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

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!


Image


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]
Carcass wrote:
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)


Is question asking for the minimum value of x?
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
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shvndry wrote:
Carcass wrote:
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)


Is question asking for the minimum value of x?



No basically we need to figure out WHEN x has a certain value to have lines parallel

The minimum value is x=10 but the question is different when asking

Quote:
[What is the minimum possible population that the least populated district could have?


Hope this helps
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
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Lets try with x and 2x ---> x +2x = 180 for line to be ||-----> 3x = 180 ---> x = 60, so options 10, 15, 20 are okay, with 10 being the 1st in the queue to satisfy the condition.

Ans is option B
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Re: In the figure above, all the marked angles are some multiple of x [#permalink]
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