It is known that a computer company has three types of computers: Type A, Type B, and Type C

Since the money spent is 100,500 yuan, one of the two models purchased must be C-type, and the number of C-type must be odd. Because if there are only types A and B, there will not be 500 yuan, that is, all the money will not be spent.

If a type X unit is purchased, then type C is (36-x) unit, so there are 6000*x+2500*(36-x)=100500, and the solution is x=3, so there are 3 units of type A and 33 units of type C; If B type Y is purchased, then C type is (36-Y) unit, there are 4000 * Y + 2500 * (36-Y) = 100500, and the solution is Y = 7, so there are 7 B type and 29 C type.

To sum up, the school has two plans, the first one: buy 3 units of type A and 33 units of type C; The second type: buy 7 units of type B and 29 units of type C.

Scheme 1: Solution: Set up type A to buy x units, and type B to buy (36-x) units.

6000x+4000(36-x)=100500x12 does not fit the topic.

Abandon it. Option 2:

Solution: Set up and buy A-type X units and C-type (36-X) units.

6000x+2500（36-x）=100500600x+9000-2500x=100500350x=100500

x=336-3=33 (sets).

Scheme 3: Solution: Set up and buy B type X units and C type (36-x) units.

4000x+2500（36-x）=1005004000x+10000-2500x=1005001500x=100500

x=736-7=29 (sets).

A: There are two plans, buy 3 units of type A and 33 units of type C. I bought 7 units of type B and 19 units of type C.

1) A type X stage, B type Y stage.

x+y=36

6000x+4000y=100500

The solution is x=, impossible.

2) Type A X stage, C type Y stage.

x+y=36

6000x+2500y=100500

The solution is x=3, y=33

3) Type B X stage, C type Y stage.

x+y=36

4000x+2500y=100500

The solution is x=7 and y=29

Two schemes: 3 units of type A and 33 units of type C; There are 7 units of type B and 29 units of type C.

There are three scenarios to consider.

1，6000x+4000y<=100500x+y=36

2，6000x+2500y<=100500x+y=36

3,4000x+2500y<=100500x+y=36

It will be found separately and you will be able to do it.

There are only two options: assuming that the requisition quantities of a, b, and c are x, y, and z, then you can list three groups according to the conditions you give.

A 2-yuan 1st order equation: that is, 6000x+4000y=100500, x+y=36, this group has no solution, 6000x+2500z=100500, x+z=36, this group x=3, z=33, that is, 3 A-type computers, 33 C-type computers.

4000Y+2500Z=100500, Y+Z=36, this group Y=9, Z=27, that is, 9 B-type computers, 27 C-type computers.

100500/6000=<36

So you have to buy Type C.

Type A: (100500-2500*36) (6000-2500) = 3 sets Type B: (100500-2500*36) (4000-2500) = 7 sets of two schemes:

1. Purchased 3 sets and 33 sets of Type A and Type C.

2. Purchased 7 sets of B and C types and 29 sets each.

Set up to buy AX Taiwan, buy BY Taiwan, buy CZ Taiwan, then.

x+y+z=36

6000x+4000y+2500z=100500 Multiply 2500 on both sides of the first formula, and then subtract with the second formula to get 3500x+1500y=10500

x=(10500-1500y)/3500

x=(21-3y)/7

Because the values of x, y, and z are all integers, y can only take the values of 0 or 7, and then find x and z: respectively

x=3 y=0 z=33

x=0 y=7 z=29

So there are two schemes, which are to buy A3 units, B not to buy, and C33 units; Or A does not buy, B7 units, C29 units.

Wipe. This is some child's homework problem, right?

Condition 1: 6000x+4000y+2500z=100500 is reduced to 12x+8y+5z=201

Condition 2: x+y+z=36

Let x=0, y=0, z=0 and bring into the system of equations to calculate when x=0, y=29, z=7

When y=0, y=3, z=33

When z=0, y=, z= is not on topic. In addition, because 100500 has a singular number, z must be singular, and z=1,3,5 ,...... can be taken31, 33 max to 33, a system of equations.

12x+8y+5z=201

x+y+z=36

Bring the z value from 1 to 33 into the calculation, round off x, and the y value is not an integer, and get the full solution.

If the money doesn't have to be spent in its entirety ... Then you are doing the math yourself, there are new schemes. 36 units are all Z, as long as 9W, you can slowly reduce it yourself.

Set up A type X station, B type Y station, C type Z station.

then we get x+y+z=36

6000x+4000y+2500z<=100500When x=0, find the value of y and z.

When y=0, find the values of x,z.

