Afaloo

1- Last week the Star Bakery made 3 kinds of cakes. One half of the cakes were made with 4 eggs each, two thirds of the rest of the cakes were made with 3 eggs each, and the remaining 54 cakes were made with 2 eggs each. What was the total number of eggs used to make all of these cakes?

2- square tiles measuring 1/2 foot by 1/2 foot are sold in boxes containing 10 tiles each. What is the least number of boxes of tiles needed to cover a rectangular floor that has dimensions 12 feet by 13 feet?
(A) 7
(B) 16
(C) 32
(D) 57
(E) 63
3- in the xy-plane, line L passes through the point (4,-5) and the vertex of the parabola with the equation: y=-2(x-2)^2+3. What is the slope of the line L?
(A) -4
(B) -1/4
(C) 0
(D) 1/4
(E) 4

rspence

1. Letting x be the number of cakes,
x/2 of the cakes had 4 eggs each
x/3 of the cakes had 3 eggs each (x/3 is "2/3 of the rest").
x/6 of the cakes had 2 eggs each (x/6 is the remaining part).

The question implies that x/6 = 54 --> x = 324. Now it's just arithmetic, which I'll leave up to you.

2. First we find the number of tiles. Since we can simply tile the whole floor without overlaps, the number of tiles is just (12*13)/(.5*.5) = 624 tiles. 624/10 = 62.4, but we round up to 63, E.

3. The vertex of the parabola occurs at x = 2, y(2) = 3. So line L passes through (2,3) and (4,-5). Find the slope.

Afaloo

@rspence, thank you for your time and response, but I have 2 problems:
1- How do I know the vertex of a parabola?
2- I did not quite understand the explanation for the first question. Could you please clarify it a bit more?

DrSteve

The following is right from my article "The Math Formulas You Should Memorize for the SAT."

The general form for a quadratic function is y = ax2 + bx + c.

The graph of this function is a parabola whose vertex has x-coordinate x = -b/2a

The parabola opens upwards if a > 0 and downwards if a < 0.

Here is an example.

Let the function f be defined by f(x) = -2x^2 - 3x + 2. For what value of x will the function f have its maximum value?

The graph of this function is a downward facing parabola, and we see that a = -2, and b = -3. So the x-coordinate of the vertex is x = 3/(-4) = -3/4 .


The standard form for a quadratic function is y - k = a(x-h)^2

The graph is a parabola with vertex at (h,k). Again, the parabola opens upwards if a > 0 and downwards if a < 0.

Here is an example.

Let the function f be defined by f(x) = 3(x - 1)2 + 2. For what value of x will the function f have its minimum value?

The graph of this function is an upward facing parabola with vertex (1,2). Therefore, the answer is 1.

Remark: Note that in this example k = 2, and it is on the right hand side of the equation instead of on the left.

rspence

The vertex of a parabola in the form y = ax^2 + bx + c occurs at x = -b/2a (this is easily proved using calculus by taking the derivative of both sides).

If the parabola in the form y = a(x-h)^2 + k, the vertex is at (h,k). Clearly, this must have some equivalence with the first statement, which it does (h = -b/2a). So the vertex of the parabola y = -2(x-2)^2 + 3 is (2,3).

Also, 1/2 of the cakes had 4 eggs each. How much is two thirds of the rest? Here, "the rest" is simply the remaining 1/2, and two thirds of 1/2 is 1/3. So 1/3 of the cakes had 3 eggs each. The remaining 54 cakes had two eggs each, 1 - 1/2 - 1/3 = 1/6, so 54 cakes comprises 1/6 of the cakes.

Agnes1025

about vertex of a parabola...
y=ax²+bx+c (a≠0)
=a[x²+(b/a)x+(b/2a)²-(b/2a)²+(c/a)]
=a[(x+b/2a)²-(b/2a)²+(c/a)]
=a(x+b/2a)²-(b²-4ac)/4a
when x=-b/2a, y=(4ac-b²)/4a (max/min)

rspence

*Optional* Another derivation of the vertex of a parabola.

y = ax^2 + bx + c. In calculus there is a certain operator called a "derivative" that basically tells you how fast a function value is changing with respect to y. For 2D functions, this is equal to the slope of a tangent line through that point (for continuous functions) and is usually denoted dy/dx for this problem (some texts use y', other notations as well).

