Posts tagged with: Plug In

Could you please explain Test 2 Calculator Active #25

Sure thing. The best way to see what’s going on with this one is to plug in. When they tell you that ab = 0 and a ≠ b, they’re telling you that a and b must be nonzero opposites (i.e., that a = –b and that a ≠ 0). So just pick any nonzero number and rock and roll!

I’ll pick 3 for a, which means I’m picking –3 for b. Therefore, the graph passes through (3, 0) and (0, –3). What do we know about its slope? (Don’t just imagine it in your head—draw it in your test book!)

That’s a positive slope! Conveniently, finding a positive slope for even one set of values of a and b is enough to eliminate choices B, C, and D: a positive slope is not negative, zero, or undefined!

If you’re nervous though, maybe see what happens if you set a = –3 and b = 3 instead. Then your line would go through (–3, 0) and (0, 3). That’s still a positive slope!

Hi Mike, Can you work the solution for Test 7, Section 3, #13? Thanks!

Absolutely. First of all, this is a great one to plug in on. Equivalent means they’ll be the same for all values of x, so pick a value of x that’s easy to work with and get crackin’. I’m going to use x = 4, because that’ll make the denominator (x – 3) equal to 1.

The original expression:

    \begin{align*}&\dfrac{x^2-2x-5}{x-3}\\\\=&\dfrac{4^2-2(4)-5}{4-3}\\\\=&\dfrac{16-8-5}{1}\\\\=&3\end{align*}

Now, which answer choice gives you 3 when you plug in 4 for x? You can cross off A and B right away because they’re going to be negative—they start with x – 5! So try C and D:

    \begin{align*}\text{C)}\qquad &x+1-\dfrac{8}{x-3}\\\\=&4+1-\dfrac{8}{4-3}\\\\=&-3\\\\\text{D)}\qquad &x+1-\dfrac{2}{x-3}\\\\=&4+1-\dfrac{2}{4-3}\\\\=&3\end{align*}

Boom! D it is.

If you don’t want to plug in, you’re going to need to do polynomial division. (In other words, you should want to plug in!)

Polynomial long division is a real pain to render in text, so please forgive the handwriting.

Step 1: You’re always looking at the lead terms in long division. The leading term of the divisor, x, goes into the leading term of the dividend, x^2, exactly x times. Therefore, put an x at the top, and then subtract the product of x and x-3 from the dividend:

Step 2: x goes into x exactly 1 time, so put a + 1 on top and then subtract the product of x-3 and 1 from the x-5 we’re working with:

Step 3: Now you’re done, because x can’t go into –2. What’s on top is your quotient, and what’s left on bottom is your remainder.

The SAT will often show quotients and remainders as it does in the answer choices of this question: in the \text{QUOTIENT}+\dfrac{\text{REMAINDER}}{\text{DIVISOR}} form. In other words, x+1-\dfrac{2}{x-3} is the answer we’re looking for.

Could you solve Test 6, Section 4, number 18 for me? Thanks!

Sure thing. You’ve got your riser-tread formula: 2hd = 25. Now you’re told that d must be at least 9 and h must be at least 5. Translating that into math: d ≥ 9 and h ≥ 5.

Of course, since 2hd must always equal 25, there are also upper bounds on the values of d and h—they can’t be too small, but they can’t be too big, either. In fact, as one gets bigger, it forces the other one to get smaller.

Let’s plug in 9 for d (that’s its minimum value) and see what we’d get for h:

2hd = 25
2h + 9 = 25
2h = 16
h = 8

What would happen if we plugged 10 in for d instead?

2h + 10 = 25
2h = 15
h = 7.5

Ahh, see? If d gets bigger, it makes h smaller. So the BIGGEST value we can have for h is 8 if we have a minimum value for d at 9.

We now know that h must be at least 5, but also cannot be greater than 8: 5 ≤ h ≤ 8 is the answer.

 

Practice test 2, section 3 Q10, the answer is C (which you can easily obtain by plugging in x = 0) but is there another way to do or it some other logic you can use?

Thank you!

Plugging in zero is smart! The “other logic” way is to recognize that although all choices have a –2 in them, in only one choice is that –2 within parentheses so that the exponent applies. (x-2)^2 will never be negative.

Knowing your function shift rules helps here: (x-2)^2 is just the x^2 parabola shifted 2 units to the right. Of course, the minimum value of x^2 is zero; shifting it 2 units right won’t change that minimum.

Could you help me with question 9 on page 38 please? I don’t understand how to solve it even using back-solving

Yeah, this one is pretty tough! 🙂

To backsolve it successfully, it’s helpful to also plug in. Say that the base and height of triangle A are both 10. Then if you try, say, answer choice C, which says that p = 23, you know that r = 18, and therefore that the base of triangle B is 12.3 (23% more than 10) and the height is 8.2 (18% less than 10). Does that give you the same area for both triangles?

