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Two problems that are giving me hell

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Two problems that are giving me hell

by yasik19 » Wed Jan 21, 2009 10:59 am
Problem #1.

A rectangular tabletop consists if a piece of laminated wood bordered by a thin metal strip along its four edges. The surface area of the tabletop is x square feet, and the total length of the strip before it was attached was x feet. If the tabletop is 3 feet wide, what is its approximate length, in feet?

(A) 12
(B) 10
(C) 9
(D) 8
(E) 6

Answer is E

I have no idea what they are asking.

Problem #2.

For all real numbers v, the operation v* is defined by the equation v*=v-v/3. If (v*)*=8, then v=

(A) 15
(B) 18
(C) 21
(D) 24
(E) 27

Answer is B

Thanks a bunch.
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by dmateer25 » Wed Jan 21, 2009 11:09 am
Here is how i solved problem 2

You know the operation is v* is defined by v – v/3

So plug in your answer choices into this first. Then the answer you get from this needs to be plugged back in a second time.

Starting with A:

A) 15 – 15/3 = 15 – 5 = 10 now plugging in a 2nd time we get 10 – 10/3 and this doesn’t equal 8.

b) 18 – 18/3 = 18 – 6 = 12 now plugging in a 2nd time we get 12 – 12/3 = 12 – 4 = 8.

Bingo, B is the answer.
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by 480ocean » Wed Jan 21, 2009 1:44 pm
Dmateer225,

I'm having a hard time reconciling this in my mind. How would we know to plug the first answer (12) into the equation again?
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by awesomeusername » Wed Jan 21, 2009 3:20 pm
Here is my stab at #1:

So here is what we know:
1. The area of the tabletop is x square feet.
2. The width of the tabletop is 3 feet.
3. The length of the metal trim is x feet

The question is asking for the length of the tabletop. We have two equations:

Area = length times width
So we then have x = L x 3

The perimeter of the tabletop is:
P = 2L + 2W
So we know that the metal trim around the perimeter of the tabletop is x.
Thus,
x = 2L + 6

Now we have a system of two equations:
x = L x 3
x = 2L + 6

3L = 2L + 6
L = 6

(E)
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by yasik19 » Wed Jan 21, 2009 3:52 pm
dmateer25 wrote:Here is how i solved problem 2

You know the operation is v* is defined by v – v/3

So plug in your answer choices into this first. Then the answer you get from this needs to be plugged back in a second time.

Starting with A:

A) 15 – 15/3 = 15 – 5 = 10 now plugging in a 2nd time we get 10 – 10/3 and this doesn’t equal 8.

b) 18 – 18/3 = 18 – 6 = 12 now plugging in a 2nd time we get 12 – 12/3 = 12 – 4 = 8.

Bingo, B is the answer.
Thanks. But isn't the answer supposed to be something to the power of that same thing?
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by awesomeusername » Wed Jan 21, 2009 4:12 pm
We know v*=v-v/3. Then (v*)* =

((v-(v/3)) - ((v-(v/3))/3 =
* Above is because we replace v-v/3 where we see a v in (v-v/3)

((3v - v)/3) - ((3v - v)/9) =

(2v/3) - (2v/9) =

((6v - 2v)/9) =

4v/9

Thus 4v/9 = 8

v = 72/4 = 18

It appears that the above explanation of plugging and playing will be the simpler way to go if you're not comfortable with algebra. I'm just showing you that there are no squared variables involved.
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by chykie » Wed Jan 21, 2009 4:32 pm
my stab @2

(v*)* = 8

let's work the outer first
substitute v* = u
so (U)*= 8

u-u/3 = 8
(3u-u)/3 = 8
3u-u= 24
2u= 24
u= 12

Let's put back the U
so
u= 12
then (v-v/3)= 12
(3v-v)/3= 12
3v-v=36
2v=36
v=18

Hope it helps.
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by yasik19 » Wed Jan 21, 2009 7:25 pm
thank you both so much, I just now realized that i'm a moron. :oops:
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For Problem 1

by gaggleofgirls » Wed Jan 21, 2009 8:43 pm
In Problem one, they have told you the area of the table top as XsqFt and the perimiter of the table being the length of the edging before it was applied as X and the width as #.

So, the Area = Width * Length= 3L
The Perimeter = 2(Width) + 2(Length) = 2*3 + 2L = 6 + 2L

Since both the area and the Perimeter = X units, then they equal each other, so

3L = 2L + 6
L=6

Answer is E


-Carrie
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by chintudave » Wed Jan 21, 2009 8:54 pm
I would stick with awesomeusername's method for the 1 problem.

Here is my stab at the second one...

V* = V-V/3

i.e. V* is nothing but two third V... 2V/3
Extending the above ... (V*)* = 4V/9

Given 4V/9 = 8, so V = 18
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