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Question states that side length are integers
Therefore, sq. root a^2 + b^2 = int
Therefore, a^2 + b^2 = int^2

But we don't know if a & b are int.

Now, on to the statements:

St1. area = 35 (1x35 or 5x7)
Therefore AO can be 1 or 35 or 5 or 7

Clearly insuff.

St2. a and b are integers but this doesn't tell us anything about their values

Clearly insuff.

Combined,
both 5 and 35 satisfy the equation a^2 + b^2 = int^2

Therefore, insuff. Answer is E.

Bunuel chetan2u is this approach correct or is my answer a fluke? Also, it took me a really long time to solve this question (I was stuck between C and E for the longest time), IS THERE A QUICKER WAY TO SOLVE THIS QUSTION?

Yes, you are correct and that is the way to do it.

What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.

Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35
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If the side lengths of rectangle ABCO are integers, What is the value of a+b?

Statement I tells us that Area is 35.
Since the sides are integers, the sides could be 7*5 or 1*35.
Possible answers: 1+35 or 5+7
Insufficient

Statement II tells us that a and b are integers.
OA is dependent on (a,b) but nothing on values of sides.
Insufficient

Combined
What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.

Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35

Both 1+35 or 7+5 possible.
Insufficient

E
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nahid78
If the side lengths of rectangle ABCO are integers, What is the value of a+b?

Statement I tells us that Area is 35.
Since the sides are integers, the sides could be 7*5 or 1*35.
Possible answers: 1+35 or 5+7
Insufficient

Statement II tells us that a and b are integers.
OA is dependent on (a,b) but nothing on values of sides.
Insufficient

Combined
What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.


Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35

Both 1+35 or 7+5 possible.
Insufficient

E


Please guide where I am wrong as I believe when taken together (a) + (b) should be sufficient:

On the basis of above discussion, we know that the length of OA has to be either 1, 35, 5, or 7

We do know that "a" and "b" both have to be integers and positive integers based on the picture given.

so Length of OA= a^2 + b^2 where "a" and "b" both have to be positive integers.

from 1 to 35 there are only 5 perfect squares that are 1, 4, 9, 16, and 25

Length of OA has to be sum of 2 perfect squares from these, therefore

Length of OA cannot be equal to 1 (1+ anything will greater than 1)
Length of OA cannot be equal to 35 (the closest we can get is 9+25 which is 34)
Length of OA cannot be equal to 7 (none of the pairs of perfect squares satisfy the equation)

The only possible value for Length of OA is 5 as 2^2+1^2 is equal to 5.

So there is only two possible situations:

a=2, b=1 i.e. a+b =3

or

a=1, b=2 i.e. a+b= 3

i.e. a +b =3 only in any case.

So both conditions combined should be sufficient.
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chetan2u
nahid78
If the side lengths of rectangle ABCO are integers, What is the value of a+b?

Statement I tells us that Area is 35.
Since the sides are integers, the sides could be 7*5 or 1*35.
Possible answers: 1+35 or 5+7
Insufficient

Statement II tells us that a and b are integers.
OA is dependent on (a,b) but nothing on values of sides.
Insufficient

Combined
What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.


Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35

Both 1+35 or 7+5 possible.
Insufficient

E


Please guide where I am wrong as I believe when taken together (a) + (b) should be sufficient:

On the basis of above discussion, we know that the length of OA has to be either 1, 35, 5, or 7

We do know that "a" and "b" both have to be integers and positive integers based on the picture given.

so Length of OA= a^2 + b^2 where "a" and "b" both have to be positive integers.

from 1 to 35 there are only 5 perfect squares that are 1, 4, 9, 16, and 25

Length of OA has to be sum of 2 perfect squares from these, therefore

Length of OA cannot be equal to 1 (1+ anything will greater than 1)
Length of OA cannot be equal to 35 (the closest we can get is 9+25 which is 34)
Length of OA cannot be equal to 7 (none of the pairs of perfect squares satisfy the equation)

The only possible value for Length of OA is 5 as 2^2+1^2 is equal to 5.

So there is only two possible situations:

a=2, b=1 i.e. a+b =3

or

a=1, b=2 i.e. a+b= 3

i.e. a +b =3 only in any case.

So both conditions combined should be sufficient.

Length of OA will not be \(a^2+b^2\), but \(OA^2=a^2+b^2\)
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chetan2u, if you are ok with revisiting this one, I am struggling why 1, 35, 5, or 7 can all be the hypotenuse. Given the image, wouldn't it be clear that OA is longer than OC? Hence, out of the 1*35 and 5*7 combination, only the shorter width, 1 and 5, are possible values for OA the hypotenuse.
chetan2u


Yes, you are correct and that is the way to do it.

What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.

Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35
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Asan92
chetan2u, if you are ok with revisiting this one, I am struggling why 1, 35, 5, or 7 can all be the hypotenuse. Given the image, wouldn't it be clear that OA is longer than OC? Hence, out of the 1*35 and 5*7 combination, only the shorter width, 1 and 5, are possible values for OA the hypotenuse.


You cannot conclude from the diagram that OA is longer than OC. On the GMAT, diagrams are not drawn to scale unless explicitly stated, so relying on the picture to compare lengths would be an assumption.

And second, you don’t need to worry about this kind of question for the GMAT Focus Edition. Geometry is no longer tested in GMAT Focus. So you can safely ignore it.
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Asan92
chetan2u, if you are ok with revisiting this one, I am struggling why 1, 35, 5, or 7 can all be the hypotenuse. Given the image, wouldn't it be clear that OA is longer than OC? Hence, out of the 1*35 and 5*7 combination, only the shorter width, 1 and 5, are possible values for OA the hypotenuse.

Hi
Bunuel has already explained it. The point on sketches not being accurate is also mentioned in OG.
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avggmatstudent23
Question states that side length are integers
Therefore, sq. root a^2 + b^2 = int
Therefore, a^2 + b^2 = int^2

But we don't know if a & b are int.

Now, on to the statements:

St1. area = 35 (1x35 or 5x7)
Therefore AO can be 1 or 35 or 5 or 7

Clearly insuff.

St2. a and b are integers but this doesn't tell us anything about their values

Clearly insuff.

Combined,
both 5 and 35 satisfy the equation a^2 + b^2 = int^2

Therefore, insuff. Answer is E.

https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel

Bunuel chetan2u is this approach correct or is my answer a fluke? Also, it took me a really long time to solve this question (I was stuck between C and E for the longest time), IS THERE A QUICKER WAY TO SOLVE THIS QUSTION?
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Bunuel chetan2u is this approach correct or is my answer a fluke? Also, it took me a really long time to solve this question (I was stuck between C and E for the longest time), IS THERE A QUICKER WAY TO SOLVE THIS QUSTION?

You don’t need to worry about this kind of question for the GMAT Focus Edition. Geometry is no longer tested in GMAT Focus. So you can safely ignore it.
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