## Posts tagged with: percents

Test 5 Section 4 Number 24

According to the table, Townsend Realty purchased the Glenview Street property for $140,000. This question tells us that$140,000 was actually a pretty heavily discounted price. They received both a 40% discount off the original price and another 20% discount for purchasing in cash. Let’s say the original price was x dollars.

A 40% discount off a price of x can be written like this: (1 – 0.4)x. Of course, we can simplify the expression in the parentheses: 0.6x.

To take another 20% off, we do a similar operation again, starting with that 0.6x: (1 – 0.2)(0.6x) = (0.8)(0.6x) = 0.48x.

So a 40% discount followed by a 20% discount off an original price of x dollars can be written as 0.48x. We know that the final price was $140,000, so we can solve for the original price: 0.48x =$140,000
x ≈ $291,667 That rounds to$291,700, which is choice B.

Putting all the math in one place for those who like it that way:

(1 – 0.4)(1 – 0.2)x = $144,000 (0.6)(0.8)x =$144,000
0.48x = $144,000 x ≈ 291,700 Test 5 Section 4 Number 22 There are a lot of words here, and many of them are necessary! They tell you that Mr. Camp’s class is representative of the 8th grade classes in that state. That means you can assume that if 34.6% of the students in Mr. Camp’s class have at least 2 siblings, then 34.6% of 8th grade students across the state have at least 2 siblings. If there are 1,800 8th grade classes in the state, with an average class size of 26, then there are 8th graders in the state. 34.6% of them have at least 2 siblings. HOWEVER, the question asks how many have fewer than 2 siblings! So we don’t need to solve for 100% – 34.6% = 65.4% of those students. That rounds to 30,600, which is choice C. Can you do Test 7 Section 4 #30? The manager projects that the percent increase from 2012 to 2013 will be double the increase from 2013 to 2014. You’re given the 2012 and 2013 numbers, so start by figuring out that percent increase. There were 5600 subscriptions sold in 2012 and 5880 sold in 2013. Percent increase is change divided by original value times 100%: If there was a 5% increase from 2012 to 2013, and the manager believes that will be double the increase from 2013 to 2014, then he projects a 2.5% increase into 2014. All we need to do from here is apply 2.5% growth to the 2013 number we know to arrive at the manager’s 2014 projection. So the answer is B. Test 5, Section 4, Questions 25 and 28. How do you get to the answers? In the future, please ask only one question per submission. I know that seems like a random request, but it makes the linking I do on the Official Test post guide easier on my end. 🙂 #25: More than 20% of all participants (all 300 of them) chose the first picture. In the first group of 150, 36 chose the first picture, and in the second group of 150, p chose the first picture; therefore, 36 + p chose the first picture overall. 36 + p must therefore be greater than 20% of 300. Translate that to math: Rearrange a tiny bit and you have choice D: #28: From the graph shown, you can see that the slope of function f is . The fastest way to get that is to count the rise and the run from (0, 3) to (2, 4). That’s all you need from the graph. From there you’re told that the slope of g is 4 times the slope of f, so the slope of g must be 2. You’re also told that g goes through (0, –4), which means its y-intercept is –4! You know enough to write the slope-intercept equation of g: From there, all you need to do is solve for : Can you explain question #24 in Practice Test #8 Section 4? The key to this one is to eliminate every choice that makes an unjustified (i.e., not directly supported by the question) assumption. You can eliminate choice A because we don’t know anything about the preferences of the viewers who didn’t vote. Generally speaking, don’t draw conclusions about what could have happened, draw conclusions about what DID happen. You can eliminate choice B because the question mentions nothing about the ages of voters. Don’t let your own assumptions about who texts and who uses social media cloud your judgment! You can eliminate choice C because it’s mathematically false (more on this below) but you should also lean towards eliminating it just because it’s using something that didn’t happen as a basis for its conclusion. As before, resist the urge to make conclusions about what could have happened when you only know what DID happen. Choice D is totally supported by the data: 70% of social media voters preferred Contestant 2, and only 40% of text message voters preferred Contestant 2. Therefore, social media voters were more likely to prefer Contestant 2 than text message voters. Now, as for why C isn’t mathematically true…let’s plug in! Say there were 100 voters. We know 30% of the votes came in via social media, so that’s 30 votes. The other 70 votes must have come in via text message. Of the 30 social media votes, Contestant 2 got 70% of them, or 21 votes. Of the 70 text message votes, Contestant 2 got 40% of them, or 28 votes. So of the 100 votes, Contestant 2 only got 21 + 28 = 49 votes! That’s not enough to win the contest. 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 . 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:

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

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

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

based on the line graph above, which of the following best approximates the percent decreases in the number of farms in the united states from 1960 to 2000?

