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Two persons start walking on a road that diverge at an angle of 120°.
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Updated on: 24 Oct 2018, 02:18
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68% (02:46) correct 32% (02:42) wrong based on 118 sessions
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Two persons start walking on a road that diverge at an angle of 120°. If they walk at the rate of 3 km/h and 2 km/h respectively. Find the distance between them after 4 hours. A. 5 B. 4√19 C. 7 D. 8√19 E. √19 Attachment:
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Originally posted by techiesam on 12 May 2016, 04:01.
Last edited by Bunuel on 24 Oct 2018, 02:18, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Two persons start walking on a road that diverge at an angle of 120°.
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12 May 2016, 05:16
techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 hi, Trignometry is NOT tested in GMAT so there has to be another solution.. SEE att image.. Draw an altitude from A to BC extende at D.. Triangle ACD is a right angle triangle, whose HYP is 2*4.. rest sides are CD4, opp 30 degree and AD  \(4\sqrt{3}\), opp 60 degree..Now take ADB.. c is hyp.. BD = 3*4 + 4 = 16 and AD = \(4\sqrt{3}\).. HYp = AC = \(\sqrt{16^2+(4*Sq Root 3})^2\) = \(\sqrt{304}\) = \(4\sqrt{19}\) B
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Re: Two persons start walking on a road that diverge at an angle of 120°.
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12 May 2016, 05:37
ronny123 wrote: chetan2u wrote: techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 hi, Trignometry is NOT tested in GMAT so there has to be another solution.. SEE att image.. Draw an altitude from A to BC extende at D.. Triangle ACD is a right angle triangle, whose HYP is 2*4.. rest sides are CD4, opp 30 degree and AD  \(4\sqrt{3}\), opp 60 degree..Now take ADB.. c is hyp.. BD = 3*4 + 4 = 16 and AD = \(4\sqrt{3}\).. HYp = AC = \(\sqrt{16^2+(4*Sq Root 3})^2\) = \(\sqrt{304}\) = \(4\sqrt{19}\) B hi, Don't be offended but isn't the maths inherent to subject correct ? if you use c^2 = a^2 + b^2  2abcos(120) = a^2 + b^2  2ab(1/2) = a^2 + b^2 + ab a = 2 x 4 = 8 b = 3 x 4 = 12 Putting these values, c^2 = 8^2 + 12^2 + (8)(12) = 64 +144 + 96 = 304 c = 4√‾19 Saying Trigo isn't tested on the GMAT isn't changing the solution, my friend Again, no offense meant. Hi, It is not meant for you .. It is for people who are preparing for GMAT so that they do not get into the area where it is NOT required.. You are most welcome to learn anything being tested or not, BUT people who are interested in just GMAT should not go away with thought that they have to do trignometry to do this Q.. I don't require to take offence or offend you, I am giving the WAY one should think about this solution
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Re: Two persons start walking on a road that diverge at an angle of 120°.
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Updated on: 12 May 2016, 05:32
hi techiesam,
can you please check if the question is correct, specifically whether one of the speeds is 3 kmph or 33 kmph Also, you can simply use the following formula for cos x to get your answer :
c^2 = a^2 + b^2  2abcos(120) = a^2 + b^2  2ab(1/2) = a^2 + b^2 + ab
Originally posted by ronny123 on 12 May 2016, 04:52.
Last edited by ronny123 on 12 May 2016, 05:32, edited 1 time in total.



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Re: Two persons start walking on a road that diverge at an angle of 120°.
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12 May 2016, 05:31
chetan2u wrote: techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 hi, Trignometry is NOT tested in GMAT so there has to be another solution.. SEE att image.. Draw an altitude from A to BC extende at D.. Triangle ACD is a right angle triangle, whose HYP is 2*4.. rest sides are CD4, opp 30 degree and AD  \(4\sqrt{3}\), opp 60 degree..Now take ADB.. c is hyp.. BD = 3*4 + 4 = 16 and AD = \(4\sqrt{3}\).. HYp = AC = \(\sqrt{16^2+(4*Sq Root 3})^2\) = \(\sqrt{304}\) = \(4\sqrt{19}\) B hi, Don't be offended but isn't the maths inherent to subject correct ? if you use c^2 = a^2 + b^2  2abcos(120) = a^2 + b^2  2ab(1/2) = a^2 + b^2 + ab a = 2 x 4 = 8 b = 3 x 4 = 12 Putting these values, c^2 = 8^2 + 12^2 + (8)(12) = 64 +144 + 96 = 304 c = 4√‾19 Saying Trigo isn't tested on the GMAT isn't changing the solution, my friend Again, no offense meant.



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Re: Two persons start walking on a road that diverge at an angle of 120°.
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19 Oct 2018, 05:31
techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 Dear Moderator, There is a typo in this question, the latter speed seems to be 3 kmph rather than 33 kmph, hope you will do the needful. Thank you.
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Re: Two persons start walking on a road that diverge at an angle of 120°.
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24 Oct 2018, 02:19
stne wrote: techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 Dear Moderator, There is a typo in this question, the latter speed seems to be 3 kmph rather than 33 kmph, hope you will do the needful. Thank you. ________________ Edited. Thank you.
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Re: Two persons start walking on a road that diverge at an angle of 120°.
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24 Oct 2018, 05:07
Longest side will be opposite to angle C. This means c should be more than 12. Also by triangle property 4<c<20 Combining, 12<c<20. By POE answer is B. PS: sqrt 19 is almost 4.5
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Two persons start walking on a road that diverge at an angle of 120°.
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20 Jul 2019, 08:34
chetan2u wrote: techiesam wrote: Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5 B. 4√19 C.7 D. 8√19 E. √19 hi, Trignometry is NOT tested in GMAT so there has to be another solution.. SEE att image.. Draw an altitude from A to BC extende at D.. Triangle ACD is a right angle triangle, whose HYP is 2*4.. rest sides are CD4, opp 30 degree and AD  \(4\sqrt{3}\), opp 60 degree..Now take ADB.. c is hyp.. BD = 3*4 + 4 = 16 and AD = \(4\sqrt{3}\).. HYp = AC = \(\sqrt{16^2+(4*Sq Root 3})^2\) = \(\sqrt{304}\) = \(4\sqrt{19}\) B Hi chetan2u, Just a small clarification required here, how can we conclude that the Hypotenuse is 8, why can't it be 12. How do we deduce that the one going along side b is the one walking at 2mph , the one walking along side b could also be walking at 3mph then the Hypotenuse could be 12. Am I missing anything? Thank you.
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Re: Two persons start walking on a road that diverge at an angle of 120°.
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22 Jul 2019, 21:12
You should modify the answers to be in the correct order of either increasing or decreasing value as per GMAT guidelines. Fastest way to do this problem is to use pythagorean theorum and find c = ~14. Since the angle is greater than 90, we know the answer must be greater than 14. 4sqrt(19) and 8sqrt(19) are your only two choices.
Besides the fact that 8sqrt(19) is way too huge, if you multiply 14 by 1.3 (~30% larger since 120 is about 30% larger) you get the the answer should be somewhere between 14 and 18, with a number closer 18. 4sqrt(19) = 17.4 = ~17.
This kind of math obviously wouldn't fly on a straight math test, but this is the fastest way to approximate an answer. And unless the GMAT decides to be extremely unforgiving and interested in testing math skills rather than reasoning  then there should only be 1 answer that fits the condition of between 14 and 18, closer to 18.




Re: Two persons start walking on a road that diverge at an angle of 120°.
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