In a class of 50 students, the number of students who offer Accounting is twice the number who offer Economic.10 students offer neither the two subjects . Illustrate the information on the venn diagram. How many students offer Accounting. How many students offer Economic. How many students offer only one subject

Answer 1

To solve the problem, let's denote the number of students who offer Accounting as \( A \) and the number who offer Economics as \( E \).

According to the information provided:

1. The total number of students = 50.
2. The number of students who offer Accounting is twice the number who offer Economics: \( A = 2E \).
3. 10 students offer neither subject.

Based on this, we can find the number of students who offer at least one subject:

\[ \text{Number of students who offer at least one subject} = 50 - 10 = 40 \]

The total number of students who offer Accounting or Economics can be expressed in terms of the individual groups:

\[ A + E - X = 40 \]

where \( X \) is the number of students who offer both subjects (the intersection).

Now substituting \( A = 2E \) into the equation:

\[ 2E + E - X = 40 \] \[ 3E - X = 40 \]

Now, to find the number of students who offer only one subject, we calculate:

1. Students who offer only Accounting: \( A - X \)
2. Students who offer only Economics: \( E - X \)

Therefore, the formula for students who offer only one subject is:

\[ (A - X) + (E - X) = A + E - 2X \]

We also have:

\[ A + E = 3E \quad \text{(because } A = 2E\text{)} \]

Thus:

\[ 3E - 2X = \text{students who offer only one subject} \]

Now we need to express \( E \) or \( A \) in terms of \( X \) to simplify our calculations.

Let’s solve \( 3E - X = 40 \) for \( X \):

1. Rearranging gives us \( X = 3E - 40 \).
2. Substitute this into \( A + E - X = 40 \).

Now substitute \( X \):

\[ A + E - (3E - 40) = 40 \]

Since \( A = 2E \):

\[ 2E + E - 3E + 40 = 40 \] \[ 40 = 40 \text{ (this is true, but does not help solve for E)} \]

Now consider substituting back into \( 3E - X = 40 \):

Using trial values is practical for this kind of problem. Let’s assume values for \( E \):

Let \( E = 10 \):

Then \( A = 2E = 20 \), thus \( 20 + 10 - X = 40 \):

We already have \( X = 3E - 40\):

Setting E = 10 gives \( X = 3(10)-40 = 30-40 = -10 \) (which does not work).

Now try \( E = 15 \):

Then \( A = 2E = 30\): Using \( 30 + 15 - X = 40\):

Hence, \( 45 - X = 40 \Rightarrow X = 5\).

In this scenario:

• Students only in Accounting: \( 30 - 5 = 25\)
• Students only in Economics: \( 15 - 5 = 10\)

Thus, we have:

• Number of Students offering Accounting: 30
• Number of Students offering Economics: 15
• Students offering only one subject: \( (25 + 10 = 35) \)

Finally, the Venn diagram would show:

• Circle for Accounting: 25 (only Accounting) + 5 (both) = 30 total
• Circle for Economics: 10 (only Economics) + 5 (both) = 15 total
• Outside both circles: 10 students (neither)

Thus confirmed with the total students: \( 25 + 10 + 5 + 10 = 50\) which is correct.

Overall:

• Number of students who offer only one subject is 35.