F.2Maths唔識做~Ratio and Rate

2012-03-28 11:38 pm
我星期六就要補數,但校內測緊驗,冇時間做,有兩條唔識~
來自New Progress in Junor Mathematics Workbook 2A
Chapter1 Ratio and Rate
10) Kevin mixed 6L of alcohol with 4L of water and Jerry mixed 3L of alcohol with 5L of water.
a) How much water should Kevin further add to his mixture so that there is 50% of alcohol?
b) They now mix their mixtures together. How much alcohol should they add to their mixture so that there is 75% of alcohol?

11) Alvin has 2 model cars A and B. Car A's speed is 2m/s and car B's speed is 3m/s. He puts the cars on the same track to travel 100m.
a) Find the difference between the times needed for car A and car B to travel 100 m.
b) Alvin increases car A's speed by 100%.Find the percentage increase in the speed of car B so that it finishes the track ahead of car A by 5 seconds.

各位除了提供答案,式& steps 外,最好解一解,THX~~

回答 (1)

2012-03-29 10:14 am
✔ 最佳答案
10) .a)
For 50% alcohol, the volume of alcohol and volume of water should be equal. Since Kelvin has 6L of alcohol and 4 L of water, Kelvin only has to add 2L of water to his mixture, so that finally he has 6L of alcohol and 6L of water.
Volume of solution 6L + 6L = 12 L
Volume of alcohol = 6L
Percentage of alcohol in solution
= (Volume of alcohol/volume of solution)x 100% = (6L/ 12L) x 100% = 50%

Kelvin should add 2L of water to the mixture. (answer)

b)
If they mix their mixture together,
Volume of alcohol = 6L + 3L = 9L
Volume of water =4L + 5L = 9L
Let y be the volume of alcohol added to make 75% alcohol
9L + y
-------- X 100% = 75%
18 + y
100(9 + y) = 75(18 + y)
900 + 100y = 1350 + 75y
100y – 75y = 1350 – 900
25y = 450
y = 450/25 = 18

18 L alcohol should be added to their mixture so that there is 75% alcohol. (answer)

11)
a)
time required = distance/speed
Time required for car A to travel 100m = 100 m/2m/s = 50 s
Time required for car B to travel 100m = 100 m/3m/s = 33 1/3 s

Difference in time = 50 s – 33 1/3 s = 16 2/3 s or 16.66 s (answer)

b)
Alvin increases car A’s speed to 4 m/s
Time required for car A to travel 100m if the new speed is 4 m/s = 100 m/4m/s = 25 s
For car B to finish the track ahead of car by 5 seconds, time required by car B to finish the track = 25 s – 5s = 20 s
The new speed of car B = distance/time = 100 m/20 c = 5 m/s

Perentacge increase in the speed of car B
= (increase in speed/original speed) X 100% =[ (5 m/s – 3 m/s)/ 3 m/s] X 100%
= [2m/s/3 m/s] x!100% = 66 2/3 % or 66.67% (Answer)


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