<Question>
The 7th and 8th grades have enrollments of 520 and 650 respectively. The two grades have a total of 18 representatives on the Student Council. How many representatives should the 8th grade have if there is to be a fair representation of the two grades?
</Question>

<think>
<Plan_1>
Okay, let's see. The problem is about figuring out how many representatives the 8th grade should have on the Student Council to have fair representation. The 7th grade has 520 students and the 8th has 650. Together, they have 18 representatives. So, I need to distribute those 18 representatives between the two grades in a way that's proportional to their enrollments.

First, maybe I should find the total number of students in both grades combined. That makes sense because if we're talking about fair representation, it's probably based on the size of each grade.

So, 7th grade is 520 and 8th is 650. Adding those together: 520 + 650.
Let me compute that. 520 plus 600 is 1120, and then plus 50 more is 1170. So total students are 1170.

Now, the total representatives are 18. To find out how many each grade should have, we can calculate the proportion of each grade's enrollment relative to the total. Then apply that proportion to the total number of representatives.
Let's start with the 7th grade. Their enrollment is 520 out of 1170. So the fraction is 520/1170. Similarly, the 8th grade would be 650/1170.

Let me simplify those fractions. Starting with 520/1170.

<prompt_1.1> Simplify the fraction 520/1170. </prompt_1.1>

For the 8th grade, 650/1170. Let's divide numerator and denominator by 130.

<prompt_1.2> Simplify the fraction 650/1170. </prompt_1.2>
-----

Alternatively, another way to approach this is by using ratios. The ratio of 7th to 8th grade enrollments is 520:650.
Simplify that by dividing both by 130, which gives 4:5. So the ratio is 4:5.

Therefore, the total parts are 4 + 5 = 9 parts. The total representatives are 18, so each part is 18 divided by 9, which is 2. Hence, 4 parts would be 4*2=8, and 5 parts would be 5*2=10.
</Plan_1>

<Execution_1>
<execution_1.1>
Both numbers end with 0, so divide numerator and denominator by 10 first: 52/117. Hmm, can this be simplified further? Let's check. 52 and 117. The factors of 52 are 2, 4, 13, 26, 52. Factors of 117 are 3, 9, 13, 39, 117. They have a common factor of 13. So dividing both by 13: 52 ÷13=4, 117 ÷13=9. So 4/9.
</execution_1.1>

<execution_1.2>
650 divided by 130 is 5, and 1170 divided by 130 is 9. So that simplifies to 5/9.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, the 7th grade should have 4/9 of the 18 representatives, and the 8th grade should have 5/9.
Let me check if 4/9 +5/9 equals 1.
Yes, 9/9 is 1. That makes sense.

Now, calculating the actual numbers. For the 7th grade: (4/9) *18. Let's do that.
4 divided by 9 times 18. Alternatively, 18 divided by 9 is 2, multiplied by 4 is 8. So 8 representatives. For the 8th grade: (5/9)*18.
Similarly, 18 divided by 9 is 2, times 5 is 10. So 10 representatives.
-----

Based on execution_1.2:
Let's check if 8 +10 equals 18.
Yes, it does.

Wait, but let me just make sure I didn't skip any steps or make any miscalculations. Let me verify again.

<prompt_2.1> Verify the calculations of representatives using the simplified fractions. </prompt_2.1>
-----

Another check: 8 and 10 have a ratio of 4:5, which matches the student enrollment ratio.
So that's consistent.

So, the answer should be 8 for 7th grade and 10 for 8th grade.
Therefore, the 8th grade should have 10 representatives.

Based on execution_1.1, execution_1.2:
But wait, just to be thorough, are there other possible ways this could be approached? For example, using decimals instead of fractions. Let me try that.

<prompt_2.2> Verify the calculations of representatives using decimal approximation. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Total students: 520 +650=1170. 7th grade fraction: 520/1170=4/9. 4/9 of 18 is (4*18)/9=72/9=8. 8th grade: 650/1170=5/9. 5*18/9=90/9=10.
</execution_2.1>

<execution_2.2>
Calculating 520/1170. Let's divide 520 by 1170. Well, 520 divided by 1170. Let me compute that. 520 ÷1170. Let's divide numerator and denominator by 10: 52 ÷117. 52 divided by 117. Let me compute this as a decimal. 117 goes into 52 zero times. Add a decimal point, add a zero: 520 divided by 117. 117*4=468. 520-468=52. Bring down next zero: 520. So it's 0.444... repeating. So approximately 0.444. Similarly, 650/1170: 650 ÷1170. Divide numerator and denominator by 10: 65 ÷117. 117 goes into 65 zero times. Add decimal, 650 divided by 117. 117*5=585. 650-585=65. Bring down zero: 650 again. So 0.555... repeating. So approximately 0.555. 

So 0.444 *18. Let's compute that. 0.444*10=4.44, 0.444*8=3.552. Adding those: 4.44 +3.552=7.992. Which is approximately 8. Similarly, 0.555*18. 0.555*10=5.55, 0.555*8=4.44. Adding: 5.55+4.44=9.99. Approximately 10. So that confirms the earlier calculation.
</execution_2.2>
</Execution_2>

<Final_answer>
Therefore, the 8th grade should have $\boxed{10}$ representatives to ensure fair representation based on their student population.
</Final_answer>
</think>