Posts tagged with: Plug In

Will you please work #20 in section 4, Test 1.

Sure! When you see percents, you should almost always think about plugging in 100. In this case, that works like a charm. Say the laptop originally cost $100. When it was put on sale for 20% off, the price went to $80. Buying an $80 laptop with 8% sales tax will cost 80+\frac{8}{100}(80)=86.4. Which answer choice gives you 100 when you plug in 86.4 for p? Only choice D.

If you want to know WHY that works, then look at the algebra. Forget about the tax for a moment. Alma bought the computer at a 20% discount. So if the computer originally cost x dollars, then the price she paid can be calculated thusly:

    \begin{align*}&x-\frac{20}{100}(x)\\=&x-0.2x\\=&0.8x\end{align*}

Now we need to add 8% sales tax to that. We do that by taking the price she paid and increasing it by 8%:

    \begin{align*}&0.8x + \frac{8}{100}(0.8x)\\=&0.8x+0.08(0.8x)\\=&0.8x(1+0.08)\\=&0.8x(1.08)\\=&(0.8)(1.08)x\end{align*}

That expression gives you the price Alma paid for the laptop—which the question calls p. So we can say:

    \begin{align*}p=(0.8)(1.08)x\end{align*}

Since the question asks you for the original price, x, in terms of p, solve for by dividing.

    \begin{align*}\dfrac{p}{(0.8)(1.08)}=x\end{align*}

Test 1 Sec 3 14 please

There are two ways to go here. First, exponent rules.

Recognize that 8=2^3, so you can rewrite \dfrac{8^x}{2^y} as \dfrac{\left(2^3\right)^x}{2^y}. Of course, that simplifies:

    \begin{align*}&\dfrac{\left(2^3\right)^x}{2^y}\\=&\dfrac{2^{3x}}{2^y}\\=&2^{3x-y}\end{align*}

Because you know that 3x-y=12, you know you’ve got 2^{12}.

The other way to go is to plug in: pick values for x and y such that 3x – y = 12. For example, you may pick x = 5 and y = 3, because 3(5) – 3 = 12. Then you can evaluate \dfrac{8^5}{2^3} and match it with an answer. However, since this is the no calculator section, you may decide that this is a suboptimal way to go because the numbers are a little too big to deal with easily unless you know your exponent rules (in which case you didn’t need to plug in in the first place).

Will you please answer question # 27, Test 3, section 4. Is it possible to use real numbers in your example? Thanks!

Sure! The thing you want to remember when you’re plugging in numbers is that you’re doing it to make your life easier, so pick numbers that will really make your life easier. In this case, that means I’m going to say that the original rectangle is a 10 by 10 square, so its area is 100. Obviously, it’s very easy to work with 100 when dealing with percents, and it’s also easy to take different percents of 10 to increase/decrease side lengths.

So, yeah. Here’s our original rectangle:

Now we need to increase its length by 10 percent (10 percent of 10 is 1, which means the length increases from 10 to 11), and decrease its width by p percent. What should we do for p? Well, why not backsolve?

Say p = 20, like answer choice C says. (I’m picking that because it’s in the middle, and also because 20 is the easiest choice to work with so why not try it first.) If p = 20, then we decrease the width by 2, taking it from 10 to 8. So here’s the new rectangle:

What’s the difference between the original area of 100 and the new area of 88? 12. And here’s where we pat ourselves on the back for picking numbers that gave us 100 as our starting area: 12 is what percent of 100? 12 percent! So the first choice we tried, C, is right. Awesome!

(Whenever I illustrate backsolving and the first choice I try is the right choice, I feel compelled to show what a wrong choice would look like. So let’s quickly look at choice B. We’re still increasing the length by 10 percent, so the length is still 11. The width is now decreased by 15 percent, making it 8.5. The new area is 11\times 8.5=93.5. Is that a 12 percent decrease from the original area of 100? No, 93.5 is too big! So if we didn’t already know the answer, we’d know that the next choice we should try should be smaller.)

If x+y=r and x-y=s, in terms of x and y, what is r^2-s^2? Please show step by step. Thanks.

