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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
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 [#permalink]
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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 [#permalink]
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 [#permalink]
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 [#permalink]
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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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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
hey adkikani Bunuel chetan2u

Can 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 [#permalink]
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Expert Reply
rahul16singh28 wrote:


Attachment:
The attachment 5033f36b-5ff1-481c-bda2-71f312ce5212.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:
angle-320-degrees.png
angle-320-degrees.png [ 3.66 KiB | Viewed 16824 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:
images-13.png
images-13.png [ 1.56 KiB | Viewed 16829 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 [#permalink]
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 [#permalink]
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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.
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
16x is an angle, so 16x<180.... x<12

I do not understand this statement. my assumption would have been that 16x<=90 not 180
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
nausherwan wrote:
16x is an angle, so 16x<180.... x<12

I do not understand this statement. my assumption would have been that 16x<=90 not 180


Hi nausherwan, I think the key is not to be distracted by how the initial diagram is being drawn (definitely not to scale). The key principle to follow here is that opposite complementary angles split by a straight line (or lines) should add up to be 180 degrees. There are different combinations that can be explored in the diagram above, so it's about finding one that works AND that is stated in the answer choice.

Hope this clarifies.

-tinytiger
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
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