Course: 7.3 混合问题 | The Way To Learn

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混合问题

Mixture Problems
::混合问题

Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in a solution, nuts and raisins, or even the change in your pocket! Let’s look at some examples of these.
::平方程式系统经常出现问题, 这些问题涉及两种化学物的混合物—— 溶液中的坚果和葡萄, 甚至你口袋里的变化! 让我们看看这些例子。

Mixture Problem: Coins
::混合问题: 硬币

Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 55 cents, how many of each coin does she have?
::珍妮(Janine)将她的钱包空空起来,发现它只包含镍(每块5美分)和硬币(每块10美分)。如果她共有7个硬币,其总价值为55美分,那么她每块硬币有多少?

Since we have 2 types of coins, let’s call the number of nickels $\begin{aligned} x \end{aligned}$ and the number of dimes $\begin{aligned} y \end{aligned}$ . We are given two key pieces of information to make our equations: the number of coins and their value.
::既然我们有两种硬币,那么让我们来计算镍x和硬币y的数量。我们得到两个关键的信息来决定我们的方程:硬币的数量及其价值。

\begin{aligned} # of coins equation: & x + y = 7 & (n u m b e r o f n i c k e l s) + (n u m b e r o f d i m e s) \\ value equation: & 5 x + 10 y = 55 & (s i n c e n i c k e l s a r e w o r t h 5 c a n d d i m e s 10 c) \end{aligned}

::硬币方程式 #:x+y=7(镍数)+(硬币数)等价:5x+10y=55(因为镍值5c和10c角)

We can quickly rearrange the first equation to isolate $\begin{aligned} x \end{aligned}$ :
::我们可以快速重新排列第一个方程式以分离 x:

\begin{aligned} x = 7 - y & n o w s u b s t i t u t e i n t o e q u a t i o n 2 : \\ 5 (7 - y) + 10 y = 55 & d i s t r i b u t e t h e 5 : \\ 35 - 5 y + 10 y = 55 & c o l l e c t l i k e t e r m s : \\ 35 + 5 y = 55 & s u b t r a c t 35 f r o m b o t h s i d e s : \\ 5 y = 20 & d i v i d e b y 5 : \\ \underline{y = 4} & s u b s t i t u t e b a c k i n t o e q u a t i o n 1 : \\ x + 4 = 7 & s u b t r a c t 4 f r o m b o t h s i d e s : \\ \underline{x = 3} \end{aligned}

::x=7-ynow 替代方程 2:5(7-y)+10y=55 分配 5:35-5y+10y=55 集合类似条件 :35+5y=55 分数 35 双侧:5y=20divide 5:y=4=4 替代方程 1:x+4=7 4 双侧:x=3_

Janine has 3 nickels and 4 dimes.
::Janine有三分四分钱

Sometimes a question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below.
::有时一个问题会要求您确定特定物质的使用量( 由浓度决定) 。所涉物质可以是像上面的硬币一样的东西, 或者可能是一种溶解的化学物质, 甚至是热。在这种情况下, 您需要知道每一部分中物质的数量。在几种常见情况下, 要获得一个方程, 您只需要添加两个给定的数量, 但要获得一个产品需要的第二个方程。下面有三个例子。

Type of mixture	First equation	Second equation
Coins (items with $ value)	total number of items $\begin{aligned} (n_{1} + n_{2}) \end{aligned}$	total *value* (item value $\begin{aligned} \times \end{aligned}$ no. of items)
Chemical solutions	total solution volume $\begin{aligned} (V_{1} + V_{2}) \end{aligned}$	amount of *solute* (vol $\begin{aligned} \times \end{aligned}$ concentration)
Density of two substances	total amount or volume of mix	total *mass* (volume $\begin{aligned} \times \end{aligned}$ density)

For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the final mixture. (A solute is the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine.) To find the total amount, simply multiply the amount of the mixture by the fractional concentration . To illustrate, let’s look at an example where you are given amounts relative to the whole.
::例如,在考虑混合化学溶液时,我们最可能需要考虑溶液在个别部分和最后混合物中的总量。 (溶液是指溶解在溶液中的化学物。溶液的例子之一是在水中添加盐以制成盐。 )要找到总量,只需将混合物的含量乘以分数浓度即可。举例来说,让我们看看一个给出与整体相对数量的例子。

Mixture Problem: Chemistry
::混合问题:化学

A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use?
::化学家需要准备浓度为15%的500毫升铜硫酸盐溶液。她希望使用高浓度溶液(60%),并用低浓度溶液(5%)稀释它,以便做到这一点。她应该使用每种溶液中的多少?

