2) Both Party A and Party B drive in the same direction from two places at the same time. Four hours later, A caught up with B in C. At this time, the two traveled 78 kilometers, and B needed 1 hour 45 minutes from A to B. What is the distance between A and B?
3) If the car speed is increased by 20%, it can arrive 1 hour earlier than the original time; If you drive 80 kilometers at the original speed and then increase the speed by 25%, you can arrive 40 minutes earlier. What's the distance between a and b?
4) The ratio of red balls to white balls in the bag is 19:: 13. After putting some red balls, the ratio of red balls to white balls is 5: 3; The number ratio after adding a few white balls is 13: 1 1. There are 40 fewer red balls than white balls. How many balls are there in the bag?
5) A rides a bike and B walks from A and B respectively. After meeting, A arrived at B in 20 minutes, but B arrived at A at 1: 20. B how many hours did it take from b to a?
6) Bus A and Bus B depart from cities A and B respectively. The speed ratio of car A and car B is 3:2. There is a city C between the two cities. The time for A and B to arrive in City C is 5: 00 a.m. and 5: 00 p.m. respectively. When did A and B meet?
1, directly set a xkg, and b is 450-x kg.
Equation 3/4 * x=7/9 * 2/3 *(450-x) exists.
That is, 240kg for Party A and 2 10kg for Party B..
2. let the speed of b be x;
AB distance =1.75x;
BC distance = 4x
* * * Drive 78km= 1.75x+2*4x (just draw a picture).
The speed of b is 8 km/h.
So AB=8* 1.75= 14km.
3. If the vehicle speed is increased by 20%, it can be 1 hour earlier than the original time, and it is easy to draw the conclusion that the original time is 6 hours.
Let the original speed be x, that is, the distance is 6x.
80/X+(6x-80)/ 1.25x = total time = 16/3.
X=30, and the distance between the two places is 180km.
4. Suppose the original red ball has 19x, then the white ball has13x; ;
Let the red ball be y, and the white ball be y+40;
( 19x+y)/ 13x=5:3
( 19x+y)/( 13x+y+40)= 13: 1 1
Get: x =15;
In other words, there are 480 balls in total.
5, let a speed x, b speed y;
Time from departure to meeting =(4/3 *y)/x =( 1/3*x)/y
Get the relationship x=2y, get the meeting time =40 minutes; So B+1 hour takes 40 minutes, and 20 minutes =2 hours.
6 It's 5 o'clock in the morning and 5 o'clock in the afternoon respectively, and you can get a message that A arrived first 12 hours;
Let the speed of A be 3x, that is, the speed of B is 2x; Let the AC distance be y.
So BC distance =(y/3x+ 12)*2x.
AB distance = y+(y/3x+12) * 2x = 5/3 * y+24x
So the meeting time =(5/3 *y+24X)/(3x+2x)=y/3x +4.8.
The time from a to c is y/3x.
So we have to wait 4.8 hours to meet, that is, 4.8 hours after 5 am.
Meet at 9: 48 a.m.
Of course, the simpler method is when a goes to c, when b goes to d,
So the distance between CDs is actually 12*2x=24x.
You can turn the problem into two places 24x apart. The speed of A is 3x, and the speed of B is 2x. How long will it take to meet?
The simple calculation is 4.8 hours.