1. (Chongqing, 20 10) There are two kinds of drinks, A and B, which contain the same fruits and vegetables, but with different concentrations. The first cup weighs 40 Jin and the second cup weighs 60 Jin. Now, a part of each beverage is poured out at the same weight, and then the poured part of each beverage is mixed with the rest of another beverage. If two mixed drinks contain the same fruits and vegetables, then

The typical concentration ratio problem is analyzed: solution concentration = solute mass/total solution mass. In this problem, the concentrations of two kinds of fruits and vegetables are unknown, but because the poured fruits and vegetables have the same quality, the total mass of the original A-type beverage is still the last 40 kg, and the total mass of the original B-type beverage is still the last 60 kg. The concentration of Class A beverage can be set as A, and the concentration of Class B beverage can be set as X kg, because the poured fruits and vegetables have the same quality.

Denominator,

Not wearing a seat belt:

Transferred items:

Merge:

So:

2. Cut a piece with the same weight from two alloys with different copper percentages, each weighing 10 kg and 15 kg respectively, and then combine each cut piece with the rest of the other piece. After smelting, the percentage of copper in the two pieces is just equal, so the weight of the cut piece is 6 kg.

Solution: let the block weight be X kg, and let the percentages of copper in the alloys of 10 kg and 15 kg be a and b,

=, finishing (b-a)x=6(b-a), x=6.

3. There are two kinds of alloys with different copper percentages, weighing 40 kg and 60 kg respectively. Cut a piece of the same weight from these two alloys, and then add another remaining alloy to each piece. After melting, the copper percentage of the two pieces is equal, so the weight of the alloy after cutting is () A.65438+D.24 kg B. 15 kg C.66

Test center: the application of one-dimensional linear equation.

Analysis: Let the copper content A be A, B be B, and the cutting weight X. According to two alloys with different copper percentages, A is 40 kg and B is 60 kg, and the copper percentages of the two alloys are equal after melting, the equation is solved.

Solution: let copper content a, b and cutting weight x.

The solution is x = 24. The cut alloy weighs 24 kg, so choose D.

A batch of goods is ready to be transported to a certain place, and there are three trucks A, B and C that can be hired. It is known that the freight volume of the three trucks A, B and C is constant every time, and the tonnage ratio of the goods transported by the two trucks A and B is1:3; If vehicles A and C jointly transport goods for the same number of times, A has transported 120 tons; If two cars B jointly transport the same number of goods, B transports 180 tons. Then the goods ***240 tons.

Solution: We assume that the gross tonnage of goods is X tons. A transports a ton at a time, B transports 3a ton at a time, and C transports b ton at a time.

, = ,

The solution is x = 240. So the answer is: 240.

5.(20 1 1 Chongqing) There are three kinds of bonsai, namely, pot A, pot B and pot C. Pot A consists of 15 red flowers, 24 yellow flowers and 25 purple flowers, and pot B consists of 10 red flowers and/kloc-.

Solution: There are three kinds of bonsai in pedestrian street: X, Y and Z.

From the meaning of the title, yes, from ①, 3x+2y+2z=580③, from ②, x+z= 150④,

Substituting ④ into ③, we can get x+2y=280, ∴2y=280-x⑤, Z = 150-X⑤ from ④. ∴4x+2y+3z = 4x+(280-x)+3( 150

A pool is equipped with a water inlet pipe and three identical water outlet pipes. Open the inlet pipe first, and then open the outlet pipe after some water is stored in the pool (the inlet pipe is not closed). If you open two outlet pipes at the same time, the pool will be empty after 8 minutes; If you open three outlet pipes at the same time, the pool will be empty in five minutes. Then the outlet pipe will open 40 minutes later than the inlet pipe.

Test site: the application of ternary linear equations.

Solution: let the outlet pipe open x minutes later than the inlet pipe, the inlet pipe speed is y, and the outlet pipe speed is z,

Yes,

Divide the two formulas to get:

Solution: x=40,

In other words, the outlet pipe opens 40 minutes later than the inlet pipe.

So the answer is: 40.

6.( 1) The original sales profit rate of a commodity is 47%. Now the purchase price has increased by 5%, but the selling price has not changed, so the sales profit rate of this commodity has become 0.40%.

(2) The current purchase price of a commodity is 20% cheaper, but the selling price has not changed, and its profit has increased by 30%, so the original profit rate is. 20%

7. A businessman deals in two commodities, A and B. The profit rate of each commodity is 40% and B is 60%. When the sales volume of commodity B is 50% more than that of commodity A, the businessman gets 50% of the total profit rate. When the sales volume of Class B goods is 50% less than that of Class A goods, the total profit rate obtained by this businessman is. 45%

Test center: the application of binary linear equation. Special topic: application questions; Equation thinking.

