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Difficulty:
85%
(hard)
Question Stats:
47% (01:54) correct
53%
(02:07)
wrong
based on 420
sessions
History
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Time
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If the side lengths of rectangle ABCO are integers, What is the value of a+b?
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
Attachments
Hard CO-Geo.png [ 24.86 KiB | Viewed 6686 times ]
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nahid78
St. 2 is insufficient
because a and b can be any integers to suffice the pythagoras theorem for triangle AO and either of the point of perpendicular at y-axis of x-axis. a and b can be 0.6 and 0.8 or 1 and 2 when OA is 1 or 5 respectively.
St. 1 is insufficient
because 35 = 5*7 OR 1*35
leading to different values of a and b(refer above)
Together 1 and 2
OA can't be 1 since that would lead a and b to be decimals. OA = 5 then a and b is 1 and 2 respectively.
SUFFICIENT.
Answer C.
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Posts: 39
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?
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?
avggmatstudent23
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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nahid78
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
Attachments
229CC05E-0B98-4D32-8B3A-D6B4935BBACD.jpeg [ 65.99 KiB | Viewed 5868 times ]
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chetan2u
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.
Sidban
Length of OA will not be \(a^2+b^2\), but \(OA^2=a^2+b^2\)
_________________
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
Asan92
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
Hi
Bunuel has already explained it. The point on sketches not being accurate is also mentioned in OG.
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Anointed
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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