Posts filed under: Q and A

A memory chip is designed to hold a number of transistors and heat sinks. There must be at least 1 heat sink for every 2000 transistors to prevent overheating. Also, each transistor has an area of 2.0 x 10^-10 mm^2, each heat sink has an area of 3.6 x 10^-6 mm^2, and the total area of transistors and heat sinks must be at most 2 mm^2. What is the approx. max. # of transistors that the chip can hold according to this design?
A. 2.78 x 10^2 B. 5.56 x 10^5 C. 1.0 x 10^9 D. 1.0 x 10^10
Ans=C. S.O.S, PLZ!

Good grief! FWIW, this is one of the Khan Academy questions that really misses the mark for me in terms of similarity to official test questions.

Anyway, let’s translate this into math. We’ll use x for the number of heat sinks and y for the number of transistors.

“There must be at least 1 heat sink for every 2000 transistors…”

    \begin{align*}\dfrac{1}{2000}&\leq\dfrac{x}{y}\\\\y&\leq 2000x\end{align*}

“each transistor has an area of 2.0 x 10^-10 mm^2, each heat sink has an area of 3.6 x 10^-6 mm^2, and the total area of transistors and heat sinks must be at most 2 mm^2”

    \begin{align*}2\times 10^{-10}y+3.6\times 10^{-6}x&\leq 2\\\\2\times 10^{-10}y&\leq -3.6\times 10^{-6}x+2\\\\y&\leq\dfrac{-3.6\times 10^{-6}}{2\times 10^{-10}}x+\dfrac{2}{2\times 10^{-10}}\end{align*}

Use your calculator to evaluate those fractions…

    \begin{align*}y\leq -18,000x+10,000,000,000\end{align*}

Now, some logic. If we’re trying to maximize the number of transistors, then we’re definitely going to use all the available space we have, and the minimal amount of heat sinks, right? So despite the wording of the question, we can avoid persisting with inequalities and just use equalities! (Another way to think about this: if we have y\leq \text{something} and we’re trying to maximize y, we can change it to y=\text{something}!)

    \begin{align*}y&=2000x\\y&= -18,000x+10,000,000,000\end{align*}

So let’s substitute.

    \begin{align*}2000x&= -18,000x+10,000,000,000\\20,000x&= 10,000,000,000\\x&= 500,000\end{align*}

Now plug that back into the first equation to get y:

    \begin{align*}y&=2000x\\y&=2000(500,000)\\y&=1,000,000,000\\y&=1\times 10^9\end{align*}

What’s the fastest way to do question 9 from practise test 2 section 3?


The fastest way might be to draw it. The first line goes through (1, 8) and has a slope of 2, which means it goes down 2 units and left 1 unit to (0, 6). The second line goes through (1, 2) and (2, 1), which means it has a slope of –1. Your drawing might look like this. (It’s OK that it’s ugly! You don’t need to draw it super carefully—just accurately enough to be able to trace the lines.)

From there, you can pretty easily use the slopes to trace the lines back to their intersection! The first line will go from (0, 6) to (–1, 4). The second line will go from (1, 2) to (0, 3) to (–1, 4). And there you have it—the intersection is (–1, 4)! Since the question asks for the sum of the coordinates, the answer is –1 + 4 = 3.

How would you do practise test 3, section 3, question 16? I plugged in 1 and happened to get the answer. But is there an algebraic way to do it?


Sure. Algebraically, you can do it in a few steps:


You can actually go further by expanding the differences of two squares if you want (below), but because the question tells you that x> 0, you know from the above that it can either be 1 or 2.


Hi Mike. Just wondering if you had video solutions for the algebraic manipulation practise questions?

I don’t have them currently, but if they’d be helpful I can prioritize doing them soon.

