SAT Tip of the Week: Math in Translation

SAT Tip of the Week - Full For those of us who grew up speaking English, we rarely find a personal benefit from translating English into another language. One big exception to this is in the math world. We may find ourselves able to understand the most erudite texts with ease, but figuring out how to interpret mathematical terms can be difficult without a little translating. Here is a quick and easy guide to help translate our language of communication into a language of computation.

 

1. Define Your Variables

2. Look for key words (“Sum” “Product” “Is added to” “Is twice” “Fixed cost” “increases __ for every increase/rotation/” etc.)

3. Replace English with an equation

4. Check with real numbers for plausibility

Let’s look at an example of how you may use these steps on a problem on the SAT.

“A number is multiplied by 4. This product is the same as the sum of this initial number and 4. What is 3 times the initial number?”

In this problem we are simply told that we start with “a number”. If the problem gives us “a number” or “some number”, it is trying to communicate that we are dealing with a variable. Let’s define our variable as m. We could call it anything. The letter is a placeholder to indicate it is unknown at this time.

Now let’s look for key words. “Is multiplied by 4” and “is the sum of” indicate multiplication and addition respectively. So the problem is stating that if you times m by 4, it should equal m plus four or:

4m = 4 + m

3m = ?

I added the “3m = ?” because I like to write down the thing that I am solving for and quantity that we need is “3 times the initial number”. From here on, it should be a piece of cake to solve this equation.

4m = 4 + m

-m -m

3m = 4

/3 /3

m = 4/3

3m = 4

Let’s look at what I call a “Taxi Problem.” Taxi problems are not always about taxis; they are essentially derived from problems where a taxi is charging some amount per mile over a certain distance. This could also be used to define phone call charges or any situation where there are costs accrued after some quantity of distance, time, or use. This example would be a classic taxi problem:

“A taxi traveling m miles charges a fixed charge of $3.50 for the first mile and then c cents for every quarter mile traveled. What is an equation that describes the cost of taxi ride over 2 miles in terms of dollars?”

We can start at the very beginning and define our variables. I’m going to call the cost we are trying to find in dollars d. This may seem obvious, but be sure that the variable you are solving for is alone on one side of the equation. An equation for d should solve for d, which means having d by itself on one side of the equation. We have defined d as the cost in dollars, and m, the number of miles, and c, the cost per quarter mile in cents, are defined for us. Let’s get to work on finding those key words.

“Fixed charge for the first mile” is a big key phrase that tells us we are going to be adding the per mile cost to $3.50. We can pretty much guarantee our equation will look something like “3.5 +[per mile charges][number of miles]”. We are also told that the cost of the first mile is included in this fixed cost, so any per mile charges should be multiplied by m - 1 (instead of m) to reflect that the charges are for all travel after the first mile. This is a detail that many students miss the first time around.

“Cents for every quarter mile” indicates that the taxi will charge c cents four times for every mile after the first, so in order to convert the cost per quarter mile to a cost per mile, we will have to multiply the cost by four. Let’s skip to step four and check for plausibility. If the cost was 25 cents per quarter mile, every mile of travel after the first would cost 4 (25 cents) or one dollar. This demonstrates another problem. c is in cents right now, so if we plug in the work we just did to our initial work we would have an equation that looks like this:

d = 3.5 + 4(c)(m-1)

If we stick with c as 25 cents, then a trip of 3 miles would cost 3.5 + 4(25)(2) = $203.5! Even in NY, that is too much money. This shows us we still need to convert c to cents, which can be done by dividing c by 100. Let’s check our work with real numbers now.

d = 3.5 + 4(c/100)(m – 1)

d = 3.5 + 4(25/100)(2) = 5.5

This answer is much more reasonable and accounts for all of key phrases in the word problem.

Though word problems can be tricky, with a little translation they can become simple equations that are easy to solve or manipulate. Just remember to go through these steps and always check with real numbers to see if the equation is plausible. Happy test taking!

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David Greenslade is a Veritas Prep SAT instructor based in New York. His passion for education began while tutoring students in underrepresented areas during his time at the University of North Carolina. After receiving a degree in Biology, he studied language in China and then moved to New York where he teaches SAT prep and participates in improv comedy. Read more of his articles here, including How I Scored in the 99th Percentile and How to Effectively Study for the SAT.

 

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