Although arithmetic is only one of the three types of math tested on the GMAT, arithmetic problems comprise about half of the total number of math questions.

 

Here are the specific arithmetic topics tested on the GMAT:

 

1. Axioms and Fundamentals (properties of integers, positive and negative numbers, even and odd). These were covered in Chapter 7.

 

2. Arithmetic Operations

 

3. Fractions

 

4. Decimals

 

5. Ratios

 

6. Percentages

 

7. Averages

 

8. Exponents and Radicals

 

In this chapter, we will first discuss the fundamentals of each topic and then show how the test-writers construct questions based on that topic.

儘管算術只是在GMAT上測試的三種數學類型之一,但算術問題約佔數學問題總數的一半。

 

以下是在GMAT上測試的特定算術主題:

 

1.公理和基本原理(整數,正負數,偶數和奇數的屬性)。 這些已在第7章中介紹。

 

2.算術運算

 

3.分數

 

4.小數

 

5.比率

 

6.百分比

 

7.平均值

 

8.指數和根號

 

在本章中,我們將首先討論每個主題的基礎,然後說明測試編寫者如何基於該主題構造問題。

ARITHMETIC OPERATIONS

There are six arithmetic operations you will need for the GMAT:

算術運算

GMAT需要進行六種算術運算:

1. Addition (2 + 2): The result of addition is a sum or total.

 

2. Subtraction (6 – 2): The result of subtraction is a difference.

 

3. Multiplication (2 × 2): The result of multiplication is a product.

 

4. Division (8 ÷ 2): The result of division is a quotient.

 

5. Raising to a power (x2): In the expression x2, the little 2 is called an exponent.

 

6. Finding a square root : 

1.加法(2 + 2):加法的結果是總和或總計。

 

2.減法(6 – 2):減法的結果是不同的。

 

3.乘法(2×2):乘法的結果是乘積。

 

4.除(8÷2):除的結果是商。

 

5.求冪(x2):在表達式x2中,小數2稱為指數。

 

6.求平方根:

Which One Do I Do First?

In a problem that involves several different operations, the operations must be performed in a particular order, and occasionally GMAC likes to see whether you know what that order is. Here’s an easy way to remember the order of operations:

The first letters stand for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. Do operations that are enclosed in parentheses first; then take care of exponents; then multiply and divide; finally add and subtract, going from left to right.

我首先要做哪一個?

在涉及幾個不同操作的問題中,這些操作必須以特定的順序執行,並且有時GMAC喜歡看您是否知道該順序是什麼。 這是記住操作順序的一種簡單方法:

首字母代表括號,指數,乘法,除法,加法和減法。 首先進行括號內的操作; 然後照顧指數; 然後乘除 最後加減,從左到右。

DRILL 2

Just to get you started, solve each of the following problems by performing the indicated operations in the proper order. The answers can be found in Part VI.

 

1. 74 + (27 – 24) =

 

2. (8 × 9) + 7 =

 

3. 2(9 – (8 ÷ 2)) =

 

4. 2(7 – 3) + (–4)(5 – 7) =

練習2

只是為了入門,請按照正確的順序執行指示的操作來解決以下每個問題。 答案可以在第六部分中找到。

 

1. 74 +(27 – 24)=

 

2.(8×9)+ 7 =

 

3. 2(9 –(8÷2))=

 

4. 2(7 – 3)+(–4)(5 – 7)=

Here’s an easy question that shows how GMAC might test PEMDAS.

 

5. 4(–3(3 – 5) + 10 – 17) =

 

–27

 

  –4

 

  –1

 

  32

 

  84

 

It is not uncommon to see a Data Sufficiency problem like this on the GMAT:

 

6. What is the value of x ?

 

(1) x3 = 8

(1) x3 = 8

 

(2) x2 = 4

 

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

 

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

 

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

 

EACH statement ALONE is sufficient.

 

Statements (1) and (2) TOGETHER are not sufficient.

 

There are two operations that can be done in any order, provided they are the only operations involved: When you are adding or multiplying a series of numbers, you can group or regroup the numbers any way you like.

 

2 + 3 + 4 is the same as 4 + 2 + 3

 

and

 

4 × 5 × 6 is the same as 6 × 5 × 4

 

This is called the Associative Law, but the name will not be tested on the GMAT.

 

Another law that GMAC likes to test states that

 

a(b + c) = ab + ac and a(b – c) = ab – ac.

 

This is called the Distributive Law but, again, you don’t need to know that for the test. Sometimes the Distributive Law can provide you with a shortcut to the solution of a problem. If a problem gives you information in “factored form”—a(b + c)—you should distribute it immediately. If the information is given in distributed form—ab + ac—you should factor it.

這是一個簡單的問題,顯示了GMAC如何測試PEMDAS。

 

5. 4(–3(3 – 5)+ 10 – 17)=

 

–27

 

  –4

 

  –1

 

  32

 

  84

 

在GMAT上看到這樣的數據充足性問題並不少見:

 

6. x的值是多少?