Percent change is calculated thusly: . For this question, then, you’d want to do the following:

The circle graph above shows the percent of 4th graders at an elementary school who have the indicated numbers of pets in their homes. If 68 of the 4th graders have at least one pet, how many have exactly two pets?

(A) 16
(B)17
(C)20
(D)33
(E)34

“At least one” means one, two, or more than two—anything but zero pets. So that’s 30 + 20 + 35  = 85 percent of the students.

If 68 students represent 85 percent of the class, how many students represent 20%? We can do a ratio:

M gallons of a p% salt solution must be mixed up with G gallons of a q% salt solution to produce an r% solution. Which of the following best describes how to find the value of r? (This question is driving me insane.)

(A) (p + g) / (M + G) = r / 100

(B) (0.01p + 0.01q) / (M + G) = r / 100

(C) 0.01p / M + 0.01q / G = r / 100

(D) (0.01M + 0.01G) / (M + G) = r / 100

(E) (0.01pM + 0.01qG) / (M+G) = r / 100

Break the parts down.

M gallons of a p% solution”:

G gallons of a q% solution”:

The r% solution will be M + G gallons, so:

The setup, then, is this:

That simplifies to :

This really isn’t a realistic SAT question, though. If you got this out of an SAT prep book, you’re being led astray.

Thanks to Hurricane Sandy, this has been a rough week for a lot of people. If you’re feeling charitable, please note that I am still giving away Math Guides for donations to the Red Cross.

Let me note, before we get into this, that these challenge questions are WAY harder than anything you’d see on the SAT. That’s why I call them challenges. They test your knowledge of concepts that appear on the SAT, sure, but my challenge questions are meant to stump you for at least a few minutes.  Please don’t freak out about these, especially if you’re taking the November SAT tomorrow. Anyway, let’s get to it.

Amy is putting together an epic 7-song playlist that she can listen to while she’s working out at the gym. She’s a bit particular about how it’s constructed. The second song must be 25% longer than the first. The third song must be 12.5% shorter than the second. The fourth song must be half a minute shorter than the third. The fifth song must be 1.5 times as long as the fourth, and the sixth song must be 30% shorter than the fifth. The seventh song must be her current favorite song, “Gangnam Style,” the mp3 for which she recently purchased from Amazon. The total playlist must last exactly 25 minutes, with no pauses between songs. How long must the first song on Amy’s playlist be?

As usual, the first correct answer in the comments will win a copy of the Math Guide.

[Full contest rules.]

UPDATE: It’s taken me a long time to write the solution to this, because, well, I kinda started thinking I had already done it. Commenter Yeana got it right first, and the commenter Peter posted a good short explanation. My full explanation below the cut.

The first thing I’d recommend doing is deciding whether you want to work with minutes in decimal form, or in seconds. My preference is for seconds (and I went to some trouble to make sure none of these songs had fractional second lengths). So yeah. My explanation will use seconds.

A 25 minute playlist will be 25 minutes × 60 seconds per minute = 1500 seconds long. And “Gangnam Style” (I specified the Amazon version because I found other online versions that differed by a second and I didn’t that to frustrate anyone) is 3 minutes and 39 seconds, or 219 seconds long. Everything else you’re going to have to figure out by reading through the problem.

Let’s work through this step by step.

