GAMSAT Tutorial: Section 3 problems (ACER Practice Test 2 - Purple Booklet): Questions 29-32

This is a free online GAMSAT tutorial made by Dr. Ferdinand, the author of the Gold Standard GAMSAT book, addressing ACER's GAMSAT sample questions from the purple booklet: Section 3, questions 29-32. This science teaching is to help students better understand the logic behind the GAMSAT exam.

Solutions are cross-referenced to the ACER's GAMSAT sample problems PDF e-book/book. There is no need for torrent or download as these videos are available for free although they also come as DVDs.

[laa]

I did 32 a slightly quicker way (which is pretty helpful for GAMSAT!)

Because I know 0.03m is already submerged, I know that I need to submerge the additional 0.07m still above the surface to achieve full submersion. Width and Length are the same so I calculated the additional volume needed as 0.07 x 2 x 0.2, which becomes 0.07 x 0.4, ending with 0.028m3 as the additional volume to be submerged. Then its a quick look at Fb = mg = Vpg to see that m = 0.028 x 1000 which is 28kg. I need not even take this step, since as soon as I got 0.028 and looked at the answers available, the only one with 28 in it is option C.

Just a method for people to also consider in the future if they watch this :) You're videos throughout have helped me so much with S3 its unreal, so thanks.

[edwardhong]

HI for question 31, why didn't you take note of the mass of the log in the calculation, but rather just the mass of the object?

[BradWillo]

Q31:
I used the mass of the log + mass of the 20kg object coming to 32kg. Which when divided by the density of water = 0.032m3, meaning that the 12kg log would only be submerged 0.012 without the 20kg addition. However this is not the case shown as it is submerged 0.03m initially. How has this discrepancy occurred? Thanks

[wayne]

Your video assumed so much. It is terrible for someone from a non-science background