To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution $\begin{aligned} (x) \end{aligned}$ and the amount of dilute solution "> $\begin{aligned} (y) \end{aligned}$ . We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml) and the final amount of solute (15% of $\begin{aligned} 500 m l = 75 m l \end{aligned}$ ). Our equations will look like this:
::要设置这个问题, 我们首先需要定义变量。我们的未知数是集中溶液( x) 和稀释溶液的数量。我们还将把百分比( 60%、 15% 和 5%) 转换成小数( 0. 6、 0. 15 和 0.05) 。关键信息有两部分是最后卷( 500 ml) 和溶液的最终数量( 500 ml= 75 ml 的15%)。我们的方程式将像这样:

Volume equation: $\begin{aligned} x + y = 500 \end{aligned}$
::音量方程: x+y=500

Solute equation: $\begin{aligned} 0.6 x + 0.05 y = 75 \end{aligned}$
::绝对方程: 0.6x+0.05y=75

To isolate a variable for substitution, we can see it’s easier to start with equation 1:
::为了分离替代的变数,我们可以看到从1等式1开始比较容易:

\begin{aligned} x + y = 500 & s u b t r a c t y f r o m b o t h s i d e s : \\ x = 500 - y & n o w s u b s t i t u t e i n t o e q u a t i o n 2 : \\ 0.6 (500 - y) + 0.05 y = 75 & d i s t r i b u t e t h e 0.6 : \\ 300 - 0.6 y + 0.05 y = 75 & c o l l e c t l i k e t e r m s : \\ 300 - 0.55 y = 75 & s u b t r a c t 300 f r o m b o t h s i d e s : \\ - 0.55 y = - 225 & d i v i d e b o t h s i d e s b y - 0.55 : \\ \underline{y = 409 m l} & s u b s t i t u t e b a c k i n t o e q u a t i o n f o r x : \\ x = 500 - 409 = \underline{91 m l} \end{aligned}

::x+y = 500 subtraction y 从两侧 : x= 500 -y = ynow 替代方程 2:0.6( 500-y)+0.05y= 75 分布0. 0. 6: 300-0.6y+0.05y= 75 类似条件 : 300- 055y= 75 subtraction 300 从两侧 : - 055y\ @ @ 225divide 双两侧乘以- 055:y= 409 ml_ 替代方程返回 x: x= 500- 409= 91 ml_ 等式

So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.
::所以化学家应该把60%溶液的91毫升和5%溶液的409毫升混合起来。

Mixture Problem: Coffee
::混合问题:咖啡

A coffee company makes a product which is a mixture of two coffees, using a coffee that costs $10.20 per pound and another coffee that costs $6.80 per pound. In order to make 20 pounds of a mixture that costs $8.50 per pound, how much of each type of coffee should it use?
::咖啡公司生产的产品是两种咖啡的混合物,一种咖啡每磅10.20美元,另一种咖啡每磅6.80美元。为了制造每磅8.50美元的20磅混合物,每种咖啡应该使用多少?

Let $\begin{aligned} m \end{aligned}$ be the amount of the $10.20 coffee, and let $\begin{aligned} n \end{aligned}$ be the amount needed of the $6.80 coffee. Since we want 20 pounds of coffee that costs $8.50 per pound, the total cost for all 20 pounds is $\begin{aligned} 20 \cdot $ 8.50 = $ 170 \end{aligned}$ . The cost for the 20 pounds of mixture is equal to the cost of each type of coffee added together: $\begin{aligned} 10.20 \cdot m + 6.8 \cdot n = 170 \end{aligned}$ .
::由于我们需要每磅8.50美元的20磅咖啡,所有20磅的合计成本为20美元8.50美元=170美元,20磅混合咖啡的成本等于每类咖啡加起来的费用:1020立方米+6.8当=170美元。

Also, the amount of each type of coffee added together equals 20 pounds: $\begin{aligned} m + n = 20 \end{aligned}$ .
::另外,每种咖啡加起来的量等于20磅:m+n=20。

The system is: $\begin{aligned} {\begin{cases} m + n = 20 \\ 10.20 \cdot m + 6.8 \cdot n = 170 \end{cases} \end{aligned}$ .
::系统是 {m+n=2010.20_m+6.8_n=170。