• Since the question asks about the first song, let’s call its length x.
• The second song’s length is 25% greater than that of the first. So it’s going to be 1.25x.
• The third song is 12.5% shorter than the second. Another way to think of this is that it’s 87.5% as long as the second. So the third song will be 0.875(1.25x) seconds long.
• The fourth song is an easy one, but at the same time continues to complicate the question. It’s 30 seconds shorter than the third song, so its length is 0.875(1.25x) – 30.
• The fifth song is 1.5 times as long as the fourth, so it’s 1.5(0.875(1.25x) – 30) seconds long.
• The sixth song must be 30% shorter than the fifth (in other words, 70% as long as the fifth). So its length is 0.7(1.5(0.875(1.25x) – 30))
• And of course, as already stated, the seventh song is 219 seconds long.
All of those add up to 1500. So let’s solve for x!
…On second thought, let’s let Wolfram Alpha do it for us. 🙂
x = 192
So the last thing we need to do is convert that to minutes and seconds. Note that seconds-to-minutes conversions are one place other than the SAT and 3rd grade where remainders are useful. The closest minute mark is 3 minutes (180 seconds), and there are 12 seconds left over. In other words, 192/60 = 3 remainder 12. So the first song is 3:12 long.

I’m going to be out of town for a few days, so I won’t be able to post a solution for this until probably Monday, but hopefully I’ll be able to check in and at least congratulate the winner, who will receive a free copy of my Math Guide.

A student is drawing concentric circles, each with a diameter one centimeter longer than that of the one before. When he tires of this random task, he has drawn 21 circles. If the area enclosed by his circles increased by 300% from the time he drew his first circle until the time he was done, what is the radius of the smallest circle?

LATE UPDATE: David’s the winner—he got the correct answer first. I’ve been buried under a bunch of other things and have been slow to get the solution posted. Sorry about that. It’s below the cut.
The first task here is to figure out how much bigger the last circle is than the first. This requires a tiny bit of care. Note that the second circle’s diameter is 1 cm bigger than that of the first. The 3rd circle’s diameter is 2 cm bigger than that of the first. And so on. So the 21st circle’s diameter will be 20 cm larger than that of the first.

Of course, this means the radius of the largest circle is 10 bigger than the radius of the smallest circle, which is what we really care about if we’re talking about areas.

Let’s call the radius of the smallest circle r. The area of the smallest circle is πr2. The area of the largest circle is π(+ 10)2.

The next tricky part is the 300% increase. An increase of 300% does not mean that the new area is 3 times bigger than the old. It means the area increased by its original size 3 times over.

This is an important insight, and it makes this problem a lot easier. I’ll include the mathy way below, because I like that kind of thing and I think it’s important to show that there are multiple ways to solve this, but note that it’s a bit more involved if you don’t know that a 300% increase means you end up with 4 times what you started with.

If the new area is 4 times the original area, we can set up a simple equation:

$4(\pi r^2)=\pi(r+10)^2$
$4\pi r^2=\pi(r^2+20r+100)$
$4r^2=r^2+20r+100$

At this point, you’re going to have to use the quadratic formula, or do some factoring. You never need the quadratic formula on the SAT, so I’d recommend factoring, but really it’s whatever makes you happy. 🙂

You know your first terms have to be 3r and r, and you know one factor must contain subtraction to get -100 in the last term, etc. A little trial and error later, you get…

$(r-10)(3r+10)=0$

So your solutions are r = 10, or r = -10/3. Since a circle’s radius can’t be negative, the answer is r = 10.

That was a bit mathy, right? Here’s the other, slightly mathier way:

If the area enclosed increased by 300%, then we can use our percent change formula to solve.

$\textup{Percent\; Change}=\frac{\textup{Amount\;of\;Change}}{\textup{Original\;Value}}\times100\%$
$300\%=\frac{\pi(r+10)^2-\pi r^2}{\pi r^2}\times100\%$

Let’s clean that up a little bit by dividing both sides by 100% and knocking the π term out…

$3=\frac{(r+10)^2-r^2}{r^2}$

And simplify a little more…

$3=\frac{r^2+20r+100-r^2}{r^2}$
$3=\frac{20r+100}{r^2}$
$3r^2=20r+100$
$3r^2-20r-100=0$

And that’s the same place we ended up at above. You’ll still get r = 10.

I went back to my old high school last night to attend the final concert of the choir director that presided over so many of my formative moments. I got to hang out with some of the same people with whom I used to have the kind of deep, meaning-of-life conversations that only happen in movies or in real life when you’re between the ages of 15 and 17. A good number of them are teachers now, as it turns out. Their students are lucky to have them.

