F.3 Maths - Ans.fyi 參考答案

F.3 Maths

2010-01-23 11:00 pm

請列出詳細步驟及答案。[Thank you very much]

1. An aeroplane flying between 2 towns takes 20 minutes more than its usual time when its normal speed is reduced by 30 km per hour, and 10 minutes less than its usual time when its normal speed is increased by 20 km per hour. Find the normal speed and distance between the towns.

2. A garden path, bounded by two circles, is gravelled at $2 per m^2(metre square), the cost being $132. Along both edges of the path is a stone edging at $3 per metre, the cost of this being $396. Find the width and inner radius of the path.
[Take 兀(pi) = 3又7分之1]

回答 (1)

用戶9112

2010-01-23 11:41 pm

✔ 最佳答案

(1) Let the normal speed be x km/h and the distance between the 2 towns be y km.
Normal time = y/x hours
When speed is reduced by 30 km/h, time = y/(x – 30)
y/(x – 30) – y/x = 20/60
30y/[x(x – 30)] = 1/3
y = x(x – 30)/90 … (1)
When the speed is increased by 20 km/h, time = y/(x + 20)
y/x – y/(x + 20) = 10/60
20y/[x(x+20)] = 1/6
y = x(x + 20)/120 … (2)
Hence x(x – 30)/90 = x(x+20)/120
4(x^2 – 30x) = 3(x^2 + 20x)
x^2 – 180x = 0
x(x – 180) = 0
x = 0 (rejected) or x = 180
From (1) y = 180*(180 – 30)/90 = 300
Normal speed = 180 km/h, distance = 300 km
(2) Let the inner radius be r m and the width be d m
Area of the path = π(r + d)^2 – πr^2 = (2πrd + πd^2) m^2
Total length of edges = 2π(r + d) + 2πr = 2π(2r + d) m
Cost of gravelling = 2(2πrd + πd^2) = 132
2rd + d^2 = 21
d(2r + d) = 21 … (1)
Cost of stone edging = 3* 2π(2r + d) = 396
2r + d = 21 …(2)
(1)/(2) => d = 1
From (2), 2r + 1 = 21 => r = 10
The width of the path is 1 m and the inner radius =10m

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