 10 more than another. The number of students who passed only in Maths and Science is 10 more than that of science only, the number of student who passed only in Science and Nepali is also 10 more than that of Maths only and the number of students who passed only in Maths and Nepali is equal to the passed students in Science only. If the number of students who passed in Science is 150 and passed students in all subjects is 50. Find the number of students who passed in Maths.

My Attempt: Let $U$ be the universal set, $M$, $N$ and $S$ be the set of students who passed in Maths, Nepali and Science respectively. Let the total number of students in the group be $x$. Then, $n(U)=x$ $n(S)=150$.

Then, How should I move on? Please help.

Thanks in Advance.

1 Answers 1

1

$$\begin{array}[ccccc]( &|&M&S&N\\ \hline x_1&|&+&0&0\\ x_2&|&0&+&0\\ x_3&|&0&0&+\\ y_1&|&+&+&0\\ y_2&|&+&0&+\\ y_3&|&0&+&+\\ z_1&|&+&+&+\\ z_2&|&0&0&0 \end{array}$$ This array explains the variables, eg. $x_2$ is number of students who passed only Science, $y_3$ - number of students who passed only science and Nepali.

We can create the following system of equations:

$$\begin{cases} x_1+x_2+x_3&=y_1+y_2+y_3+z_1+z_2+10\\ y_1&=x_2+10\\ y_3&=x_1+10\\ y_2 &= x_2\\ x_2+y_1+y_3+z_1&=150\\ z_1&=50\\ x_1+y_1+y_2+z_1 &=a \end{cases}$$

where $a$ is our unknown number of students who passed math.

In matrix representation we have: $$\left[\begin{array}[cccccccccc] ( 1&1&1&-1&-1&-1&-1&-1&|&10\\ 0&-1&0&1&0&0&0&0&|&10\\ -1&0&0&0&0&1&0&0&|&10\\ 0&-1&0&0&1&0&0&0&|&0\\ 0&1&0&1&0&1&1&0&|&150\\ 0&0&0&0&0&0&1&0&|&50\\ 1&0&0&1&1&0&1&0&|&a \end{array}\right]$$

After 5 steps of gaussian elimination we obtain a matrix with row $$\left[\begin{array}[cccccccccc] ( 0&0&0&0&0&0&0&0&|&a-140 \end{array}\right]$$ which indicates equation $$0=a-140$$ So the number of students who passed math is $140$.