## Posts tagged with: averages

I turn 30 years old tomorrow. 30 f’ing years old. CLING TO EVERY SECOND OF YOUR YOUTH. SQUEEZE EVERY OUNCE OF JOY OUT OF IT LIKE JUICE FROM YOUR FAVORITE FRUIT. SOMEDAY YOU WILL BE OLD LIKE ME.

The prize for the challenge: Free access to the PWN the SAT Math Guide Beta.

Week 1: Served customers numbered a through b
Week 2: Served customers numbered c through d
Week 3: Served customers numbered e through f
Week 4: Served customers numbered g through h
Week 5: Served customers numbered i through j

The proprietor of a deli is trying to project how many customers he usually has on Mondays, so that he can order enough roast beef, but not order too much because nobody likes rotten roast beef. At the deli, customers take numbers before they are served, so he plans to collect data over the next 5 Mondays in the format of the list above, and is looking for an expression for average number of customers that he can plug numbers into as he collects them. What is the expression for the average number of customers (in terms of a through j) that are served in the deli on those 5 Mondays?

Good luck, children. I’ll post the solution Monday.

UPDATE: Congrats, Serrilius. I hope you enjoy the book.

Solution below the cut.

As Collin so deftly pointed out in the comments, a little plugging in does wonders here. If, for example, the deli serves customers numbered 35 through 39, how many customers were served? 35, 36, 37, 38, 39. 39-35 = 4, but 5 customers were served. So when you want to know how many customers passed through on a day when numbers a through b were served, it’s b – a + 1.

So then, to find the average, you find the sum, and then divide by 5.

[(b a + 1) + (dc + 1) + (fe + 1) + (hg + 1) + (ji + 1)]/5
= (-a + bc + de + fg + hi + j + 5)/5
= (-a + b – c + d – e + f – g + h – i + j)/5 + 1

Something you don’t often see in SAT prep materials is a pictograph, but when you actually look at a real SAT, pictographs are all over the place!

It’s not a conspiracy by a cabal of prep writers and SAT writers; pictograph questions are usually just really easy, and so prep writers don’t pay them much mind. This morning I decided to see if I could write a pictograph question that’s worthy of being called a Weekend Challenge. I guess you guys can judge whether or not I succeeded.

The prize this week: Same as the last few weeks. First correct response in the comments gets access to the PWN the SAT Math Guide Beta Program. One note: if you comment anonymously, I have no way of contacting you, or verifying who you are if you try to claim it was you later, so you can’t win. If you want to win, don’t post anonymously.

Sam is the kind of guy who keeps track of meaningless things in unnecessarily complicated charts. The chart above depicts the average number of times Sam finds himself laughing out loud during 3 distinct parts of the day. Sam has been keeping this chart for 4 days, and calculates that he will have to be made to LOL 10 more times than usual on the 5th night when his gassy uncle comes to family dinner in order to necessitate a change of one added face on his chart. On average, how many times a day has Sam been been made to LOL in school?

UPDATE: Nice work, Kira! I hope you enjoy the Math Guide Beta.

Everyone else: The solution is below the cut.
So this might be a pictograph question, but it’s also a weighted average question, so the solution will involve the average table. If you’re a longtime reader, you should be nodding your head like yeah. (‘sup Miley?)

Basically, Sam averages 4n LOLs per family dinner, and it’ll take him 10 more LOLs than usual at dinner on the night in question to get his average up to 5n LOLs. Luckily, his flatulent uncle is on the way! Here’s the setup:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 4 nights 4n LOLs per night 16n LOLs + 1 night 4n + 10 LOLs + 4n + 10 LOLs = 5 nights 5n LOLs per night = 20n + 10 LOLs= 25n LOLs

There are two ways to fill in the bottom right field in the table: 1) add the right column and get 20n + 10 LOLs, or 2) multiply the bottom row to get 25n LOLs.

