A, B and C transported 40% of the total amount of wheat, 150 kg more than B. the weight ratio of B to C was 3:2 How many kilos?

A, B and C transported 40% of the total amount of wheat, 150 kg more than B. the weight ratio of B to C was 3:2 How many kilos?

According to experts, the total number of wheat transported by Party B is (1-40%) × 3 / (3 + 2) = 9 / 25, the total number of wheat transported by Party C is (1-40%) × 2 / (3 + 2) = 6 / 25, the total number of wheat transported by Party A is (3750 × 40%) = 1500 (kg), and the total number of wheat transported by Party B is (3750 × 9 / 25)
Party B transported (1-40%) × 3 / (3 + 2) = 9 / 25 of the total
The proportion of the total amount transported by C is (1-40%) × 2 / (3 + 2) = 6 / 25
This batch of wheat has: 150 (40% - 9 / 25) = 3750 (kg)
Total wheat transported by Party A: 3750 × 40% = 1500 (kg)
Wheat transported by Yi: 3750 × 9 / 25 = 1350 (kg)
C total transport of wheat: 3750 × 6 / 25 = 900 (kg)
Party B transported (1-40%) × 3 / (3 + 2) = 9 / 25 of the total
The proportion of the total amount transported by C is (1-40%) × 2 / (3 + 2) = 6 / 25
This batch of wheat has: 150 (40% - 9 / 25) = 3750 (kg)
Total wheat transported by Party A: 3750 × 40% = 1500 (kg)
Wheat transported by Yi: 3750 × 9 / 25 = 1350 (kg)
C total transport of wheat: 3750 × 6 / 25 = 900 (kg)
Pile B coal is 24 tons more than pile a coal. After 3 / 4 of pile a coal is transported, the rest is equal to 1 / 5 of pile B coal. How many tons of pile B coal?
Let a have X and B have 24 + X
X(1-3/4)=(24+X)*1/5
5X=4(24+X)
5X=96+4X
X=96
So B has 96 + 24 = 120 tons
A: B has 120 tons of coal
A + 24 = b
A (1-3 / 4) = 1 / 5 B
A + 24 = b
5 A-4 B = 0
5 A-4 (a + 24) = 0
A = 96
B = a + 24 = 96 + 24 = 120 tons
A and B drive from a to B at the same time, with the speed of 42 km / h and 38 km / h respectively. A returns immediately after arriving at B, and meets B 20 km away from B. how far is ab from B?
When meeting, a is more than B: 2 * 20 = 40 km
The meeting time is 40 / (42-38) = 10 hours
AB distance: 42 * 10-20 = 400 km
Total driving time: 20 * 2 / (42-38) = 10 hours
Distance between two places: 10 * 38 + 20 = 400 km
After driving for t hours
The whole journey is equal to 38t + 20km = 42t-20
The solution is t = 10
So it's equal to 400x120 + 38km
A walks 42-38 kilometers more per hour than B;
A walked 20 + 20 = 40 kilometers more than B;
40 / 4 = 10 hours.
The distance is 38 * 10 + 20 = 400 km.
The driving time of the two vehicles is the same. A travels 40 km more than B, and a travels 42-38 km faster than B = 4 km / h. The extra mileage must be 4 km / h faster than car a. In this way, we can calculate the travel time of two cars, which is 40 / 4 = 10 hours. The distance between natural AB: 1. Calculated by car a: 42 * 10-20 = 400; 2. Calculated by car B: 38 * 10 + 20 = 400 (km). ... unfold
The driving time of the two vehicles is the same. A travels 40 km more than B, and a travels 42-38 km faster than B = 4 km / h. The extra mileage must be 4 km / h faster than car a. In this way, we can calculate the travel time of two cars, which is 40 / 4 = 10 hours. The distance between natural AB: 1. Calculated by car a: 42 * 10-20 = 400; 2. Calculated by car B: 38 * 10 + 20 = 400 (km). Put it away
A, B and C transported a pile of wheat together. A transported 40% of the total amount, 152 kg more than B. B transported 1.5 times as much as C. how many kg of wheat did each of the three transport
analysis:
B = 1.5 C, a = B + 152
So a = 1.5 C + 152
So suppose C x kg, a 1.5x + 152, B 1.5x
1.5x+152=（x+1.5x+152+1.5x）×40%
1.5x+152=（4x+153.5）×0.4
x=912
A = 1.5 * 912 + 152 = 1520
B = 1.5 × 912 = 1368
B and C transport 60% of the total
Suppose B 1.5, then C 1
Party B shipped 1.5 / 2.5 * 60% = 36% of the total
1 / 2.5 * 60% = 24% of the total
152 / (40% - 36%) = 3800 kg
A: 3800 * 40%
B: 3800 * 36%
C transported 3800 * 24%
Just do it yourself
(1-40%) / (1 + 1.5) = 24%
B: 24% × 1.5 = 36%
Total weight of wheat: 152 △ 40% - 36% = 152 △ 4% = 3800 (kg)
Weight transported by Party A: 3800 × 40% = 1520 (kg)
Weight of shipment B: 1520-152 = 1368 (kg)