What’s the best and fastest way to do practise test 3, section 3, question 15? How would you interpret the equation? (which I find very confusing and difficult)

Thank you

If the equation is confusing for you, try plugging in! The three statements you’re evaluating are about 1 degree increases, so plug in 0 and 1 for F and see what happens to C. Then plug in 0 and 1 for C and see what happens to F.



That tells you that a 1 degree increase in F results in a \dfrac{5}{9} increase in C.



Because \dfrac{9}{5}=1.8, that tells you that a 1 degree increase in C results in a 1.8 degree increase in F.

That’s enough to choose choice D.

Hello, I know you’ve already solved practise test 2, section 4 question 29 (by either using your graphic calculator or by looking at the equation of a parabola) but how would you use backsolving? Lets say I try in option C and im getting y as 3 (which means my equations do NOT have 2 real solutions), how do I know whether to try out option B or D next?

Thank you so much!

You wouldn’t really know whether to try B or D next without knowing that you’re playing with whether the parabola opens up or down and its y-intercept when you change a and b, respectively. This is important, though: don’t waste more than a couple seconds trying to figure out which way to go! If you don’t know which choice to try next right away, pick any one. If you’re backsolving simply by changing a graph on your calculator, it shouldn’t take more than a couple seconds to try each choice until one works.

Hi Mike! For practise test 2, section 3 Q6, how exactly could I use backsolving to solve this? Lets say I start with C and I plug in 8. My gradient of line l is 2/5. If I plug in p as 8, I’m getting gradient of line k as 4/8. Do I now compare the fractions? How do I know if I should try plugging in a bigger or smaller number to get closer towards 2/5 (initial gradient)?


Right, you’re comparing the fractions. (Of course, this requires you to know that parallel lines have the same slope.)

To decide which direction to go, recognize that \dfrac{4}{8}=\dfrac{1}{2} is bigger than \dfrac{2}{5}. You need your fraction to get smaller. Since the value you’re playing with is p, which is in the denominator, you make the fraction smaller by increasing the denominator.

A couple notes:

  1. Generally speaking, don’t spend too much time trying to figure out which answer choice to try next if it’s not obvious. It’s probably faster just to guess at which direction to go and see if you get closer or farther away from where you want to be!
  2. Just because I say a problem CAN be solved by backsolving doesn’t mean I’m saying that backsolving is the best way for that problem. I just want you to train yourself to recognize how often it’s possible to backsolve so that when you encounter a problem that can be backsolved on test day, you’re not blind to the possibility.

Hi. Please, explain the solution to this question:

If the slope of a linear equation f(x) is 2 and the y intercept is -2, then what is the y intercept of 3f(x-2)-3

If you know the slope and the y-intercept, then you know all you need to put the f(x) into slope-intercept form (ymxb, where m is the slope and b is the y-intercept). In this case, you have f(x)=2x-2.

To go from there to 3f(x-2)-3, your first step should be to get the f(x-2) out of there. To do that, just put in x-2 everywhere the original function has x:

If f(x)=2x-2, then f(x-2)=2(x-2)-2, which simplifies to f(x-2)=2x-6.

Once you’ve got that, substitute and simplify:


That’s in mxb form, so the y-intercept is –21.



Hi. Please can you explain test 3,section 4, number 35? Thank you

Just did that one the other day! See here.

official practice test 3 section 4 how to solve?

The question tells you that the store’s average after 10 ratings is 75. That means that the first 10 ratings have a sum of 750, because \text{average}=\dfrac{\text{sum}}{\text{number of values}}, \text{average}\times\text{number of values}=\text{sum}. In this case, 75\times 10=750.

You need the first 20 ratings to have an average of 85, which means you need the sum of the first 20 ratings to be 20\times 85=1700. So far so good?

Now, we know that the first 10 ratings added to 750, so we need the next 10 ratings to add to 1700-750=950.

Since we want to know the smallest possible value for the 11th rating, let’s assume that the 12th-20th ratings are the highest they can be, 100. In other words, let’s assume there are 9 perfect 100 ratings, and set the 11th rating equal to x.



