Step-by-Step 4/15/14

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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

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“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.”

Ugh.

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:

q*p=c

Why not use variables that make sense?

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

p=1.25*(100-18)%/100%=1.25*.82=1.025

c=24,000*1.025=24,600

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.

quantQ
 Translating Directly from Word Problem to Algebra

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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.

quantQ
 Algebraic Expressions

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On the GRE and GMAT, word problems are composed of encrypted expressions and equations, and nothing more. When you face a word problem, your highest priority should be to translate it directly into equations. Only after you have properly transformed the words of a word problem into equations should you pursue problem-solving.

What are the expressions that compose the following word problem?

Question of the Day

A car covers a distance of 39 km in 45 minutes. It moves at a speed of x km/h for the first 15 minutes, then moves at double the speed for the next 20 minutes, and finally moves at its original speed for the rest of the journey. What is the value of x in km/h?

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This is a tough problem, but gets easier as you translate one equation at a time. Once we’re able to completely view the description of the problem in terms of equalities and expressions, the problem instantly becomes solvable. The first and most important step of solving any word problem is going straight from words to math!