What that tells us is:

20n + 10 = 25n
10 = 5n
2 = n

If n = 2, then we know how many times Sam has been laughing out loud in school, on average. The pictograph tells us there are 6n LOLs per day, which means Sam’s friends must be hilarious. He averages 12 LOLs a day in school.

I found this awesome graphic at Amazin’ Avenue, and it inspired me to write a baseball-themed challenge question this week. I’m a huge Mets fan, and my favorite player on the team right now is R.A. Dickey. He faces the hated Phillies tonight. I am so pumped to watch.

The prize this week: A free (imaginary) R.A. Dickey bobblehead doll! Best prize ever!

In the 2011 season, R.A. Dickey has thrown 1758 pitches and batters have completed 494 plate appearances against him as of July 15. What is the least number of batters he would need to face to have a pitch per plate appearance average under 3.50?

 Let’s go Mets!

UPDATE: solution below the cut.

Don’t get hung up on the terminology! Even for baseball fans, this isn’t a stat that comes up in conversation. The SAT won’t ask about something so culturally anchored as baseball, but it very well might throw a question like this at you in another context.

We know that Dickey has thrown 1758 pitches through 494 plate appearances. So right now he has a pitch per plate appearance average of 1758/494, about 3.56.

If we’re looking for the minimum number of batters he must face to get that average under 3.5, we need to figure out the minimum number of pitches that can occur in a plate appearance. That number is 1. A batter can see one pitch and, say, get a hit. He could also ground out on that pitch. Pitch per plate appearance, as you see, doesn’t tell much of a story on its own.

So the minimum number of batters he must face must be the SAME as the number of additional pitches he must throw, since it’s one pitch per batter.

$\textrm{Pitch\:Per\:Plate\:Appearance\:Average}=\frac{1758+x\:\textrm{Pitches}}{494+x\:\textrm{Plate\:Appearances}}$

From here, you can either throw the equation into your trusty graphing calculator and use the table to scroll down until you see the average drop under 3.5, or you can solve algebraically. Either way, you should see that after 11 additional plate appearances and pitches, his average is just over 3.5, and after 12, it’s just under.

So the answer I’m looking for here is 12.

I wanted to take it easier on you guys after last week’s hard-as-hell question, so here’s one that could actually appear on an SAT.

This week’s prize: you and your future random college roommate will have remarkably similar tastes in music. Trust me — it matters.

1. After m days, the average (arithmetic mean) temperature for the month of June in the city of Riverside was is 87° Fahrenheit. If the temperature the next day was 99° and the average temperature for June rose to 89°, what is m?

UPDATE: Excellent job, Guilherme. Solution below the cut.

To solve this question, you’re going to want…bum bum BUMMMM….the AVERAGE TABLE!!!

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 m days 87° 1 day 99° 99° m + 1 days 89°

The power of the average table is that it makes it very easy to deal with sums; sums end up being the path to success for most average questions on the SAT. Above I’ve filled in all the given values (the average and sum temperature on the 1 day are, obviously, the same). We’re going to solve by filling the rest of the table in. The sum of the temperatures for the first m days will be 87m. The sum for the m + 1 days will be the average (89) for those days times m + 1, but it will ALSO be the sum of the first m days plus the temperature of the 1 day:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 m 87 87m 1 99 99 m + 1 89 87m + 99 or 89(m + 1)

Important enough to repeat even though I kinda already said this: I can make two separate expressions in the right-hand corner. One comes from adding up the first two entries in the column on the right, and one comes from multiplying the first two values in the last row. If you’re not following the way this table works, please see my original post on the table, which should answer your questions.

Here’s the thing: now I can solve for m because I have two expressions containing m, which equal each other!

87m + 99 = 89(m + 1)
87m + 99 = 89m + 89
87m + 10 = 89m
10 = 2m
5 = m

Oh hell yes.

Same prize as last weekend for whoever gets this first (either on the blog, or on Facebook; you’re competing against each other here). Any \$5 album from Amazon.com. Ready?