For this parabola, dy/dx = 2ax + b. The vertex of the parabola occurs when the slope of the tangent line at a point (x0, y0) is zero. If we set dy/dx = 2ax + b = 0, we obtain one solution, x = -b/2a, which turns out to be the x-coordinate of the vertex of the parabola.

The advantage of knowing the derivative is that this works for higher-degree polynomials, as well as any differentiable function in general. But you won't need this for the SAT...

RandomHSer

A non-calculus derivation of the vertex
let f(x)=ax^2+bx+c. We wish to see if a value k exists such that f(k+x)=f(k-x) for all x. If the graph has symmetry about k, then k must be either a max or min. Doing some algebra we get
a(x+k)^2+b(x+k)+c=a(k-x)^2+b(k-x)+c, cancelling c's and expanding:
ax^2+2akx+ak^2+bx+bk=ak^2-2akx+ax^2+bk-bx. We can cancel ax^2,ak^2, and bk
leaving us with 2akx+bx=-2akx-bx, 4akx=-2bx, k=(-2bx)/(4ax) so k=-b/(2a), the result rspence got using calculus.

rspence

^ The second question on that list, how is that in any way tough for an SAT? Ans is just 4! = 24.

GaleSkylex

3. If you know what a vertex is, i.e the lowest or highest point, then you should know that what you're looking for is a highest or lowest value for Y. For this sort of a value to exist there must be some sort of a limit set, and these are mostly set by what is being added or subtracted. Note that 3 is being added in the equation, so now you just need to figure out if there is a highest or lowest for the other part. X-2 becomes 0 if you put X=2, and thus that whole part becomes 0 (and you get y=3). Then to make sure this is the value check if you can get a lower value if you use a different value of X, a highest since the whole variable part if being squared it is impossible for the that side of the equation to equal a negative number, thus it can be assumed that whatever value of X is given the value of Y will not drop below Y.

Hope this helps.

pckeller

For the record, I have never seen an SAT problem that required a student to find the vertex of a parabola in standard form. So if you know that x=-b/2a it can't hurt but is not necessary. Same thing for the calculus.

What you DO want to be able to do is look at a parabola such as: y = -2(x-2)^2 + 3

and see that compared to a basic y=x^2 parabola...

i. it is "upside down" because of the -2 in front
ii. it has been shifted two units to the right because of the (x-2)
iii. it has been shifted up 3.

So the vertex is a (2,3 )

If anyone has an example of a college board SAT problem that requires you to find the vertex by using -b/2a, calculus, completing the square, etc, I'd like to see it.

bhph11

i found the easiest way to do it is factor -2(x-2)^2+3 out and get -2x^2+8x-5. plug into graphing calculator and you get a downward facing parabola with vertex (2,3) simply plug in the formula rise/run=slope and you get (3+5)/(2-4) (y1-y2)/(x1-x2) you get the answer as a the slope of line l = -4. hope that helps

bhph11

2. is quite simple the area of the rectangular floor is length x width. 12 x 13 = 156 feet if each tile is 1/2 sq foot then double the area is needed in tiles (set up a proportion if you're unsure) multiply 156 x 2= number of tiles, then divide by the amount per package (10) to receive the number of packages (32)

bhph11

in #1 you are left with a remainder that is a fraction of the whole. 1/2 of the total is 4 eggs, 2/3 of the remaining half is 3 eggs, and the remainder of the rest is 54 cakes with 2 eggs. i split the fraction into 6ths and was left with 3/6 of total is 4, 2/6 of total is 3, 1/6 of total =54. multiply 54 x 6 to receive the total number of eggs. after that the answer is arithmetic (162 x 4) + (108 x 3) + (54 x 2)=1080 eggs total