One Yuan One Application Problem Rush Hurry Hurry 11 1!

On the day off, my brother and my mother set off from home to go to my grandmother's house together, and after they walked for 1 hour, my brother found that he had forgotten the gift he brought to his grandmother, so he immediately took the gift and chased it at a speed of 6 kilometers per hour, if my brother and my mother traveled 2 kilometers per hour, it would take them 1 hour and 45 minutes to get from home to my grandmother's house, and asked my brother if he could catch up with my brother and mother before they went to my grandmother's house.

1 solution: Set the time for my brother to catch up with x hours, which can be obtained according to the topic.

6x=2(1+x)

Solution: x=The younger brother and mother set off from home to go to grandma's house together, and walked a total of 1+x= that is, 1 hour and 30 minutes before being caught up by the brother.

Since it takes 1 hour and 45 minutes for my brother and mother to go to my grandmother's house together, my brother can catch up 15 minutes before my brother and mother go to my grandmother's house.

4. A and B are 240 kilometers apart, and a slow train comes out of station A at a speed of 80 kilometers per hour; An express train departs from Station B at a speed of 120 km/h. Q: If two cars are going in the same direction (the fast train is behind), how long does it take for the fast train to catch up with the slow train?

Solution: After X hours, can the fast train catch up with the slow train? It can be obtained according to the title.

80x+240=120x

Solution: x=6

5. Students from the Municipal Experimental Middle School walked to the suburbs. Class (1) students form a front team with a walking speed of 4 km, and (2) class students form a back team with a speed of 6 km. The front team set off one hour before the back team set off, and at the same time the back team sent a liaison officer to communicate back and forth between the two teams on a bicycle without interruption, and he rode at a speed of 12 km/h.

1) How long does it take for the back team to catch up with the front team?

2) What is the distance traveled by the liaison officer during the time when the back team catches up with the front team?

3) When are the teams 3 km apart?

4) When did the two teams get 8 kilometers apart?

When the rear team caught up with the front team, the liaison officer walked a distance of 2*12 24 kilometers.

There are two scenarios when the two teams are 3 kilometers apart.

Catch-up is 1 (6-4) hours.

After that, it is 2+3 (6-4) hours.

8+4) (6-4) 6 hours apart by 8 km.

Question 1: Let the time when the elder brother catches up with them is t, 6t = 2 (t+1).

t = hours.

Hours 45 minutes so it is possible to catch up.

When the basic price of 130 yuan is set at the base price of x yuan, the product can be sold every day, and each piece makes a profit of 10 x yuan.

Set at the base price of 130 yuan** price x yuan.

x1=100，.x2=30

Sales price: 130 + 100 = 230 yuan, or 130 + 30 = 160 yuan.

o( o Thank you, hope.)

Solution: If the single digit is x, then the ten digit is (x 4).

x²－4）－2＝x

x²-x-6=0

x-3）（x+2）=0

Solve x=3 or x=-2 (do not match the question and round it off).

Ten digits: 3+2=5

A: The number is 53

Solution: (1) When the price of each product is 170 yuan, it is 40 yuan higher than the price of 130 yuan per product, that is, 170-130 = 40 (yuan), (1 point).

Then 30 goods can be sold per day, that is, 70-40 = 30 (pieces), (2 points) the daily profit of the shopping mall is (170-120) 30 = 1500 (yuan) (3 points) Answer: 30 goods can be sold every day, and the daily profit of the shopping mall is 1500 yuan 2) When the daily profit of the shopping mall reaches 1600 yuan, the price of each commodity is x yuan, then each product is higher than 130 yuan (x-130) yuan, and each piece can be profitable (x-120) yuan (4 points).

The daily sales of goods is 70-(x-130)=200-x (pieces) (5 points) according to the equation (200-x) (x-120) = 1600 (6 points), and x2-320x+25600=0, that is, (x-160)2=0 (7 points).

The solution is x=160 (9 points).

Answer: When the price of each item is 160 yuan, the daily profit of the mall reaches 1600 yuan (10 points).

Set up to buy x pair of socks.

221*10+（x-10）*26=（221*10+26*x）*

Solution, get x=15

Suppose the unit wants to buy x pairs of socks, then the first plan needs to pay 2210 + 26 * (x-10) The second plan needs to pay.

Now you have to pay the same, i.e. 2210 + 26 * (x-10) = get x = 15 pairs.

Let the raw material of A need to be x, and the raw material of B needs to be y

x+ solution yields x=28 y=30

Later, you will go to the math know-it-all to ask, and there is a professional teacher who is answering, and it is very timely. I'm also a regular visitor there. Hey.

When X grams of raw materials for A and Y grams for raw materials for B are just enough to meet the needs of patients. According to the title, there are:

x+ biquint, y=30, x=28

That is, 30 grams of raw material A and 18 grams of raw material B are just enough to meet the needs of the patient.

Suppose raw material A needs to be x, and raw material B needs to be y

x+ solution yields x=28 y=30

Set A x grams, B y grams.

Rule. x+ solution yields x=28 y=30

That is, 28 grams for A and 30 grams for B.

Set X grams of raw materials for A and Y grams for raw materials of B. Get x+

solution, x=28, y=30

That is, 28 grams of raw material A and 30 grams of raw material B are required.

Solution: Set up x grams of raw material A and y grams of raw material B in each meal, according to the topic, get.

x+solutionx=28 y=30

Let A be x and B be y, and we get the system of equations x+

Finding the system of equations yields x=28 and y=30

Set a x gram, B y gram.

By the title:

x+ 2- gets: y=30

then x=28 so A needs 28 grams and B 30 grams.