Solution: If the buying price of A is A yuan, the selling price is 1.4a yuan; The buying price of B is B yuan, so the selling price is 1.6b yuan; If x pieces are sold, 1.5x pieces are sold.

=0.5,

The solution is a= 1.5b,

∴ When the sales quantity of Class B goods is 50% less than that of Class A goods, when the quantity of Class A goods is Y, the quantity of Class B goods is 0.5y 。

The merchant's total profit rate = = = 45%.

So the answer is 45%.

8. A shopping mall sells a batch of TV sets. /kloc-the gross profit of each TV set in October is 20% of the selling price (gross profit = selling price-buying price). In February, the mall lowered the price of each TV set by 65,438+00% (the purchase price remained unchanged). Therefore, the sales volume of TV sets increased by 65,438+020% compared with 1 month, so the ratio of gross profit in February to gross profit in 1 month. 1 1: 10

Solution: If the selling price of 1 month is X and the sales volume is Y, then the buying price is X × (1-20%) = 80% X.

1 month gross profit is x× 20% x y = February selling price is x (1-10%) = 90% x.

The gross profit of each set is 90%x-80%x= 10%. The number of units sold in February is y× (1+120%) = 220% y.

So the gross profit in February is 10%x×220%y=22%xy.

The ratio of February gross profit to 1 monthly gross profit is 22%: 1 1: 10.

9.(20 1 1 No.1 Middle School March Test) A beverage produced by a company is made of two raw materials, A and B, in which the cost of raw material liquid A is 15 yuan /kg, and the cost of raw material liquid B is 10 yuan /kg. If it is sold at the current price, the profit rate will be high. The raw material liquid B increased by 65,438+00%, and the total cost after preparation increased by 65,438+02%. In order to expand the market, the company intends to invest 25% of the total cost in advertising. If you want to keep the profit per kilogram unchanged, the profit margin of this beverage is 50% at this time.

Analysis: After calculating the price increase according to the meaning of the question, A original price 18 yuan and B original price 10% become 1 1 yuan. It is concluded that the total cost has increased by 12%, that is, the cost per100kg before the price increase and the cost per100kg after the price increase. Then find the value of x to get the answer. Solution: Solution: Cost price of stock solution A 15 yuan /kg, and cost price of stock solution B 10 yuan /kg. After the price increase, the original price of A rose by 20% to 18 yuan; B It increased by 65,438+00% to become 65,438+065,438+0 yuan, and the total cost increased by 65,438+02%. If every 65,438+000 kilograms of finished products, two kinds of raw materials A account for X kilograms, and B accounts for (65,438+000-x) kilograms, then before the price increase, it is 65,438+00 kilograms.

18x+ 1 1( 100-x)=[ 15x+ 10( 100-x)]？ ( 1+ 12%),

The solution is: x = 100-x= 600/7 7kg, 100-x = 600/7kg, that is, their ratio is: A: B = 1: 6.

Then before the price increase, the cost per kilogram is 15/7+ 60/7= 75/7 yuan, the price is 127.57 yuan, and the profit is 7.5 yuan. After the price increase of raw materials, the cost per kilogram becomes 12 yuan, 25% of the cost is 3 yuan, and the guaranteed profit is 7.5 yuan, so the profit rate is 7.5 ".

10. "Energy saving and emission reduction, low-carbon economy" is the future development direction of China. A car manufacturer produces large, medium and small displacement cars. Under normal circumstances, small displacement cars account for 30% of the total output. In order to actively respond to the call of the state and meet the consumer demand of the public, it is planned to increase the production of small-displacement vehicles. Due to the adjustment of its production structure, the output of large and medium displacement vehicles is only 90% of that under normal circumstances. 48.3%

Analysis: To ask the percentage that the output of small-displacement vehicles should increase compared with normal conditions, we must first set an unknown X, and then read it to understand the meaning of the question. The equivalent relationship of this problem is that the total output of cars with three displacements remains unchanged after adjustment. For the convenience of doing the problem, we can set the total before adjustment to a.

Solution: Let's assume that the percentage of output of small-displacement vehicles is X, and the original total number of vehicles is A. 。

Then we can get the equation: 30% a (1+x)+70% a× 90% = (1+7.5%) a, and the solution is x ≈ 48.3%, so fill in 48.3.

1 1. Company A sells three products, A, B and C. In last year's sales, the sales amount of high-tech product C accounted for 40% of the total sales amount. Due to the impact of the international financial crisis, the sales amount of products A and B will be reduced by 20% this year, so high-tech product C is the focus of sales this year. If the total sales this year is the same as last year, then this year's.

