节: 应用中的理性数字 | 2.4 应用中的理性数字

章节大纲

Rational Numbers in Applications
::应用中的理性数字

Let's use the skills we learned in the last concept to solve some real-world problems.
::让我们利用我们在最后一个概念中学到的技能解决一些现实世界的问题

Real-World Application: School Trip
::现实世界应用:校外旅行

Peter is hoping to travel on a school trip to Europe. The ticket costs $2400. Peter has several relatives who have pledged to help him with the ticket cost. His parents have told him that they will cover half the cost. His grandma Zenoviea will pay one sixth, and his grandparents in Florida will send him one fourth of the cost. What fraction of the cost can Peter count on his relatives to provide?
::彼得希望去欧洲的学校旅行。机票费用为2400美元。彼得有几个亲戚承诺帮助他支付机票费用。他的父母告诉他,他们将支付一半的费用。他奶奶泽诺维亚将支付六分之一的费用,而他在佛罗里达的祖父母将支付四分之一的费用。彼得可以指望他的亲戚提供哪部分费用呢?

The first thing we need to do is extract the relevant information. Peter’s parents will provide $\begin{aligned} \frac{1}{2} \end{aligned}$ the cost; his grandma Zenoviea will provide $\begin{aligned} \frac{1}{6} \end{aligned}$ ; and his grandparents in Florida $\begin{aligned} \frac{1}{4} \end{aligned}$ . We need to find the sum of those numbers, or $\begin{aligned} \frac{1}{2} + \frac{1}{6} + \frac{1}{4} \end{aligned}$ .
::我们需要做的第一件事就是提取相关信息。彼得的父母将提供12个成本;他奶奶泽诺维亚将提供16个;他的祖父母将在佛罗里达14。我们需要找到这些数字的总和,即12+16+14。

To determine the sum, we first need to find the LCD. The LCM of 2, 6 and 4 is 12, so that’s our LCD. Now we can find equivalent fractions:
::为了确定总和,我们首先需要找到液晶体,2、6和4的LCM是12,所以这就是我们的液晶体。现在我们可以找到相等的分数:

$\begin{aligned} \frac{1}{2} & = \frac{6 \cdot 1}{6 \cdot 2} = \frac{6}{12} \\ \frac{1}{6} & = \frac{2 \cdot 1}{2 \cdot 6} = \frac{2}{12} \\ \frac{1}{4} & = \frac{3 \cdot 1}{3 \cdot 4} = \frac{3}{12} \end{aligned}$

Putting them all together: $\begin{aligned} \frac{6}{12} + \frac{2}{12} + \frac{3}{12} = \frac{11}{12} \end{aligned}$ .
::612+212+312=1112。

Peter will get $\begin{aligned} \frac{11}{12} \end{aligned}$ the cost of the trip, or $2200 out of $2400, from his family.
::Peter会从他的家人那里拿到1112美元的旅费, 也就是2400美元中的2200美元。

Real-World Application: Property Management
::实际世界应用:财产管理

A property management firm is buying parcels of land in order to build a small community of condominiums. It has just bought three adjacent plots of land. The first is four-fifths of an acre, the second is five-twelfths of an acre, and the third is nineteen-twentieths of an acre. The firm knows that it must allow one-sixth of an acre for utilities and a small access road. How much of the remaining land is available for development?
::一家财产管理公司正在购买几块土地,以建造一小块公用公寓,它刚刚购买了三块相邻的土地,第一块是一英亩的五分之四,第二块是一英亩的五十二,第三块是一英亩的十九分二十。该公司知道它必须允许一英亩土地的六分之一用于公用事业和一条小的出入道路。剩下的土地有多少可以用于发展?

The first thing we need to do is extract the relevant information. The plots of land measure $\begin{aligned} \frac{4}{5}, \frac{5}{12}, \end{aligned}$ and $\begin{aligned} \frac{19}{20} \end{aligned}$ acres, and the firm can use all of that land except for $\begin{aligned} \frac{1}{6} \end{aligned}$ of an acre. The total amount of land the firm can use is therefore $\begin{aligned} \frac{4}{5} + \frac{5}{12} + \frac{19}{20} - \frac{1}{6} \end{aligned}$ acres.
::我们首先需要的是提取相关信息。45,512和1920英亩土地的地块,公司除了16英亩外,可以使用所有这些土地。因此,公司可以使用的土地总数是45+512+1920-16英亩。

We can add and subtract multiple fractions at once just by finding a common denominator for all of them. The factors of 5, 9, 20, and 6 are as follows:
::我们可以通过为所有这些分母找到一个共同的分母来同时增减多个分数。

$\begin{aligned} 5 5 \\ 12 2 \cdot 2 \cdot 3 \\ 20 2 \cdot 2 \cdot 5 \\ 6 2 \cdot 3 \end{aligned}$

