课程： 2.7 应用中的混合数字

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选择节应用中的混合数字

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应用中的混合数字
Mixed Numbers in Applications
::应用中的混合数字

Real-World Application: Chemistry
::现实世界应用:化学

In the chemistry lab there is a bottle with two liters of a 15% solution of hydrogen peroxide $\begin{aligned} (H_{2} O_{2}) \end{aligned}$ . John removes one-fifth of what is in the bottle, and puts it in a beaker. He measures the amount of $\begin{aligned} H_{2} O_{2} \end{aligned}$ and adds twice that amount of water to the beaker. Calculate the following measurements.
::在化学实验室里,有一瓶装有两升氢过氧化氢(H2O2)15%溶液的瓶子,约翰将瓶子里的五分之一脱去,放入烧杯中。他测量了H2O2的含量,并将水量增加了两倍。计算以下测量量。

a) The amount of $\begin{aligned} H_{2} O_{2} \end{aligned}$ left in the bottle.
:a) 瓶中残留的H2O2数量。

To determine the amount of $\begin{aligned} H_{2} O_{2} \end{aligned}$ left in the bottle, we first determine the amount that was removed. That amount was $\begin{aligned} \frac{1}{5} \end{aligned}$ of the amount in the bottle (2 liters). $\begin{aligned} \frac{1}{5} \end{aligned}$ of 2 is $\begin{aligned} \frac{2}{5} \end{aligned}$ .
::为了确定瓶中残留的H2O2数量,我们首先确定被去除的数量。这一数量是瓶中数量(2升)的15分。 2中的15分是25分。

The amount remaining is $\begin{aligned} 2 - \frac{2}{5} \end{aligned}$ , or $\begin{aligned} \frac{10}{5} - \frac{2}{5} = \frac{8}{5} \end{aligned}$ liter (or 1.6 liters).
::其余为2-25,或105-25=85升(或1.6升)。

There are 1.6 liters left in the bottle.
::瓶子里还剩1.6升

b) The amount of diluted $\begin{aligned} H_{2} O_{2} \end{aligned}$ in the beaker.
:b) 油箱中稀释的H2O2含量。

We determined that the amount of the 15% $\begin{aligned} H_{2} O_{2} \end{aligned}$ solution removed was $\begin{aligned} \frac{2}{5} \end{aligned}$ liter. The amount of water added was twice this amount, or $\begin{aligned} \frac{4}{5} \end{aligned}$ liter. So the total amount of solution in the beaker is now $\begin{aligned} \frac{2}{5} + \frac{4}{5} = \frac{6}{5} \end{aligned}$ liter, or 1.2 liters.
::我们确定,去除的15%的H2O2溶液是25升。增加的水量是这一数量的两倍,或45升。因此,烧杯中的溶液总量现在是25+45=65升,或1.2升。

There are 1.2 liters of diluted $\begin{aligned} H_{2} O_{2} \end{aligned}$ in the beaker.
::烧杯里有1.2升稀释的H2O2。

c) The concentration of the $\begin{aligned} H_{2} O_{2} \end{aligned}$ in the beaker.
:c) 烧烤箱中H2O2的浓度。

The new concentration of $\begin{aligned} H_{2} O_{2} \end{aligned}$ can be calculated.
::可以计算出新的H2O2浓度。

John started with $\begin{aligned} \frac{2}{5} \end{aligned}$ liter of 15% $\begin{aligned} H_{2} O_{2} \end{aligned}$ solution, so the amount of pure $\begin{aligned} H_{2} O_{2} \end{aligned}$ is 15% of $\begin{aligned} \frac{2}{5} \end{aligned}$ liters, or $\begin{aligned} 0.15 \times 0.40 = 0.06 \end{aligned}$ liters.
::约翰开始使用 25 升的15%的H2O2溶液, 所以纯H2O2的含量是 25 升的15%, 即0.15x0. 40=0.06 升。

After he adds the water, there is 1.2 liters of solution in the beaker, so the concentration of $\begin{aligned} H_{2} O_{2} \end{aligned}$ is $\begin{aligned} \frac{0.06}{1.2} = \frac{1}{20} \end{aligned}$ or 0.05. To convert to a percent we multiply this number by 100, so the beaker’s contents are 5% $\begin{aligned} H_{2} O_{2} \end{aligned}$ .
::在他添加水后,烧烤箱中有1.2升溶液,所以H2O2的浓度是0.061.2=120或0.05。要转换成一个百分比,我们乘以100,因此烧烤箱的含量是5% H2O2。

Real-World Application: Splitting Candy
::真实世界应用程序: 分割糖果

Anne has a bar of chocolate and she offers Bill a piece. Bill quickly breaks off $\begin{aligned} \frac{1}{4} \end{aligned}$ of the bar and eats it. Another friend, Cindy, takes $\begin{aligned} \frac{1}{3} \end{aligned}$ of what was left. Anne splits the remaining candy bar into two equal pieces which she shares with a third friend, Dora. How much of the candy bar does each person get?
::另一个朋友辛迪拿了剩下的13块。安妮把剩下的糖果棒分成两块,她和第三个朋友朵拉分享。每人有多少糖果棒?

