kitchen table math, the sequel: the brick problem

Friday, June 11, 2010

the brick problem

The pattern shown above is composed of rectangles. This pattern is used repeatedly to completely cover a rectangular region 12L units long and 10L units wide. How many rectangles of dimension L by W are needed?

answer choices:
A) 30
B) 36
C) 100
D) 150
E) 180

solution here and here

The Official SAT Study Guide


Catherine Johnson said...

I love the internet.

SteveH said...

When I first look at a problem like this, I don't necessarily come up with the easiest solution. Often, when people give solutions to problems like this, they are carefully constructed and not the first approach they took. They present the cleaned up solution as if it is something everyone should come up with on their first attempt. If you are talking about a timed test, then studying an optimal solution may not be helpful for the next completely different problem you see.

In my teaching days, I used to have students give me new problems to do so that I could explain all of my missteps and thinking processes. It wasn't very helpful. They didn't have my knowledge or experience, and they couldn't learn it by listening to me.

Niels Henrik Abel said...

My initial solution was essentially the same as reply #3 at College Confidential - to me, it provides a more transparent approach to the problem.

Catherine Johnson said...

I got 200.


Tomorrow I will attempt to come up with 180.

VickyS said...

I got 200 too! What's wrong with us?

VickyS said...

Oh I know...we got lulled by the visual into thinking thet 2W=L (it looks like that along the bottom) instead of trusting the data which says 3W=2L! Clearly time to go to bed...

Barry Garelick said...

SteveH is correct. First attempts are clumsy. Here's my clumsy approach. I calculated the area of the rectangle made up of five bricks and got 2L(W + L).

Since W = (2/3)*L, then expanding the above expression and substituting, you get the area as (10/3)*L^2.

Area of the 12L by 10L rectangle is 120L^2, so I divide that by (10/3)L^2 and get 36. There are 36 5-brick rectangles in the large rectangle, so 36 x 5 = 180.

Not pretty but it works.

SteveH said...

I always hated explanations to problems that made it seem like that's what I should have thought of the first time I attempted the problem. They made me feel stupid.

My first reaction was that this was one of those packing or fitting problems where the number of bricks was going to be greater than the area required - some bricks might have to be cut. (or that it was a complex arrangement of the blocks)This led me away from an equation approach to the problem and to a pattern/picture approach. (However, it didn't say that the bricks couldn't be cut.)

In any case, I approached it as a (no-cut) fitting problem where I had to visualize how I could fit the bricks evenly into the area. The layout pattern in the picture was also misleading. It led me to believe that I should look at repeating that pattern over the full area. I soon dropped that idea to see how the W's would line up with the 10L or 12L side. After that, the answer came quickly.

How much time is allocated for this problem on the test?

Does anyone have any comments about questions like this in terms of speed of solution? I noticed that I spent some time on determining how complex the problem is. I know that my son overthinks many standardized test problems.

SteveH said...

I forgot to mention that even though they said that the pattern should be repeated, I decided to look only at the number of bricks I needed for the area. That could have been a critical mistake, but I was trying to find an answer fast.

What if you repeated the pattern, but one side only allowed room for part of the pattern. You wouldn't have an integer number of patterns, but you might have an integer number of bricks.

ChemProf said...

SteveH, for speed I do think you want to think of it as a unit. I looked at it and said
W = 2/3 L
So the unit is 2 L by 1 2/3 L.

The area to be covered is 12 L by 10 L, and 12 L = 6*2L while 10 L = 6 * 1 2/3 L. So we need 36 squares. Each one contains 5 blocks, so that's 180.

I admit I figured that it was going to be whole numbers, so checked what 10 / (1 2/3) was.

Barry Garelick said...

Hey Chem Prof, I think you and I solved it the same way! (My method is a few posts above.)

VickyS said...

I focused at first on the dimensions of the area to be covered. Since one side was 10L, looking at the unit pattern, I figured I'd extend the unit pattern 5X in the vertical direction to get 10L. Seemed logical, since 5x2=the 10L dimension. But, turns out that was the wrong way to go about it. Then (here is were I compounded my mistake) the bottom of the unit pattern looked like 1.5L, so I could extend the pattern horizontally 8X to get the 12L side. That gave me 40 unit patterns which gave me the wrong answer of 200 bricks.

To get the correct answer, actually the pattern should be extended vertically for a total of 6 units to get the 12L side, and horizontally also for a total 6 units (since the true dimension of the bottom is is 5/3 L) to get the 10L side. You get a 6x6 unit pattern, 36 units, each unit has 5 bricks, gives you 180.

It is a good thing 200 wasn't one of the answers on the test as I would for sure have gotten it wrong. But seeing that my answer wasn't among the options, I went back and did it algebraically and got the right answer.

I've always been a bit of a visual/spatial learner as it is now called, but I've also come to see the severe limitations of that mode of analysis. Too often it leads you in the wrong direction. Better to focus on the data!

palisadesk said...

