Tuesday, June 5, 2012

this week's MGRE Math Beast Challenge

From here:

An item originally cost p dollars, where p > 0.

y – x > x – y


Quantity A
The price of the item if the original price were decreased by x%, increased by 35%, and then increased by y%

Quantity B
The price of the item if the original price were increased by x%, decreased by y%, and then increased by 35%


(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

Go to it! My own answer will eventually appear in the comments.


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2 comments:

Kevin Kim said...

I'm pretty sure the answer to this is (D)-- cannot be determined. Here's why I think so.

First, let's figure out how to express Quantity A.

"the original price [p] decreased by x%..."

p - (x/100)p, or p - (xp/100)

"...increased by 35%..."

1.35(p - (xp/100))

"...and then increased by y%..."

((100 + y)/100)•[1.35(p - (xp/100))]

Yikes. But bear with me.

Next, let's try to express Quantity B.

"...original price [p] increased by x%..."

p + (xp/100)

"...decreased by y%..."

((100 - y)/100)•(p + (xp/100))

"...and then increased by 35%..."

1.35•[((100 - y)/100)•(p + (xp/100))]

Double yikes.

Let's set Quantities A and B opposite each other, once again using "Q" to mean "undefined relationship-- greater than, less than, equal to... we don't know."

A Q B, or, more exactly:

((100 + y)/100)•[1.35(p - (xp/100))]

Q

1.35•[((100 - y)/100)•(p + (xp/100))]


We can divide by 1.35 on both sides, leaving us with this relationship:

((100 + y)/100)•(p - (xp/100))

Q

((100p - yp)/100) + ((100xp - xpy)/10,000)


We can then multiply by 100 on both sides to eliminate the "100" in the denominator:

(100 + y)(p - (xp/100))

Q

(100p - yp) + ((100xp + xpy)/100)


Using FOIL, Quantity A becomes:

100p - xp + yp - (xpy/100);

Quantity B, meanwhile, can be rewritten as:

100p - yp + xp + (xpy/100)

--so the relationship is now:

100p - xp + yp - (xpy/100)

Q

100p - yp + xp + (xpy/100)


We can subtract 100p from both sides, then factor out p and eliminate it:

y - x - (xy/100) Q x - y + (xy/100)

At this point, or perhaps a step or two before, you may have noticed that Quantity B is the negative of Quantity A, and since we were given that

(y - x) > (x - y) [i.e., y > x],

we will be tempted to assume that, since Quantity B is always the negative of Quantity A, Quantity A is always greater. The problem is that, if Quantity A is negative, then B will be positive, making B greater than A.

To be sure, though, we can try plugging in some numbers for x and y. I'll use coordinate pairs to express these values, with "QuantA" and "QuantB" representing the results of plugging those values in.

Scenario 1: (0,1)
QuantA = +1
QuantB = -1
∴ A > B

Scenario 2: (-1,0)
QuantA = +1
QuantB = -1
∴ A > B

Scenario 3: (-2,-1)
QuantA = +0.98
QuantB = -0.98
∴ A > B

Scenario 4: (-100, 100)
QuantA = +300
QuantB = -300
∴ A > B

Scenario 5: (-200,-100)
QuantA = -100
QuantB = +100
∴ A < B UH-OH!!!!

Scenario 5 is the kicker. Since it's possible to find values that shift the balance such that B is greater than A, then all bets are off. I think the answer is (D).

If you beg to differ, and/or if you think you have a more economical way to solve this problem, I'm all ears.

Kevin Kim said...

Note, too, that I've taken the liberty of assuming that x and/or y could be negative. The problem offers no evidence to the contrary. Essentially, then, the problem boils down to this:

x Q -x

--and without knowing whether x is positive or negative, there's no way to define Q as > or < or =.