Hi Mike. I bought your math guide about a month ago, and it’s great! I do have a question about the daily PWN #5. The explanation says that solutions exist in all quadrants except quadrant IV. But how this helps you come up with the conclusion that a>0 and b>0?

Thank you!

Here’s a link to that question.

The conclusion isn’t that a > 0 and b > 0. The conclusion is that IF a > 0, THEN b > 0. That’s a really important difference! Here’s the same graph from the solution to that question, but with a little added detail (in yellow):

The full solution set is that purple-ish region that is the overlap of the blue and the red. The part highlighted in yellow is the only part of the solution set where the x-values are positive. Therefore, you know that if (ab) is in the solution set and IF a is greater than zero, THEN the value of b MUST also be greater than zero.

Does that help?

Hello, I am having trouble with the explanation for Angles, Triangles, and Polygons problem #2 on page 343. To get the answer to this problem, I multiplied 180*4, as there are four sets of angles that each make up a straight line, and the question asks for their sum. The solution seems to be getting at this, but instead of saying “180° x 4=720°” it says “180°.4 x 180° = 720°”. (I can send you a picture if you want.) Is this a typo, or am I misunderstanding?

That’s not a typo—the period you see after 180° is the end of the sentence. Then the next sentence is all math: 4\times 180^\circ=720^\circ. Does that help?

Hello, I am a little confused on the solution given on page 332 for problem #6 of “Percents and Percent change”. The first two steps to solving the problem are listed as

1. 7=x/100*56

2. 5/56=x/100

I’m not sure how to get from step 1 to step 2 – or where the 7 in step 1 came from. It seems to me that the 7 in step 1 should be a 5 – as the problem asks for babies born on Thursday, not Tuesday. Is this a typo, or am I missing something?

You’re not missing anything—that’s totally a typo! The 7 in the first step should be a 5.

…so embarrassed. :/

Hi, Mike! Can you explain the second way we can approach question number 6 from the Parabola chapter? (The two points where the higher y-coordinate is also farther from the line of symmetry.) It would be great if you can provide an example. Thank you.

Sure. Here’s the question:

The key, as I say in the solution, is to look for points that are consistent with what you know about the parabola given the equation: that it opens upwards and that its line of symmetry is at x = 6. Because parabolas get wider and wider, points in a parabola that opens up must get higher and higher the farther they get from the line of symmetry.

Therefore, you’re looking for either

  1. Two points with the same y-coordinate that are the same distance from the line of symmetry, or
  2. Two points that are otherwise consistent with the shape of an upwards-opening parabola that’s symmetrical about x = 6.

Choice D does the first one—x = 2 and x = 10 are each 4 away from the line of symmetry at x = 6, and both points have the same y-coordinate of –5.

No choice satisfies the second one (otherwise there’d be more than one possible right answer!) but a choice like (4, 10) and (9, 15) would. (4, 10) is closer to x = 6 and has a lower y-value; (9, 15) is farther from x = 6 and has a higher y-value:

How would you do question 11 from practise test 3, the no calculator section? (using the plugging in method?)

Thank you!

It involves a little trial and error. The basic idea here is that it’s easier to work with real numbers and the geometrical constraints (i.e., vertical angles must be congruent, straight lines must equal 180°) than it is to write a bunch of equations. Note that my figure isn’t to scale, but then again, neither is the one in the original question.

Start by plugging in for y, which means you are also plugging in for u because of vertical angles.

Now you’re going to plug in for x, which means you’re also going to be picking t (again, because of vertical angles). But wait! The question also tells you that xyuw. Given what we’ve already plugged in, that means:

x + 100 = 100 + w
= w

So when you’re picking x, you’re not only picking t, but you’re also picking w and z! Better pick something that’s going to make all those lines add up to 180°!

So there you have it. Once you chose a number for y, all the rules in the question gave you no choice for any other variable.