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# Plz help to solve this ps

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Plz help to solve this ps [#permalink]

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09 Dec 2009, 00:07
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tricky ps

If equation encloses a certain region on the coordinate plane, what is the area of this region?

|x|/|2|+|y|/|2| =5

20
50
100
200
400

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40
56
72
81
104

If a circle is inscribed in a square, then the area outside the circle is what percent of the total area of the square (approximately)?

14%
18%
22%
28%
30%
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Re: Plz help to solve this ps [#permalink]

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09 Dec 2009, 06:01
raghavs wrote:
tricky ps

If equation encloses a certain region on the coordinate plane, what is the area of this region?

|x|/|2|+|y|/|2| =5

20
50
100
200
400

if x = 0 y = 10
if y = 0 x = 10
if x = 0 y= -10
if y = 0 x = -10

four triangles with base 10, height 10 area = 50
50*4 = 200
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Re: Plz help to solve this ps [#permalink]

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09 Dec 2009, 06:11
raghavs wrote:
tricky ps

If a circle is inscribed in a square, then the area outside the circle is what percent of the total area of the square (approximately)?

14%
18%
22%
28%
30%

let side of square be 2a then radius of circle will be a. Area of square = 4a^2 and Area of Circle = πa^2

So area outside circle is 4a^2 - πa^2

% of area outside = (4a^2 - πa^2)/a^2 *100 = 22% approx.
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Re: Plz help to solve this ps [#permalink]

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10 Dec 2009, 03:12
Thanks Lagomez and Kp1811 for the solution.
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Re: Plz help to solve this ps [#permalink]

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11 Dec 2009, 18:56
2)

The first digit can only be 8 or 9.

If it is 8, the second digit can assume 9 values, 5 of which are odd and 4 of which are even. The third digit must be odd, hence it can assume 5 values at the most. If the second digit is even, we can pick any of these 5 values, hence we have 4(#of suitable even second digits)*5(#number of suitable odd third digits)=20. If the second digit is odd, the third has still to be odd, but it cannot assume the same value as the second, hence we have 5(#of suitable odd second digits)*4(# of suitable odd third digits)=20. The total in the 800-899 range is 40.

If the first digit is 9, the second digit can assume 4 even values and 5 odd values. If the second digit is even, the third digit can be any odd digit but 9, hence we have 5(# of suitable even second digits)*4(# of suitable odd third digits)=20. If the second digit is odd, the third digit cannot be 9 or equal to the second digit, and it still has to be odd, hence 4(# of suitable odd second digits)*3(# of suitable odd third digits)=12. The total in the 900-999 range is 32.

The total number of odd numbers with different digits above 800 is then 72.
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Re: Plz help to solve this ps [#permalink]

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12 Dec 2009, 04:07
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Re: Plz help to solve this ps [#permalink]

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12 Dec 2009, 05:26
ans for 1) 200...
take 4 cases where both x n y are +,both -, and two cases where each are opposite to each other...
ull get a square with sides given by <(0,10)(10,0)> ,<(0,-10)(10,0)> , <(0,10)(-10,0)> and <(0,-10)(10,0)>...
the diag of sq will be 10+10=20...so each side is 10(2)^1/2....
2)
kp1811 wrote:
raghavs wrote:
tricky ps

If a circle is inscribed in a square, then the area outside the circle is what percent of the total area of the square (approximately)?

14%
18%
22%
28%
30%

let side of square be 2a then radius of circle will be a. Area of square = 4a^2 and Area of Circle = πa^2

So area outside circle is 4a^2 - πa^2

% of area outside = (4a^2 - πa^2)/a^2 *100 = 22% approx.

shud be 18%... as mentioned above but while dividing it has to be 4a^2 and not a^2
3) 72....
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Re: Plz help to solve this ps [#permalink]

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10 Jan 2010, 19:55
chetan2u wrote:
ans for 1) 200...
take 4 cases where both x n y are +,both -, and two cases where each are opposite to each other...
ull get a square with sides given by <(0,10)(10,0)> ,<(0,-10)(10,0)> , <(0,10)(-10,0)> and <(0,-10)(10,0)>...
the diag of sq will be 10+10=20...so each side is 10(2)^1/2....
2)
kp1811 wrote:
raghavs wrote:
tricky ps

If a circle is inscribed in a square, then the area outside the circle is what percent of the total area of the square (approximately)?

