Three math problems that are a little difficult, and three math problems that are more difficult

1.There are X people in the fifth grade and Y people in the sixth grade.

x+y=336, 5/7*x+3/7*y=188x=154, y=182

There were 154 students in the fifth grade and 182 in the sixth grade.

2.There are x first prizes, y second prizes, and z third prizes.

54z=2y, y=2x

543.Let this batch of *** machines have x units, the purchase price of *** machine is y, and the selling price of *** machine is x*(, x-10)*

x=150,y=40

So there are 150 *** machines in this batch.

1.Suppose the fifth and sixth grades are drawn 5 7, and the following will be drawn:

336 5 7 240 (person).

More than the actual number of people drawn:

240 188 52 (person).

Sixth-graders are:

52 (5 7 3 7) 182 (person).

Fifth-graders are:

336 182 154 (person).

pieces, 120x50%=60 pieces, the ratio of the first prize, the second prize, and the third prize is 1:2:(2x2)=1:2:4

1+2+4=7, so, get.

The number of first, second and third prize works should be in the range of 54 to 60, and at the same time it is a multiple of 7, then only 56 is compliant, and the first prize: 56 x 1 7 = 8 pieces.

2nd prize: 56x 2 7 = 16 pieces.

3rd prize: 56x 4 7 = 32 pieces.

3.There are x sets of *** machines in this batch.

Unit price: (6000 20%) x=30000 x6000*(1-96%)] 10=(30000 x)*(1+20%-90%)=9000 x

24=9000/x

x=375There are 375 *** machines in this batch.

1. There are x students in the fifth grade and y students in the sixth grade, according to the topic.

x+y=336

5/7x+3/7y=188

x=154 y=182.

2. The first prize is set to be x, according to the title, the second prize is 2x, and the third prize is 4x.

Because the total number of entries is 120 and the probability of winning the prize is between 45% and 50%, there are.

45%*120 x+2x+4x 50%*120 solution:

Because the prizes are integer, the first prize is 8, the second prize is 16, and the third prize is 32.

3. There are x units in this batch, and the purchase price is Y yuan.

Depending on the title, there is.

Divide by 2.

Substituting 2 formulas, you can get y = 1200 yuan, so as to get x = 25 units.

There are a total of 25 *** machines in this batch.

The answer to the third question seems to be wrong, in short, I calculated 200, neither 375 nor 150I'm also asking for verification, so let's ask the teacher around me. It's not the road to success yet.

120 times 45% equals 54, and 120 times 50% is 60 pieces.

Solution: Set the first prize x pieces.

x+2x+4x=y (y takes the value between 54 and 60), and 1+2+4=7,54 between 60 and the common multiple of 7 is 567x=56x=8

Question 2).

1. A and B are xy

x+y=72

x/6+y/8=11

x=48 y=24

They were 48 and 24 respectively.

The answer should be 1

3. Let's look at it separately, after all the processes, the total amount of solution in B should be 800 + 800 2-(800 + 800 2) 2 + [800-800 2+(800 + 800 2) 2] 2 1100 grams.

Looking at the alcohol again, half of A to B> the alcohol in B is 800 2 2 200, and then half of it is poured back to A, and 200 2 100 is left in B, and the amount of alcohol in A is 800 2 + 100 300. Go back to the general B is 300 2 150, plus the original 100, the total alcohol is 100 + 150 250 grams, the problem is not very clear, B is 250 grams of alcohol, 1100 grams of solution, and 850 grams of water, so the answer is 250 1100 5 22 (accounting for the total) 250 850 5 17 (the ratio of alcohol and water) 250 400 5 8 (the proportion of alcohol in B to the total alcohol) 250 (800 + 800) 5 32 (the proportion of alcohol in B to the total solution) If you want the specific answer, you can look at it.

1 Class A 48 Class B 24

Let A be x and B be y

x+y=72, x 6+y 8=11 so we get x=48, y=24

The numerator denominator can be eliminated.

3 Original A 400 Alcohol 400 Water B 0 Alcohol 800 Water.

After the first time, A 200 alcohol 200 water B 200 alcohol 1000 water.

After the second time, A 300 alcohol 700 water B 100 alcohol 500 water.

After the third time, A 150 alcohol 350 water B 250 alcohol 850 water.

So alcohol accounts for 250 (250+850)=5 22

1 Class A 48 Class B 24

Let A be x and B be y

x+y=72, x 6+y 8=11 so we get x=48, y=24

So in the end there is 1 left

3 After the first time, A 200 alcohol 200 water B 200 alcohol 1000 water.

After the second time, A 200 + 100 alcohol 200 + 500 water B 100 alcohol 500 water.

After the third time, A 150 alcohol 350 water B 250 alcohol 850 water.

So alcohol accounts for 250 (250+850)=5 22

Question 2 The last remaining number is 1

The first inequality can be reduced to.

x-2)(x+1)>0

Solution. x>2 or x<-1

The second inequality can be reduced to.

2x+5)(x+k)<0

When -k>-5 2, i.e., k<5 2, the solution set is -5 2 -1, i.e., k<1.

When the solution set of the original inequality group is -5 25 2, the solution set is -k k, and the value range of k k is k<1

The first square of y is the opposite of each other <=1 5)x+1+(2x+1) 4=0 The answer is to do the math yourself;

2.9 eggs + 5 duck eggs + 3 goose eggs = yuan.

2 eggs + 6 duck eggs + 8 goose eggs = yuan.

11 eggs + 11 duck eggs + 11 goose eggs = yuan.

1 egg + 1 duck egg + 1 goose egg = yuan.

3.The denominator is = 8, so when the denominator is the smallest, the algebraic formula is the largest, that is, when x=1, the maximum value is 3;

Because there is no maximum value for the denominator, there is no minimum value for algebraic formulas.

When the maximum value is 3 x = 1, the denominator is the smallest, and the value of the algebraic formula is the largest.

1.From the problem, it can be obtained: the first square of y + the second square of y = 0

That is: (1 5) x + 1 + (2 x + 1) 4 = 0 x = -5 14

Yuan. 3.The numerator is 8 when the value is maximum, i.e., x=1 (because the quadratic square of 3(x-1) is 0.).The minimum value is 0, so the minimum value of the molecule is 8).

There is no minimum value for algebraic formulas.

1 Let the first power of y = the second power of negative y, and the solution is x=-25 142 Let 9x+5y+3z=

2x+6y+8z=

then i.e. 11x 11y 11z

x+y+z＝

3 In this fraction, the denominator is a variable, so the magnitude of the fractional value depends on the size of the denominator. When the denominator is larger, the fractional value is smaller, and vice versa. When x=1, the denominator has a minimum value of 8, and the maximum value of the denominator is 3; There is no maximum value for the denominator, so there is no maximum value for the fractional value.