48-a-(2a+3)-(3a+3)=42-6aWhen a=3cm, 42-6a=24

So 2a+3=9, 3a+3=12

So this one is still a quadrilateral.

When a = 7 cm, 42-6a = 0

So you can't get a quadrilateral.

The first side is a centimeter long.

The second side length is: 2a+3

The third length is: a+2a+3=3a+3

The length of the fourth side is: 48-(a+2a+3+3a+3)=42-6aWhen a=3 cm, the length of the 4 sides is:

The length of the first side = 3 cm.

The second side length = 2a + 3 = 9 cm.

The length of the third side = a + (2a + 3) = 12 cm.

The length of the fourth side = 42-6a = 42-6a = 24 cm because: the sum of the first three sides = (3 + 9 + 12) cm = 24 cm = the length of the fourth side So: at this time, the figure is a straight line that coincides with the first three sides and the fourth edge.

To form a quadrilateral, as long as the longest side is less than the sum of the other three sides.

When a = 7 cm, the length of the 4 edges is:

The length of the first side = 7 cm.

The length of the second side = 2a + 3 = 17 cm.

The length of the third side = a + (2a + 3) = 24 cm.

The length of the fourth side = 42-6a = 42-6a = 0 cm because: the sum of the first three sides = (7 + 17 + 24) cm = 48 cm = circumference, and the length of the fourth side is 0, that is: there is no fourth edge.

So: at this time the quadrilateral (or figure. Because the triangle is not true, at most it is a straight line with 3 sides overlapping) does not hold.

So 3

, you can get a quadrilateral; a=7cm, you can't get a quadrilateral.

Total price: (4400+1200) (1-44%)x10=10000 (yuan).

Total original price: 10000-1200=8800 (yuan).

The purchase price of the original product: 8800 100 = 88 (yuan).

Let the purchase unit price be x and the pricing price be y

100x=100y+1200

100*44%=100y-4400

x=88,y=100

The cost of making the price of x 100 items is yRule.

100x=y+1200

100x*44%=y-4400

Get x=100

y=8800

The unit price = 88 yuan.

The set price is y yuan, and the cost price is x yuan, then.

y-x=1200 100---1)44%y-x= -4400 100 --2) consists of (1), (2) to form a system of equations.

y=100 --3)x=100-12=88

Let the purchase unit price of the original product be X yuan.

1200+100x）

The solution is x=88 answer——

Pricing x original price y

100*x-100*y=1200

The solution is x=100 and y=88

Total price: (4400+1200) (1-44%)x10=10000 (yuan).

Total original price: 10000-1200 = 8800 (yuan) The purchase unit price of the original commodity: 8800 100 = 88 (yuan).

Set the purchase unit price to be $ x and the ** price to be $ y.

List binary linear equations:

100y-100x=1200

The solution is x=88 and y=100

That is, the purchase unit price is 88 yuan.

To lay out a square, the side length must be both a multiple of 45 and a multiple of 30, and "at least" requires the least common multiple of 45 and 30.

The least common multiple of 45 and 30 is 90

90 45 = 2 (blocks) 2 blocks are needed for the long side.

90 30 = 3 (blocks) 3 blocks are required for a wide side.

2 3 = 6 (block).

A: At least 6 such bricks can be used to pave a square.

Solution: Find the least common multiple of 45 and 30, that is, 90, the side length is 90 cm, then one side should be 90

45 = 2 (block), 90 30 3 (block) on one side, 2 3 6 (block). Answer: 6 pieces.

The six bricks are equal in length because the sides of the square.

So 45x=30y

At the very least, take x=2

y=3 excludes x=y=0.

That's 3*2=6

1.If there are x people in the 2nd workshop, then the 1st workshop is.

x-10）*

x = 250 is 170 for the first workshop and 250 for the second workshop

2.Solution: The average speed from Beijing to Tianjin at the time of this test run is x kilometers per hour.

The average speed from Tianjin to Beijing is x+40 km/h.

List the equations according to the title:

x+40)*

Solution x=200

That is, the average speed from Beijing to Tianjin was 200 kilometers per hour during this test run.

1. Solution: If there are x people in the second workshop, then the first workshop has (people.

Solution x=250

Answer: 2. If the average speed from Beijing to Tianjin is x kilometers per hour during this test run, the average speed from Tianjin to Beijing is X + 40 kilometers per hour.

x+40)*

Solution x=200

A: The average speed from Beijing to Tianjin was 200 kilometers per hour.

！100% correct!!

1.Set the number of people in the first workshop x, then the number of people in the second workshop 4 5 * x - 30 column x+10 = (4 5 * x-30-10) * 3 42Set this test run from Tianjin to Beijing at a speed of x kilometers per hour.

Then Beijing to Tianjin speed X-40

30 60*x=(30+6) 60*(x-40), then the average velocity = 2*(30 60*x) [30+36) 60].

1. The second workshop x people first workshop 4 5x-30

3 4 (4 5x-30 + 10) = 4 5x-30-10x = 2501 workshop is 170, 2 workshop is 250

2. The distance between Tianjin and Beijing is x kilometers.

x (36 60) + 40 = x (30 60) x = 120 average speed = 2x [(36 + 30) 60] = km h).

1;Let the second workshop be m people, and the column equation is: 4 5 m-30 + 10 = (m-10) x3 4 m = 250 is the number of people in the second workshop, and the first workshop is 170 people.

1.Set the number of people in the second workshop to x, and the first workshop to y

4/5x-30=y

x-10=3/4x

Solution: x=40, y=2

So the first is 2 people, and the second is 40 people.