14%
18%
22%
28%
30%

let side of square be 2a then radius of circle will be a. Area of square = 4a^2 and Area of Circle = πa^2

So area outside circle is 4a^2 - πa^2

% of area outside = (4a^2 - πa^2)/a^2 *100 = 22% approx.

shud be 18%... as mentioned above but while dividing it has to be 4a^2 and not a^2
3) 72....

Hi.. plz help me with this calculation..

% of area outside = (4a^2 - πa^2)/4a^2 *100 = 18% , how do you calculate this to get 18%
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Re: Plz help to solve this ps [#permalink]

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10 Jan 2010, 22:01
Question c)
----------

Let d represent the length of the square's side. Therefore, the radius of the circle is d/2.

Area of square = $$d^2$$
Area of circle = (pi)$$(d/2)^2$$ = pi*(d^2)/4

The area of of the square less the inscribed circle is:
$$d^2$$ - pi*(d^2)/4
= $$\frac{d^2*(4-pi)}{4}$$

As a percentage of the square's area, we get

= $$\frac{d^2*(4-pi)}{4}$$ * 1/d^2

Simplifying, we get $$\frac{4-pi}{4}$$
= 1-pi/4
= 1-(22/7)/4
= 1-11/14
= 3/14

Since 3/14 > 3/15, and 3/15 is 20%, we get 22% as the answer.
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Re: Plz help to solve this ps [#permalink]

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11 Jan 2010, 05:16
gmatJP wrote:

Hi.. plz help me with this calculation..

% of area outside = (4a^2 - πa^2)/4a^2 *100 = 18% , how do you calculate this to get 18%

(4a^2 - πa^2)/4a^2 *100 =a^2(4-pi)/4a^2*100=(4-22/7)/4*100=(6/7)*100/4=600/28=22%..
and sorry i meant 22% when i wrote 18%
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Re: Plz help to solve this ps [#permalink]

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12 Jan 2010, 16:09
for Q1, why can't we take the coordinates to be(+-8,+-2) and why only integers?
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Re: Plz help to solve this ps [#permalink]

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12 Jan 2010, 17:34
$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

After solving you'll get equation of four lines:

$$y<0$$ and $$x<0$$ --> $$y=-10-x$$
$$y>0$$ and $$x<0$$ --> $$y=10+x$$
$$y>0$$ and $$x>0$$ --> $$y=10-x$$
$$y<0$$ and $$x>0$$ --> $$y=x-10$$

If you draw these four lines (line segments) you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin.

Diagonal of this square would be 20, so the $$Area=\frac{20*20}{2}=200$$. Or the $$Side= \sqrt{200}$$, $$Area=\sqrt{200}^2=200$$.

Hope it's clear.
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Re: Plz help to solve this ps [#permalink]

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30 Apr 2010, 00:59
Bunuel wrote:
$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

After solving you'll get equation of four lines:

$$y<0$$ and $$x<0$$ --> $$y=-10-x$$
$$y>0$$ and $$x<0$$ --> $$y=10+x$$
$$y>0$$ and $$x>0$$ --> $$y=10-x$$
$$y<0$$ and $$x>0$$ --> $$y=x-10$$

If you draw these four lines (line segments) you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin.

Diagonal of this square would be 20, so the $$Area=\frac{20*20}{2}=200$$. Or the $$Side= \sqrt{200}$$, $$Area=\sqrt{200}^2=200$$.

Hope it's clear.

I still dont understand how area becomes 200..
First, the question says certain region then how do you know its square or rectangle and not triangle..
Second, if each side of the length of square is 10 then area should be 100 no?

Re: Plz help to solve this ps   [#permalink] 30 Apr 2010, 00:59
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