majorie asked in 科學及數學數學 · 9 years ago

# 2條中三數學 (20點!)

1.

in the figure, there is a metal ball with the radius of 5cm inside a container in the shape of right prism. The base of the container is a rectangle of dimensions 15cm x 12cm. Wather is poured into the container until the metal ball is just covered by water.

a) find the volume. ( ans: 1280 cm^3 (corr. to 3 sig fig))

b) now, 10 more metal balls with diameters of 2.4 cm each are put into the container. Assume that the metal balls are fully immersed in the water and the water dose not overflow, how much does the water level rise?

( ans: 0.402cm (corr. to 3 sig fig))

2.

For a batch of raffle tickets, the probability of winning a prize is 1/30.

if 240 more tickets are added without prizes, the probability of winning a prize becomes 1/90. Find the original number of raffle tickets.

( ans: 120 )

Rating
• c
Lv 5
9 years ago

1.

(a)

the volume of water poured

=the volume of the right prism with the base 15cm x 12cm and the height 5cm x2=10cm

-the volume of the ball

=15cm x 12cm x 10cm - (4*pi/3)*[(5 cm)^3]

=1276.401224 cm^3 (corr. to 6 decimal places)

=1280 cm^3 (corr. to 3 sig. fig.) (b)

the volume of 10 balls added

=10*(4*pi/3)*{[(2.4 cm)/2]^3}

=23.04*pi cm^3 let the length of the rise of water level be h cm.(h cm)*[(15 cm)*(12 cm)]=23.04*pi cm^3

h*15*12=23.04*pi

h=0.402123859 (corr. to 9 decimal places)

h=0.402 (corr. to 3 sig. fig.) the water level rises 0.402 cm2.

let the original number of raffle tickets be a.

let the number of prizes be b.

So,b/a=1/30

b=a/30 if 240 more tickets are added without prizes,

then the number of raffle tickets is (a+240)

the number of prizes is still b.b/(a+240)=1/90

b=(a+240)/90

a/30=(a+240)/90

(a/30)*90=[(a+240)/90]*90

3a=a+240

2a=240

a=120the original number of raffle tickets is 120

• 9 years ago

第一題要cover個metal ball既長方體既volume係15x12x10, 減去個metal ball就可以得到所需既水既volume

1a)

(15x12x10) - (4x兀x5^3)/3

=1800 - 523.59877

=1280 cm^3 (cor. to 3 sig fig)

1b)

{[(4x兀x1.2^3)/3]x10} / (15x12)

=72.38229 / 180

=0.402 cm (cor. to 3 sig fig)

第二題我會用linar eauation (我唔知有無更好既方法)

2)

let x be the number of ticket that can get a prize

let y be the number of ticket that can not get a prize

x/y = 1/30 ------------------ 1

x/(y+240) = 1/90 --------- 2

1. x/y = 1/30

x = y/30

sub 1 into 2

(y/30) / (y+240) = 1/90

(y/30) x [1/(y+240)] = 1/90

y/(30y+7200) = 1 /90 <-- 呢度交叉相乘到下一步

90y = 30y+7200

y(90-30)=7200 <-- 抽common factor

60y=7200

y=120

之後d step你識做

Source(s): 我盡量用中文解釋啦, 努力讀書 !!!