Hi there,
I took the Veritas Prep test this morning, which included the following question on the IR section:
For triangle JKL, angle JKL measures 90 degrees, and side JL has a length of 260 centimeters. If side JK > side KL, which of the following could be a combination of the lengths of sides JK and KL?
A) JK:
50
100
120
200
240
B) KL:
50
100
120
200
240
Answer for A): [spoiler] 240
[/spoiler]
Answer for B): [spoiler] 100
[/spoiler]
Since the question says that JK > KL, I knew JK could not be 50 and KL could not be 240  I knocked down this combination. Then, I used the rule "the sum of any two sides of a triangle must be greater than the third side" to cross off A)/B) combinations 120/50, 120/100, and 200/50.
That left the following options:
A could be 240, in which case Y could be 50/100/120/200
A could be 200, in which case Y could be 50/100/120
At this point, I began using Pythagorean Theorem to figure out the answer but it just became too lengthy and timeconsuming. How would I proceed to answer this question in another, more efficient way?
Thank you.
"Triangle JKL measures 90 deg"  Veritas Prep Test
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 krithika1993
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Hi Krithika
You have broken it down to a good extent already and I dont see a way out of pythagoras from where you are. Just one thing, for A = 200, B cannot be 50.
From hereon, you can ignore unit's digit for ease since all of the unit's digits are 0.
We know that 26^2 is 676. Hence you now have to work with only two scenarios. When A is 20 (0 ignored), you need B to be sqrt(676400) and when A is 24, you need B to be sqrt (676  576)  this one has the answer for you.
I guess there is no way out of pythagoras here, but you can simply the things by dropping 0.
Cheers
You have broken it down to a good extent already and I dont see a way out of pythagoras from where you are. Just one thing, for A = 200, B cannot be 50.
From hereon, you can ignore unit's digit for ease since all of the unit's digits are 0.
We know that 26^2 is 676. Hence you now have to work with only two scenarios. When A is 20 (0 ignored), you need B to be sqrt(676400) and when A is 24, you need B to be sqrt (676  576)  this one has the answer for you.
I guess there is no way out of pythagoras here, but you can simply the things by dropping 0.
Cheers
Gaurav
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Another little trick here: be on the lookout for classic pythagorean triples, in this case, 5x: 12x :13xkrithika1993 wrote:Hi there,
I took the Veritas Prep test this morning, which included the following question on the IR section:
For triangle JKL, angle JKL measures 90 degrees, and side JL has a length of 260 centimeters. If side JK > side KL, which of the following could be a combination of the lengths of sides JK and KL?
A) JK:
50
100
120
200
240
B) KL:
50
100
120
200
240
Answer for A): [spoiler] 240
[/spoiler]
Answer for B): [spoiler] 100
[/spoiler]
Since the question says that JK > KL, I knew JK could not be 50 and KL could not be 240  I knocked down this combination. Then, I used the rule "the sum of any two sides of a triangle must be greater than the third side" to cross off A)/B) combinations 120/50, 120/100, and 200/50.
That left the following options:
A could be 240, in which case Y could be 50/100/120/200
A could be 200, in which case Y could be 50/100/120
At this point, I began using Pythagorean Theorem to figure out the answer but it just became too lengthy and timeconsuming. How would I proceed to answer this question in another, more efficient way?
Thank you.
When you see 260, hopefully you recognize that it's a multiple of 13. If 13x = 260, x = 20, 5x = 100, and 12x = 240.