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In the figure above, all the marked angles are some multiple of x
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Updated on: 27 Jan 2018, 23:03
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34% (01:36) correct 66% (01:53) wrong based on 393 sessions
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Originally posted by rahul16singh28 on 27 Jan 2018, 20:07.
Last edited by Bunuel on 27 Jan 2018, 23:03, edited 2 times in total.
Renamed the topic and edited the question.




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Re: In the figure above, all the marked angles are some multiple of x
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27 Jan 2018, 20:48
rahul16singh28 wrote: Please give Kudos, if it helps you in Preparing for GMAT. An immediate observation... 16x is an angle, so 16x<180.... x<12 Only 8 and 10 are possible.. x=10 can be found from 2nd line and 5th line. B Let's check all.. Let number lines 1, 2,3,4,5 from left.. 1) let's see 1 PARALLEL to others. 1 ll 2.... x+2x=180...x=60... Not given 1 ll 3.....x=4x....not possible 1 ll 4....x+8x=180...x=20..E 1 ll 5...x=16x...not possible 2) 2 2 ll 3...2x+4x=180...x=30.. not given 2 ll 4 ...2x=8x...not possible 2 ll 5 ...2x+16x=180..x=10..B 3) 3rd 3 ll 4..4x+8x=180....x=15..C 3 ll 5...4x=16x ...not possible 4) 4th 4 ll 5..16x+8x=180... Not possible as X will not be an integer So possible values of x are 10,15 and 20 But possible is only 10.
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Re: In the figure above, all the marked angles are some multiple of x
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28 Jan 2018, 09:22
chetan2u wrote: rahul16singh28 wrote: Please give Kudos, if it helps you in Preparing for GMAT. An immediate observation... 16x is an angle, so 16x<180.... x<12 Only 8 and 10 are possible.. x=10 can be found from 2nd line and 5th line. B Let's check all.. Let number lines 1, 2,3,4,5 from left.. 1) let's see 1 PARALLEL to others. 1 ll 2.... x+2x=180...x=60... Not given 1 ll 3.....x=4x....not possible 1 ll 4....x+8x=180...x=20..E 1 ll 5...x=16x...not possible 2) 2 2 ll 3...2x+4x=180...x=30.. not given 2 ll 4 ...2x=8x...not possible 2 ll 5 ...2x+16x=180..x=10..B 3) 3rd 3 ll 4..4x+8x=180....x=15..C 3 ll 5...4x=16x ...not possible 4) 4th 4 ll 5..16x+8x=180... Not possible as X will not be an integer So possible values of x are 10,15 and 20 But possible is only 10. How did you know that 16x is an angle, so 16x<180.... x<12 What is the reasoning behind your solution? Your help is much appreciated



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Re: In the figure above, all the marked angles are some multiple of x
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31 Jan 2018, 19:41
Quote: How did you know that 16x is an angle, so 16x<180.... x<12
What is the reasoning behind your solution?
Your help is much appreciated Hi, Because the greatest angle is 180 (a line) => 16x is an angel (which is obviously shown in the figure) => 16x must be < 180 Hope it's clear



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Re: In the figure above, all the marked angles are some multiple of x
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09 Feb 2018, 03:19
Hello, why 15 and 20 are not possible? Thanks



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Re: In the figure above, all the marked angles are some multiple of x
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20 Feb 2018, 09:07
I am getting confused here. With x=15, the lines with 4x and 8x will also be parallel.
So, why are we bothered about any other line?



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In the figure above, all the marked angles are some multiple of x
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24 Apr 2018, 04:21
I did this,
for lines with angles 4x and 8x to be parallel => 4x + 8x = 12x = 180, x = 15 for lines with angles 2x and 16x to be parallel => 2x + 16x = 18x = 180, x = 10 (minimum so far) tried with x = 8, found none to be parallel.
So answer (B)



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In the figure above, all the marked angles are some multiple of x
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Updated on: 30 Aug 2018, 17:58
Corrected: I made table for all values of x and tried searching for equal values in columns A,C and D or in columns B, and E; or Supplementary angles one from either columns A,C, and D; and one from either B, and E.
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Originally posted by Blackishmamba on 23 Aug 2018, 17:23.
Last edited by Blackishmamba on 30 Aug 2018, 17:58, edited 1 time in total.



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Re: In the figure above, all the marked angles are some multiple of x
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23 Aug 2018, 17:41
hey adkikani Bunuel chetan2uCan you tell why x = 15, 18 or 20 not possible? Technically all of them can form an angle with the line (x  axis). Any angle greater than 180 will change the quadrant. Let me know what is incorrect with my understanding here
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Re: In the figure above, all the marked angles are some multiple of x
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23 Aug 2018, 21:46
rahul16singh28 wrote: Attachment: The attachment 5033f36b5ff1481cbda271f312ce5212.jpg is no longer available Tricky question and here is why  shown angles are angles between two lines (not line segments). So think about this  how will you show a 320 degrees angle? Attachment:
angle320degrees.png [ 3.66 KiB  Viewed 1596 times ]
Note that two intersecting lines make 4 angles around the point of intersection such that vertically opposite angles are equal. Can one of the angles be 320 degrees? The other needs to be 320 degrees too. If a is 320, c needs to be 320 too! Attachment:
images13.png [ 1.56 KiB  Viewed 1600 times ]
Hence, the options (C), (D) and (E) are not possible. Now it is just about looking for a pair of supplementary angles. If x = 10, 2x + 16x = 18x = 180 So 2x = 20 degrees and the angle supplementary to 16x would be 20 degrees too making these two lines parallel. Answer (B)
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Re: In the figure above, all the marked angles are some multiple of x
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08 Sep 2018, 08:24
IS THIS from gmatprep? pls, shot the screen



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In the figure above, all the marked angles are some multiple of x
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31 Dec 2018, 00:42
Hmmn my approach was to not be distracted by the picture and was similar to what was discussed above.
1) Look at the properties of parallel lines  for 2 lines to be parallel, there must be a perpendicular line joining these two lines (whether it's  or crossing both lines i.e. ).
2) Any angle subtended by the perpendicular line, would have be 90 degrees.
3) With this, I tried adding the different x's together with the following: a) x + 2x = 180, x = 60 (not in answer choice, immediately calibrate approach by trying combination of x's that can give me 180 degrees) b) 2x + 16x = 180 > x = 10
Answer: B.




In the figure above, all the marked angles are some multiple of x &nbs
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31 Dec 2018, 00:42






