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33% got 5 right

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Comments (5)

Hey Mike,
In your SAT Diagnostic Test Ques 19, I’m doing it as x-(x/3 + 20 + x/6 +10) = 76
This qives an answer of 212. But the answer given is D) 240. How?
Thanks

One third of the attendees at a concert left the show before the encore. Twenty more people left during the encore. After the encore, half as many people as had left before the encore left. If 76 people remained in the theater after the concert was over to try to get autographs from the band, how many people, in total, attended the concert?

The answer is (C), actually. The reason you’re not getting that is that you’ve got a 10 in your math that’s not necessary–you don’t need to cut that 20 in half because those people left during the encore, not before. Does that help?

For this problem: The equation of a certain parabola is f(x) = ax2 + bx + c, where a, b, and c are constants. If the minimum of the parabola is at x = 5 and its y-intercept is 12, then all of the following must be true EXCEPT…

Why can’t the answer be E? How do you know that b=0 for sure? Thank you! I am confused because -b/2a is the x value of the vertex and how do you know if b definitely equals 0?

Careful—you know that b definitely does NOT equal 0. If b were equal to 0, then the parabola would be of the form ax^2 + c, which would mean its vertex would be somewhere on the y-axis. To understand why that’s true, refer to the rules for graph translation. If f(x) = ax^2, then ax^2 + c is just f(x) + c—it can only move up or down.

## Comments (5)

Hey Mike,

In your SAT Diagnostic Test Ques 19, I’m doing it as x-(x/3 + 20 + x/6 +10) = 76

This qives an answer of 212. But the answer given is D) 240. How?

Thanks

One third of the attendees at a concert left the show before the encore. Twenty more people left during the encore. After the encore, half as many people as had left before the encore left. If 76 people remained in the theater after the concert was over to try to get autographs from the band, how many people, in total, attended the concert?

(A) 150

(B) 165

(C) 192

(D) 240

(E) 270

The answer is (C), actually. The reason you’re not getting that is that you’ve got a 10 in your math that’s not necessary–you don’t need to cut that 20 in half because those people left during the encore, not before. Does that help?

For this problem: The equation of a certain parabola is f(x) = ax2 + bx + c, where a, b, and c are constants. If the minimum of the parabola is at x = 5 and its y-intercept is 12, then all of the following must be true EXCEPT…

Why can’t the answer be E? How do you know that b=0 for sure? Thank you! I am confused because -b/2a is the x value of the vertex and how do you know if b definitely equals 0?

Careful—you know that b definitely does NOT equal 0. If b were equal to 0, then the parabola would be of the form ax^2 + c, which would mean its vertex would be somewhere on the y-axis. To understand why that’s true, refer to the rules for graph translation. If f(x) = ax^2, then ax^2 + c is just f(x) + c—it can only move up or down.

Ohh ok this makes so much more sense now! Thanks so much!!! I think I misread the question but now it makes sense :P.