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Inequalities

Algebra: Inequalities

Read the following SAT test question and then select the correct answer.

Always read the question carefully, identifying the bottom line.  Assess your options for reaching the bottom line and use the most efficient method to attack the problem.  When you have an answer, loop back to verify that your answer matches the bottom line.

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On the line above, if AB < BC < CD < DE, which of the following must be true?

Bottom Line: wotf must be true = ? (which of the following)


Assess your Options:  For a “wotf” question, you will have to look at the answer choices.  Most students will start with “A,” so Knowsys recommends that you start with “E.”  You may also find that this is a good problem to use the strategy of plugging in numbers.

Attack the problem:  Take a look at your answer choices:
(A) AC < CD
(B) AC < CE
(C) AD < CE
(D) AD < DE
(E) BD < DE

(E) BD < DE  Look up at the figure.  On the figure, does BD look smaller than DE?  No!  It looks slightly larger.  You know that the figure is not drawn to scale, but the figure does give you one possible depiction of the rule.  Use the figure!  If it is possible for BD to be bigger than DE, then this answer is incorrect because you are looking for something that must be true.  Eliminate this choice.

(D) AD < DE  Look up at the figure.  The figure shows you that it is possible for AD to be larger than DE.  Eliminate this choice.

(C) AD < CE  These lengths are very similar on the line.  Break each length down into the parts that compose it so that you can make a precise comparison.  For example, AD contains AB + BC + CD.  CE contains CD + CE.  You now have: AB + BC + CD < CD + DE.  When you have the same thing on both sides of an equation, it cancels.  Eliminate the CD.  You now have AB + BC < DE.
You cannot come to a conclusion about these lengths.  If you want to prove this, try plugging in numbers.  Suppose AB starts at 10 and each portion along this line gets larger by 1.  AB = 10, BC = 11, CD = 12, DE = 13.  Is 10 + 11 < 13?  No.  Eliminate this choice.
(B) AC < CE  This one looks like it could be true, based on the figure.  See if you can prove it.  Break it down into its parts just as you broke down the last answer choice.  AC contains AB + BC.  CE contains CD + DE.  At first it seems as if you cannot compare these either because all of the numbers are different.  Try plugging in the same values as you used before: AB = 10, BC = 11, CD = 12, DE = 13.  Is 10 + 11 < 12 + 13?  Yes!  Will this work for all numbers?  Yes!  You are adding a small number plus a medium number and comparing it to a big number plus an even bigger number.  The former will always be smaller than the latter.  Once you know this, you do not even need to check (A).

(A) AC < CD You can tell from the figure that this does not have to be true.

Loop back:  You solved for what must be true, so you should select the answer you found.

The correct answer is (B).


On sat.collegeboard.org, 68% of the responses were correct.

To get help preparing for the SAT exam, visit www.myknowsys.com!

ACT Question of the Day:

If you have gone 4.8 miles in 24 minutes, what was your average speed, in miles per hour?

Your bottom line here is in miles per hour.  That would be miles over hours.  Your distance (miles) is in the correct unit, but your time (minutes) is not.  You know that there are 60 minutes in an hour.   Find the fraction of an hour that was spent traveling. Take your minutes and put them over the total minutes in an hour:


Now you know that you went 4.8 miles in .4 hours.  How many miles per hour was that?  Divide 4.8 by .4 and you will see that the answer is 12. 

Note: You can do this in your head if you realize that this is the same thing as dividing 48 by 4.  This whole problem can be done in seconds if you know your times table all the way up to 12. 

Look down at your answer choices.

(A)  5.0
(B) 10.0
(C) 12.0
(D) 19.2
(E) 50.0

The correct answer is (C).


For the ACT Question of the Day, visit http://www.act.org/qotd/.

To get help preparing for the ACT exam, visit 
www.myknowsys.com!