First, let’s look at this problem visually.
::首先,让我们看一看这个问题。

Anne starts with a full candy bar.
::安妮从一个完整的糖果棒开始

Bill breaks off $\begin{aligned} \frac{1}{4} \end{aligned}$ of the bar.
::比尔打破了14个酒吧。

Cindy takes $\begin{aligned} \frac{1}{3} \end{aligned}$ of what was left.
::Cindy拿走了剩下的13块

Dora gets half of the remaining candy bar.
::多拉得到一半剩下的糖果棒

We can see that the candy bar ends up being split four ways, with each person getting an equal amount.
::我们可以看到糖果棒最终被分成四条路, 每个人得到的金额相等。

Each person gets exactly $\begin{aligned} \frac{1}{4} \end{aligned}$ of the candy bar.
::每个人都得到14个糖果棒。

We can also examine this problem using rational numbers. We keep a running total of what fraction of the bar remains. Remember, when we read a fraction followed by of in the problem, it means we multiply by that fraction.
::我们也可以使用合理数字来研究这个问题。我们保留了条形数的总数。记住, 当我们读到问题的一个部分之后, 这意味着我们乘以这个部分。

We start with 1 bar. Then Bill takes $\begin{aligned} \frac{1}{4} \end{aligned}$ of it, so there is $\begin{aligned} 1 - \frac{1}{4} = \frac{3}{4} \end{aligned}$ of a bar left.
::我们从一个酒吧开始,然后Bill要14个, 所以还有1 -14=34个酒吧。

Cindy takes $\begin{aligned} \frac{1}{3} \end{aligned}$ of what’s left, or $\begin{aligned} \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4} \end{aligned}$ of a whole bar. That leaves $\begin{aligned} \frac{3}{4} - \frac{1}{4} = \frac{2}{4} \end{aligned}$ , or $\begin{aligned} \frac{1}{2} \end{aligned}$ of a bar.
::Cindy拿走了剩下的13块,或者整个酒吧的1334=14块。剩下34 –14=24块,或者12块。

That half bar gets split between Anne and Dora, so they each get half of a half bar: $\begin{aligned} \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \end{aligned}$ .
::那半个酒吧在安妮和朵拉之间分开所以他们每人得到半个酒吧:1212=14

So each person gets exactly $\begin{aligned} \frac{1}{4} \end{aligned}$ of the candy bar.
::所以每个人都得到14个糖果棒。

Extension: If each person’s share is 3 oz, how much did the original candy bar weigh?
::扩展名:如果每个人的份额是3oz,那么原来的糖果棒的重量是多少?

Real-World Application: Newton's Law
::现实世界应用:牛顿法

Newton’s second law $\begin{aligned} (F = m a) \end{aligned}$ relates the force applied to a body in Newtons $\begin{aligned} (F) \end{aligned}$ , the mass of the body in kilograms $\begin{aligned} (m) \end{aligned}$ and the acceleration in meters per second squared $\begin{aligned} (a) \end{aligned}$ . Calculate the resulting acceleration if a Force of $\begin{aligned} 7 \frac{1}{3} \end{aligned}$ Newtons is applied to a mass of $\begin{aligned} \frac{1}{5} k g \end{aligned}$ .
::牛顿的第二项法律(F=ma)将牛顿(F)体体的强度、体重(m)公斤和加速度以每秒平方(a)米计。如果将713牛顿的力用于15公斤的重量,计算产生的加速度。

First, we rearrange our equation to isolate the acceleration, $\begin{aligned} a \end{aligned}$ . If $\begin{aligned} F = m a \end{aligned}$ , dividing both sides by $\begin{aligned} m \end{aligned}$ gives us $\begin{aligned} a = \frac{F}{m} \end{aligned}$ . Then we substitute in the known values for $\begin{aligned} F \end{aligned}$ and $\begin{aligned} m \end{aligned}$ :
::首先,我们重新排列我们的方程以分离加速度, a。如果F=ma, 以m为单位将两边除以m 给我们一个=Fm。然后我们用已知的F和m值替代F和m:

$\begin{aligned} a = \frac{7 \frac{1}{3}}{\frac{1}{5}} = \frac{22}{3} \div \frac{1}{5} = \frac{22}{3} \times \frac{5}{1} = \frac{110}{3} \end{aligned}$

::a=71315=2223=15=223×51=1103

The resultant acceleration is $\begin{aligned} 36 \frac{2}{3} m / s^{2} \end{aligned}$ .
::由此产生的加速度为每秒3623米/秒。

Example
::示例示例示例示例

Andrew is driving down the freeway. He passes mile marker 27 at exactly mid-day. At 12:35 he passes mile marker 69. At what speed , in miles per hour, is Andrew traveling?
::12: 35时,他经过69英里,每小时以多少英里的速度,安德鲁正在旅行?