\text{Area}_A=\dfrac{1}{2}(10)(10)=50

\text{Area}_B=\dfrac{1}{2}(12.3)(8.2)=50.43

Not quite, but you’re close! For that reason, probably makes sense to try the next closest choice, D. Sure enough, that one works.

I sorry I am very confused about why the answer question 5 on page 28 is A would you help me please?

Sure. Here’s the question:

This question appears in the plugging in chapter, so the solution (page 301) approaches it that way. I also plug in when I solve this question in the video solutions available to Math Guide Owners. So in this post, I’ll approach the question algebraically.

If you use the Pythagorean theorem, you know that:

    \begin{align*}3^2+a^2&=AC^2\\9+a^2&=AC^2\\\sqrt{9+a^2}&=AC\end{align*}

Since the value of \sin A will be opposite leg over hypotenuse, it will be equal to \dfrac{a}{AC}. Substituting for AC based on the Pythagorean theorem calculation above gives you choice A.

    \begin{align*}\sin A&=\dfrac{a}{AC}\\\\\sin A&=\dfrac{a}{\sqrt{9+a^2}}\end{align*}

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.

C=\dfrac{5}{9}(0-32)\\\\C=\dfrac{-160}{9}

C=\dfrac{5}{9}(1-32)\\\\C=\dfrac{-155}{9}

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

0=\dfrac{5}{9}(F-32)\\\\0=F-32\\\\32=F

1=\dfrac{5}{9}(F-32)\\\\\dfrac{9}{5}=F-32\\\\\dfrac{9}{5}+32=F

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.

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:

xyuw
x + 100 = 100 + w
x
= 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.

For the official practise test 2, section 3, question 4 is listed as a “plugging in” question.
So when I tried it out, I substituted 1 for both variables a and b. That easily allows me to pick option A.
However, one of your rules was to never plug in the values 0 and 1, yet 1 was a viable option for this question.
So how can I know when plugging in 1 will be okay and when it’s a bad idea? Also, is my method correct for this question or would you suggest another way? Thanks so much!

The reason you generally want to avoid plugging in 1 or zero is that doing so may result in more than one answer being correct. This is also the reason you generally should not plug the same number in multiple times in the same question. As you point out, though, you got away with doing both things here: setting both a and b equal to 1 in this case worked very nicely, eliminating all choices but the right one.

Recognize that plugging in other numbers would also work here—that’s the beauty of plugging in. In fact, you could even have plugged in the other number to avoid, zero, for one variable and gotten away with it! (Plugging in zero for both variables wouldn’t eliminate any choices, though.)

As for how you can know when plugging in 1 will be OK, if you want to plug in 1 just make sure you check every answer choice (which you should be doing anyway when you plug in). If plugging in 1 left more than one choice, plug in a new number and try the choices you didn’t eliminate again.

Hi Mike, I find these Qs confusing and I lose valuable time trying to think my way through them (usually get them wrong anyway!) What’s a good stepwise approach? Thanks!

If 6 < |x-3| < 7 and x < 0, what is one possible value of |x| ?

I think it’s helpful on questions like this to plug in—to think in terms of actual numbers as much as possible. In this case, I’d look at the inequality, which says that |x-3| must be between 6 and 7, and I’d simplify that by saying that 6.5 is between 6 and 7, so why not just say |x-3|=6.5?

From there, you can say that x-3=6.5, in which case x=9.5, or x-3=-6.5, in which case x=-3.5. Since the question says that x<0, we must choose x=-3.5, which means |x|=3.5. So that’s what we grid.

Algebraically this is a bit trickier, but if you want to see a step-by-step solution, you have to begin by noting that removing the absolute value brackets from the original inequality range results in two possible inequality ranges:

    \begin{align*}6&<x-3<7\\&-or-\\-6&>x-3>-7\end{align*}

Again, we know that x<0, so we only need to deal with the second inequality. Solve for x by adding 3 to each part:

    \begin{align*}-6+3&>x>-7+3\\-3&>x>-4\end{align*}

So x must be between –3 and –4. That means the absolute value of x must be between 3 and 4.

Can you do Test 6 Section 3 #15?

Sure can. I’ll even do it two ways. First, I’ll plug in: let’s say a=3 and b=4. Then we can say that \left(a+\dfrac{b}{2}\right)^2=\left(3+\dfrac{4}{2}\right)^2=25. When you plug those values into the answer choices, which one gives you 25? Only choice D will: a^2+ab+\dfrac{b^2}{4}=3^2+(3)(4)+\dfrac{4^2}{4}=9+12+4=25.

