Gmat Pret-PS-Series, Circle.

Question

Can anyone help me in this.

GMAT TIGER · Answer

neeraj.kaushal · Answer

I could solve first one, but second is tricky as by all means it looks like as if value of s has to be square root of 3. The only way s can come out to be 1 is if x and y coordinates flip over . But cannot justify how.

Second is as follows.

First term = 1/2
second term -1/4
third term = 1/8
forth term = -1/16
fifth term = 1/32 
sixth term= -1/64
seventh= 1/128
eighth = -1/256
ninth= 1/512
tenth = -1/1024

add and solve u get appx 0.3 which is between 1/4 and 1/2

greenoak · Answer

For the problem with the circle and the right triangle: the answer is 1.

Solution.

First, don’t fall into a trap: the triangle is NOT necessary symmetrical relatively y axis.
I think we need coordinate geometry approach here.

1) From PO=OQ => s^2+t^2=4
2) From PO^2 + OQ^2 = PQ^2 => (s+√3)^2 + (t-1)^2 = 8

Now, we have system from two equations: s^2+t^2=4 and s^2+t^2 + 2√3s – 2t = 4
To solve it, first substitute 4 for s^2+t^2 in the second equation => √3s=t. Then, from the first equation, we have 4s^2=4 =>s^2 = 1. Since Q is in the first quadrant, s>0 => s=1.

greenoak · Answer

For the sequence problem:

While the solution suggested by the other poster is correct, I think that you might be interested in more general approach.

Notice that the series is geometric progression with the first term a1= ½ and the common ratio r=-1/2. Thus, we can use the formula for the sum of geometric progression: 
S(n) = a1*(1- r^n)/(1-r). 
For S(10) this gives us ½*(1-1/1024)/(3/2) = (1-1/1024)/3 which is a bit less than 1/3.
So the answer is D.

maratikus · Answer

second problem: a(n) converges to (1/2)/(1-(1/2))=1/3
error term is below: 1/2^10

a(10) is between 1/3-1/2^10 and 1/3+1/2^10 -> D

greenoak · Answer

I don't

This is my stupid typo. It should be PO=OQ, of course. I'll change the post with explanation. Thanks, fresinha12!

alpha_plus_gamma · Answer

second problem:

answer B.1.

calculate the slope of line P0 = -1/sqrt(3)
PQ is perpendicular to P0 so slope of PQ = sqrt(3)

equation of PQ, y = sqrt(3)x + 0 = > t= sqrt(3)* s ................(1)

PO=PQ (radii)

PO= (sqrt(3))^2 + 1^2 = 2 = PQ

(PQ)^2= 2^2 = s^2 + t^2

4= s^2 + (sqrt(3)*s)^2 = 4*s^2

s^2 = 1

s=1

Gmat Pret-PS-Series, Circle.

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