To find the speed, we need the distance traveled and the time taken. If we want our speed to come out in miles per hour, we’ll need distance in miles and time in hours .
::为了找到速度,我们需要所走的距离和所花费的时间。如果我们想要以每小时英里的速度走出来,我们需要以英里和时数的距离。如果我们想要以每小时英里的速度走出来,我们需要以英里和时数的距离。

The distance is $\begin{aligned} 69 - 27 \end{aligned}$ or 42 miles. The time is 35 minutes, or $\begin{aligned} \frac{35}{60} \end{aligned}$ hours, which reduces to $\begin{aligned} \frac{7}{12} \end{aligned}$ . Now we can plug in the values for distance and time into our equation for speed.
::距离是 69 - 27 或 42 英里。时间是 35 分钟, 或 3560 小时, 减到 712 。现在我们可以将距离和时间的值插入速度的方程式。

$\begin{aligned} Speed = \frac{42}{\frac{7}{12}} = 42 \div \frac{7}{12} = \frac{42}{1} \times \frac{12}{7} = \frac{6 \cdot 7 \cdot 12}{1 \cdot 7} = \frac{6 \cdot 12}{1} = 72 \end{aligned}$

::速度=42712=42712=42712=421×127=6712=6121=72

Andrew is driving at 72 miles per hour.
::安德鲁每小时开72英里

Review
::回顾

For 1-8, perform the operations of multiplication and division.
::对于1-8,进行乘法和分法的操作。

$\begin{aligned} (3 - x) \end{aligned}$
:3-x)

$\begin{aligned} \frac{1}{13} \times \frac{1}{11} \end{aligned}$

$\begin{aligned} \frac{7}{27} \times \frac{9}{14} \end{aligned}$

$\begin{aligned} {(\frac{3}{5})}^{2} \end{aligned}$

$\begin{aligned} \frac{1}{2} \div \frac{x}{4 y} \end{aligned}$
::12x4y

$\begin{aligned} (- \frac{1}{3}) \div (- \frac{3}{5}) \end{aligned}$

$\begin{aligned} \frac{7}{2} \div \frac{7}{4} \end{aligned}$

$\begin{aligned} 11 \div \frac{- x}{4} \end{aligned}$
::11x4

For 9-11, solve the real-world problems using multiplication and division.
::9-11,使用乘法和分法解决现实世界的问题。

The label on a can of paint says that it will cover 50 square feet per pint. If I buy a $\begin{aligned} \frac{1}{8} \end{aligned}$ pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less or the same as that stated on the label?
::油漆罐上的标签表示它覆盖每品脱50平方英尺。如果我购买18品脱样品,它覆盖一平方2英尺长3英尺高的面积。我得到的覆盖是否比标签上所说的多、少或相同?

The world’s largest trench digger, "Bagger 288", moves at $\begin{aligned} \frac{3}{8} \end{aligned}$ mph. How long will it take to dig a trench $\begin{aligned} \frac{2}{3} \end{aligned}$ mile long?
::世界上最大的战壕挖掘者“Bagger 288 ” ( Bagger 288 ) , 移动38米。挖一条23英里长的战壕需要多长时间?

A $\begin{aligned} \frac{2}{7} \end{aligned}$ Newton force applied to a body of unknown mass produces an acceleration of $\begin{aligned} \frac{3}{10} m / s^{2} \end{aligned}$ . Calculate the mass of the body using $\begin{aligned} F = m \cdot a . \end{aligned}$
::对未知质量体施用27牛顿力,加速度为310 m/s2.,用F=ma计算体质量。

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

常规

应用中的混合数字

Mixed Numbers in Applications ::应用中的混合数字

Real-World Application: Chemistry ::现实世界应用:化学

Real-World Application: Splitting Candy ::真实世界应用程序: 分割糖果

Real-World Application: Newton's Law ::现实世界应用:牛顿法

Example ::示例示例示例示例

Review ::回顾

Review (Answers) ::回顾(答复)

Mixed Numbers in Applications
::应用中的混合数字

Real-World Application: Chemistry
::现实世界应用:化学

Real-World Application: Splitting Candy
::真实世界应用程序: 分割糖果

Real-World Application: Newton's Law
::现实世界应用:牛顿法

Example
::示例示例示例示例

Review
::回顾

Review (Answers)
::回顾(答复)