2 GMAT PREP, Trick

[gmat78854p]

Can the wonderful people on this board show me the way to answer these 2. The first one seems like a trick question. thanks

[Trouble]

the best way to solve Q30 is if u thought of point p as a reflection of point Q so point s=1 (B)...What's the OA?

I am not sure of 31 but I think it's E. what's the OA?

[GMAT-HELP]

Q1)
Distance between points O and P = 2 (= sqrt(3+1) )
Distance between points O and Q = 2 (since OPQ is an isosceles triangle)
angle OPQ = OQP = 45 degrees.. (180 -90)/2

We have 2 eqns and 2 variables:
Distance between P and Q = 2*(sqrt2) from the property of right angle isosceles triangle(OPQ).

s^2+t^2 = 4 (distance between points O and Q)
(s+sqrt(3) )^2 + (t-1)^2 = (2*sqrt2)^2

Ans B.


Solving them, we get S =1 and t = sqrt3..

[gmat78854p]

OA for 30 is B. thanks guys

[GMAT-HELP]

Q 2)
Stmt 1.
No info about Marsha.. Insufficient to determine who drove the greatest distance..

Stmt 2:
No info on the other 2. Marsha drove 450 miles..
insufficient..

Combining them, still insufficient to figure out who drove the greatest..
Ans E.

Thanks,
GMAT-HELP

[wilsonucon]

I struggled with q2 in quantifying statement 1. It helped me to do it out mathematically to realize that there are an infinite number of situations where Al and Pablo can drive within an hour of each other and within 5 miles per hour of each other.

[kook44]

Q1)
Distance between points O and P = 2 (= sqrt(3+1) )
Distance between points O and Q = 2 (since OPQ is an isosceles triangle)
angle OPQ = OQP = 45 degrees.. (180 -90)/2

We have 2 eqns and 2 variables:
Distance between P and Q = 2*(sqrt2) from the property of right angle isosceles triangle(OPQ).

s^2+t^2 = 4 (distance between points O and Q)
(s+sqrt(3) )^2 + (t-1)^2 = (2*sqrt2)^2

Ans B.


Solving them, we get S =1 and t = sqrt3..

this question drives me nuts every time it comes up. Where do you get the part in blue?

[GMAT-HELP]

this question drives me nuts every time it comes up. Where do you get the part in blue?

Distance between P and Q.

[manish8109]

Kook44,
Since OP & OQ are the radius of the semicircle, means
OP=OQ

OP= sqrt[(0-sqrt(3))^2 + (0-1)^2] = 2

OP=OQ=2
so,
OQ= sqrt[(s-0)^2 + (t-0)^2]= s^2+t^2 =4 ---------(1)

PQ is the hypotenuse as OPQ is a right angled triangle, so

PQ^2 =OP^2+OQ^2
= 2^2+2^2 =8
=> PQ=2sqrt(2)

Now distance of PS can also be written as,

(s+sqrt(3) )^2 + (t-1)^2 = (2*sqrt2)^2

s^2+ 3+2sqrt(3)*s + t^2 +1- 2t = 8
Since s^2+t^2=4 , substituting in above eqn,

s*sqrt(3) = t--------(2)

Substituting the value of (2) in (1)

s^2+ t^2 = 4
=> s^2 + s^2 * 3 =4
=> 4s^2 =4
=> s^2 =1
=> s=1 or -1 as s lies in first quadrant do +ve value is to be considered, so s=1 (ANS)


Thanx

=

[jjaacc]

just wondering, can't we do the mirror image thing? thats how I"ve been solving such questions actually.