Sorry, I will not be able to answer all of my semi-promised exercises. The UPCAT sked is so toxic, I am writing this midnight and I have to be at UP at 530 am! suicide, huh. haha
Anyway, I’ll just work on some examples of my own and just read between the lines why.
An inlet pipe can fill the pool in 3 hours while a drain pipe will drain the pool in 8 hours. If the inlet pipe is turned on and the outlet pipe is accidentally also turned on after one hour that the inlet pipe gushing, how many more minutes would it take for the pool to be filled halfway?
Soln: rate of inlet pipe is 1/3
rate of outlet pipe is -1/8
Since the inlet pipe has been gushing for 1 hour, it would have done 1 times 1/3 or 1/3 of the work. The equation would look like this:
1 (1/3) + (1/3 + -1/ 8 ) t = 1/2
where t is the time wherein the pool becomes halfway full. You know of course how to solve for this.
Find the value of s such that the perimeter of the the equilateral triangle with side s is equal to the area of the same equilateral triangle with side equal to s.
Soln: We know that the perimeter of an equilateral triangle with side equal to s is 3s and you may verify that the area of the triangle with side equal to s is sqrt(3) s^2 divided by 4. Equate the two and you would get the correct answer.
August 3, 2008 at 10:01 pm |
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August 4, 2008 at 10:16 am |
Thanks sir!
August 4, 2008 at 2:03 pm |
hay.. Godbless na lng po smen tom.. thanks po sir!
August 4, 2008 at 10:02 pm |
Thank you, sir. Epal na UPCAT. Hahah just kidding
August 4, 2008 at 10:13 pm |
TT.TT
pero, thanks na rin poh.,,
^_^
August 12, 2008 at 6:15 pm |
I found your site on technorati and read a few of your other posts. Keep up the good work. I just added your RSS feed to my Google News Reader. Looking forward to reading more from you down the road!