Another poorly written word problem

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t construct sentences that can be understood by elementary students (or their parents).

poorwordproblem

 

Another poorly written word problem

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this gem that was assigned to Texas 5th graders is a doozy.

Source: https://www.facebook.com/photo.php?fbid=10203673212963557&set=o.121316371311714&type=1&theater

I get what the textbook wants the student to do: rounding to the nearest $10 and developing the skill of approximating a sum without actually laboriously computing the sum exactly. According to this logic, $54.26 gets rounded to $50 and $34.34 gets rounded to $30. So Fran spends about $80, and (according to this logic) so she has about $20 left. So the textbook wants the student to answer B.

But this is wrong on so many levels only destined to confuse parents and children alike.

First, the actual answer, without using approximations, is $11.40. Undeniably, $5 (answer C) is closer to $11.40 than $20 (answer B).

Second, it’s entirely reasonable and appropriate for students to approximate to either the nearest dollar or else the nearest $5. Indeed, nothing in this problem says that the rounding must occur to the nearest $10… I’d imagine that this could only be inferred from the context of other problems on the page. By rounding to the nearest dollar, Fran would have about $12 left. By rounding to the nearest $5, Fran would have about $10 left. And there’s nothing “wrong” with either of these approximations.

Third, in real life, Fran would not say that it would cost about $80 to buy the sneakers and shirt. In real life, Fran would always round up to be sure that she has enough money to complete the transaction. If Fran keeps rounding to the nearest $10, she’ll end up short of money at the cash register sooner or later. So while rounding up or down may be appropriate for some problems, it probably shouldn’t be advocated for the sake of financial literacy.

In short, this problem does little except confuse students and get them to hate math. I do advocate that children should be able to estimate a sum without finding it. This is one of the standards for teaching Texas 5th graders mathematics:

Number, operation, and quantitative reasoning. The student estimates to determine reasonable results. The student is expected to use strategies, including rounding and compatible numbers to estimate solutions to addition, subtraction, multiplication, and division problems. Source: http://ritter.tea.state.tx.us/rules/tac/chapter111/ch111a.html

That said, this particular problem is an exceptionally poor way of determining whether students have acquired that skill. It’s hard to believe that this problem survived the proofreading process before the textbook was published.

See also: https://meangreenmath.com/2013/06/20/acceleration/

Acceleration

The following two questions came from a middle-school math textbook. The first is reasonable, while the second is a classic example of an author being overly cute when writing a homework problem.

  1.  A car slams on its brakes, coming to a complete stop in 4 seconds.  The car was traveling north at 60mph. Calculate the acceleration.
  2. A rocket blasts off. At 10 seconds after blast off, it is at 10,000 feet, traveling at 3600mph.  Assuming the direction is up, calculate the acceleration.

For the first question, we’ll assume constant deceleration (after all, this comes from a middle-school textbook). First, let’s convert from miles per hour to feet per second:

60 ~ \frac{\hbox{mile}}{\hbox{hour}} = 60 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 88~ \frac{\hbox{feet}}{\hbox{second}}

The deceleration is therefore equal to the change in velocity over time, or

\frac{-88 ~ \hbox{feet/second}}{4 ~ \hbox{second}} = -22 ~\hbox{ft/s}^2

Now notice the word north in the statement of the first question. This bit of information is irrelevant to the problem. I presume that the writer of the problem wants students to practice picking out the important information of a problem from the unimportant… again, a good skill for students to acquire.

green line

Let’s now turn to the second question. At first blush, this also has irrelevant information…  it is at 10,000 feet. So I presume that the author wants students to solve this in exactly the same way:

3600 ~ \frac{\hbox{mile}}{\hbox{hour}} = 3600 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 5280 ~ \frac{\hbox{feet}}{\hbox{second}}

for an acceleration of

\frac{5280 ~ \hbox{feet/second}}{10 ~ \hbox{second}} = 528 ~\hbox{ft/s}^2

The major flaw with this question is that the acceleration of the rocket completely determines the distance that the rocket travels. While middle-school students would not be expected to know this, we can use calculus to determine the distance. Since the initial position and velocity are zero, we obtain

x''(t) = 528

x'(t) = \int 528 \, dt = 528t + C

x'(0) = 528(0) + C

0 = C

\therefore x'(t) = 528t + 0 = 528t

x(t) = \int 528t \, dt = 264t^2 + C

x(0) = 264(0)^2 + C

\therefore x(t) = 264t^2 + 0 = 264t^2

Therefore, the rocket travels a distance of 264 ~ \hbox{feet/second}^2 \times (10 ~ \hbox{second})^2 = 26400 ~ \hbox{feet}. In other words, not 10,000 feet.

As a mathematician, this is the kind of error that drives me crazy, as I would presume that the author of this textbook should know that he/she just can’t make up a distance in the effort of making a problem more interesting to students.