We need a 5, two 2’s, and a 3 in our LCD. $\begin{aligned} 2 \cdot 2 \cdot 3 \cdot 5 = 60 \end{aligned}$ , so that’s our common denominator. Now to convert the fractions:
::我们需要5,2,2,2,3,2,2,2,2,2,2,2,2,3,5=60,这就是我们的共同分母。现在将分数转换为:

$\begin{aligned} \frac{4}{5} & = \frac{12 \cdot 4}{12 \cdot 5} = \frac{48}{60} \\ \frac{5}{12} & = \frac{5 \cdot 5}{5 \cdot 12} = \frac{25}{60} \\ \frac{19}{20} & = \frac{3 \cdot 19}{3 \cdot 20} = \frac{57}{60} \\ \frac{1}{6} & = \frac{10 \cdot 1}{10 \cdot 6} = \frac{10}{60} \end{aligned}$

We can rewrite our sum as $\begin{aligned} \frac{48}{60} + \frac{25}{60} + \frac{57}{60} - \frac{10}{60} = \frac{48 + 25 + 57 - 10}{60} = \frac{120}{60} \end{aligned}$ .
::我们可以重写我们的数字为4860+2560+5760-1060=48+25+57-1060=12060。

Next, we need to reduce this fraction. We can see immediately that the numerator is twice the denominator, so this fraction reduces to $\begin{aligned} \frac{2}{1} \end{aligned}$ or simply 2. One is sometimes called the invisible denominator , because every whole number can be thought of as a rational number whose denominator is one.
::接下来,我们需要减少这个分数。我们可以看到分子是分母的两倍, 所以这个分数会减少到21或者简单2。一个有时被称为无形分母, 因为每个整数都可以被看作一个理性数字, 其分母是一个。

The property firm has two acres available for development.
::该地产公司有两英亩土地可供开发。

Evaluate Change Using a Variable Expression
::使用变量表达式评估变化

When we write algebraic expressions to represent a real quantity, the difference between two values is the change in that quantity.
::当我们写代数表达式以表示实际数量时,两个值之间的差异是该数量的变化。

The intensity of light hitting a detector when it is held a certain distance from a bulb is given by this equation :
::这个方程式给出了在与灯泡保持一定距离时击中探测器的光强度:

$\begin{aligned} I n t e n s i t y = \frac{3}{d^{2}} \end{aligned}$

::密度=3d2

where $\begin{aligned} d \end{aligned}$ is the distance measured in meters , and intensity is measured in lumens . Calculate the change in intensity when the detector is moved from two meters to three meters away.
::计算探测器从两米移动到三米之外时强度的变化。

We first find the values of the intensity at distances of two and three meters.
::我们首先在两米和三米的距离找到强度的值。

$\begin{aligned} Intensity (2) & = \frac{3}{(2)^{2}} = \frac{3}{4} \\ Intensity (3) & = \frac{3}{(3)^{2}} = \frac{3}{9} = \frac{1}{3} \end{aligned}$

::密度(2)=3(2)2=34 密度(3)=3(3)2=39=13

The difference in the two values will give the change in the intensity. We move from two meters to three meters away.
::这两个值的差异将改变强度。我们从两米移到三米。

$\begin{aligned} Change = Intensity (3) - Intensity (2) = \frac{1}{3} - \frac{3}{4} \end{aligned}$
::变化=密度(3)-密度(2)=13-34

To find the answer, we will need to write these fractions over a common denominator.
::为了找到答案,我们需要用一个共同的分母来写这些分母。

The LCM of 3 and 4 is 12, so we need to rewrite each fraction with a denominator of 12:
::3 和 4 的 LCM 是 12, 所以我们需要重写每个分数, 分母是 12 :

$\begin{aligned} \frac{1}{3} & = \frac{4 \cdot 1}{4 \cdot 3} = \frac{4}{12} \\ \frac{3}{4} & = \frac{3 \cdot 3}{3 \cdot 4} = \frac{9}{12} \end{aligned}$

So we can rewrite our equation as $\begin{aligned} \frac{4}{12} - \frac{9}{12} = - \frac{5}{12} \end{aligned}$ . The negative value means that the intensity decreases as we move from 2 to 3 meters away.
::因此我们可以重写我们的方程式为 412 - 912\\\\\512。负值意味着当我们从2米到3米之间移动时强度会下降。

When moving the detector from two meters to three meters, the intensity falls by $\begin{aligned} \frac{5}{12} \end{aligned}$ lumens.
::当探测器从2米移动到3米时,强度下降512磅。

Guided Practice
::实践指南实践指南实践指南实践指南实践指南实践指南实践指南

Example 1
::例1

Elsa baked a small cake for her family. First her sister ate one quarter and her mom ate one third. How much was left for Elsa?
::爱尔莎为家人烤了一个小蛋糕首先她姐姐吃了四分之一,她妈妈吃了三分之一

The whole cake is represented by 1. To solve this problem, we subtract the fraction that each person ate.
::1. 为了解决这个问题,我们减去每个人所吃的分数。

$\begin{aligned} 1 - \frac{1}{4} - \frac{1}{3} \end{aligned}$ .

