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Compound Inequalities problem

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by Jim@StratusPrep » Mon May 12, 2014 10:48 am
The answer is c. You simply subtract 5! from each of the 3 different expressions.
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by Matt@VeritasPrep » Mon May 12, 2014 11:30 am
GenEdMBA wrote:Concept: Compound inequalities
Question: 5! + 3 < a < 5! + 8

a) 3 < a < 8
b) 0 < a < 5
c) 3 < a- 5! < 8
This can also be solved by actually computing 5! (5 * 4 * 3 * 2 * 1, or 120).

Then you have

120 + 3 < a < 120 + 8

or

123 < a < 128

From this you can tell that the first two inequalities won't work, while the last one is just

3 < a - 120 < 8

which is the same thing as the red inequality above, with 120 subtracted from all three parts.
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by GenEdMBA » Tue May 13, 2014 7:26 am
Answer: 3 < a-5! < 8

Don't make the mistake of subtracting 5! From only first and last term and choosing option (a) as answer. Subtract 5! From all 3 terms, so

5!+3<a<5!+8
Subtracting 5! From all 3 terms

3!< a- 5! < 8

Mantra: For compound inequalities (inequalities with more than one inequality sign) , perform operations on all terms together, not on first and last term alone.
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