The answer should be seven.

The OA??

The answer is 4. I don't understand why it gets 4.

We have x+y>3 AND x^2-y^2<8. Therefore x-y< 8/3 (nearly equals 2.63) since x^2-y^2 = (x+y)(x-y).
x and y are integers, plus 0 < y < x then x-y equal z which is an integer, lower than 2.63. We only have (1,2)
Since the smallest number of x and y that product x-y = 2 is x=3 and y=1. But x^2-y^2=8 which is wrong.
Then only 1 case left which is x-y = 1. We can create a table of x and y from then to see the greatest value of x is 4

Is this really a PP2 Question. ? So much harder.

No.

The tag was wrong because the user used it. All and only the questions from powerprep are located here

https://gre.myprepclub.com/forum/gre-power ... -3118.html

If you see a question with the same tag but NOT from the directory above, then it is NOT a pp question. Absolutely.

Regards

Thanks, Although this question basically i found in Magoosh Very Hard section. Its' from Magoosh.

Given : x^2-y^2<8
x+y>3

(x+y)*(x-y)<8
since x+y>3
case 1 : x+y = 4 -> x-y = 1 -> x=2.5 not an integer
case 2 : x+y = 5 -> x-y = 1 -> x=3 and y=2
case 3 : x+y = 6 -> x-y =1 ->x=3.5 not an integer
case 4 : x+y=7 -> x-y=1 -> x=4 and y=3.
As question asks for max value of x, hence 4

OA answer should be 4 not 7.

Hi,

I tried this one using trial and error.

x^2-y^2= (x+y)(x-y)
What we can try to understand here is we can minimize the value of x-y to get the maximum value of x+y. As x^2-y^2<8 the maximum value it can take is 7. Thus we can make x+y=7 and similarly minimize x-y=1. The two number, which we can use to satisfy the eqn are 4 & 3.

Thus x=4

Hope this helps!

In the question, its clearly given that y<x. Therefore, you cannot take values of y more than x. I think someone above in the comments has termed this question as a bad one, but I think the questions states things clearly.

1. x^2-y^2<8
2. x+y>3
3. 0<y<x (positive integers)

Solution:

x^2-y^2<8
(x+y) (x-y) < 8 (Using property)

- we are given than x+y >3, therefore we can conclude that (x-y) has to be = 1. (Since, if (x+y >3 and are positive integers, we take the value to be 4. If x+y = 4 , then x-y cannot be 2 since (x+y) (x-y) < 8, therefore x-y = 1 (Next less positive integer before 2)

If x-y=1, then x=y+1 giving us that x and y are consecutive positive integers.

We check for (x+y> 3) and take x=3 and y=2, this holds true and we check for next set till one of the conditions above do not satisfy.

We get, x=4 and y=3, which holds true. We then check x=5 and y=4, this does not hold x^2-y^2<8 as x^2 - y^2 = 9.

Therefore, x has a maximum value of 4.

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