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Manager
Joined: 27 Sep 2006
Posts: 70
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09 Nov 2006, 22:52
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi,

I posted 5 queries under the Challenge category, but no reply yet. Anyone here have explanations? Thanks much!

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)
2002: S(1-x/100)^2
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.

How?

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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 clock at 12:24= 0+12 = 12 deg.

Similarly, 14 = 60 degrees +
30 degrees = 60 mins therefore 36 mins =?
(30 * 36)/60 = 18
60+18 = 78

therefore distance between 2 times is 78-12=66 degrees.

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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;
then??
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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)

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22. f(x) satisfies f(x)=f(x^2)
solution: f(6) - f(-2)=0

I don't get this problem. Help please!

THANK YOU!
Manager
Joined: 03 Jul 2005
Posts: 192
Location: City
Followers: 1

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09 Nov 2006, 23:32
If you post each question seperatly then you should get some answers.
09 Nov 2006, 23:32
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