Qn1. The answer should be D

let x= L PRS
y= L PQR

we need to find x-y

by stmt 1
L QPR=30

By properties of triangle=> external angle = sum of opposite internal angle)

x=y+30=> sufficient

By stmt 2
y+180-x=150
=> sufficient

Answer is D

Qn 2
t/s*(-1/root(3))=-1---> since lines OP and OQ are perpendicular

also let
s^2+t^2=r^2=3+1=4--->(since P lies on the same circle and O is origin)

solve for s

s^2+3s^2=4=>s=+/-1

Qn3
arc length=r*angle(in radians)
since ABC is equilateral therefore angle subtended by the arc=4*pi/3
we get r=18/pi
d=36/pi=(36/22)*7=126/[email protected]

thanks, OAs are D, B and C.

LevelOne wrote:Please see attachments.

My thoughts:

1) Stmt (1) & (2) give exactly the same info that L QPS = 30. How do we go from here?

2) I dropped a perpendicular line from point P to x-axis to create a point M. So it looked like OM = -√3. As point Q is symmetrical to P, I thought s should be equal to √3. Where am I going wrong here?

3) I used this formula and it worked: length of arc = 24 = 240/360*2pi*r. Any alternative solutions?

2) I dropped a perpendicular line from point P to x-axis to create a point M. So it looked like OM = -√3. As point Q is symmetrical to P, I thought s should be equal to √3. Where am I going wrong here?

GOLDEN RULE which I follow:

Never assume anything on the figures unless it is mentioned or marked on the figure, unless mentioned assume P can be anywhere on the circle.