Okay, let's see. The problem is about fairly representing the 7th and 8th grades on the Student Council. The enrollments are 520 for 7th and 650 for 8th. There are a total of 18 representatives, and we need to figure out how many should come from 8th grade. Hmm, fair representation usually means that the number of representatives is proportional to the size of the group, right?

First, maybe I should check if the total number of students is needed here. Let's add the two enrollments: 520 + 650. That equals 1170 students total. So there are 1170 students across both grades. The total representatives are 18. So if we want proportional representation, the number of representatives each grade gets should be based on their proportion of the total student body.

So for 7th grade, the proportion would be 520/1170, and for 8th grade, it's 650/1170. Then multiply each proportion by the total number of representatives, which is 18. Let me calculate that.

First, simplify the fractions. 520 divided by 1170. Let's see, both numbers can be divided by 10, so that's 52/117. Hmm, 52 and 117. Let's check if they have a common divisor. 52 is 2*26, 117 is 13*9. No common factors, so 52/117 is the simplified fraction. Similarly, 650/1170 simplifies to 65/117, since both divided by 10 again. Wait, 650 divided by 10 is 65, 1170 divided by 10 is 117. So 65/117. Wait, 65 and 117 have a common factor? 65 is 5*13, 117 is 9*13. Oh, they both have 13. So divide numerator and denominator by 13: 65 ÷13=5, 117 ÷13=9. So 5/9. So 65/117 simplifies to 5/9. Similarly, 52/117: 52 and 117. 52 is 4*13, 117 is 9*13. So divide both by 13: 52 ÷13=4, 117 ÷13=9. So 4/9. So 52/117 is 4/9, and 65/117 is 5/9. That makes sense because 4/9 +5/9 =9/9=1, which is the total proportion.

So if we apply these fractions to the total representatives, which is 18, then 7th grade should get 4/9 of 18, and 8th grade 5/9 of 18. Let's compute that. 4/9 *18 = (4*18)/9 = (4*2)=8. Similarly, 5/9 *18 = (5*18)/9 =5*2=10. So 7th grade would have 8 representatives, 8th grade 10. That adds up to 18, which is correct.

Wait, but let me verify this. Let's check if 8 +10 equals 18, which it does. And the fractions of the total enrollment. So 520/1170 *18 = (520*18)/1170. Let's compute that. 520*18=9360. 9360 divided by 1170. Let's divide numerator and denominator by 10: 936/117. 936 ÷117: 117*8=936. So 8. Similarly, 650*18=11700. 11700 ÷1170=10. So yes, that's correct. So 8 and 10. Therefore, the 8th grade should have 10 representatives.

But wait, maybe there's another way to approach this problem. For example, using ratios. The ratio of 7th to 8th grade enrollments is 520:650. Let's simplify that. Divide both by 130: 520 ÷130=4, 650 ÷130=5. So the ratio is 4:5. Therefore, the number of representatives should be in the ratio 4:5. So total parts are 4+5=9. 18 representatives divided by 9 parts is 2 per part. Therefore, 4 parts for 7th grade: 4*2=8, and 5 parts for 8th grade:5*2=10. Same result. So that's another way to see it.

Alternatively, if we didn't simplify the ratio, we could have set up the proportion as 520/650 = x/18, where x is the number of representatives for 8th grade. Wait, no. Wait, actually, if we let x be the number for 8th grade, then 7th grade would have 18 -x. The ratio of their enrollments should equal the ratio of their representatives. So 520/650 = (18 -x)/x. Solving for x. Let's see:

520/650 = (18 -x)/x

Cross-multiplying: 520x =650(18 -x)

520x =11700 -650x

520x +650x =11700

1170x =11700

x =11700 /1170=10

So again, x=10. So that's another method. So three different ways all leading to 10. So that seems solid.

Therefore, the answer should be 10 representatives for 8th grade.
Final Answer:
The 8th grade should have \boxed{10} representatives.
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