So I got the Q & A services after taking the May SAT and I’m currently reviewing all the problems I got wrong, some answers I look at the answer and understand how they go there and the small mistake I made but others I have no clue how they got their answer. Can someone explain to me how to get the answer to the following:
x I f(x)
0 l 18
2 l 12
3 l 9
7 l -3
Some values of the linear function f are shown in the table above. What is the value of f(6) ?
A) 0
B) 1
C) 2
D) 3
E) 4
Much appreciation,
ECP
f(6) = 0 --> A.
A quick way is to note that f(0) = 18, f(3) = 9, and 3 is halfway between 0 and 6. Because f is linear, f(3) is also halfway between f(0) and f(6).
You know the function is linear, so it has to have a constant slope and is in the form of mx+b. Since f(0) is 18, we can assume that the function is of the form f(x) = mx + 18. Since we know that f(2) = 12, and we know that f(x) = mx+18, plug in 2 for x and then solve the equation. You get m = -3. From there just plug in 6 for x, getting f(6) = 0.
Alternatively, just see that for each time x increases by 1, f(x) or y decreases by 3. Intuitively it should make sense then that if you start at 18 and decrease by 3 each time, when you increase x by 6 you decrease f(x) by 18, giving you f(x) = 0.
Edit: Damn, @MITer94 – you beat me by two minutes 
Haha thank you guys, both were very helpful!
@MITer94 @micmatt513 What about this one. A fence consist of sections built in a row. Each section is identical and consist of two hoizontal boards between consecutive vertical posts. Each posts cost $8 and each board costs $6. i the toal cost of the boards and posts was $208, how many posts are in the fence?
A) 10
B) 11
C) 20
D) 26
E) 31
I assumed this was a system of operations and went about this by trying to isolate a variable and so I “found” B and tried to plug it back in but it took me in a circle. So, how should I go about this one?
I’m not sure I understand what this question is asking.
Is it supposed to be like |- -|- -|- -| for the boards, or is it supposed to be like |-|-|-|.
I was going to do this a different way (more algebraically at first), but this is the easier way. I’m assuming it’s the first diagram btw.
The first complete section costs $28 ($8 + $6 + $6 + $8). After that, each subsequent section costs $20 because you only need 1 post and 2 boards. Subtract the $28 from the first section and you get that you have $180 left. That’s 9 extra sections. Since you already had 2 posts and now you’ve added 9 more posts, you get 11 total. Hope that explains it well enough!
Algebraic solution:
If there are p posts, then there are 2(p-1) boards. Then 8p + 6*2(p-1) = 208 <==> 20p - 12 = 208 <==> p = 11.
Thanks @MITer94 and @micmatt513, yeah it was very similar to the first diagram actually it was more like this l=l=l a *(but that docen’t affect the answer it seems. The answer indeed was 11! Thanks a ton, you two are da bomb 