Question

(a) In a class of 40 students, 18 passed
Mathematics, 19 passed Accounts, 16
passed Economics, 5 Mathematics and
Accounts only, 6 Mathematics only, 9
Accounts only, 2 Accounts and Economics
only. If each student offered at least one of
the subjects.
i. How many students failed in all the
subjects?

Answer 1

Answer:

4

Step-by-step explanation:

Let's denote:

• $n(U)$ as the total number of students (40 in this case).
• $n(E)$, $n(M)$, and $n(A)$ as the number of students who passed Economics, Mathematics, and Accounts, respectively.
• $n(ME)$, $n(AE)$, and $n(AM)$ as the number of students who passed both Mathematics and Economics, both Accounts and Economics, and both Mathematics and Accounts, respectively.
• $n(M)only$, $n(A)only$, and $n(E)only$ as the number of students who passed only Mathematics, only Accounts, and only Economics, respectively.
• $r$ as the number of students who failed in all subjects.

Now, we are given the following information:

$\begin{aligned} n(U) & = 40 \\\\n(E) & = 16 \\\\n(M) & = 18 \\\\n(ME) & = 5 \\\\n(A) & = 19 \\\\n(AE) & = 2 \\\\n(M)only & = 6 \\\\n(A)only & = 9 \\\\\end{aligned}$

From this information, we can find the values of $x$, $y$, and $z$ as follows:

1. For Accounts:

$x + 5 + 2 + 9 = 19$

$x + 16 = 19$

$x = 19 - 16$

$x = 3$

2. For Mathematics:

$y + 6 + x + 5 = 18$

$y + 6 + 3 + 5 = 18$

$y + 14 = 18$

$y = 18 - 14$

$y = 4$

3. For Economics:

$y + x + z = 16$

$4 + 3 + z = 16$

$z + 7 = 16$

$z = 16 - 7$

$z = 9$

Now, we need to find the number of students who failed in all subjects ($r$):

$40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7$

$40 - r = 36$

$r = 40 - 36 = 4$

So, the number of students who failed in all subjects is 4.