Tough(?) Algebra(?) Problem!

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Tough(?) Algebra(?) Problem!

by razorback » Thu Oct 06, 2011 2:09 pm
From MGMAT...Their explanation is just raw number crunching. Is there a way to set this up algebraically?

As a bicycle salesperson, Norman earns a fixed salary of $20 per week plus $6 per bicycle for the first six bicycles he sells, $12 per bicycle for the next six bicycles he sells, and $18 per bicycle for every bicycle sold after the first 12. This week, Norman earned more than twice as much as he did last week. If he sold x bicycles last week and y bicycles this week, which of the following statements must be true?

I. y > 2x

II. y > x

III. y > 3

A) I only
B) II only
C) I and II
D) II and III
E) I, II, and III

OA: D

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by bhavsk9 » Thu Oct 06, 2011 2:50 pm
Norman's salary can be algebraically expressed as:
20 + 6(6) + 12(6) + 18(total - 12)
where total = the total number of bicycles sold during that week.

The first three terms in the above equation are constant. So, the only way to double Norman's income is to increase the sales.

Considering statement III, y > 3 has to be true because y > 12. That leaves us with answer options D and E.

Consider x and y as the sales of previous and current week respectively as mentioned in the question.
The variable part can be represented as

18(y - 12) > 2[18(x - 12)]
y - 12 > 2(x - 12)
y + 12 > 2x

OR y + k > 2x

Also, x and y cannot be negative. Therefore, y > 2x doesn't hold true.

That leaves us with answer D.

Moreover, if x = 18 (for example), y + 12 > 36 OR y > 24. => y > x.

Hope this helps!

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by razorback » Thu Oct 06, 2011 10:48 pm
Yup. thanks!