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rahul16singh28
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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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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Hello, why 15 and 20 are not possible? Thanks
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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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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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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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rahul16singh28


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 18495 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 18487 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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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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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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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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DO these questions still appear in Focus GMAT? KarishmaB Bunuel
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DO these questions still appear in Focus GMAT? KarishmaB Bunuel

I'd say it's unlikely because of the geometry content.
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