V(x,y) is a point on the line segment from P(0,20) to : PS Archive
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# V(x,y) is a point on the line segment from P(0,20) to

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V(x,y) is a point on the line segment from P(0,20) to [#permalink]

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02 Aug 2006, 10:25
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V(x,y) is a point on the line segment from P(0,20) to Q(30,0) and a rectangle is constructed so that OV is one of its diagonals, where O has coordinates (0,0). If x is an integer, which of the following could not be the area of the rectangle?

(A) 54 (B) 108 (C) 126 (D) 144 (E) 150
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Re: PS: Area of Rectangle [#permalink]

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02 Aug 2006, 11:38
kevincan wrote:
V(x,y) is a point on the line segment from P(0,20) to Q(30,0) and a rectangle is constructed so that OV is one of its diagonals, where O has coordinates (0,0). If x is an integer, which of the following could not be the area of the rectangle?

(A) 54 (B) 108 (C) 126 (D) 144 (E) 150

Damn this is brutal..took me way more than 2 mins to solve, but here it is:

Equation of line PQ is:

y= - 2/3x + 20

=> V can be represented as (x, -2/3x+20)

If this is the diagonal, then the area is x*y = 20x -2/3x^2
Our requirement is to find x such that x is a +ve I.

From a,b,c,d,e we get 5 equations:

a. 20x - 2/3x^2 = 54
or 10x -x^2/3 = 27
or 30x - x^2 = 81
or x^2 - 30x + 81 = 0
x^2 -27x -3x +81 = 0
x= 27 or x=-3

possible.

b. 20x - 2/3x^2 = 108
or 10x-x^2/3 = 54
or 30x - x^2 = 162
or x^2 - 30x + 162 = 0

There are no whole roots for the above equation, therefore B is the answer. Look no further.
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02 Aug 2006, 12:22
B...

eqution of line passing thru the points : y = -2/3x +20

Since the diagonal is a line connecting the origion and a point on this line. the sides of the rectangle will be y and x.

So basically area is x* (-2/3x +20) =?

Now if you see each of the options in the question is a whole number, so which means that the value of x has to be a multiple of 3 ( -2/3*X^2 +20 is a whole no. )

To save time I listed down all multiple of 3 between 0-30 (range given in q) and did a calculation to get the areas.
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Re: PS: Area of Rectangle [#permalink]

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02 Aug 2006, 17:20
F(x) = -(2/3)X + 20
(X, -(2/3)X+20)

-(2/3)X+20 = integer.
X is multiple of 3. X=3m (m is an integer, 0<m<10)
Area = X(-2/3X+20) = 3m*(20-2m)

plug in m => 3m*(20-2m)
m=1 : 3*(20-2) = 3*18 = 54
m=2 : 6*(20-4) = 6*16 = 96
m=3 : 9*(20-6) = 9*14 = 126
m=4 : 12*(20-8) = 12*12 = 144
m=5 : 15*(20-10) = 15*10 = 150

Thus B
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02 Aug 2006, 23:59
Great work!

Also for a given area A x=(30 +- sqrt(900-6A))/2

Only if 900-6A is the square of an even integer will x be an integer.
02 Aug 2006, 23:59
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