Two math application problems and two application problems for junior high school

1.If the first pile is doubled, the second pile is increased by one, the third pile is reduced by two, and the fourth pile is doubled to x, then there is x-1+x+2+2x+x 2=19

x=4 So 2 in the first pile, 3 in the second pile, 6 in the third pile, and 8 in the fourth pile.

Hours 40 minutes = 5 3 hours.

The speed of the ship is V, and the distance between the two places is S kilometers.

s/（v+12）=5/3 s/（v-12）=33s=5v+60 s=3v-36

9s=15v+180 5s=15v-1809s-5s=360

s = 90 km.

So you can finish the fight.

4.(1) 400 students can pass in 2 minutes, and 200 students can pass in 1 minute.

Set up a partial door to pass x students.

x+x+40=200 x=80

So the side door passes 80 people.

The main entrance is 120 minutes.

2) A total of 45·4·6 = 1080 students.

The efficiency is only 80%, so the partial door passes 64 people.

The main entrance passes 96 minutes.

A total of 96 + 64 + 64 = 224 people.

1080 224 minutes. Plausible!

It's not just two, it's two.

Categories: Teaching Nuclear Zen Education Academic Exam >> Study HelpProblem Description:

1. There are 3 teacups on the table with the rims facing up, and 2 cups are flipped each time, can they be flipped several times to make them turn all the cups down? What about 7 cups with the rim facing up, 3 flipped at a time? If "+1" and "1" are used to indicate different directions of the cup, can you use the operation of rational numbers to explain the reason?

2. In front of the 12 numbers on the clock face, add a positive or negative sign appropriately, so that their sum is 0 urgent, everyone help me, thank you, thank you.

Analysis: 1. There are 3 teacups on the table with the rim facing up, 2 of them are flipped each time, can they be flipped several times to make them all face down? What about 7 cups with the rim facing up, 3 flipped at a time?

If "+1" and "1" are used to indicate different directions of the cup, can you use the operation of rational numbers to explain the reason?

1. This is not feasible.

Remember that the cup is "+1" for Tong's hand upwards and "-1" for downwards

At the beginning, it is 1, and the three are 3, and two cups are flipped at a time, and there are several cases:

Two "1s" become two "-1s", and the value is reduced by 4

Two "1s" become two "1s" and the value increases by 4

A "1" and a "1" become a "1" and a "-1", and the value increases to 0, so no matter how you flip it, the sum of the values of the three cups divided by 4 is always the remainder of 3, which means that the value is only 3, 1, two possibilities, so it is impossible to flip them all over.

2. For the second case is possible, the cup situation is as follows:

Flip 4 times and you're good to go.

2. In front of the 12 numbers on the clock face, add a positive or negative sign appropriately, so that their sum is 0 and 0, only half of them need to be positive and half of them are negative.

1) 3, 4, 6, 7, 9, 10 are negative.

2) 8, 9, 10, 12 are negative.

3) 1, 8, 9, 10, 11 are negative.

①.Solution: Let : A be x, then (x+2x+3x) 2+(x+3x), solution: x=1 23;

Answer: Limin store** has 20 baskets.

1) 1. B is 10% more expensive than A, and B** closure is.

b=a*(1+10%)=2409 yuan.

2. The power consumption difference between B and A is.

degrees are less than a per day).

Refer to 2, b to a above

365 days per year for 10 years to save electricity**p for.

p = degree * yuan degree * 365 days * 10 years = 657 yuan.

3. To sum up, B's **minus the **p saved in 10 years, that is: 2409-657 = 1752 yuan.

When A's ** is less than 1752 yuan, it can be used for 10 years or less, and A is cost-effective.

A's ** discount is 1752 divided by A's original price = 1752 Shen J 2190 = that is, A is 8% off or lower, and it is cost-effective if it is used for 10 years or within the sedan car.

Earn back the cost after the year.

That is, the fourth year begins to make a profit.

2) 32*15-120+20=3.8 million.

3,800,000 15 = 10,000 yuan.

The average annual profit in these 15 years is 10,000 yuan.

1.Set up a power outage for x minutes.

1-x/120=2(1-x/60)

1-x/120=2-x/30

x/40=1

x=402.Set the original plan for x days.

40*x/3+80(2/

160/3x-120=80/3x

80/3x=120x=

Let the burning candle length be x

1-1/2x=2(1-x)

x = 2 3 power outage 2 3 hours.

That's 40 minutes.

A total of x-sides are made.

2/3x）/40-（2/3x）/80=2/3x=180

1.Solution: If X staff members of type B are recruited, then A is 30-x.

600 (30-x) + 1000x 22000 solve: x 10

A: A maximum of 10 Type B staff members will be recruited.

2.Solution: The remaining fruits can be discounted at least x according to the original sales price**1000 2 * (11-7) + 1000 2 * (11x-7) 2900 solution: x

A: The remaining fruits can be discounted by as little as 20% off the original sales price**.

1 A x and B y.

According to the inscription, there is: x + y=30 1 formula 600x +1000y<=22000 2 formula is known by 1 formula: x=30 - y

Bringing in 2 formula has: 18000 + 400y<=22000 solution: y<=10

That is, up to 10 people can be recruited.

2. Set x discount.

1000/2*（11-7）+500*（11x-7）=29002000+500（11x-7）=2900

55x-35=9

55x=44x=4/5

The remaining fruit can be discounted by as little as 20% off the original sale price**.

