FOR THE FIRST QUESTION

from the main statement:
AB=OC. OC=Radius=OB. This means that ∆ABC, and ∆BOC are isosceles.
Angle BAC=angle BOD, angle BCO=angle CBO. Angle BAO=x degrees
(1)
Angle COD=60 degr. Thus, ang COA=120.
∆ABC in isosceles so ang BAC=ang BOA=x
Ang ABO = 180-x-x=180-2x. angl CBO=180-(180-2x)=2x
∆BOC is isosceles so angl CBO=ang OCB=2x.
For ∆ACO we have x+2x+120=180, 3x=60, x=20.
Angle BAO=x degrees=20 degrees
From this we can found that ang OCA= 40 degrees
SUFFICIENT

(2)
This is the same that (1) is – from knowing that ang OCA= 40 degrees we can find Angle BAO=x degrees=20 degrees
You can use the same reasoning.
SUFFICIENT

So, the answer is D

Hope it will help

Thank u

FOR THE SECOND QUESTION:


From the main statement we deduce that a and b are either both positive or both negative. Only in this case (-a;b) and (-b;a) can be in one quadrant – in II or in IV

(1)
xy>0
x and y can both negative and positive
(x positive)(y positive)>0 will be in the same quadrant
(x negative)(y negative)>0 will NOT be in the same quadrant
It is INSUFFICIENT

(2)
ax>0
a and x can both negative and positive
(a positive)(x positive)>0
(a negative)(x negative)>0
But we don know is a positive or negative!
It is INSUFFICIENT

(1) and (2)
This means that a, b, x and y are either ALL positive or all negative.
Here I substituted positive and negative meanings and found that taking (1) and (2) I can answer that (-a;b), (-b;a) and (-x;y) will always be in the same quadrant.
ANSWER - C

Hey,
Sorry I didnt understand ur explanation for the 2nd question. Can u please explain it in any other way?

Thanks

I've made some corrections. may be it will help.
May be someone else will find another explanation - I dont know what to add.
You can plug in some numbers to try to solve it - this can help

In this problem, we basically have to prove that for the points (-a,b), (-b,a) & (-x,y) to lie in the same quadrant, it is necessary that a,b,x & y, all have the same sign (All positive or all negative)
This can b proved only by combining the 2 statements

Thanks...

4meonly wrote:FOR THE SECOND QUESTION:


From the main statement we deduce that a and b are either both positive or both negative. Only in this case (-a;b) and (-b;a) can be in one quadrant – in II or in IV

(1)
xy>0
x and y can both negative and positive
(x positive)(y positive)>0 will be in the same quadrant
(x negative)(y negative)>0 will NOT be in the same quadrant
It is INSUFFICIENT

(2)
ax>0
a and x can both negative and positive
(a positive)(x positive)>0
(a negative)(x negative)>0
But we don know is a positive or negative!
It is INSUFFICIENT

(1) and (2)
This means that a, b, x and y are either ALL positive or all negative.
Here I substituted positive and negative meanings and found that taking (1) and (2) I can answer that (-a;b), (-b;a) and (-x;y) will always be in the same quadrant.
ANSWER - C

Plz. help me understand this.
I am not clear how(-a,b) and(-b,a) can be only in II and IV.
According to me slope is coming positive so it should be in I or III
Is this the only method to solve this question.
Also when we say (x negative)(y negative)>0 why they are not in same quadrant. III quadrant contains x and y both negative.
Sorry I am not able to understand the concept here.