G12_MATH_P1A2_13

10a) Average speed = 30km/3hr = 10km/hr
10bi)Time spent to reach C = (15/20 + 4/3)hrs = 25/12 hrs = 125mins
So Tommy reach C 125mins after 10:00, i.e. 12:05
10bii)Normal time spent from C to A = 11/15 hr = 44mins
So Tommy rest for (55-44)mins = 11mins
I guess is like that not sure....!!!

12a)Let f(x) = 2x^3-ax^2+27x+b
f(3)=2(3)^3-a(3)^2+27(3)+b=0 --> -9a+b=-135..(1)
f(1)=2(1)^3-a(1)^2+27(1)+b=-2 --> -a+b = -31..(2)
Solve (1) and (2), a=13 b=-18
so f(x) = 2x^3-13x^2+27x-18

12b) Coordinates of P= (h, 6h^2-39h+81)
Hence, area of OKPH = h x (6h^2-39h+81) = (6h^3-39h^2+81h) sq units
12c) From b, (6h^3-39h^2+81h)= 54
6h^3-39h^2+81h-54=0
2h^3-13h^2+27h-18=0
(h-3)(2h^2-7h+6)=0
(h-3)(h-2)(2h-3)=0
h=3/2, h=2, h=3
Coordinates of P = (3/2, 36)
(2, 27)
(3, 18)
13a) Mean = 7.7
Median = (7.8+7.8)/2=7.8
13bi) For mean unchange, (p+q)/2 = 7.7 = 15.4
For median unchange, Let p>q
then p will be larger or equal to 7.8
and q will be smaller or equal to 7.8
一頭一尾咁加個median 就不變
Fulfil the above conditions (p=9.4, q=6)...etc.
13bii)The total rating for the 14 audience = 14x7.7=107.8
Assume all the other 6 audience give the highest rating, i.e 10
The average rating of these 20 audience = (107.8+10x6)/20=8.39 < 8.6
So it is impossible

2015-01-06 10:39:22 補充：
Total distance of the journey = 30km
It means A to B and then from B to A.
And tommy walks the same route.
So AB=BA = 30/2 = 15km
Distance of AB=15km

20km/hr 係(a) 答案x2

我估係咁, 唔肯定,有錯勿_