Three math problems, help me

1) The subtracted number is equal to the subtraction plus the difference, so it is equal to 5:3:2 Answer: The subtracted number is 30, and the difference is 12.

A: The new brine has 30 grams of water.

3) From 12 noon to 4:12 a.m., after about 16 o'clock, 16:24 2:318 (2 3) 12 (min).

4:12 12:12 4:00.

A: The actual time is 4 o'clock.

1. Minus difference minus 60 minus.

So, subtract 30:

Minus: Difference 3:2 (30 Difference): Difference.

So, the difference is 10

2. (35-2 water): water 1:10

So, water 330 11 30 grams.

So, the exact time is 1440 minutes 1440 60 16 points.

1.Let the subtracted number be x, the subtraction be y, and the difference be z

x-y=zx+y+z=60

y/z=3：2

x=30，y=18 z=12

The raw brine weighs 35-2=33 grams.

Water = 33x (10 (1+10)) = 30 grams.

The hour is fast 18 minutes.

is fast per hour.

18-24 minutes.

Twelve noon to 4:12 am

16 hours and 12 minutes.

18 hours faster, 12 minutes faster.

So now it's 4 o'clock.

Because x-y=z x+y+z=60

So x=30 y+z=30

again y 3=z 2 so y = 18 z = 12

Let the salt xx + 10 x + 2 = 35 so x = 3 so the water in the new brine has 10 * 3 = 30

24 hours faster than 18 minutes, that is, the speed ratio.

Actual clock = 24 * 60 (24 * 60 + 18) so: 24 * 60 (24 * 60 + 18) = x minutes (12 + 4) * 60 + 12

x = 960 minutes = 16 hours.

So the actual time is four o'clock in the morning.

1) Because x-y=z x+y+z=60

So x=30 y+z=30

again y 3=z 2 so y = 18 z = 12

2) The total weight of the brine after adding 2 grams of salt is 35 grams, then the total weight of the brine before adding 2 grams of salt is 33 grams, and the ratio of salt to water is 1:10, you can know that there are 30 grams of water, and the water in the new brine has not changed, so it is still 30 grams.

3), a day and night 18 minutes faster can know that the hour is faster, the actual time from 12 noon to 4 o'clock in the morning has passed 16 hours, this clock should be 16 * minutes faster, exactly 4 hours 12 minutes of the clock now, so the actual time is 4 o'clock.

1.The minus is 18 and the difference is 12

2.3 grams of salt, 30 grams of water, 59 minutes and 51 seconds.

All right. 1.Let the substance be x tons, then , then x=130

00 when the wheel turns 3000 weeks = 2 r3000 = d 3000 =, you can rush.

3.Originally payable 360 280 = 640, actually paid 360 120 = 480, 480 640 =, played seven or five.

1) C moved: 5*3=15 units.

Then c means -5+15=10

2) Let the number represented by b be x, then ab distance x-(-5) = x + 5 units.

Encounter time of A and C: (x+5) (3+2)=(x+5) 5 Encounter time of A and B: (x+5) (3+1)=(x+5) 4 The equation obtained from the problem: (x+5) 4-(x+5) 5=1 is solved x=15

So b means 15

3.Assuming that it exists, the distance between C and B is: (x+5)-t(1+3)=20-4tA-C distance: (x+5)-t(2+3)=20-5t20-4t=(20-5t)*2

The solution yields t=10 3

The meeting time of A and C is: 20 5 = 4

So t exists, for 10 3 seconds.

25、v=1/3*sh=1/3*1/2*ab*a=1/6a^2b26、sina+cosa=（2t+1）/5sinacosa=（t^2+t)/25

sina + cosa) 2 = 1 + 2 sinacosa solution : t = 4

t=-6 (rounded).

27. The straight line l passes through (-1 2, 3 2) and is perpendicular to the straight line at the center of the circle (0, 1) and (-1 2, 3 2).

k=-1/[（1-3/2）/（1/2）]=1（y-3/2）/（x+1/2）=1

y-3/2）=（x+1/2）

y=x+2

(1) Let the speed of the train be V and the length of the train be X

From the question (v+

and (so v=,x=918 km.

2) The speed of the car is 110 km/h, and the speed of the truck is 100km/h. , the relative velocity of the two is 10 kilometers per hour, and the relative distance between the two is 12 meters, so the time is hour.

3) The speed of the two ships A and B is 10 kilometers downstream, and 5 kilometers when the speed is against the water. Let the ab distance be x

If point C is between AB.

x／10）+（x-10）／5=4

So x=20. Time t=2

Ship A sailed 20 kilometres away from place B.

If point A is between BC.

x／10）+（x+10）／5=4

So x=20 3. Time t = 10 3

Ship A sailed 100 3 km from place B.

(1) A walked 15m

B walked 17m

The speed of the train remains the same.

x-15)/15=(x+17)/17

x-15)/(x+17)=15/17

15x+255=17x-255

x=255m

2)110km/h=

100km/h=

It takes x seconds.

x=(3) Let B go from b to c in x hours.

Then A and B took 4-x hours to get from A to B.

AB length (BC length (

ab-bc=ac

10(4-x)-5x=10

x=2, then A leaves place B (

1) It is a problem of encounter with A, and it is a problem of pursuit with B. Let the vehicle length l m, and the vehicle speed v m l (v+1)=15

l/(v-1)=17

The simultaneous equation yields: l 17 + 1 = v = l 15 - 1l = 16

2)s=4+12=16m

Relative velocity v = 110-100 = 10 km h = 10 t = s v = 16*

3) There are two cases.

If A is upstream of C, let the BC distance be X km and the AB distance be (10+X)km.

10+x)/(

x=10B from b to c t=x (

A leaves b:s=(

If A is downstream of C, let the BC distance be X km and the AB distance be (X-10)km.

x-10)/(

x=50/3

B from b to c t=x ( h

A leaves b:s=( km=

Question 1: 255m

Question 2: Question 3: 20km