HI Mike…Thanks for all your help ! Here’s another question

HI Mike…Thanks for all your help ! Here’s another question:

When a buffet restaurant charges $12.00 per meal, the number of meals it sells per day is 400. For each $0.50 increase to the price per meal, the number of meals sold per day decreases by 10. What is the price per meal that results in the greatest sales, in dollars, from meals each day?

A) $16.00
B) $20.00
C) $24.00
D) $28.00

QAS March 2021, Section 3, # 20.

Hi Mike… QAS March 2021, Section 3, # 20. Since “no solution,” can you consider each side a separate parallel line & use corresponding coefficients to confirm that k must equal 1/2? Or, what’s the best way to solve?

1/2x + 5 = kx + 7

In the given equation, k is a constant. The equation has no solution. What is the value of k?

March 21 QAS Sec 3 #11

Hi Mike …can you solve & explain? (from QAS March 2021)
Section 3; #11.
y = (x-1)(x+1)(x+2)
The graph in the xy plane of the equation above contains the point (a,b). If -1 < or = a < or = 1, which of the following is NOT a possible value of b?
A) -2
B) -1
C) 0
D) 1

In this equation, k is a constant…

Some help please, Mike?

x^2 – 12x +k = 0 In this equation, k is a constant. For which values of k does the equation have only one solution? I know I can set the discriminant to zero and solve for k. But is there another way to solve? Thanks!

Test 7 Section 4 Question 6

Hi Mike…SAT 7, Section 4, Q6: I now see the shortcut here (that both sides of the equation are perfect squares,) but if I did expand and FOIL the left side, wouldn’t I still get the correct “a” values even though it takes longer? I can’t get it to work !! Can you please show the alternate path math steps? Or is recognizing the perfect squares the ONLY way to solve this one ? Thanks!

SAT 8, Section 3, Number 7

Hi Mike, I’m asking about SAT 8, Section 3, Number 7. Is it always best to immediately plug in answer choices on questions like this? For the algebraic practice, I rewrote the equation as a quadratic and solved for (x=5) and (x= -1). Then sub’d each value back into the given equation to find that only (x=5) worked. OK…but what a time-killer at #7 out of 20. Any other solution path to consider? Thanks!