Solution: There are A people in Class A, so there are 85-a people in Class B..
[a+(85-a)× 1/ 1 1]:(85-a)×( 1- 1/ 1 1)= 9:8
8a+8/ 1 1×(85-a)= 90/ 1 1×(85-a)
88a+680-8a=7650-90a
170a=6970
a=4 1
There are 4 1 person in class a.
37. Class A donated two-thirds of the sum of Class B and Class C, and Class B donated two-fifths of the sum of Class A and Class C. If Class A and Class B donated 144 yuan, how much did Class C donate?
Solution: class a donation a, class b donation144-a.
Class c donation (144-a)/(2/5)-a=360-7/2a.
According to the meaning of the question
a=( 144-a+360-7/2a)×2/3
3a=288-2a+720-7a
12a= 1008
A=84 yuan.
Then Class C donated 360-7/2×84=66 yuan.
38. The simplest fractional sums of two identical molecules are 1 and 18, and the ratio of the two denominators is 2: 3. What are the two scores?
Solution: Let denominators be 2a and 3a respectively.
therefore
1/(2a)+ 1/(3a)= 25/ 18
5/(6a)=25/ 18
a=3/5
Then these two scores are 5/6 and 5/9 respectively.
39. There is a bus with a body length of 12 meters in the street, which runs from east to west at a speed of 18 kilometers per hour. On the sidewalk, two people ran towards each other. At some point, the bus caught up with A and left A six seconds later. After 90 seconds, the car met Run B, which took 1.5 seconds. The bus left B. How many seconds before A and B meet?
Solution:18 km/h = 5 m/s.
Car and A are catching up, with speed difference =12/6 = 2m/s.
The speed of A is 5-2 = 3m/s.
When the car meets B, the speed sum is =12/1.5 = 8m/s.
The speed of B is 8-5 = 3m/s.
Let the distance between Party A and Party B be s meters.
When the car meets B, the a * * line s-5×6-3×6=s-48.
According to the meaning of the question
(5+3)×90=s-48
s-48=720
S = 768m.
After the car leaves Party B, the distance between Party A and Party B is 768-(3+3) × (6+90+1.5) =183m.
Then meet again after 183/(3+3)=30.5 seconds.
40. The construction team repaired the road. Built 60 meters on the first day. The next day, the remaining third was repaired. At this time, what has been repaired is equal to what has not been repaired. Find out the total length of the road.
Solution: Finally, whether 1: 1 is fixed or not.
Then the unrepaired accounts for 1/2 of the total.
That is, 1- 1/3=2/3 after it was repaired the next day.
So after the first day of repair, (1/2)/(2/3)=3/4 remains.
Then the total length = 60/( 1-3/4) = 240m.
What has been repaired equals what has not been repaired, that is to say, what has been repaired accounts for 1/2.
So, let the total length = x.
60+(X-60)* 1/3=X/2
Get x = 240.
I.e. total length = 240m.
involve