Let’s do this two ways. First, with substitution. If r=x+y and s=x-y, then we can rewrite r^2-s^2 as (x+y)^2-(x-y)^2. Then we can simplify that:

    \begin{align*}&(x+y)^2-(x-y)^2\\=&(x^2+2xy+y^2)-(x^2-2xy+y^2)\\=&x^2+2xy+y^2-x^2+2xy-y^2\\=&4xy\end{align*}

The other way to go is to plug in. Say x=3 and y=2. That makes r=5 and s=1. Now you’re looking for r^2-s^2=5^2-1^1=24. Which answer choice gives you 24 when you plug in x=3 and y=2? 4xy does: 4(3)(2)=24.

T2 Sec 3 #15

If you ask me, this one is begging for plug in. Say that x = 2. Then \dfrac{5x-2}{x+3} becomes \dfrac{5(2)-2}{2+3}=\dfrac{8}{5}. Which answer choice also equals \dfrac{8}{5} when x = 2?

A) \dfrac{5-2}{3}=1

B) 5-\dfrac{2}{3}=\dfrac{15}{3}-\dfrac{2}{3}=\dfrac{13}{3}

C) 5-\dfrac{2}{x+3}=5-\dfrac{2}{2+3}=5-\dfrac{2}{5}=\dfrac{25}{5}-\dfrac{2}{5}=\dfrac{23}{5}

D) 5-\dfrac{17}{x+3}=5-\dfrac{17}{2+3}=5-\dfrac{17}{5}=\dfrac{25}{5}-\dfrac{17}{5}=\dfrac{8}{5}

Only choice D equaled \dfrac{8}{5}, so D is the answer!

The mathy way, because I know someone’s going to want to see it, is to do polynomial division:

That’s 5 with a remainder of –17, which can also be written as 5-\dfrac{17}{x+3}.

Hey Mike, I am pretty strong in word probs, and on this one, I keep getting an answer of 7 chips, but correct answer says 11 chips. I don’t get it !

Erin and Amy are playing poker. At a certain point in the game, Erin has 3 more chips than Amy. On the next hand, Erin wins 4 chips from Amy. Now how many more chips does Erin have than Amy?
A) -1
B) 1
C) 7
D) 11
E) 14

I think plugging in might help you see what’s going on here.

Say Erin has 13 chips and Amy has 10. That way, Erin has 3 more than Amy, just like the question says. Now Erin wins 4 chips from Amy. Erin ends up at 17 chips, right? But she got those 4 chips from Amy, so Amy’s chip total has to go from 10 to 6.

17 – 6 = 11

 

The C cars in a car service use a total of G gallons of gasoline per week. If each of the cars uses the same amount of gasoline, then, at this rate, which of the following represents the number of gallons used by 5 of the cars in 2 weeks?
A) 10cg
E)10g/c
Thanks to one of your methods, I first tried using c=1 and randomly chose 5 gallons to plug in. But then I got A, which was wrong. I later used c=2 and g=5, but then I got E, which is the right answer? Why doesn’t c=1 work? Thank you

Read my method more carefully—I’m very clear that you should never plug in 1, for exactly this reason. Look at the two choices you landed on! They’re exactly the same when c = 1.

When you plug in, remember that you must check every choice, and that if two choices work, you must plug in different numbers to get down to just one choice. You can avoid this happening to you most of the time if you don’t plug in 1 or 0.

Try using 20 for c and see what happens.

if 0 < x < y and x/y=r which of the following must be equal to x+y/x
(A) 1 over (r + 1)
(B) r over (r + 1)
(C) (r +1) over (r minus 1)
(D) (r + 1) over r
(E) r plus 1
can you please explain how the answer is D.

First, please always use parentheses when you enter fractions. You did in the answer choices, but not the question itself. I have seen this question before so I know it should be (x + y)/x, but that’s not what you typed.

Anyhoo, manipulate the fraction:

\dfrac{x+y}{x}=\dfrac{x}{x}+\dfrac{y}{x}=1+\dfrac{y}{x}

Since you know that \dfrac{x}{y}=r, you know that \dfrac{y}{x}=\dfrac{1}{r}, so you can simplify the above:

1+\dfrac{y}{x}=1+\dfrac{1}{r}

If you combine that into one fraction by getting a common denominator, you land on choice D:

1+\dfrac{1}{r}=\dfrac{r}{r}+\dfrac{1}{r}=\dfrac{r+1}{r}

Don’t like all that algebra? This is also a great question to plug in on! 🙂

A train started at station S and has been traveling r miles per hour for t hours. In terms of r and t , how many miles from station S had the train travelled 3t/4 hours ago ?
(A) rt/12
(B) 3rt/4
(C) rt
(D) 4rt/3
(E) rt/4

Plug in! Say r = 10 and t = 4—those will be nice, easy numbers to work with. \dfrac{3t}{4} becomes simply 3, so the question asks, given that the train has been traveling for 4 hours, how far the train had traveled 3 hours ago. In other words, how far had the train traveled 1 hour after it left?