It seemed like a simple ratio-type problem to me. I didn't need to use algebra or write anything down. I looked at the pattern, noted that 3W's equaled 2 L's, so to make the pattern unit have a width of 10L, I needed to line up 6 of those patterns horizontally. 3 of the patterns would measure 5 L (the 3 L's plus another 2 L's for the 3 W's). So, 2 sets of 5L and there's one axis with 6 patterns across. Then to get a length of 12L, I just need to stack up 6 more, given that the side dimensuion is 2 L each. Total, 36 of the pattern, and 5 rectangles for each pattern, and I came up with 180 total.

I could kind of see it in my head as I figured it out, but I did it a couple of times to make sure I came up with the same answer, because my mental math isn't that great.

Anonymous said...


I read this problem and didn't immediately see how I had enough info to solve it.

If I were taking the SAT, I'd skip it. I would teach anyone taking the SAT that they should *know the strategy* to answer the question by the time they've finished reading it, or they should move on and go back to it later.

The goal of SAT prep and math courses, generally, should be to study enough math that nearly all of the problems are that immediately clear to you--you know by the time you've finished reading them how you're going to solve them.

GoogleMaster said...

First thing to notice, as other people pointed out, is that 2L = 3W, so W = 2L/3. If I were doing this under SAT time constraints, I'd decide that working with fractions all the way through is "icky" and slows you down.

So let's call L = 3M, making W = 2M. The pattern area is (2L) x (L+W), or 6M x 5M. The larger area we're trying to cover is 12L x 10L, or 36M x 30M.

We need (36M x 30M)/(6M x 5M) = 36 patterns to fill the larger area. Each pattern has 5 rectangles, so we need 36 x 5 = 180 rectangles to fill the area.

No way would I have skipped this on the SAT. Although, truth be told, I probably would have rushed it and forgotten to multiply by 5 at the end!

ChemProf said...

Barry -- yeah, we did! When I read over yours at first I didn't quite see it.

I'd also agree with Allison that I'd probably skip it on the SAT (but I am crummy at spatial stuff -- can't pack a trunk either). If you do all the problems that you can do immediately, then you'll have time at the end for those that you have to think about. You need to average about a minute per problem, but some problems should take you much less than that.

I also thought that W was 1/2 of L at first, just looking at the picture.

lgm said...

I solved this one similarly to palisadesk. With a length of 12L, and each quilt block 2L long, figured it would be 12/2= 6 quilt blocks long. Now to find out how many fit across. Observe 2L=3W. So, do it the easy way, avoiding fractions, and work with multiples of 3W. 3 quilt blocks is w+l+w+l+w+l= 3w+3L =2L+3L=5L. It needs to be 10L wide, so that's 6 quilt blocks needed. 6 long by 6 wide is 36 q blocks altogether. 5 rectangles per block means 36x5 or 180 rectangles needed.

The key is to keep it simple and straightforward, don't assume tricks...use easy multiples for mental math and don't rotate anything unless the problem can't be solved oriented as written.

Catherine Johnson said...



The ALEKS geometry course has SOOOO many of those problems I immediately thought that.

Catherine Johnson said...


I did notice that but still didn't get the right answer.

Anonymous said...

This problem, probably by design, is stated in such a way as to lead you down the garden path to an unwieldy solution. What is being asked for is just a ratio of areas, not a particular way to tile a large rectangle with smaller rectangles. The area of the large rectangle is (12L)(10L), which equals 120LL. The area of the small rectangle is LW. The asked-for ratio is therefore 120LL/LW, which equals 120(L/W). The diagram accompanying the problem shows that 2L=3W, which upon solving for L yields L=3W/2; dividing both sides by W yields L/W = 3/2. So the answer is 120(L/W), which is 120(3/2), which is 180. There is no need to count how many rectangles fit in vertical or horizontal directions, or worry about what units of length to use, etc. If it were not possible to tile the large rectangle with the 5-rectangle pattern, then none of the answer choices would be correct. And if there were several ways to do the tiling, there would still be only one answer, the above-mentioned ratio of areas.

I think it might be useful to come up with simple, "elegant" solutions to the practice problems, in the hope that under the pressure of the real test you might be less likely to go down the garden path.

The above is just an explication of Barry Garelick's solution, which he called "clumsy" and "not pretty", but it seems to be the simplest.

ChemProf said...

"The ALEKS geometry course has SOOOO many of those problems I immediately thought that."

That is an important point. We solve new math problems by fitting them into categories, and choosing the wrong category can really slow you down.

In figuring out categories for the SAT, it is useful to remember that it tends toward "whole number" problems.

Catherine Johnson said...

Good lord.

I have just this moment realized I misread the problem.

That's another thing: lack of fluency **seriously** impairs my ability simply to READ a word problem.

Ed was complaining last week that C. isn't reading the problems correctly, which he attributed to 'careless error' and possibly 'test anxiety.'

Now that I'm finding myself not reading problems correctly (this has happened enough that it's a pattern), I realize that misreading problems is a symptom of the fundamental problem: lack of fluency.

I read text - including word problem text - fluently when I'm already fluent in the subject.