A small town has 3 theaters in it. Last Saturday night, the average age in the first theater was 29, and the average age in the second theater was 24. If the overall average theatergoer on that night was 36 years old and the ratio of attendees for the three theaters was 2:3:10, respectively, then what was the average age of the attendees in the third theater?

Good luck, folks. I’ll post the answer and contact the winner (if there is one) on Monday. Remember: you can’t win if I can’t contact you.

UPDATE: Congrats to Codi, who nailed it on Facebook. Answer below the cut.

This is an average question, so let’s use the average table! Start by filling in what the questions tells us (and plugging in obvious values for numbers of people in the theaters based on the given ratio):

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people (1st theater) 29 years per person 3 people (2nd theater) 24 years per person 10 people (3rd theater) 15 people (all theaters) 36 years per person

Now fill in the sums:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people(1st theater) 29 years per person 58 years 3 people(2nd theater) 24 years per person 72 years 10 people(3rd theater) 15 people(all theaters) 36 years per person 540 years

Hopefully you see that the sum of the ages in the first and second theaters is 58+72=130. Therefore, in order for the sum of everyone’s ages to be 540, the sum of the ages of the people in the third theater has to be 540-130=410. If there are 10 people in a room whose ages add up to 410, the average age in the room has to be 41! To finish the table:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 2 people(1st theater) 29 years per person 58 years 3 people(2nd theater) 24 years per person 72 years 10 people(3rd theater) 41 years per person 410 years 15 people(all theaters) 36 years per person 540 years

Note that the average table allows you to check your work to make sure you didn’t make any mistakes. In the left-hand column 2+3+10=15, and in the right-hand column 58+72+410=540. That’s a beautiful thing. I literally weep when I see it.

Pretty easy question here, so I’ll attach a prize of fairly small value, but that might still be fun: the first person to comment with the answer gets to name a character in a future word problem on this site, and what they do/sell/wear/eat. For example: “Rita is the diaper changer at a daycare center that feeds the kids nothing but corn.” Of course, I have no way of contacting an anonymous commenter, so you’re going to have to identify yourself. Cool? Let’s goooo.

Four kids are in a room; their average age is 8 years old. Then an adult enters the room, and the average age becomes 16. A tense conversation quickly escalates, culminating in one of the children screaming “You’re not even my real dad!” and leaving the room, but the average age in the room stays the same. A few minutes later, another kid leaves the room in search of a sandwich. If the last kid to leave was 4 years old, then the adult is how much older than the average age of the people remaining in the room, awkwardly staring at each other?

MOAR UPDATEZ and SOLUTION: Congratulations to “Amy” who answered correctly, albeit with some degree of uncertainty. Good enough, kid. Email me to claim your prize.

Solution after the jump.

##### Solution

We’re going to solve this bad boy with the average table. Because it is awesome and if you don’t love it then you probably are incapable of love. Let’s start by finding the age of the adult. Here’s what the problem tells us:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 4 people 8 years/person +1 person 5 people 16 years/person

In order to find the adult’s age, just fill in the sums of the ages in the first and third rows, and figure out the difference. That is to say: if we know the sum of the ages when it was just the 4 kids in the room was 32, and then the sum of ages became 80 when the adult walked in, how old must the adult have been?

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 4 people 8 years/person 32 years +1 person 48 years/person +48 years 5 people 16 years/person 80 years

OK, so if the adult is 48 years old, we have half of what we need. Now let’s figure out what happened after the adult walked in. We know two people left the room, and we know enough about each to figure out the average age in the room when the dust eventually settles. Picking up from where we left off, let’s deal with the first kid having a little hissy fit and leaving:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 5 people 16 years/person 80 years -1 person 4 people 16 years/person

This looks a lot like the table above did, so we’ll fill it in the same way. Note that I filled in the age of the kid who left, although it doesn’t actually matter at all to the solution of the problem.

