The first year of junior high school application questions should be detailed in the process The pro

!! made with inequalities

Solution: Suppose the second count is x times.

5x+2≤55

5x≤53x≤

x is a positive integer and 5x is around 55.

x is only) Answer: A total of 52 eggs.

The mantissa that satisfies the division by 5 is 0,5, so the mantissa must be , and the estimation of the full basket must only be 52, of course, 22,37 is also correct.

My answer needs to be detailed.

Because 5 numbers leave 2, and only the single digits of 0 and 5 are divisible by 5, the single digit of the total number of eggs is 2 or 7And the total number of eggs is less than or equal to 55, so the range is locked. There are 9 numbers of 12, 22, 32, 42, 52, 17, 27, 37, and 47 that meet the above.

And because there is 1 left in 3, so the above 9 numbers are subtracted from 1 and divided by 3, and if it can be divisible, it is eligible. The calculations that meet are 22, 37 and 52Yes, because the question tells that it is full of blue, so it is 52 eggs.

Friendly reminder: (tips for numbers divisible by 3) to see if the numbers on each digit of this number add up to be divisible by 3, if it is divisible by 3, this number must be divisible by 3, if not, it must not be divisible by 3.

Is this an application problem?

The students of the two travel agencies are 20% off, and the teachers of the A travel agency are free, and the teachers of the B travel agency are charged at 20% off, so the B travel agency charges more for the teacher, and if there are 3 teachers to participate, the B travel agency will definitely charge more. It can't be equal. No need to calculate.

Your question is wrong, both are 20% off, then it must be A is cheap, assuming A is 8% off, B is 7% off, and the answer is as follows.

Suppose there is a student A who charges $B:

Then: A travel agency charges = a times multiplied by b B travel agency charges = (a + 3) multiplied by b when A and B charges are equal.

A multiplied by b = (a+3) multiplied by b

The solution is a=21

When the number of students is equal to 21, both families are charged the same.

When a = 45 times get A = 36 b, B = b can be seen A travel agency is cheaper.

If I assume that the number of discounts is different, then I can enter it.

1.(20a+30(a-2))(1+20%)=2.Schoolgirl x

x+1=2(x-1) x=3 Students have 3+3+1=7

x=2000

4.1st class car X

64x+92(6-x)=496 x=2 second-class car 6-2=4

5.Set to buy Yali xkg

5x+3(8-x)=28

x=23*2+5*(8-2)-28=8 yuan.

6.Set the truck X speed, the bus Y speed.

4(x+y)=x(

x=75 y=60

The distance between the two places is 4 (75 + 60) = 540

1. (20a+30(a-2)) x (1+20%)=60a-72 The store sells a total of 60a-72 yuan.

2. Set x boys, because each boy sees as many white and red hard hats, so the number of girls = x-1

And because: the white helmet that each girl sees is twice as red as the red one, then x (x-2) = 2 solves the equation to obtain: x = 4

So there are 4 boys and 3 girls.

3. Set rough processing x kilograms.

then x+3x+2000=10000

x = 2000 kg.

Rough machining 2000 kg.

4. If the first-class carriage is set up x sections, the second-class carriage is 6-x sections, and the equation can be listed

64x+92（6-x）=496

x = 2 2 first-class cars, 4 second-class cars.

5. If the pear is x kg, the grape is 8-x

5x+3（8-x）=28

x = 2 actual amount of money to sell = 3 x 2 + 5 x 6 = 36 yuan, the bear lost 36-28 = 8 yuan.

The topic is simple, too wasted time 1, [20a 30 a 2 ] 1 20

:How much is a share on Wednesday**? 27+4+

How many yuan is the most ** per share this week? How much is the lowest price per share? 27+4+

It is known that Wang Gang bought ** and paid the handling fee of the transaction amount, and when he sells, he needs to pay the handling fee of the transaction amount and 1 transaction fee, if Wang Gang sells ** before Saturday**, what is his income?

