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"Triangle JKL measures 90 deg" - Veritas Prep Test

This topic has 1 expert reply and 1 member reply
krithika1993 Junior | Next Rank: 30 Posts
Joined
23 Jul 2016
Posted:
16 messages

"Triangle JKL measures 90 deg" - Veritas Prep Test

Post Sat Sep 10, 2016 3:10 pm
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): 240


Answer for B): 100


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 time-consuming. How would I proceed to answer this question in another, more efficient way?

Thank you.

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Post Sun Sep 11, 2016 7:52 am
krithika1993 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): 240


Answer for B): 100


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 time-consuming. How would I proceed to answer this question in another, more efficient way?

Thank you.
Another little trick here: be on the lookout for classic pythagorean triples, in this case, 5x: 12x :13x

When you see 260, hopefully you recognize that it's a multiple of 13. If 13x = 260, x = 20, 5x = 100, and 12x = 240.

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Gaurav Mittal Senior | Next Rank: 100 Posts Default Avatar
Joined
06 Apr 2016
Posted:
33 messages
Followed by:
1 members
Upvotes:
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GMAT Score:
740
Post Sat Sep 10, 2016 6:27 pm
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(676-400) 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

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