We can isolate one variable and use substitution to solve the system:
::我们可以分离一个变量,使用替代来解决系统问题:

$\begin{aligned} m = 20 - n \end{aligned}$
::m=20-n

Now solve for $\begin{aligned} n \end{aligned}$ .
::N. 现在解决 N.

\begin{aligned} 10.20 (20 - n) + 6.8 n & = 170 \\ 204 - 10.20 n + 6.8 n & = 0170 Distributive Property \\ 204 - 3.4 n & = 170 Add like terms. \\ - 3.4 n & = - 34 Subtract 204. \\ n & = 10 Divide by - 3.4. \end{aligned}

::10.20(20-n)+6.8n=170204-10.20n+6.8n=0170分配性财产204-3.4n=170Add like terms.-3.4n\*34 减号204.n=10除以-3.4。

Since $\begin{aligned} n = 10 \end{aligned}$ , we can plug that into $\begin{aligned} m + n = 20 \end{aligned}$ .
::从n=10开始,我们可以将它插入 m+n=20。

$\begin{aligned} m + 10 = 20 \Rightarrow m = 10 \end{aligned}$ .
::m+10=20m=10。

The coffee company needs to use 10 pounds of each type of coffee in order to have a 20 pound mixture that costs $8.50 per pound.
::咖啡公司需要每类咖啡使用10磅,才能混合20磅,每磅8.50美元。

Example
::示例示例示例示例

Example 1
::例1

A light green latex paint that is 20% yellow paint is combined with a darker green latex paint that is 45% yellow paint. How many gallons of each paint must be used to create 15 gallons of a green paint that is 25% yellow paint?
::黄色涂料是20%的浅绿色胶面涂料,与黄色涂料是45%的深绿色胶面涂料相结合。每涂料中必须用多少加仑来制造15加仑的黄色涂料,即25%的黄色涂料?

Let $\begin{aligned} x \end{aligned}$ be the number of gallons of the 20% yellow paint and let $\begin{aligned} y \end{aligned}$ be the number of gallons of the 40% yellow paint. This means that we want those two numbers to add up to 15: $\begin{aligned} x + y = 15 \end{aligned}$
::x 应该是20%黄色涂料的加仑数, y 应该是40%黄色涂料的加加仑数。这意味着我们希望这两个数字加起来最多15: x+y=15

Now if we want 15 gallons of 25% yellow paint, that means we want $\begin{aligned} 0.25 \cdot 15 = 3.75 \end{aligned}$ gallons of pure yellow pigment. The expression $\begin{aligned} 0.20 \cdot x \end{aligned}$ represents the amount of pure yellow pigment in the $\begin{aligned} x \end{aligned}$ gallons of 20% yellow paint. The expression $\begin{aligned} 0.45 \cdot y \end{aligned}$ represents the amount of pure yellow pigment in the $\begin{aligned} y \end{aligned}$ gallons of 45% yellow paint. Combing the last two adds up to the 3.75 gallons of pure pigment in the final mixture:
::现在,如果我们想要15加仑的黄色涂料25%,这意味着我们需要0.251515=3.75加仑的纯黄色色素。 0. 20xx表示x加仑的纯黄色涂料含量为20%。 0. 45+y表示的纯黄色涂料含量为45%。将最后两加起来, 在最后混合物中加起来的纯色含量为3. 75加仑:

$\begin{aligned} 0.20 x + 0.40 y = 3.75 \end{aligned}$
::0.20x+0.40y=3.75

The system is: $\begin{aligned} {\begin{cases} x + y = 15 \\ 0.20 x + 0.45 y = 3.75 \end{cases} \end{aligned}$ .
::系统是 { x+y=150.20x+0.45y=3.75。

We can isolate one variable and use substitution to solve the system:
::我们可以分离一个变量,使用替代来解决系统问题:

$\begin{aligned} x = 15 - y \end{aligned}$
::x=15-y

Now solve for $\begin{aligned} y \end{aligned}$ .
::现在为y解决问题。

\begin{aligned} 0.20 (15 - y) + 0.45 y & = 3.75 \\ 3 - 0.20 y + 0.45 y & = 3.75 Distributive Property \\ 3 + 0.2 y & = 3.75 Add like terms. \\ 0.25 y & = 0.75 Subtract 3. \\ y & = 3 Divide by 0.25. \end{aligned}

::0.2045y+0.45y=3.7553-0.20y+0.45y=3.75分配性财产3+0.2y=3.75Add like tems.0.25y=0.75 减号3.y=3除以0.25。