Weight of C transportation: 1368 △ 1.5 = 912 (kg)
Because a transported 40% of the total,
So 60% of the second and third
60% divided by (1 + 1.5) × 1.5 = 36% -- B
152 divided by (40% - 36%) = 3800
Let C transport x, and the total number is y
According to the meaning of the title, we can get two equations: 1
The total number is y = a + B + C = 1.5x + 152 + 1.5x + X
The number of a-transports is 0.4y = 1.5x + 152
A 1520 B 1368 C 912 can be obtained from the two equations
Without equation, it can be analyzed as follows:
A carried 40% of the total, so B and C carried 60% of the total
The ratio of B to C is 1.5 times that of C, so the ratio of B to C is 1.5:1 = 3:2
So Party B transported 60% of the total amount * 3 / (3 + 2) = 36%
40% - 36% = 4% of the total, 152 kg
As a result, three people transported wheat 152 / 4% = 3800 (kg)
Without equation, it can be analyzed as follows:
A carried 40% of the total, so B and C carried 60% of the total
The ratio of B to C is 1.5 times that of C, so the ratio of B to C is 1.5:1 = 3:2
So Party B transported 60% of the total amount * 3 / (3 + 2) = 36%
40% - 36% = 4% of the total, 152 kg
As a result, three people transported wheat 152 / 4% = 3800 (kg) away
A 1520
B 1368
C 912
Because a transported 40% of the total,
So 60% of the second and third
60% divided by (1 + 1.5) × 1.5 = 36% -- B
152 divided by (40% - 36%) = 3800, ---- A
3800-152 = 3648 --- B
3648 / 1.5 = 2432 -- C
Three people transported x tons.
40%x+40%x-152+（40%x-152）÷1.5=x
The solution is x = 3800
40%x=1520
40%x-152=1368
（40%x-152）÷1.5=912
A: 1520 tons for Party A, 1368 tons for Party B and 912 tons for Party C.
Pile a coal is 24 tons more than pile B coal. After 3 / 4 of pile B coal is transported away, the remaining coal is equal to 1 / 5 of pile B coal. How much is pile a coal?
There is a mistake in the title. Please correct it
A. The distance between two places is 169 km. Car a drives from place a to place B at the speed of 42 km / h. After 20 minutes of departure, it needs to stop for repair due to the fault. At this time, car B drives from place B to place a at the speed of 39 km / h. it takes 10 minutes for car a to clear the fault. How many hours has car B met car a after departure?
Car B meets car a after X hours
20 minutes = 1 / 3 hour
10 minutes = 1 / 6 hour
42*(x-1/6)+42*(1/3)+39x=169
42x-7+14+39x=169
81x=162
X=2
I hope you can understand, you can understand and agree
20 minutes = 1 / 3 hour
10 minutes = 1 / 6 hour
[169-42×（1/3-1/6)]÷（42+39）
=162÷81
=2 hours
Suppose B meets a x hours after departure
From the meaning of the title
39X+42(x-1/6)=169
39x+42x-6.5=169
81x=175.5
X = 2 and 1 / 6
X = 2 hours 10 minutes
Two fifths of the total amount of wheat transported by Party A is 152kg more than that transported by Party B. the ratio of wheat transported by Party B and Party C is 3:2. How many kg of wheat are transported by each of the three people
Let B transport 3x kg, then a transport x + 152 (kg), C transport 2x (kg) 3x + X + 152 + 2x = (x + 152) △ 2 / 56x + 152 = (x + 152) × 5 / 2, both sides multiply by 212x + 304 = 5 (x + 152) 12x + 304 = 5x + 76012x-5x = 760-3047x = 456x = 456 / 7, that is: B transport 3 × 456 / 7 = 1368 / 7 (kg), then a transport 456 / 7
A freight car and a passenger car leave 413km apart at the same time. After 3.5 hours, the two cars meet. The speed of the freight car is 65km / h,
What is the speed of the bus in km / h
Suppose the speed of the bus is x km / h
3.5*(X+65)=413
X+65=118
X=118-65
X=53
The speed of the bus is 53 km / h
Fifty-three
(413-65 * 3.5) / 3.5 = 53 km / h
10. Pile B coal is 24 tons more than pile a coal. After three fourths of pile a coal is transported away, the remaining coal is one fifth of pile B coal. How many tons of pile a coal?
Score calculation,
nail
24x1/5÷(1-3/4-1/5)
=24x1/5÷1/20
=96 tons
The speed ratio of a and B is 5:4. When they meet, a travels 35 kilometers more than B. how far is ab?
Let the total distance be 9s
Then a line of 5S, B line of 4S
The distance from a to B is 35km more, that is 5s-4s = 35
So s = 35,
Then, the total distance is 9s = 35x9 = 315km