A few math problems. It should be done with the solution ratio.

1. The total number of people in the two workshops: 156 * 2 =, 312 * (7 12) = 182, A 130, B 182

2. Senior grades: x=150 (lower grades), 3x=450 (middle grades) + 15=15a+7a+8a, a=15, the original imported oranges = 15 * 7-15 = 90

2500=a/

36*150=18*24*a，a=200.4*，a=192

1600=240/a，a=96

6=a/（a+24），a=48,48*5/4=60,48*6/4=72

1.A 130, B 182

Hope it helps.

1. (a+b) 2=156, a b=5 7, a130, b182-200=3a+a, a=150 (lower grade), 3a=450 (middle grade) + 15=15a+7a+8a, a=15, original imported oranges = 15*7-15=90

2500=a/

36*150=18*24*a，a=200.4*，a=192

1600=240/a，a=96

6=a/（a+24），a=48,48*5/4=60,48*6/4=72

Summary. 9 a.m. plus 1 p.m. 45 minutes is 10 a.m. 45 minutes.

A: Arrive at place B at 10:45 minutes.

Help to solve a math problem with the proportions.

Hello dear? This question.

Okay, wait a minute. It is necessary to use the method of solving the proportion.

Uh-huh. A: Arrive at place B at 10:45.

The parsing is being sorted out, please wait.

Good. Analysis: Let the intercavity distance of Chi Shan Shi Zheng in the two codes of ab is xcmSo 1:3000000=:xx=1050000010500000cm=105 km.

And then what about the dear. When needed, the old mold room is: 105 60 = small and with return time. Minute. So the time required is 1 hour and 45 minutes to wake up the hunger bell.

Ok thank you for the thumbs up.

You're welcome. 9 a.m. plus 1 p.m. 45 minutes is 10 a.m. 45 minutes. A: Arrive at place B at 10:45 minutes.

Let the total number of fish in the pond be x, and the forty fish caught for marking account for 40 x of the total. If you know how many of the 60 fish were marked, the total number of fish can be derived from 40 x x60 = (the number of marked fish in 60).

It's a matter of probability. If there are 2 marked fish in 60, then the total number of fish in the pond is 40*60 2=1200,.

The answer is simple: if 2 of the 60 fish are marked, then the following sentence can be used as the center of the scale equation: "If there are 2 marks for 60 fish, how many will have 40 markers?" "60 2 x 40, that is to say, there must be 40 marked fish in the whole fish pond, this ratio is equal to the proportion of 60 fish with 2 markers, and the calculation result should be 1200.

Solution: Suppose the spring of an object weighing 5kg is x cm long.

54-4x=1

x=53/4x=

Let the original length of the spring be xcm and the coefficient of elasticity is k, and the spring length of the object of 5kg is ycmx+2k=

x+6k=x=12

k=y=12+

Solution: Let the original length of the spring be x cm, weigh the object of 5kg, and the spring length is ycm( 2 =(y- x) (5-0).

Solution: x=12cm, y=

A: The spring length is .

Find out how many centimeters each kilogram of spring elongation, (divide by (6-2)=, and then find the original spring length as.

Then the five-kilogram spring length is 12+5*

The number of gear teeth is inversely proportional to the number of revolutions per minute.

20 x=40 60 solution gives x=30

The number of teeth of the driven wheel should be 30

Let this bundle of wires be xm long

2500/250=x/5

x = 50 This bundle of wires is 50m long

Since the mass of the wire is directly proportional to the length.

So if you set the length of the wire to be xm

then there is 2500:x=250:5

Get x=50 so the length is 50m

Wire Quality Wire length = unit length of wire mass.

Here the mass of the wire is directly proportional to the length of the wire.

Set the length of the wire to be x meters.

Kilogram = 2500 grams.

2500/x=250/5

250x=2500×5

250x=12500

x=50

Let the bundle be x-long

x x = 50 meters.

Let the distance between the two places be x kilometers.

x：x+78=1：7/9+80%

x+78=71x/45

26x/45=78

x=135 The distance between the two places is 135 km.

Solution: If the distance between the two places is x kilometers, then there is 7x 9+

Simplification yields 26x 45=78

The solution is x=135

So the distance between the two places is 135 kilometers.

Solution: Let each person be assigned x, y, and z parts.

x:1/6=y:1/5=z:1/

x+y+z=1590

Solution: x=450

y=540z=600

Answer: A allocates 450 parts, B allocates 540 finger-resistant parts, and C allocates 600 parts.