Math problem On the eve of June 1st , a toy dealer used 2,350 yuan

Topic: On the eve of "June 1st", a toy dealer used 2,350 yuan.

Reward points: 0 - There are 9 days and 23 hours before the end of the problem 1, on the eve of "June 1st", a toy dealer used 2350 yuan to buy A, B, C three new types of electric toys a total of 50 sets, and the purchase of three kinds of toys are not less than 10 sets, set to purchase a kind of toy set, B kind of toy set, three kinds of electric toys The purchase price and selling price are shown in the table on the right, model. ab

c purchase price (yuan set).

The selling price (yuan set).

The algebraic expression of containing and expressing the number of sets of C toys purchased;

2) Suppose that all the three toys purchased can be sold, and in the process of purchasing and selling such toys, you need to spend an additional 200 yuan on various expenses. 1.Find the profit p (yuan) and (set) 2Find the maximum value of the profit and write out how many sets of each of the three toys there are at this time.

Answer: 1) 50-x-y

2）y=2x-30

3) p=10x=25y, 15(50-x-y)-200, bring y=2x-30 into this formula, and then do it according to the increase or decrease.

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1) The number of sets of C toys purchased: 50 x Y (or 47 x Y).

2) From the meaning of the question, 40x+55y+50(x-y)=2350 is sorted out, and y=2x-30y

3) Profit, sales revenue, purchase price, and other expenses.

y=2x-30y to get p=15x+250

The number of units of type C electric toys purchased is: 50-x-y=50-x-(2x-30=80-3x.)

According to the group of inequalities in the title, the range of x is , and x is an integer x and the maximum value is 23

In p=15x+250, k=15 0 p increases with the increase of x.

When X takes the maximum value of 23, P has a maximum value, and the maximum value is 595 yuan At this time, 23 sets, 16 sets, and 11 sets of toys were purchased A, B, and C respectively

Buy 24 copies, get 6 free.

Saved $36.

Let the number of sets of toys purchased in B be y, and the number of sets of toys purchased in c is Z;

From the meaning of the title: 30 x 10, 30 y 10, 30 z 10, and x+y+z=50, 40x+55y+50z=2350

So: *50- gets:y=2x-30;

55- gets: z=-3x+80

Profit w = 10x + 25y + 15z = 10x + 25 (2x-30) + 15 (-3x + 80) - 200 = 15x + 250

So when x is larger, the higher the profit w.

Since the values of x, y, and z must be greater than or equal to 10 pieces, y=2x-30 10, which is simplified to x 20;

z=-3x+80 10, simplified x 70 3 23;

i.e. 20 x 23

Therefore, when x=23, the maximum profit w is 595 yuan.

So y=16, z=11.

Analysis: (1) According to the purchase of A, B, C three new types of electric toys a total of 50 sets, the number of C toys can be expressed;

2) According to the purchase of three kinds of toys should be 2350, list the inequalities, can be expressed as the functional relationship between y and x;

3) Profit = total sales - total purchase price - expenses paid, list the function relationship; According to the purchase of three kinds of toys are not less than 10 sets, list the inequality group for solution Answer: Solution: (1) It is known that a total of 50 sets of three new types of electric toys A, B and C have been purchased, so the number of sets of C toys purchased is:

50-x-y；

2) 40x+55y+50(50-x-y)=2350 from the title, and y=2x-30;

3) Profit = sales revenue - purchase price - other expenses, therefore: p = (50-40) x + (80-55) y + (65-50) (50-x-y) - 200, and y = 2x-30, sorted out p = 15x + 250, the number of sets of purchased c electric toys is: 50-x-y = 50-x-(2x-30) = 80-3x, according to the title of the inequality group x 102x-30 1080-3x 10, the solution is 20 x 703, and the range of x is 20 x 703, and x is an integer, so the maximum value of x is 23, in p = 15x + 250, k = 15 0, p increases with the increase of x, when x takes the maximum value of 23, p has a maximum value, the maximum value is 595 yuan At this time, 23 sets, 16 sets, and 11 sets of toys are purchased A, B, and C respectively

Summary. A shopping mall 61 toys** all toys buy 5 get 2 free, a toy per piece, obviously bought back 14 pieces of yuan. Because buy five get two free, buy 10 and get 10 4 equals 14.

So I bought a total of 10 toys and gave four toys away. And then a toy . Just use 10 equals.

A shopping mall 61 toys** all toys buy 5 get 2 free, a kind of toy per mu is yuan, and it costs 14 yuan to buy back. Because buy five get two free, so buy 10 and get 10 4 is equal to blind and unbearable 14. So I bought a total of 10 toys and gave four sedan chair toys.

And then a toy . Just use 10 equals.

So I bought back 14 toys and spent it.

Set the purchase price of the first set to be $x.

The first entry is 2500 x

The number of the second entry is.

3750(x+10)/x=4500

3750x+37500=4500x

750x=37500

x = 50, that is, the first batch of each set of purchase price of 50 yuan.

The first batch is in quantity x, and the second batch is.

i.e. 4500-15x=3750

15x=750

x=502500 50=50 yuan.

Solution: (1) Let the purchase price of each set of the first batch of toys be x yuan, 2500 x, x=50, and the test of x=50 is the solution of the fractional equation, which is in line with the topic

Therefore, the purchase price of each set of the first batch of toys is 50 yuan;

2) Set the price of each set is at least Y yuan, 2500 50 (sets).

50Y+75Y-2500-4500 (2500+4500) 25%, Y 70, then the price of each set is at least 70 yuan

1 Set the first purchase x pcs, 2500 x + 10 = 4500

x=50 yuan.

2. Set each set to sell x yuan, from the first question can be known: the first purchase of 50, the second purchase of 75 (x-50) * 50 + (x-60) * 75 = (2500 + 4500) * x = 70 yuan.

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Solution: (1) Let the purchase price of each set of the first batch of toys be x yuan and the quantity is a ax = 2500 ......

Get. x ： = 5 ： 9

x = 50

A: The purchase price of the first batch of toys is 50 yuan per set.

2) (2500+4500) (1+25%) (50+75) = 7000 125 = 70 yuan.

A: The price of each set is at least 70 yuan.

Solution: Let the purchase price of each set of the first batch of toys be x yuan, 2500x, x = 50, and the test x = 50 is the solution of the fractional equation, which is in line with the topic

Therefore, the purchase price of each set of the first batch of toys is 50 yuan;

2) Set the price of each set is at least Y yuan, 2500 50 (sets).

50Y+75Y-2500-4500 (2500+4500) 25%, Y 70, then the price of each set is at least 70 yuan

5a（x+10）=

5（x+10）=5

9x=50A: Yes.

x;(50+75)

1:1 solution: The purchase price of each set of the first batch of toys is 50 yuan.

70 yuan. Answer: (1) The purchase price of each set of the first batch of toys is X yuan.

The quantity is a ax=

14 (5+2)=2 groups.

5 2 yuan.

Answer: Lele spent yuan.