Application of one-dimensional linear equation. Special topic: growth rate.

Solution: Suppose that the sales amount of high-tech product C this year will increase by X compared with last year. According to the meaning of the question, it is 0.4 (1+x)+(1-40%) (1-20%) =1,and the solution is x=30%.

So fill in 30.

1 1. (Chongqing Nankai Middle School 20 1 1, the second half of the ninth grade) Beaker A contains m liters of salt water with a concentration of a%, and beaker B contains m liters of salt water with a concentration of b% (A >: B). Now put the salt water in A/kloc- It is estimated that the brine in A will be reduced to m liters, so the ratio of the difference between the pure salt content in the beakers of A and B after mixing with the pure salt content before mixing in the beakers of A and B is _ _ _ _ _ _ .3/5.

According to the fact that beaker A contains M liters of physiological saline with a concentration of a% and beaker B contains M liters of physiological saline with a concentration of b% (A > B), the difference of pure salt content between the two beakers is obtained, and then the physiological saline in A is poured into B, and the answer is obtained after being mixed evenly.

Solution: ∫ Beaker A contains m liters of saline with concentration of a%, and Beaker B contains m liters of saline with concentration of b% (A > B).

∴ The difference of pure salt content between two beakers is: ma%-mb%=m(a%-b%),

∵ Pour the brine in A into B, mix it evenly, and then pour it back into A from B,

∴ After the brine is poured into B, the concentration of beaker B is: =,

Then return from b to a according to uniform mixing,

After returning to A, the salt content of A is: ma%+ × m= ma%+ b%,

The salt content of b is: m,

∴ The difference of pure salt content between A and B beakers after mutual doping is m(a%-b%).

∴ The ratio of the difference of pure salt content in beakers A and B after mutual mixing to the difference of pure salt content in beakers A and B before mutual mixing is:

So the answer is:

12. (Chongqing Bashu Middle School 20 1 1 the second half of grade) A tea beverage on the market is made up of tea stock solution and purified water in a certain proportion, of which 20 tons of purified water can be bought with the money for buying one ton of tea stock solution. Due to the continuous drought in Yunnan Province, the purchase price of tea stock solution has increased by 50%, and the price of pure water has also increased by 8%, resulting in a 20% increase in the cost of this tea beverage. What is the ratio of tea stock solution to pure water in this tea beverage?

Analysis: If the ratio of tea stock solution to purified water in this tea beverage is A: B, and the price of purchasing one ton of purified water is X, then the price of purchasing tea stock solution is 20x. According to the increase of 50% in the purchase price of tea stock solution and 8% in the price of purified water, the cost of preparing tea drinks has increased by 20%, and the proportion can be obtained by listing the equations. Solution: If the ratio of tea stock solution to pure water in this tea beverage is A: B, then the price of buying a ton of pure water is.

So the answer is: 2: 15.

13. Chongqing Changan Automobile Company distributes luxury cars, intermediate cars, intermediate cars and compact cars. In last year's sales, the sales of compact cars accounted for 60% of the total sales. Due to the impact of the international financial crisis, the sales of luxury cars, mid-size cars and mid-size cars will be reduced by 30% this year, so compact cars are the focus of sales this year. If the total sales this year is the same as last year.

Analysis: Suppose that the sales volume of four grades of cars last year was ***a yuan, of which the sales volume of compact cars was 60%a yuan, and the sales volume of luxury cars, middle-high cars and intermediate cars was ***( 1-60%)a yuan; Assuming that the percentage of the sales amount of compact cars this year will increase by X compared with last year, the sales amount of compact cars this year will be 60%( 1+x)a yuan, and the sales amount of luxury, medium-high and intermediate-grade cars will be * * (1-60%) (1-30%) A yuan. According to this year's total sales and

According to the meaning of the question: 60% (1+x) a+(1-60%) (1-30%) a = a,

Solution: x = 0.2 = 20%.

A: The sales of compact cars this year should increase by 20% compared with last year.

14. A fruit and vegetable beverage is made of fruit juice, vegetable juice and purified water according to a certain mass ratio. The price ratio of pure water, fruit juice and vegetable juice is 1: 2: 2. Due to market reasons, the price of fruit juice and vegetable juice increased by 15%, and the price of pure water decreased by 20%, but it did not affect the cost of beverages.