Now we can plug in $\begin{aligned} y = 3 \end{aligned}$ into $\begin{aligned} x + y = 15 \end{aligned}$ :
::现在我们可以插入 y= 3 到 x+y=15 :

$\begin{aligned} x + y = 15 \Rightarrow x + 3 = 15 \Rightarrow x = 12 \end{aligned}$ .
::xy=15*x+3=15*x=12。

This means 12 gallons of 20% yellow paint should be mixed with 3 gallons of 45% yellow paint in order to get 15 gallons of 25% yellow paint.
::这意味着12加仑的黄涂料(20%)应该与3加仑的黄涂料(45%)混在一起,以便获得15加仑的黄涂料(25%)。

Review
::回顾

I have $15 and wish to buy five pounds of mixed nuts for a party. Peanuts cost $2.20 per pound. Cashews cost $4.70 per pound.
1. How many pounds of each should I buy?
  ::我每人要买多少磅?
2. If I suddenly realize I need to set aside $5 to buy chips, can I still buy 5 pounds of nuts with the remaining $10?
  ::如果我突然意识到我需要花5美元买薯片我还能用剩下的10美元买5磅坚果吗?
3. What’s the greatest amount of nuts I can buy?
  ::我能买到的坚果数量最多吗?
::我有15美元,想买五磅混合坚果来参加派对。花生每磅2.20美元。腰果每磅4.70美元。我应该买多少磅?如果我突然意识到我需要5美元来买薯片,我还能买5磅坚果和剩下的10美元吗?我能买多少?

A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and 35%.
1. How many liters of each should be mixed to give the acid needed for the experiment?
  ::每一升应混合多少升,以提供实验所需的酸?
2. How many liters should be mixed to give two liters at a 15% concentration?
  ::有多少升应该混合在一起以15%的浓度给两升?
::化学实验需要一升浓度为15%的硫酸,但供应室只储存浓度为10%和35%的硫酸。每种硫酸的含量应该混合多少升才能提供实验所需的酸?应该混合多少升才能给出浓度为15%的两升?

Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc. The jeweler told her that the volume of silver in the bracelet was 10 cc and the volume of gold was 20 cc. Find the combined density of her bracelet.
::Bachelle想知道她的手镯的密度,这是金和银的混合体;密度是总质量除以总量;金的密度为19.3克/cc,银的密度为10.5克/cc。珠宝商告诉她,手镯中的银量为10cc,金的含量为20cc。

Tickets to a show cost $10 in advance and $15 at the door. If 120 tickets are sold for a total of $1390, how many of the tickets were bought in advance?
::演出票价预付10美元,门前15美元,如果售出120张罚单共计1390美元,预购票数是多少?

A light purple latex paint that is 40% blue paint is combined with a blue latex paint that is 100% blue paint. How many gallons of each paint must be used to create 15 gallons of a dark purple paint that is 60% blue paint?
::浅紫色乳胶涂料为40%蓝色涂料,与蓝色涂料为100%蓝色涂料结合。每涂料中必须用多少加仑来制造15加仑深紫色涂料,即60%蓝色涂料?

In 6-10, the multiple-choice questions on a test are worth 2 points each, and the short-answer questions are worth 5 points each.
::在6-10年中,测试的多选题各值两点,短题各值五点。

If the whole test is worth 100 points and has 35 questions, how many of the questions are multiple-choice and how many are short-answer?
::如果整个测试值100点,并有35个问题,其中有多少问题是多重选择,有多少是简短回答?

If Kwan gets 31 questions right and ends up with a score of 86 on the test, how many questions of each type did she get right? (Assume there is no partial credit.)
::如果关问对了31个问题,最后在测试中得分86分,那么她每类回答的问题有多少? (假设没有部分信用。 )

If Ashok gets 5 questions wrong and ends up with a score of 87 on the test, how many questions of each type did he get wrong? (Careful!)
::如果Ashok发现5个问题是错误的,最后在测试中得分为87,那么他犯了多少个问题?

What are two ways you could have set up the equations for part c?
::你用什么两种方法来设置C部分的方程?

How could you have set up part b differently?
::你怎么能用不同的方式建立B部分呢?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

混合问题

Mixture Problems ::混合问题

Mixture Problem: Coins ::混合问题: 硬币

Mixture Problem: Chemistry ::混合问题:化学

Mixture Problem: Coffee ::混合问题:咖啡

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)