To get this algebraically, all you need to do is FOIL. Just be super careful with the fraction.

    \begin{align*}&\left(a+\dfrac{b}{2}\right)\left(a+\dfrac{b}{2}\right)\\\\=&a^2+\dfrac{ab}{2}+\dfrac{ab}{2}+\left(\dfrac{b}{2}\right)^2\\\\=&a^2+\dfrac{2ab}{2}+\dfrac{b^2}{4}\\\\=&a^2+ab+\dfrac{b^2}{4}\end{align*}

PSAT/NMSQT practice test #1 section 4 question 18

This is a great one to plug in on. Say the height of the trapezoid is 2, the lower width is 5, and the upper one is 3. That way, assuming you don’t know the formula for the area of a trapezoid (which exists—see below—but which you don’t need to memorize for SAT purposes) you can break the trapezoid neatly into two right triangles and a rectangle thusly (note that my drawing is not to scale):

The rectangle has an area of 3\times 2=6, and the triangles each have an area of \dfrac{1}{2}(1)(2)=1. Total area of the trapezoid: 1+6+1=8.

Now if you do the manipulations the question asks you to do, the height gets cut in half and the bases are doubled:

The areas of the triangle are still \dfrac{1}{2}(2)(1)=1. The area of the rectangle is also unchanged: 6\times 1 = 6. Therefore, the area of the trapezoid is still 1+6+1=8.

The answer is C: the area does not change.

Since this post will be a reference point now, you can also do this using the area of a trapezoid formula:

A=\dfrac{b_1+b_2}{2}h

If you double both bases and cut the height in half, you get:

A=\dfrac{2b_1+2b_2}{2}\dfrac{h}{2}

A=\left(b_1+b_2\right)\dfrac{h}{2}

Of course, that’s equivalent to the original formula. Therefore, doubling the bases and halving the height won’t change the area of any trapezoid.

I want to stress, though, that the lesson you should take from this question is not that you need to memorize the trapezoid area formula. Rather, the lesson should be that you can 1) plug in, and 2) break more complex shapes into things like triangles and rectangles that you already know how to work with.

I’m not sure you are answering questions about Practice Test 6 yet. If so, would you please work # 12 in section 3 of that test?

Sure I am—I just haven’t been tracking those answers yet. I’ll get that set up soon.

This one is a straight up plug in question. Say x = 2:

    \begin{align*}&\dfrac{4x^2+6x}{4x+2}\\=&\dfrac{4(2)^2+6(2)}{4(2)+2}\\=&\dfrac{16+12}{8+2}\\=&\dfrac{28}{10}\\=&2.8\end{align*}

Which answer choice simplifies to 2.8?

Obviously not A, B, or C (C is 2 minus something positive, so it can’t be 2.8).

Confirm D:

    \begin{align*}&2+1-\dfrac{2}{4(2)+2}\\=&3-\dfrac{2}{10}\\=&3-0.2\\=&2.8\end{align*}

You could also do the long division…but why? 🙂

That tells you that the quotient is x + 1 with a remainder of 2, just like choice D says.

how can you use plug-in for #13 on page 607?

To get this one by plugging in, pick values for all the variables on the right side of the equals sign: tv, and k. For example, say t = 2, v = 3, and k = 5. Use your calculator to see what that gives you for h:

    \begin{align*}h&=-16(2)^2+(3)(2)+5\\h&=-53\end{align*}

Now plug –53 in for h, 2 in for t, and 5 in for k in each answer choice. The one that gives you 3 (our original value for v) is the right answer.

Choice D does the trick:

    \begin{align*}v&=\dfrac{h-k}{t}+16t\\3&=\dfrac{-53-5}{2}+16(2)\\3&=-29+32\\3&=3\end{align*}

Note that I’m not saying you need to do the question this way, just that it CAN be done this way. On a more complicated algebraic manipulation question, you might be happy to have a method like this in your back pocket.

Because I know someone will ask, here’s the algebra:

    \begin{align*}h&=-16t^2+vt+k\\h+16t^2-k&=vt\\\dfrac{h-k+16t^2}{t}&=v\\\dfrac{h-k}{t}+\dfrac{16t^2}{t}&=v\\\dfrac{h-k}{t}+16t&=v\end{align*}

how would you use substitution to figure out #26 on test 4 page 722?

Start by plugging in the first numbers that occur to you that work. How about x = 2 and y = 3? That works, because –3 < 2 < 3.

Does that eliminate any of the Roman numerals?

I. |2| < 3
II. 2 > 0
III. 3 > 0

Hmm…it does not. The next step is to think about whether you can make any changes to the numbers you chose that would still make the main condition true, but that might break one of the Roman numerals. This just requires a bit of creativity and the experience of having done questions like this before .The Roman numerals give it away a bit: you need to test if x or can be negative.

As it turns out, x totally can: x = –2 and y = 3 works fine: –3 < –2 < 3. That shows that Roman numeral II doesn’t have to be true.

To really be sure that III is true, recognize that the question is telling you that –y < y. Can that be true if y is negative? No, it cannot. Therefore, y must be greater than 0, and the answer is C.