Analysis: Let the prices of pure water, fruit juice and vegetable juice be A, 2a and 2a, and let the mass ratio of pure water, fruit juice and vegetable juice be X: Y; Z, according to market reasons, the price of fruit juice and vegetable juice increases by 15%, while the price of pure water decreases by 20%, but it does not affect the cost of beverages (only considering the purchase cost), so the equation can be listed and solved.

Solutions; Let the prices of purified water, fruit juice and vegetable juice be A, 2a and 2a, and the mass ratio of purified water, fruit juice and vegetable juice be x: y: z, AX+2ay+2az = AX (1-80%)+2ay (1+15%).

0.2x=0.3(y+z)，(y+z): x = 2: 3。 So the answer is 2: 3.

15.(20 10 Bashu) Every time a bag of certain sucrose is sold in the supermarket, the profit can reach 20%. Due to the recent continuous drought in the southwest producing areas, the purchase price of this sucrose has increased by 25%. The supermarket will raise the price of this sucrose to ensure the profit per bag remains unchanged, and the profit rate after the price increase is 16%.

Analysis: According to the meaning of the question, y-x( 1+25%)=x? 20%, but to the value of y, there is a profit rate = (selling price-buying price)/buying price to get the answer.

Solution: Assuming that the original purchase price of each bag of sucrose is X, and it is Y after the purchase price rises, then the meaning of the problem is:

Profit rate = (1+25%) x+20% x (1+25%)/x (1+25%) =16%.

16. (Bashu 2010 —— 2011) If you buy a few pieces of goods in a shopping mall, you can sell 65% of all the goods at a higher price in 30 yuan, and then lower the price by 10%, so you can still make a profit of 65438 per piece.

Solution: Let the purchase price be X yuan, (1-10%) × (x+30) = x+18x = 90, and let the selling price of the remaining commodities be no less than Y yuan.

(90+30)×65%+(90+ 18)×25%+( 1-65%-25%)×y≥90×( 1+25%)y≥75

The price of surplus commodities shall not be lower than that of 75 yuan.

17. (Chongqing No.3 Middle School Grade 20 1 1 Grade 9 May exam) Xiaofeng rode at a constant speed on the ring road. Every 5 minutes, a bus runs backwards from the opposite side, and every 20 minutes, a bus runs forwards from the back. If the bus also runs at a constant speed,

Test site: the application of ternary linear equations.

Solution 1: Let the distance between adjacent cars be L, the vehicle speed be V 1, the bicycle be V2, and the interval time be T. 。

According to the meaning of the question, you must

From, we get V 1= V2, ④ Substitute ① and ④ into ②, and the solution is t = 8. So the answer is: 8.

Solution 2: Let the bicycle speed be X, the bus speed be Y, and the interval time be A. Every 5 minutes, a bus passes in the opposite direction. That is to say, when this bus meets it, the distance between the next bus and it is 5 (x+y), that is, 5y+5x=ay. Similarly, 20 (y-x) = ay. Both of the above formulas are acceptable. This problem is mainly to draw a sketch, that is, the time and speed axis, so that the relative position of the car is intuitive.

Solution 3: This is a catch-up problem: the departure interval of the father-in-law car is unchanged, and this problem can be solved by grasping this invariant. Assuming that the distance between two workshops is s, the single car speed of small wind is V 1, the bus speed is V2, and the bus spacing is t, then there is s = (V2-v 1) × 20 = (v1+v2) × 5, and the relationship between v1and v2 is obtained.

Walking along the street at a constant speed, Xiao Wang found that every six minutes, a 18 bus passed behind him, and every three minutes, a 18 bus came from the front. Assuming that each 18 bus runs at the same speed, the bus terminal of 18 bus departs at intervals of four minutes.

Solution: Suppose the speed of the car is A, the speed of the person is B, there is a bus every t, and the distance between the two cars is at.

Overtaking a car from behind is a catching-up problem, and so is the distance between people and cars: at then: at=6(a-b)①.

The coming car is a meeting problem, then: at=3(a+b)②.

① ②, we get: a=3b, so: at=4a t=4, that is, there is a bus every 4 minutes.

Xiao Wang rode his bike at a constant speed on the ring road. Every 6 minutes, a bus passes by the opposite side, and every 30 minutes, a bus passes by the back. If the bus also travels at a constant speed, how many minutes does the bus stop leave, regardless of the time when passengers get on and off?

Suppose the speed of the bus is A and the speed of Xiao Wang is B. If a bus leaves every N minutes, then

The distance between every two buses is a, a>b.

an/(a+b)=6……①

an/(a-b)=30……②

When you divide the two formulas, you get (a+b)/(a-b) = 5 ∴ A/b = 3/2...③ Bring ③ back to ① and get n= 10, so you drive every 10 minute.