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Pls help.........

Hey Binish,

I am not sure abt the logic. But is the OA (D)?

Statement (1):
Let us draw a perpendicular from b to extended AC to the left and let us name the intersection point as H.

Then,
Area of (triangle ADb) = 1/2 * AD * bH
Area of (DCb) = 1/2 * DC * bH

Now according to the statement, both areas are equal. Hence the heights are equal and so are the sides AD and DC.

Hence, AD = DC and D is the midpoint of the side AC. Now in AbC, we have AD=DC, bD is common and the areas are equal.

Hence, bA=bC.
”AbC is isosceles. (Sufficient)

Statement (2):

This is same as statement 1 where the perpendicular is now inside the AbC.
Hence we can say that,
Area of (ADb) = Area of (DCb) = 1/2 * AC * bD

Hence, bA=bC.
”AbC is isosceles. (Sufficient)

Answer (D)

Would request Bhavin to clarify.

For the other DS question, here goes my way of doing the same:

i) if a is in S, then –a is in S, and
ii) if each of a and b is in S, then ab is in S.

Statement (1) :
1 is in S. Then -1 is in S
S ={-1,1}
(Nothing to say abt -4)...Insufficient

Statement (2):
2 is in S. Then -2 is in S
S ={-2,2}

Per rule 2 above,

New S = {.....,-4,-2,2,.....}
I believe that S here would be an infinite set as integers would increase per iteration as per rule 2 stated above.

Sufficient
Hence (b)


Sorry for the confusion b4.

rchuck21 wrote:
Hey Binish,

I am not sure abt the logic. But is the OA (D)?

Statement (1):
Let us draw a perpendicular from b to extended AC to the left and let us name the intersection point as H.

Then,
Area of (triangle ADb) = 1/2 * AD * bH
Area of (DCb) = 1/2 * DC * bH

Now according to the statement, both areas are equal. Hence the heights are equal and so are the sides AD and DC.

Hence, AD = DC and D is the midpoint of the side AC. Now in AbC, we have AD=DC, bD is common and the areas are equal.

Hence, bA=bC.
”AbC is isosceles. (Sufficient)

Statement (2):

This is same as statement 1 where the perpendicular is now inside the AbC.
Hence we can say that,
Area of (ADb) = Area of (DCb) = 1/2 * AC * bD

Hence, bA=bC.
”AbC is isosceles. (Sufficient)

Answer (D)

Would request Bhavin to clarify.

hey rahul...statement in red is incorrect...
your assumption implies 2 triangles with equal areas are congruent...do u think this is correct ??

There are infinite triangles with equal areas and not congruent...
From the figure itself we know 2 triangles have equal areas....ratio of the areas of 2 triangles with same _base_ is proportional to their height....and since 2 triangles have common vertex, hence heights are also equal...hence information in statement 1 is redundant...

[color=#008000]equal areas does not imply congruency [/color]

If BD is perpendicular to AC and D is midpt, hence BD is perpendicular bisector of AC...and any vertex on perpendicular bisector generates isosceles triangle...

Ans B..

Cheers !!

If x=-y, is (x-y)/(x+y) >1?

1.x>0
2.y less than zero

This is a question from OG and OA is E............but i think its B.......can anyone validate my reasoning...

cross multipling x-y>x+y
= 2y less than zero
=y less than zero

Hey Binish,

I think the OA in OG is correct. It is option (E).

In fact, both the statements give the same information.
Lets take Statement (1):
x > 0

If x = -y, and for x to be greater than 0, y has to be negative (y is less than 0).
Second statement gives exactly the same information.

So the information that we got is x is +ve while y is negative.
Lets take an example.

If x = 2, then y = -2
Put the values in the equation (x-y)/(x+y)
The solution is undefined as denominator is 0.

Whatever the values you put, the denominator is always 0 because of the condition, x = -y
Hence the solution is undefined.

Hence option (E) is the answer.