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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]
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24 Jan 2017, 04:42
neeraj609 wrote: anceer wrote: A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?
A. 15 B. 57 C. 93 D. 109 E. 121 So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121. I am pretty sure it cant be that simple, what am i missing here How do you figure that we are looking for (a + b)^2? We are looking for the value of (a + b) only.
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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]
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24 Jan 2017, 06:07
VeritasPrepKarishma wrote: neeraj609 wrote: anceer wrote: A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?
A. 15 B. 57 C. 93 D. 109 E. 121 So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121. I am pretty sure it cant be that simple, what am i missing here How do you figure that we are looking for (a + b)^2? We are looking for the value of (a + b) only. Actually i reading (a+b)? wrongly as (a+b)^2. Thanks alot!



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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]
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24 Jan 2017, 07:38
A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?
Case 1: \(a^2 + b^2 = 11^2 = 121\) (unit digit 1)
Perfect squares end with any one of the following: 0; 1; 4; 9; 6; 5. To have unit digit 1, we have to use either of the combos: 0 and 1; 6 and 5. But none satisfies. X
Case 2: \(a^2 + 121 = b^2\)
Subcase (a) => \(a^2 = b^2  121\) => \(a^2 = (b+11) (b11)\) Since, \((b+11)\) is not equal to \((b11)\)and also \((b+11) > (b11)\)therefore \(b11 = 1\) and \(b+11 = a^2\) \(b=12\) and \(a^2=23\) (a is not a integer) X
Subcase (b)
=> \(121 = b^2  a^2\) => \((ba) (b+a) = 121\) Therefore, \(ba=1\) and \(b+a=121\)



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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]
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12 Sep 2017, 08:30
GMATinsight wrote: anceer wrote: A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?
A. 15 B. 57 C. 93 D. 109 E. 121 The trick for this question is If the side of a right angle triangle is a prime number then other two sides will be
Second Side = [(Prime)^2 + 1]/2
Third Side = [(Prime)^2  1]/2i.e. for One side = 11 Second Side = (11^2 1)/2 = (1211)/2 = 60 Third Side = (11^2 +1)/2 = (121+1)/2 = 61 I hope this helps! I personally find it an UNSUITABLE question for GMAT... GMAT Doesn't expect such tricks from students. Bookmarking this post and making a mental note. Definitely worth knowing!
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Re: A right triangle has sides of a, b, and 11, respectively,
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