F.3maths

2009-06-28 5:51 pm
1)
There are mistakes in the the calculation. Make necessary corrections.

If the radius of a spherical balloon is increased by 10%, what is the percentage increase in surface area and percentage increase in the volume of the balloon?

Solution: Percentage increase in surface are= (10%)^2= 1%
Percentage increase in volume= (10%)^3= 0.1%

2)
A helicopter is 100m above the ground. Fanny observes it at an angle of elevation of 60 degree. If the helicopter descends to half of its original height, is the angle of elevation of the helicopter from Fanny halved?

3)
When we observe a distant object, the angle of elevation (or depression) is smaller than that when we observe a nearer object. Explain why.

4)There are 5 blue balls and some green balls in a box. Two balls are drawn at the same time at random from the box. If the probability of getting a blue ball and a green ball is greater then 0.5, find the original number of green balls in the box.

PLEASE do as much as you can, thank you very much!!!^^

回答 (1)

2009-06-28 6:49 pm
✔ 最佳答案
1) If the radius of a spherical balloon is increased by 10%, what is the percentage increase in surface area and percentage increase in the volume of the balloon?

Original surface area = 4丌r^2
New surface area= 4丌(r*1.1)^2=1.21*4丌r^2
Percentage increase =(1.21*4丌r^2-4丌r^2)/(4丌r^2) * 100% = 21%

Original volume = 4/3丌r^3
New volume=4/3丌(r*1.1)^3 = 1.331*4/3丌r^3percentage increase in volume = (1.331*4/3丌r^3-4/3丌r^3)/(4/3丌r^3)*100% = 33.1%

2) A helicopter is 100m above the ground. Fanny observes it at an angle of elevation of 60 degree. If the helicopter descends to half of its original height, is the angle of elevation of the helicopter from Fanny halved?

Let horizontal distance of helicopter from Fanny = Xm
100/X = tan 60
X = 100/tan60 = 57.74

New height of helicopter = 100/2 = 50m
Let new angle of elevation be y
50/X = tany
tany = 50/57.74 = 0.866
y = 40.9degree, it is not halved.

3) When we observe a distant object, the angle of elevation (or depression) is smaller than that when we observe a nearer object. Explain why.

Let the angle of elevation or depression be @.
Let the horizontal distance of object from observation be d.
Let the height of object above or below the observer be h.
The relationship among these parameters is
tan@=h/d

Since h is not changed but d is reduced when we come near an object, tan@ increases. Accordingly @ increases.

4)There are 5 blue balls and some green balls in a box. Two balls are drawn at the same time at random from the box. If the probability of getting a blue ball and a green ball is greater then 0.5, find the original number of green balls in the box.

Let the number of green ball be G.
Probability to get 1 green and 1 blue ball is
[5/(5+G)]*[G/(5-1+G] + [G/(5+G)]*[5/(5+G-1)]
=10G/(G^2+9G+20) >1/2
=> 20G>G^2+9G+20
=> 0>G^-11G+20
=> 0>(G-11/2)^2+20-121/4
=>11.25>(G-5.5)^2
So sqrt(11.25)>G-5.5 and sqrt(11.25)>5.5-G
8.85>G and G>2.15
Since G is an integer, therefore 8>=G>=3


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