I went back with a few of the people who helped make me into me, to honor a man who helped shape all of us. I saw a fantastic concert. I had a lot of hugs and handshakes. I found a plaque that still has my name on it. And then I drove home listening to a record I listened to a lot back then. Now I’m sitting in my apartment back in New York, wistful and weary. HOLD ON TO THESE MOMENTS WITH BOTH HANDS. DRINK FULLY AND RICHLY FROM THE CUP OF YOUTH. AND RETURN FOR SECONDS.

Anyway, here’s a challenge question! First correct response gets a free copy of the Math Guide.

Louis and Rebecca each had x dollars on Monday morning. On Monday afternoon Louis paid Rebecca 20% of his money for a computer that she was selling. On Tuesday morning, Rebecca paid Louis 20% of her money for a lawnmower that he was selling. On Wednesday morning, both Louis and Rebecca paid 20% of their money to Steven for…something. I don’t know. Steven was selling something. Assuming they did not spend or receive any other money in that period, how much money did Steven receive, in total, from Louis and Rebecca in terms of x?

Leave your answers in the comments. I’ll post a solution on Monday (assuming someone gets it by then).

In order to win a book, you must not comment anonymously. I need to be able to get in touch with you to get your shipping info, etc. Also, while you can still win a free book if you don’t live in the US, you’re going to have to pay for shipping. So the book is only sorta free if you’re international.

Good luck!

UPDATE: Congratulations to Jason, who nailed it with alarming speed, and to whom a Math Guide will be en route. Solution below the cut.

There’s a fast way to do this problem, and then there’s a very fast way to do this problem. I’m gonna give you both.

First, the fast way. Plug in. Say x = 100. Here’s how the money flows:

 Day and time Louis Rebecca Steven Monday morning $100$100 Monday afternoon $80$120 Tuesday morning $104$96 Wednesday morning $83.2$76.8 $20.80+$19.20=$40$40 is 40% of x (which, as you’ll recall, we said was 100), so the answer is 40% of x, or 0.4x.

As you look at the table, you might notice that each row adds up to $200. Interesting, no? Which brings us to the very fast way to do the question. Regardless of the money that changes hands between Louis and Rebecca, the total amount of money they have together never changes. So we can just, from the get-go, say that the amount of money that gets handed to Steven is 20% of the total money Louis and Rebecca begin with. They start with, in total, 2x dollars. 20% of 2x is 0.4x. I’m going to tell you a secret: I don’t sleep well at all the night before SAT scores are released. It’s not like I even took it. I just get so excited for all the kids I’ve worked with that I can’t sleep. So last night, after a sleepless night followed by a bunch of good news, I slept like a rock. And I woke up energized, and I got some much-needed work done on my book, and now I’m posing this weekend challenge in timely manner. I’m feeling very productive today. Yay me! By now you know the drill I’m sure. First correct (and non-anonymous) comment gets access to the Math Guide Beta (to which I just added some more hard questions this very morning). Adina keeps track of the number of rats she sees while she waits for the subway every morning. She noticed that there was a roughly 46% increase in the amount of rats she spotted between Tuesday and Wednesday. If she saw 19 rats on Wednesday, what will be the percent decrease in rat sightings if she sees the same number Thursday that she did Tuesday? Round your answer to the nearest percent. Good luck, y’all. And have a great weekend! I’ll post the solution early next week. UPDATE: Nice work, Jessica. Enjoy the Beta. Solution is below the cut. To solve this, you really just need to remember the basic formula for percent change: What you’re given in this problem is the percent change (roughly 46%) and the final value (19). You can use that to find the original value (since it’s the number of rats Adina saw Tuesday, let’s call it t. .46 = (19 – t)/t .46t = 19 – t 1.46t = 19 t = 13.01… A rat is a rat, there’s no partial rats, so we say Adina saw 13 rats on Tuesday, and thus will see 13 rats on Thursday. From here, it’s just plug-and-chug to find the answer. She’ll see 6 fewer rats on Thursday than she did on Wednesday, so the change is 6 rats. The original value is 19, because now we’re starting from Wednesday. Percent Change = 6/19 × 100% Percent Change = 31.57…% Round, like the question said to, and you end up with 32%. If you’re taking the test tomorrow morning, you should ignore this and REST YOUR BRAIN. If you’re not taking it tomorrow, though, then this is just another regular weekend for you, and you should work your brain HARD. The prize, as it has been every week for weeks and weeks, is access to the coveted PWN the SAT Math Guide Beta. If you wanna win, you have to comment in such a way that I can contact you. Preferably using Facebook, Twitter, or Gmail. A warning: this one is pretty tough. In a hand in his weekly poker game, Mike won money from both John and Sean. Mike’s percent gain on the hand was equal in magnitude to John’s percent loss. John bet twice as much as did Sean. Mike and John together held$200 in chips before the hand began. If Sean bet \$20 in chips on the hand, what was Mike’s chip total after the hand was over?

