My experience on GR9768

[amateur]

Today I tried GR9768 (I had already did it a few months back though not timed). Despite my best efforts I could manage to solve only 59 questions in the stipulated time.

I think one problem with me (and with some of us other test-takers) is that we tend to overspend time with the easy questions in the beginning and by the time we reach the 3rd hour of the test we are exhausted and have very little energy left for doing the more time consuming questions.

How about doing the test in the reverse order i.e. Try with Qs 50-66 first, then attempt Qs 25-50 and end the test with Qs 1-25?

[Kastro]

I take a little while to get my brain in gear for some reason, and tend to make silly mistakes on the first couple of questions, which I have to go back and correct later. Attempting to warm-up on the harder ones wouldn't work well at all for me.

[Nameless]

How about doing the test in the reverse order i.e. Try with Qs 50-66 first, then attempt Qs 25-50 and end the test with Qs 1-25?

Are you kidding ?
I thinks the test is not computer adaptive test so try to solve the easiest questions first, mark the questions you cannot answer at first and come back later if you have time

But if you try with the most difficult questions first and what will you do if you get stuck with one of them? I guess you even cannot stay focus ----->

[amateur]

I'll give this method a try with GR8767 (haven't tried it so far except Q 20 which was discussed in this forum a lot) and 9367 (tried a few forum posted Qs only) and report you back my experience.

[Nameless]

Today I worked on problem 16 : what is the volume of the solid formed by revolving about x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation : y=1/[sqrt(1+x^2)]

I tried by the following way :

V= \Pi*\Int_{0}^{\infinity}y^2dx and I got V=\Pi^2/4 not (\Pi^2) / 2

you guys have any explanation? thanks

[amateur]

How did you get Pi^2/4? Notice that the question says that the region to be revolved is in the first quadrant (Forming a sort of depressed triangle). It does not mean that the solid thus formed is also in the first quadrant (or octant if you like) only.

I assumed that the solid formed includes a portion of the 4th quadrant also.

[octave145]

y=1/[sqrt(1+x^2)]
V= \Pi*\Int_{0}^{\infinity}y^2dx = \Pi* [ arctan(\infinity) - arctan(0) ]

tan(x) goes to infinity when x goes to \Pi/2 and arctan(0) = 0
So V = (\Pi^2) / 2

[Nameless]

How stupid am I, I thought arctan(0)=pi/4 :d
I spent almost 3 hours to solve the test, It is not too difficult but I still made a lot of stupid mistakes. I finished 10 minutes earlier but not totally satisfied with stupid mistakes I made. Hope that it will be good on the real exam