To complete this problem, we must give the terms common denominators. Since the denominators do not share any factors, we simply multiply them together: $\begin{aligned} 4 \cdot 3 = 12 \end{aligned}$ .
::为了解决这个问题,我们必须给出共同的分母。由于分母并不存在任何共同因素,我们只需将它们相乘:4+3=12。

$\begin{aligned} 1 - \frac{1}{4} - \frac{1}{3} Start with the original expression. \\ = & 1 \cdot \frac{12}{12} - \frac{1}{4} \cdot \frac{3}{3} - \frac{1}{3} \cdot \frac{4}{4} Give each term a common denominator. \\ = & \frac{12}{12} - \frac{3}{12} - \frac{4}{12} Simplify. \\ = & \frac{12 - 3 - 4}{12} = \frac{5}{12} \end{aligned}$

::1 - 14 - 13 以原表达式开始。=11212 - 1433 - 1344 给每个术语一个共同分母。=1212 - 312 - 412 - 412 简化。=12-3 - 412=512

There is $\begin{aligned} \frac{5}{12} \end{aligned}$ of the original cake left for Elsa.
::原蛋糕还有512块留给Elsa

Review
::回顾

Which property of addition does each situation involve?
::每种情况涉及哪些附加财产?

Whichever order your groceries are scanned at the store, the total will be the same.
::不管你的杂货在店里被扫描到哪家, 总数都是一样的。

However many shovel-loads it takes to move 1 ton of gravel, the number of rocks moved is the same.
::无论需要多少铲子才能移动1吨砾石,但移动的岩石数量相同。

If Julia has no money, then Mark and Julia together have just as much money as Mark by himself has.
::如果朱莉娅没有钱那么马克和朱莉娅在一起和马克一样有钱

In 4-7, practice your addition and subtraction skills.
::4 -7,练习你的附加和减法技能。

$\begin{aligned} \frac{7}{15} + \frac{2}{9} \end{aligned}$

$\begin{aligned} \frac{5}{19} + \frac{2}{27} \end{aligned}$

$\begin{aligned} \frac{5}{12} - \frac{9}{18} \end{aligned}$

$\begin{aligned} \frac{2}{3} - \frac{1}{4} \end{aligned}$

Ilana buys two identically sized cakes for a party. She cuts the chocolate cake into 24 pieces and the vanilla cake into 20 pieces, and lets the guests serve themselves. Martin takes three pieces of chocolate cake and one of vanilla, and Sheena takes one piece of chocolate and two of vanilla. Which of them gets more cake?
::伊莲娜为派对买了两个大小相同的蛋糕。她把巧克力蛋糕切成24块,香草蛋糕切成20块,让客人自食其力。马丁拿了三块巧克力蛋糕和一个香草蛋糕,希娜拿了一块巧克力和两块香草。他们谁拿了更多的蛋糕?

Nadia, Peter and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They figure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the total cost was tax?
::Nadia, Peter 和 Ian 正在集资购买一加仑的冰淇淋。 Nadia 是最老的, 并且得到了最大的津贴。她贡献了一半的费用。 Ian 是下一个最老的, 贡献了三分之一的费用。 Peter是最小的, 得到了最小的津贴, 贡献了四分之一的费用。他们认为这笔钱足够了。当他们去结账时, 他们意识到他们忘记了销售税, 担心没有足够的钱。令人惊讶的是, 他们有准确的金额。总成本的哪一部分是税?

The time taken to commute from San Diego to Los Angeles is given by the equation $\begin{aligned} t i m e = \frac{120}{s p e e d} \end{aligned}$ where time is measured in hours and speed is measured in miles per hour (mph). Calculate the change in time that a rush hour commuter would see when switching from traveling by bus to traveling by train, if the bus averages 40 mph and the train averages 90 mph.
::从圣地亚哥到洛杉矶的通勤时间用方程式时间=120速度给出,其中时间以小时计,速度以每小时英里(mph)计。如果公共汽车平均为40mph,火车平均为90mph,计算时速通勤者从公共汽车转向火车时看到的时间变化。

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。

章节大纲

Rational Numbers in Applications ::应用中的理性数字

Real-World Application: School Trip ::现实世界应用:校外旅行

Real-World Application: Property Management ::实际世界应用:财产管理

Evaluate Change Using a Variable Expression ::使用变量表达式评估变化

Guided Practice ::实践指南 实践指南 实践指南 实践指南 实践指南 实践指南 实践指南

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Rational Numbers in Applications
::应用中的理性数字

Real-World Application: School Trip
::现实世界应用:校外旅行

Real-World Application: Property Management
::实际世界应用:财产管理

Evaluate Change Using a Variable Expression
::使用变量表达式评估变化

Guided Practice
::实践指南实践指南实践指南实践指南实践指南实践指南实践指南

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)