Good luck!

UPDATE: Nobody got this yet, so I’m gonna leave it up for now unanswered. If you solve it, book access is yours.

SECOND UPDATE: OK, Eowyn got it. Nice. Solution below the cut.
There’s a lot going on here, and at first blush it may seem as though you have too many variables to work with. If you write down everything you know, though, and play around with it enough, you’ll see that you can actually get it down to just one variable in one equation!

$percent\:&space;change&space;=&space;\frac{change}{original\:&space;value}\times&space;100\%$
$\frac{Mike's\:gain}{Mike's\:original\:value}\times&space;100\%&space;=&space;\frac{John's\:loss}{John's\:original\:value}\times&space;100\%$

Of course, we can tidy this up a bit:

$\frac{Mike's\:gain}{Mike's\:original\:value}&space;=&space;\frac{John's\:loss}{John's\:original\:value}$

Now we need to start figuring out the numbers. We know that Sean bet 20, and that John bet twice as much as Sean. So John bet 40, which means John’s loss was 40. We also know that Mike’s gain was 60, since he won all the money that was bet in the hand (John’s 40 + Sean’s 20).

$\frac{60}{Mike's\:original\:value}&space;=&space;\frac{40}{John's\:original\:value}$

And we can be a bit clever here if we want to jam all the info into one equation and say that, since Mike’s and John’s chips added up to 200 before the hand, Mike’s prior chip total can be m, and John’s prior chip total can be 200 – m.

$\frac{60}{m}&space;=&space;\frac{40}{200-m}$

Solve that for m

$40m&space;=&space;60(200-m)$
$40m&space;=&space;12000-60m$
$100m&space;=&space;12000$
$m&space;=&space;120$

…and you’ll see that before the hand began, Mike had 120 dollars in chips. Since he won 60, he has 180 after the hand.

When I was in high school, I weighed 125 pounds fully clothed and soaking wet. I couldn’t do anything to change it, either. That was the worst part. I yearned to play varsity baseball, but at my weight, I just straight up wasn’t big enough.

College was mostly the same, although I filled out a little. I’d say my average weight in college reflected the “freshman 15,” but for me it was a welcome change.

Then I got a job, spent 4 years sitting on my ass all day, eating large fast food meals, and not getting any exercise. I weigh about 170 now.

So over the years, I’ve put on 45 pounds. Damn. What’s the percent change in my weight from high school to now?

Here’s the general formula for this kind of question:

$\fn_jvn&space;Percent\,&space;Change=\frac{Amount\,&space;of\,&space;Change}{Original\,&space;Value}\times&space;100\%$

So plug my values in:

$\fn_jvn&space;Percent\,&space;Change_{Mike's\:&space;weight}=\frac{45&space;lbs}{125&space;lbs}\times&space;100\%$

And would you look at that? The percent change in my weight is 36%. Holy balls. I need to go on a diet.

So now say I go on some crazy workout plan, and I lose 45 pounds, so I’m right back where I started at 125. What will be my percent change then? Note: if the answer was just going to be 36% again I probably wouldn’t be wasting your time with this.

$\fn_jvn Percent\, Change_{Humpilates\:diet}=\frac{45lbs}{170lbs}\times100\%$

Whoa. So it was a 36% gain, but now it’s only about a 26.5% loss? How is that fair? A sorta strange truth about percents (this is true about ratios, too) is that the bigger the numbers are, the less difference a difference makes.

If you’re 16 and your little brother is 13, you’re about 123% his age. But when you were 4 and he was 1, you were 400% his age. The older you both get, the smaller the percent difference will be, but you’ll always be 3 years older than him.

This is also why it’s socially acceptable if your dad is 10 years older than your mom, but it would have been pretty creepy if they met when he was 26 and your mom was 16.

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