27 4 31 on Monday, 31 on Tuesday, 32 6 26 on Friday, 26 28 on Saturday, $1 per share on Wednesday **; This week's highest per share, the lowest is 26 yuan; Wang Gang got 1000 (28 27) 1000 28 (yuan) when he sold on Saturday.

Solution: Suppose the average speed from Beijing to Tianjin is x kilometers.

1 2 + 6 60) x = 1 2 (x + 40) The original equation can be reduced to 3 5x = 1 2x + 20

Shift the term to get 3 5x-1 2x=20

Combine similar terms to get 1 10x=20

The coefficient of transformation is 1, and x=200 is obtained

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

Solution: Suppose the average speed from Beijing to Tianjin is x kilometers.

Solution: (1 2+6 60)x=1 2(x+40)3 5x=1 2x+20

3/5x-1/2x=20

1/10x=20

x=200A: The average speed from Beijing to Tianjin is 200 kilometers per hour.

Here's how the problem is solved.

Because the distance is the same, the total distance = 40*

So the average speed from Beijing to Tianjin = 20 (

It can be thought of as x-(-1) +x-2

This means that the minimum sum of the distances from x to -1 and to 2 is of course between -1 and 2.

Therefore, the corresponding x value range is -1 x 2, and the minimum value is 3

If the first discount is set to be the X yuan of commodity A, then the equation of the column of commodity B (84-5x): 6x+3(84-5x)=108

Solution x=16

Before the discount, product A is 16 yuan, and product B is 4 yuan.

If there is no discount, 50 pieces of product A and 50 pieces of product B will cost 50 (16 + 4) = 1000, so it will cost 40 yuan less after the discount.

Question 1 (340*4-20*4) 2 The speed of the car is formed m s, and the second question is set to be a x yuan and b is y yuan.

5x+y=84

6x+3y=108

Solution x=16 y=4 It is calculated that it takes a total of 1,000 yuan to buy 50 pieces of product A and 50 pieces of product B before the event, that is, 40 less money

5. The total workload completed by the 4 workers in Group A in March was 20 pieces more than 4 times the per capita quota of the month, and the total workload completed by the 5 workers in Group B in March was 20 pieces less than the 6 times the per capita quota of the month.

1) If the average workload of the two groups of workers is equal for the month, what is the quota per person for the month?

2) If the average workload of the workers in Group A is 2 more than that of Group B, what is the per capita quota for this month?

3) If the average workload of the workers in Group A is 2 less than that of Group B, what is the per capita quota for this month?

The per capita quota is x

1)（4x+20）/4=（6x-20）/5

2)（4x+20）/4-（6x-20）/5=2

3)6x-20）/5-（4x+20）/4=2

Assuming that all of them are 6% annual interest rate, the interest is 68*6% = 10,000 yuan, and the overpayment is 10,000 yuan. The reason for the overpayment is because the B loan is also calculated at 6%, and the interest is overcalculated. Type B loan is 10,000 yuan, and type A loan is 120,000 yuan.

Column: Type B loan (68*6% 10,000 yuan.

Type A loan 68-56 = 120,000 yuan.

Let the loan of A be $0,000 and the loan of B be $y,000.

x+y=68 ①

Braces) 6% x+

y=68-x is obtained by

Substituting into , you get 6%x+

6%x+x=12

Substituting x=12 into , we get y=56

x=12 (curly braces) is the solution of the original system of equations (don't forget to write the answer) y=56

Do you want to verify. By the way, you went to middle school in **.

This is the knowledge of the middle stage.

If A is 10,000 yuan, then B is (68-x) 10,000 yuan.

x*6%+（68-x）*x=12

Answer: A paid 120,000 yuan, B borrowed 560,000 yuan.

Let A x, B y

x+y=68 (10,000 yuan).

6% x + 10,000 yuan).

So, x=12; y=56