harsh.champ wrote:The position of the line segment AB at time t = 0 is shown in the figure given below, where 0° < x < 90°.
The point A is fixed and the line segment AB is rotating at a uniform rate in the X-Y plane in an
anticlockwise direction. In which of the four quadrants will the line segment AB lie at time t = 30
seconds?


A: The angular distance covered by the line segment AB in every 5 seconds is 615°.
B: The angular distance covered by the line segment AB in every 10 seconds is 640°.

[Figure provided in the attachment]

I don't know why your attachment appears blank to me.

NOW OK

Each statement alone is sufficient if the line segment AB is rotating at a uniform rate. [spoiler]D[/spoiler] is my answer. But, why the two statements contradict? Have I missed something in hurry?

IMO A
with x least and max value for s1) it lies in 3rd Q,..........suff
and s2) it lies in 4th and 1st q...suff

harsh.champ wrote:The position of the line segment AB at time t = 0 is shown in the figure given below, where 0° < x < 90°.
The point A is fixed and the line segment AB is rotating at a uniform rate in the X-Y plane in an
anticlockwise direction. In which of the four quadrants will the line segment AB lie at time t = 30
seconds?


A: The angular distance covered by the line segment AB in every 5 seconds is 615°.
B: The angular distance covered by the line segment AB in every 10 seconds is 640°.

[Figure provided in the attachment]

A - Sufficient [ angular distance covered in 30 seconds - 3 690 , it takes 360 degrees for one full rotation so, it will be 90 degrees anticlockwise from initial position, ie 90+45 =135 degrees ie in 2nd quadrant]
B - Sufficient [angular distance covered = 1920 sufficient to conclude that at t=30 segment will be in 2nd quadrant]

D for me

Hey IMO its E. can you please explain in details why D?

raisethebar wrote:Hey IMO its E. can you please explain in details why D?

But we would be loving to read why E from you first.

It's A. Please do not forget 0<x<90 here. To simplify, assume AB start at 0 degree on Ax line traveling counterclockwise.
1) 615*(30/5) = 3690. That is 3600+90 or 360*10+90. That tells us AB travels 10 times the circle and then 90 degree more. It will stop at Ay. But because it started with x degree, it actually stops at 90+x degree position -> II quadrant. -> Sufficient
2) 640*(30/10) = 1920. That is 360*5+120. That tells us AB travels 5 times the circle and then 120 degree more (somewhere in II quadrant). But because it started with x degree, it actually stops at 120+x degree position. If x is small, 120+x can be in II quadrant. If x is big, 120+x can be in III quadrant -> Insufficient

tienvunguyen wrote:It's A. Please do not forget 0<x<90 here. To simplify, assume AB start at 0 degree on Ax line traveling counterclockwise.
1) 615*(30/5) = 3690. That is 3600+90 or 360*10+90. That tells us AB travels 10 times the circle and then 90 degree more. It will stop at Ay. But because it started with x degree, it actually stops at 90+x degree position -> II quadrant. -> Sufficient
2) 640*(30/10) = 1920. That is 360*5+120. That tells us AB travels 5 times the circle and then 120 degree more (somewhere in II quadrant). But because it started with x degree, it actually stops at 120+x degree position. If x is small, 120+x can be in II quadrant. If x is big, 120+x can be in III quadrant -> Insufficient

Excellent!!

[spoiler]But I still didn't get why the two statements contradict.[/spoiler]

Agree with ans A