720 (Q47, V41) - overall good, unhappy with Quant

First of all, a BIG THANK YOU to all on this forum. I visited this site to get help with my prep and it really paid off! I gave the GMAT the second time today. Got 720 (Q47, V41). !! :D :D The only issue is that on the first attempt I got 690 (Q47, V38) Can anyone let me know if the rather low Q sco...

We can also do this problem by picking nos. i) Since t/7 gives remainder 6 and t is a postive no, choose t= 13, 20, .. Substitute in original equation from the stimulus and we get a consistent value of 2 for r. ==> sufficient. ii) Since (t^2)/7 gives remainder 1, choose t= 6, 8 (try for atleast 2-3 ...

Looking at the co-ordinates of P, we can deduce that OP=2 (Draw a line perpendicular to the x-axis from P to point B and use the pythogoras theorem). If OP=2, PB=1 and OB=|root 3| we can deduce triangle OPB to be a 30-60-90 triangle with angle POB = 30. Draw a perpendicular line from Q to the x-axis...

Clarification for the part: H cannot be 2 since all three vertices lie on the circle. If b were 2, then the diameter would be the base and that is not possible for a triangle. For a triangle with the center of a circle as one vertex, the diameter cannot be the base of the triangle (since all three v...

This is good solution. Another way to do it:

y yards -- > 11 sec
x yards --> A secs

Cross multiplying ==> A= 11x/y

Of course this works only because speed is constant so we can use the principle of proportion.