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 5 people 16 years/person 80 years -1 person 16 years/person -16 years 4 people 16 years/person 64 years

Ok, so since the average age stayed the same after that kid left, the sum of the ages in the room is now 64. When one more kid (age 4) leaves, we’re left with 3 people in the room, aged a total of 60 years. What’s the average age in the room? 20 years:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

 4 people 16 years/person 64 years -1 person 4 years/person -4 years 3 people 20 years/person 60 years

The adult is how much older than the average age in the room?

48 – 20 = 28.
Sweet.
 source

The SAT loves to ask a particular kind of question about averages that can pretty confusing without a nice, easy way to organize your information. Enter The Average Table. KNEEL WHEN IT ENTERS THE ROOM, KNAVE! Seriously, this thing kicks ass.

To build it, just remember what you have known for a long time about averages: how to calculate them. If I gave you a set of 5 test scores and asked you to average them, what would you do? You’d add them up, and then divide the total by 5. That’s because…

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;Average\:&space;of\:&space;Values&space;=&space;\frac{Sum\:&space;of\:&space;Values}{Number\:&space;of\:&space;Values}$

That’s just how averages work. But what if we multiplied both sides of that equation by [Number of Values]?

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$

You can use this to set up a very handy little table, which will help you solve even the hairiest looking average questions. I’m going to use colors to help you see how! Aww yiss.

Let’s illustrate with a problem that looks like it sucks:

1. A delivery truck is loaded with seven packages weighing an average of 30 pounds. At his first stop, the delivery man drops off three packages weighing a total of 60 pounds. He also picks up one package weighing 15 pounds. He makes one more stop to deliver two more packages, which weigh 42 and 48 pounds. What is the average weight, in pounds, of the packages that remain on the truck?

(A) 15
(B) 17
(C) 19
(D) 25
(E) 30

OK. So let’s set up the average table, using the colors in the problem above to show what came from where (if you’re colorblind I’m so sorry):

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 7 packages 30 pounds -3 packages -60 pounds +1 package +15 pounds -2 packages -90 pounds

See where everything’s coming from? When we have an average (30 pounds for the initial 7 packages) we put it in the average column. When we have a sum, we put it in the sum column. We keep track of whether the packages are being delivered or picked up with + and – signs. Now let’s fill in the rest of the table just to see how everything works together (calculated values are in bold type…make sure you understand where they come from):

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 7 packages 30 pounds 210 pounds -3 packages 20 pounds -60 pounds +1 package 15 pounds +15 pounds -2 packages 45 pounds -90 pounds

We filled everything in just for practice, but for the next step we’re only going to need the values in the outer columns. So we know we started with a total weight in the truck of 210 pounds. We dropped off 3 packages weighing 60 pounds, picked up 1 package weighing 15 pounds, and dropped off 2 more weighing 90 pounds. Using our table, we can easily see that the number of packages left on the truck is 7-3+1-2 = 3, and the total weight on the truck is 210-60+15-90 = 75. So the average weight of the 3 packages left on the truck is 25 pounds! That’s choice (D)!

$\tiny&space;\inline&space;\dpi{300}&space;\fn_cm&space;[Number\:&space;of\:&space;Values]\times&space;[Average\:&space;of\:&space;Values]&space;=&space;[Sum\:&space;of\:&space;Values]$
 7 packages 30 pounds 210 pounds -3 packages 20 pounds -60 pounds +1 package 15 pounds +15 pounds -2 packages 45 pounds -90 pounds = 3 packages 25 pounds = 75 pounds

Some last notes about the average table before I give you a few more practice problems:

• You can only add or subtract up and down the outer columns. Try adding and subtracting averages and you’ll get all screwed up. You can only use the middle column for
• calculating the sum by multiplying the number by the average, or
• calculating the average by dividing the sum by the number.
• This will work with questions that have variables instead of numbers, as long as you follow the rules (but it’s a good idea to substitute real numbers to make your life easier whenever possible).
##### Try these two for practice:

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