A maximum of x number of staff members can be recruited.

1000x + 600(30-x) = 220001000x + 18000 - 600x = 22000400x = 4000

x = 10 names.

2.Set a minimum of x discount on the original sales price.

1000×(11-7)/2 + 500×(11x/10 -7) = 2900

2000 + 550x - 3500 = 2900550x = 4400

x = 8 (fold).

1 If there are x staff members of type A and y names of staff members of type B, there will be x+y=30 ......a 600x+1000y<=22000……b

Substituting x=30-y from formula A to formula B.

30-y）+1000y<=22000

i.e. y<=10

It can be known that up to 10 Type B staff members can be recruited.

2. Half of the sales is to make a profit, which is 500 (11-7) = 2000 yuan, so it is necessary to make a profit of 900 yuan, and the selling price is x yuan.

then there is 500 (x-7) > = 900

The solution is x>=

So the discount number is (

That is, 20% off**.

1.If you recruit x people for type B staff, then 30-x people for staff A can be listed according to the topic: 1000x+600(30-x) 22000 solution x 10, so you can recruit up to 10 type B workers.

2.Set the minimum selling price of the remaining fruits at $x.

According to the meaning of the problem, the inequality (1000 2)*(11-7)+500(x-7) 2900 can be solved to x

Therefore, the remaining fruits can be discounted by as little as 20% according to the original sales price**.

1.Untie; A maximum of x staff members of type B can be recruited, and y staff members of type A can be recruited.

x+y=30 ①

1000x+600y=22000 ②

by , x=30-y

Substituting into , we get 1000 (30-y) + 600y = 2200030000-1000y + 600y = 2200030000-400y = 22000

400y=30000 -22000

400y=8000

y=20 substituting y=20 into , we get x=30-20

x=10 A: A maximum of 10 Type B staff can be recruited.

2.Solution: The remaining fruits can be discounted at least x according to the original sales price**(1000 2 (11-7) (1000 2 (11-7) 10 x=2900

300x=2900

x=2900÷300

x=9 3 2 (nine and two-thirds).

A: The remaining fruits can be discounted at a minimum of 9 and 2/3 of the original sales price**.

The first path is set to be X and B Y

x+y is less than or equal to 30

600x+1000y is less than or equal to 10,000.

Then do the math yourself.

Set up a move X

600x + 1000（30-x)<=220008000<=400x

x=>20

n B < = 10

Set up a Y discount. 11*500+500*11*y=>2900+70005500*（1+n）=>9900

n "The minimum discount is 20%.

Set A and B two kinds of staff are x y :

x+y=30

600x+1000y=22000

y=10A: A maximum of 10 Type B staff can be recruited.

x+1000y<=22000 2, x+y=30 Bring 2 into 1 and eliminate x to get an inequality that can be calculated. (x for A, y for B) to solve y<=10

2. Half of the start of sales has already made a profit of 2000, so the remaining 500 kilograms only need to earn 900. then there is an inequality.

11x-7) * 500 > = 900 can be solved to solve the discount (such as 7 discount multiplied.)

1.A company wants to recruit 30 staff members A and B, with a monthly salary of 600 yuan for a type of staff and a monthly salary of 1,000 yuan for a type of B. It is now required that the monthly salary cannot exceed 10,000 yuan, and the maximum number of B staff can be recruited?

A: A maximum of 9 Type B staff members can be recruited.

1. x+y=30 600x+1000y=22000 calculates the value of x y.

500*(11-7)=900 yuan 500*Calculate the value of x.

Solution: The sum of the areas of the three circles is 56, the area of the circle A and the circle c and the area of the circle b are 3 5, a+ c = 3 5 b

b＋3／5⊙b=56

b=35⊙a＋⊙b=21 35÷（1／7）=51／6⊙a＋1／3(21-⊙a)=5

a=12，⊙c=9

Answer: The areas of circles a, b, and c are respectively.

2. When x=1, its value is 5:x=2, and its value is 7:x=-2: its value is 11 and is substituted into ax 2+bx+c:

a+b+c=5 4a+2b+c=7 4a-2b+c=11 - obtained: 4b=-4 b=-1

Obtain: 8a+2c=18, 4a+c=9 c=9-4a substituting b=-1, substituting a=-1 gets: a-1+9-4a=5 a=1 substituting b=-1, a=1 into obtain:

1-1+c=5 c=5Answer: a=1,b,=-1,c=5.

1.There is a problem with the question.

b=-1 c=5

Happy New Year, thank you.

Question 1: Incompetent.

2.When x=1, a+b+c=5,

When x=2, 4a+2b+c=7,

When x=-2, 4a-2b+c=11, 8a+2c=18, 4b=-4

b = -1 substituting b = -1 to get a + c = 6

a=6-c8（6-c）+2c=18

c=5, substitution a=1

1.First of all, it is easy to calculate that the area of b is 56 times five-eighths to get 35, and let the area of a be x, there is 1 6x+1 3(21-x)=5

So a12, b35, c9

4a+2b+c=7,4a-2b+c=11, and the simultaneous solution is a=1, b=-1, c=5

1.There is a total of m of water in the pool, and it takes x hours for the water to come out, and because the water outlet pipe is not closed, the water for two hours is equal to the amount of water for one hour of water inflow. m 5 2m x x x 10

2.Set to take t hours to catch up, then.

80t 60t 448 t hours.