Well, that’s easy. It’s going 10 miles per hour, so in 1 hour it went 10 miles.

Which answer choice simplifies to 10 when r = 10 and t = 4? Only E.

Would you say that for the hard rated questions on math with variables a general go-to strategy to do is plug in? I’ve just been looking over the Blue-Book and it seems like all the hard ones with variables are susceptible to plugging in.

It’s certainly very common that plugging in works on very hard questions. It’s a pretty good go-to if you’re stuck.

If x and y are positive and sqrtx = y, which of the following must be equal 0?

A) x-y
B) x- sqrty
C) y-2x
D) y- x^2
E) y^2 -x

Square both sides:

\left(\sqrt{x}\right)^2=y^2

x=y^2

From there, all you need to do is subtract!

0=y^2-x

You could also plug in! Say x = 49 and y = 7. Then it’s true that the square root of x equals y, and you can try each answer choice to see which equals 0. Only E will.

a=x^2
b=(x+1)(x-1)
c=(x+1)^2

If x is less than or equal to 0 in the three equations above, what is the ordering of A, B and C

a) A<B<C

b) B<A<C

c) B<C<A

d) C<A<B

e) C<B<A

This is either a bad question or you transcribed it wrong, because if x = 0 or x = –1, you get different answers than if x = –2 or less.  So let’s assume this question actually says x is less than –1, and then let’s PLUG INNNNN!!!!!!!!!!!

Say x = –2.

a=(-2)^2=4

b=(-2+1)(-2-1)=(-1)(-3)=3

c=(-2+1)^2=(-1)^2=1

There: c < b < a

You can also graph to solve this one, and graphing will show you precisely why the question as you submitted it doesn’t work. See how the top-to-bottom order of the graphs changes when –1 ≤ x ≤ 0?

s=t
g=h-5

If h is 6 less than t in the equations above, then g is how much less than s?

A)1
B)5
C)6
D)9
E)11

Plug in! Say h = 10, which will make t = 16. Now s = 16 as well, and g = 5. 5 is how much less than 16? 11 less.

Hi Mike, I scored a 660 on the math section on the SAT. I did not answer the last two or three math questions from each section so I wouldn’t risk getting them wrong and so I would have more time for the easy/medium questions. My main question is how do I deal with the hardest math questions on the SAT? Are there certain strategies that work better for the harder questions (i.e. backsolving tends to work better on the harder questions)? I do own a copy of your Math Guide, by the way.

Skipping the last few questions is a good move if your goal is to have more time to get the earlier ones right for sure, but you’re pretty much maxing out your score at 660 if you’re skipping that many (on average, skipping 3 per section and getting EVERYTHING ELSE RIGHT will get you about 660). So yeah, if you want to go higher, you’re going to have to answer more questions.

If you look at the Blue Book Breakdown at the back of my book, you’ll see that the hardest questions come in many forms. A fair amount of them are susceptible to plugging in, so if you’re looking to, say answer 1 out of the last 3 in each section, you might start that process by scanning for questions you can plug in on. There’s no guarantee that you’ll always find one, but it’s a good way to start.

Which of the following is an equation of the line with x-intercept a and y-intercept b, where ab is not equal to 0?
A) y= ax+b
B) y=bx +a
C) y= (a/b)x +a
D) y= (b/a)x +b
E) y= (-b/a)x =b

You have to reverse engineer this one a bit. You know, based on the question, that the points (a, 0) and (0, b) are on the graph. From that, you can find the slope:

\dfrac{b-0}{0-a}=\dfrac{b}{-a}=-\dfrac{b}{a}

Armed with that slope, and with the y-intercept of b, you can write your slope-intercept equation:

y=-\dfrac{b}{a}x+b

Note that this is also a question that’s quite susceptible to plugging in, if that helps you visualize it. Pick values for a and b, then plug those into each answer choice, graph them, and see which line actually goes through the points it’s supposed to. 🙂