Garbage In, Garbage Out


I recently read an article that proclaimed Washington D.C. as the nation’s fittest city. Having lived in DC for three years, I was curious, so I read through the article. The American College of Sports Medicine apparently decided that it would be interesting to figure out which is America’s fittest city. The people running the survey, not wanting to introduce their own personal biases figured that they would come up with an objective list of factors, collect the data, and declare a winner.

One of the factors that seemed to carry heavy weight is spending on parks. The ACSM reasoned that heavy spending on parks makes them attractive and safe places to spend time, and thus would likely lead to a more fit population. They suggest that municipalities target approximately $100 per capita in parks spending. They note to Washington’s great credit that the city spends $398 per capita.

Stop and think about that for a minute. Does that necessarily mean that DC is filled with many beautifully manicured parks where citizens can safely enjoy exercise? No. In fact, as a former resident of the area, I remember many times when we had to scramble to fit some suitable space for a softball game or some pick-up soccer. Very rarely could you find a space big enough to accommodate a good game without being right up against other groups of people or having to deal with concrete walkways cutting through the field.

So how could DC have such great park spending and not have great recreational facilities? Simply speaking, DC’s parks aren’t parks, they’re monuments. The National Mall is a huge park, but it’s not made for fitness and it’s not really made for locals. It’s made for the thousands of tourists that flock to the city. Spending money to repair the Washington Monument doesn’t make DC fitter, but it does go into DC’s park budget.

You get the idea here. The point is that when you put garbage information into your formula, you get garbage out. When you’re working through a math problem making a simple mistake in transcribing the information is just as deadly as a gap in math knowledge. Be diligent, be careful and make sure you’re putting the right information into your formulas, because if it’s not right going in, it’s garbage coming out.

 Step-by-Step 4/15/14


Today we’ll examine a word problem found in the GMAT Official Guide, 13th Edition, Section 5.3, Number 79.

“After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?”

Step 1: Dissect the information in the word problem- There’s a lot of information here and it’s essential to make sure you get it all figured out. A sketch can often help. Even though I lack artistic skills, this diagram shows you that you don’t need to be an artist to organize information.

Step 2: Figure out what you need to solve for- In this case we’re asked to find how much further south Bob can run. That’s the section of the purple line that goes south to the turn around point.

Step 3: Set up an equation to solve- How could we isolate the variable we need to solve for in this situation? Here’s one approach:

Additional distance south = Total distance south – 3.25m.

Total distance south = 1/2 total distance

Step 4: Solve your equation- Now that we know we can find our desired variable if we can find the total distance or the total distance south, we can make a second equation and solve. Here we know that Bob runs at a constant rate of 8 minutes per mile, and that the purple section of our graph takes him 50 minutes to run. So, we get this equation:

1 mile/ 8 minutes * 50 minutes = 6.25 miles

That tells us that the purple segment of our line is 6.25 miles. All that’s left to do is isolate the piece of the line we want. We can eliminate the 3.25 mile segment at the top, leaving 3 miles. All that’s left is the turn around which is equal parts: south and north. So, we divide that 3 mile segment in two and we get our answer: 1.5 miles (A).

Keep practicing and good luck!


 A Variable By Any Other Name


“Spacely Sprockets sells its widgets through a preferred buyer program. Buyers get a 1% discount off of the base price for each thousand sprockets they bought in the previous calendar year. If Ricky’s Rockets ordered 18,000 sprockets last year, and the base price per sprocket is $1.25, how much will Ricky spend on sprockets if he uses 50% more sprockets than he did during the previous year?”

First of all, if you look at that and see a jumble of words and start to panic. Don’t! We can turn this word problem into a manageable equation or set of equations if we carefully walk through it step by step.

Second, the way that you go about doing that can make the difference between an easy solve and a more challenging one. Here’s how many people would start…

“So Ricky uses x sprockets at a price of y for a total cost of z.”


What magical power does x have? Who came along and said, “When we teach children algebra, the first variable we give them must be x”? The answer, sadly, is probably that x was chosen because it was good for so little else. If we want a completely nondescript variable that isn’t likely to be confused with anything else, x is a good choice. Unfortunately, very few of the problems that we face are algebra for algebra’s sake where using generic variables makes sense. Take this problem for example.

We need to figure out the quantity and the price to figure out the total cost. So here’s a crazy idea:


Why not use variables that make sense?

q=18,000*150%/100 = 24,000



When you use variables that actually relate to the pieces of the problem, it’s going to make more sense how they fit together. Plus, you’re less likely to get them mixed up when you have an idea of what each variable means!

Don’t just throw an x in for an unknown because it’s x and it has some kind of magical powers. Use variables that make sense in your word problems, and be amazed at how it reduces the number of silly mistakes that you make.

 Translating Directly from Word Problem to Algebra


It goes without saying that your ability to transform word problems into equations and expressions is incredibly valuable on the GRE and GMAT. But don’t overthink it! Many of these translations can be done almost automatically. How would you translate the following word problem into expressions and equations you can use?

Question of the Day

If the square root of the product of and 4 is 6, and the arithmetic mean of  and  is , what is the value of ?

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Hopefully you are familiar with the word problem translation chart:

Words Math
More/Less Than Add To/Subtract From Next Term
Increased/Decreased Add/Subtract from original
Of/Times Multiply
Percent /100
What/A Number Variable
Is/Is Equal To/Equals =
Remaining/Left/Difference Subtract from original Quantity
Part Divide Part by Total
Total/Sum Add

Being able to translate words into expressions using this chart is not just necessary to get word problems right, it is also a key to doing them quickly! Using the values from this table allow you to bypass any heavy lifting for these problems and proceed directly to solving.