Let E represent Economics, M represent Mathematics, and G represent Geography.

Question

ScanSolve AI Tutor · Accepted Answer

Here's the solution to the problem.

Let $E$ represent Economics, $M$ represent Mathematics, and $G$ represent Geography.
The total number of students is $n(U) = 70$ .
All students offer at least one subject, so $n(E \cup M \cup G) = 70$ .

We are given the following information:

$n(E \text{ only}) = 6$
$n(E \cap M^c) = 18$ (Economics but not Mathematics)
$n(E \cap G) = 38$ (Economics and Geography)
$n(E) = 53$ (Economics)
$n(G) = 50$ (Geography)
$n(M \cap G) = 34$ (Mathematics and Geography)

Let's denote the number of students in each region of the Venn diagram as follows:

$a = n(E \text{ only})$
$b = n(M \text{ only})$
$c = n(G \text{ only})$
$d = n(E \cap M \cap G^c)$ (Economics and Mathematics only)
$e = n(E \cap G \cap M^c)$ (Economics and Geography only)
$f = n(M \cap G \cap E^c)$ (Mathematics and Geography only)
$g = n(E \cap M \cap G)$ (Economics, Mathematics, and Geography)

Using the given information to find the values for each region:

From (1), $a = 6$ .
From (2), $n(E \cap M^c) = a + e = 18$ .
Since $a=6$ , we have $6 + e = 18 \implies e = 12$ .
From (3), $n(E \cap G) = e + g = 38$ .
Since $e=12$ , we have $12 + g = 38 \implies g = 26$ .
From (4), $n(E) = a + d + e + g = 53$ .
Since $a=6, e=12, g=26$ , we have $6 + d + 12 + 26 = 53 \implies d + 44 = 53 \implies d = 9$ .
From (6), $n(M \cap G) = f + g = 34$ .
Since $g=26$ , we have $f + 26 = 34 \implies f = 8$ .
From (5), $n(G) = c + e + f + g = 50$ .
Since $e=12, f=8, g=26$ , we have $c + 12 + 8 + 26 = 50 \implies c + 46 = 50 \implies c = 4$ .
Since all students offer at least one subject, $n(E \cup M \cup G) = a + b + c + d +

Let E represent Economics, M represent Mathematics, and G represent Geography.

Step-by-step solution

From (1), $a = 6$ .

From (2), $n(E \cap M^c) = a + e = 18$ .

From (3), $n(E \cap G) = e + g = 38$ .

From (4), $n(E) = a + d + e + g = 53$ .

From (6), $n(M \cap G) = f + g = 34$ .

From (5), $n(G) = c + e + f + g = 50$ .

Since all students offer at least one subject, $n(E \cup M \cup G) = a + b + c + d +

Still stuck on this one?

More Geography Questions

Let E represent Economics, M represent Mathematics, and G represent Geography.

Step-by-step solution

From (1), a=6a = 6a=6.

From (2), n(E∩Mc)=a+e=18n(E \cap M^c) = a + e = 18n(E∩Mc)=a+e=18.

From (3), n(E∩G)=e+g=38n(E \cap G) = e + g = 38n(E∩G)=e+g=38.

From (4), n(E)=a+d+e+g=53n(E) = a + d + e + g = 53n(E)=a+d+e+g=53.

From (6), n(M∩G)=f+g=34n(M \cap G) = f + g = 34n(M∩G)=f+g=34.

From (5), n(G)=c+e+f+g=50n(G) = c + e + f + g = 50n(G)=c+e+f+g=50.

Since all students offer at least one subject, $n(E \cup M \cup G) = a + b + c + d +

Still stuck on this one?

More Geography Questions

From (1), $a = 6$ .

From (2), $n(E \cap M^c) = a + e = 18$ .

From (3), $n(E \cap G) = e + g = 38$ .

From (4), $n(E) = a + d + e + g = 53$ .

From (6), $n(M \cap G) = f + g = 34$ .

From (5), $n(G) = c + e + f + g = 50$ .