If a point P(x,y) inside the above square ABCD is chosen at random and

I'd say (1) is sufficient and allows to say that the probability is equal, because there is abcolutely no topologic difference between x and y, so x/y < 5 is as much probable as y/x < 5. (2) is irrelevant.

Still, I'd be happy to see a 'proper' solution, mine is more of a guess.

Also, I'm not sure if we can consider AB parallel to X based on the picture; if not, (1) is insufficient.

IMO graphic approach can be used to solve the problem. Represent the inequalities as lines- y/x<5 =>y/x=5, y=5*x and y/x=1/5 => x=5y.
Now the values on the right of the first line are <5, and the values below the second line are <1/5.
Both areas are equal so A

A...

It's more to do with symmetry

Only visualization can help solve this quickly...

Answer is (C)

A thing one has to realise is that everything that is < 5 of something is ALSO < 0,2 of the same thing.

Thus the probability that y/x is < 5 is only = the probability that y/x is < 0,2 if no points equal the sum of 0,2 or more.

Since the first statement is that 0,0 is in the middle and the second statement what constitutes the vertices of the box you just have to draw the box and try for yourself.

The only points that makes y/x < 0,2 are those where either x or y are negative or zero (50% of the box).

The only points that are NOT < 5 are (y=5, x= 1) and (y=-5, x=-1) since they = 5

Seems to be a very tough Q. Forget the probability first, to calculate Y/X we need to know the point. If we can fix the point (x,y) to a single point or to a set of points then we can calculate the probability

(1) talks about midpoint of AD. No idea bout co ordinates of any four vertices. Insuff

(2) talks about area. ie 100 SO each side = 10 and diagonal = 10 sqrt 2

Can we arrive at more than one set of vertices such that the side is 10? I think so. to arrive at 100 (square of 10) we can use (8,6) (5 sqrt 2, 5 sqrt 2)? Probably Insuff

Together we can fix the (x,y) co ordinates and can start calculating Probability, which probably is unnecessary as we will get a yes/no

OA & OE?

Looking for a good explanation

The answer is A

From the information provided, we draw two lines: y=5x and y=1/5x

From (1) we know that square ABCD is symmetrical through point 0(0,0). So O is the centre of square ABCD. If P(x,y) is inside the square,
The probability that y>5x = (the square of the the quadrilateral formed by line y=5x and the square ABCD - on the right side line y=5x) / the square of ABCD = 1/2
The probability that y>1/5x = (the square of the the quadrilateral formed by line y=1/5x and the square ABCD) / the square of ABCD = 1/2
The probability that y>5x / The probability that y>1/5x = 1

From (2) alone, we don't have enough data to answer the question.

SO THE ANSWER IS A

Please find attached the graph for your information

IMO both are needed. I used the areas method. But there is a problem it contains. We are not dealing with equalities but dealing with inequalities. Think the line y=x/5. When x and y is positive, the points between the line and the x axis gives us the (x,y) combinations in which y/x<1/5. But when we are going to the negative side. Everything changes. Think the point (-4,-1) It is under the y=x/5 line. But y/x is greater than 1/5. So the point combinations which satisfy y/x<1/5 are the points that lie between the x axis and the line. This explanation works for the line y=5x. So we can not know the exact measures of the areas that lie between the lines and the x axis only with the coordinate of midpoint. So both are needed

Let I make this right. At first, I'm sorry for making some mistakes in the figure.

Infact, the quadrilateral MBCN represents the position of P(x,y) in the square ABCD so that x>1/5y (or y/x<5) not x>5y as in the figure
Similarly for the quadrilateral QCDT represents the position of P(x,y) in the square ABCD so that x>5y (or y/x<1/5) not x>1/5y as in the figure

So, the question is the probability that y/x <5 (x>1/5y) is how much greater than the probability that y/x < 1/5 (x>5y)

From (1), we have that, no matter how small or large the square ABCD, as long as it is symmetrical through center O
+ The probability that y/x <5 <--> (x>1/5y), given P(x,y) in square ABCD, equals (area of MBCN / area of ABCD) = 1/2
+ The probability that y/x <1/5 <--> (x>5y), given P(x,y) in square ABCD, equals (area of QCDT / area of ABCD) = 1/2

Please be advised that the question does not ask us to find the point P(x,y) satisfying both inequalities y/x<5 and y/x<1/5. But it asks whether we could calculate the probability that y/x<5, given that P (x,y) is in ABCD and probability that y/x<1/5, given that P(x,y) is in ABCD.

So I think that data provided in (1) is sufficient to answer the question. And (2) alone, is not sufficient.

Therefore, the answer is A.

-points in the quadrants (x>0 and y<0) and (x<0 and y>0) are all satisfy y/x<5 and y/x < 1/5 (e.g. points (-1,3) and (2,-3) )
-points that lie over the line y=x/5 and lie under x axis are all satisfy y/x<1/5 (e.g. point (-1,-0.1) )
-points that lie over the line y=5x and lie under x axis are all satisfy y/x<5 (e.g. point (-0.1, -1) )

These do not seen in your figure anonym There is a problem in your solution.

both are needed

For me the best explanation was provided by BG.

With (1) we know that the area under y=<5x is equal to the area under y=(1/5)x

Therefore, the probability that y=<5x is equal to the probability that y=(1/5)x.

Answer is A.

The size of the box is irrelevant. Absolute values don't matter here.

I'd be very happy if somebody would prove me wrong.

Cheers

+1 for 'A'.

Size of the square does not matter. Prob is -> Area of Sqare/2 in both case.

I am going with A as well. Anyone has the OA and OE?

From what you explained A makes a lot of sense.

The probability would be greater than 1/2, because all the points in the second quadrant in the square would satisfy the inequality. Though the answer would remain same because the same would apply to the other inequality as well.

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