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I posted 5 queries under the Challenge category, but no reply yet. Anyone here have explanations? Thank you!
here were my questions: Challenge 25: #6; 12; 18b; 19; 22
#6: my solution seems different from the answer provided.
This is what I got.
2001 : S(1-x/100)
2003: S(1-x/100)^3 = T
S(1-x/100)^3 = T, therefore (1-x/100)^3 = T/S or (1-x/100) = cube root of (T/S)
cost in 2002 therefore is S(1-x/100)^2 or S*(cube root of (T/S))^2
The above checks out if depreciation is 10% for example: 100 in 2000, 90 in 2001, 81 in 2002 and 72.9 in 2003.
Challenge answer is sqrt S*T
12. degrees between 12:24 and 14:36 hour hand.
Can someone please explain the solution to me? I don't understand. I've seen this kind of solution before and not understanding worries me. I got it right though by doing the following and am not sure if the method is right.
At 12, 0degrees.
360/12=30 degrees for minutes hand.
30 degrees = 60 mins therefore 24 mins = ?
(24 * 30)/60 = 12
therefore distance between 2 times is 78-12=66 degrees.
18B. 3 times as many pupils in 2nd as in 1st. Solution says y=3x therefore x cancels out so result can be calculated. If x cancels out, what is the answer?
I got (12.2x+13.1y)/x+y; or 12.2x+13.1*3x/x+3x; = 51.5x/4x;
19. The given equation gets simplified to x+y=10. And this encloses a coordinate plane.
solution: 10^2+10^2= sqrt of 200
BUT what is the assumption in arriving at this solution.
I was at a loss so decided to do (x+y) (x+y) = 10 * 10 = 100 as area, though wasn't sure why I was doing that either! (and it appears to be incorrect anyway)
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...