When z=0, find the value of x,y.

Let a be x, b be y, and c be 36-x-y

Then make 2 random equations.

6000x+4000y+（36-x-y）2500=100500100500÷x+100500÷y+100500÷（36-x-y）=4000+2500+100500

Maybe the equation column is a bit annoying.

Solution: It can be considered in three situations:

1) Only Type A PCs and Type B PCs will be purchased.

Purchased X type A and B type (36-x) units.

6000x+4000(36-x)=100500 solution x= undesirable.

2) Purchase only Type A PCs and Type C PCs.

Purchased X type A and C type (36-x) units.

6000x+2500(36-x) =100500 The solution is x=3,36-x=33

1: Solution: Set up to buy x units from brand A, and purchase (36-x) units from brand B;

X+2000 (36-x) 100000 solution to X less than or equal to 7, that is, buy 7 A type A computers and 29 B E computers;

x+2000 (36-x) 100000 solution x 14, that is, buy 14 type A B type computers and 22 B type computers;

X+5000 (36-x) 100000 solution x 32, that is, buy 32 Type C computers and 4 Type D computers.

Reason: Spend 100,000 yuan to buy as many computers as possible.

2. If the purchase of brand A computers is type A: then the school has purchased 7 type A computers.

The first thing to do is to determine which two models to buy, to meet the number of 36 units, C is definitely there, because all buy C has 40 units, and all buy the other two can not achieve the number of 36, AB mix can not achieve this goal, because the limit of all A and all B is not good, and the two are naturally not good when mixed.

Scheme 1, C buys 33 units, A3 units, and the equation is as follows: 2500*C+6000*(36-C)=100500

Scheme 2, C buys 29 units, B 7 units, equation: 2500C + 4000 (36-C) = 100500.

The reason I said, it's easy to understand, this shouldn't be counted as a computer problem, it's just a calculation problem.

1. According to the topic, first use the list method or the tree diagram method to list all possible outcomes, and then find the probability of the event according to the probability formula;

There are 6 possible outcomes: (a,d),(a,e),(b,d),(b,e),(c,e),(c,d),(c,e);

2. From the above, it can be seen that when the scheme (A, D) is elected, the purchase of A model and D model computers are X, Y respectively, according to the topic, x+y=36,6000x+5000y=100000, and the solution is x=-80 after testing does not meet the topic, and it is discarded;

When the scheme (a, e) is selected, the purchase of a model and e model computers are x, y respectively, according to the topic, x+y=36,6000x+2000y=100000 to obtain x=7

So Hope Middle School purchased 7 A-type computers

There are 6 possible outcomes: (a,d),(a,e),(b,d),(b,e),(c,e),(c,d),(c,e);

2) Because there are 2 options for selecting model A computers, namely (A, D) (A, E), THE PROBABILITY OF MODEL A COMPUTERS BEING SELECTED IS;

3) From (2), it can be seen that when the scheme (A, D) is selected, the purchase of A model and D model computers are X and Y respectively, according to the title.

The solution is not in line with the topic after testing, and it is discarded;

When the scheme (A, E) is selected, the A model and E model computers are purchased as X and Y respectively, and they are obtained according to the theme.

So Hope Middle School purchased 7 A-type computers

Six kinds of 2 6=

x+y=36

x*6000+y*(d=5000)=100000 or x*6000+y*(e=2000)=100000

Reason d=5000 calculates y=116 so it is excluded. So the scheme is e=2000 and y=29 x=7

Everything is ok

Analysis: There are three situations: one is to buy a + b = 36, the unit price of a quantity + the unit price of b = 100500;The second is to buy a + c = 36, the unit price of a quantity + the unit price of c = 100500;The third is to buy B + C = 36, the unit price of B quantity + C unit price quantity = 100500 Answer:

Solution: If the computer company purchases X sets of type A computers, Y sets of type B computers, and sets Z of type C computers from the computer company, it can be considered in the following three situations:

1) Only purchase A type computer and B type computer, and the system of equations can be 6000x+4000y=100500x+y=36 according to the topic

The solution x = is not on topic, and should be discarded

2) Only purchase A type computer and C type computer, and the system of equations can be 6000x+2500z=100500x+z=36 according to the problem

The solution is x=3z=33

3) Only purchase B type computers and C type computers, and the system of equations can be 4000y+2500z=100500y+z=36 according to the problem

The solution yields y=7z=29

A: There are two options for the school to choose from, the first option is to purchase 3 A-type computers and 33 C-type computers;

The second option is to purchase 7 B computers and 29 C computers