 

(1)x3 = 8

(1)x3 = 8

 

(2)x2 = 4

 

陳述(1)僅是足夠的,但僅陳述(2)是不夠的。

 

陳述(2)單獨就足夠了,但是僅陳述(1)不夠。

 

兩條語句TOGETHER都足夠,但是NEITHER語句ALONE就足夠了。

 

每個語句ALONE就足夠了。

 

語句(1)和(2)不夠。

 

只要是涉及的唯一操作,就可以按照任何順序執行兩個操作:在對一系列數字進行加或乘運算時,可以按照自己喜歡的方式對數字進行分組或重新分組。

 

2 + 3 + 4與4 + 2 + 3相同

 

 

4×5×6與6×5×4相同

 

這被稱為《聯合法》,但該名稱不會在GMAT上進行測試。

 

GMAC喜歡測試的另一條法律指出:

 

a(b + c)= ab + ac,a(b – c)= ab – ac。

 

這就是所謂的分配法,但是,再次,您不需要知道該測試。有時,分配法可以為您提供解決問題的捷徑。如果問題以“分解形式”(a(b + c))為您提供信息,則應立即分發。如果信息以分佈式形式(ab + ac)給出,則應將其分解。

DRILL 3

If the following problems are in distributed form, factor them; if they are in factored form, distribute them. Then do the indicated operations. Answers are in Part VI.

 

1. x 2 + x

 

2. (55 × 12) + (55 × 88)

 

3. a(b + c – d)

 

4. abc + xyc

 

A GMAT problem might look like this:

 

5. If x = 6, what is the value of ?

 

–30

 

    6

    8

 

  30

 

It cannot be determined from the information given.

 

It is not uncommon to see a Data Sufficiency problem like this on the GMAT:

 

6. If ax + ay + az = 15, what is x + y + z ?

 

(1) x = 2

 

(2) a = 5

 

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

 

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

 

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

 

EACH statement ALONE is sufficient.

 

Statements (1) and (2) TOGETHER are not sufficient.

鑽3

如果以下問題是分佈式的,請將其分解; 如果它們是分解形式,則分發它們。 然後執行指示的操作。 答案在第六部分。

 

1. x 2 + x

 

2.(55×12)+(55×88)

 

3. a(b + c – d)

 

4. abc + xyc

 

GMAT問題可能看起來像這樣:

 

5.如果x = 6,則值是多少?

 

–30

 

     6

     8

 

   30

 

無法根據給定的信息確定。

 

在GMAT上看到這樣的數據充足性問題並不少見:

 

6.如果ax + ay + az = 15,x + y + z是多少?

 

(1)x = 2

 

(2)= 5

 

陳述(1)僅是足夠的,但僅陳述(2)是不夠的。

 

陳述(2)單獨就足夠了,但是僅陳述(1)不夠。

 

兩條語句TOGETHER都足夠,但是NEITHER語句ALONE就足夠了。

 

每個語句ALONE就足夠了。

 

語句(1)和(2)不夠。

Challenge Question #2

 

A device calculates the worth of gemstones based on quality such that a gem with a quality rating of q – 1 is worth 5 times more than a gem with a quality rating of q, and a gem with a quality rating of q – 4 is worth 625 times more than a gem with a quality rating of q. According to this device, the worth of a gem with a quality rating of p – r is how many times greater than that of a gem with a rating of p ?

 

p5 – r5

 

r5

 

(p – r)5

 

5r

 

5r

挑戰問題2

 

設備根據質量來計算寶石的價值,以使質量等級為q – 1的寶石的價值是質量等級為q的寶石的5倍,質量等級為q – 4的寶石的價值 是質量等級為q的寶石的625倍。 根據此設備,質量評級為p – r的寶石的價值是評級為p的寶石的價值的幾倍?

 

p5 – r5

 

55

 

(p – r)5

 

5r

 

5r

FRACTIONS

Fractions can be thought of in two ways:

 

A fraction is just another way of expressing division. The expression  1/2 is exactly the same thing as 1 divided by 2. x/y is nothing more than x divided by y. In the fraction x/y , x is known as the numerator and y is known as the denominator.

 

The other important way to think of a fraction is as part/whole . The fraction 7/10 can be thought of as 7 parts out of a total of 10 parts.

 

分數

分數可以通過兩種方式考慮:

 

分數只是表達除法的另一種方式。 表達式1/2與1除以2完全相同。x/ y無非是x除以y。 在分數x / y中,x被稱為分子,而y被稱為分母。

 

考慮分數的另一種重要方法是part / whole。 分數7/10被認為是10總數中的7。

Adding and Subtracting Fractions with the Same Denominator

To add two or more fractions that have the same denominator, simply add the numerators and put the sum over the common denominator. For example:

使用相同分母添加和減去分數

要添加兩個或多個具有相同分母的分數,只需將分子相加並將總和放在公共分母上即可。 例如:

Subtraction works exactly the same way:

減法的工作方式完全相同:

Adding and Subtracting Fractions with Different Denominators

Before you can add or subtract two or more fractions with different denominators, you must give all of them the same denominator. To do this, multiply the numerator and denominator of each fraction by a number that will give it a denominator in common with the others. If you multiplied each fraction by any old number, the fractions wouldn’t have their original values, so the number you multiply by has to be equal to 1. For example, if you wanted to change  into sixths, you could do the following:

用不同分母添加和減去分數

在可以添加或減去具有不同分母的兩個或多個分數之前,必須為所有分數賦予相同的分母。 為此,將每個分數的分子和分母乘以一個數字,使其與其他分母相同。 如果您將每個分數乘以任何舊數字,那麼分數將沒有其原始值,因此乘以的數字必須等於1。例如,如果要更改為六進制,則可以執行以下操作:

We haven’t actually changed the value of the fraction, because  equals 1.

If we wanted to add:

我們實際上沒有更改分數的值,因為它等於1。

如果我們要添加:

The Bowtie

The Bowtie method has been a staple of The Princeton Review’s materials since the company began in a living room in New York City in 1981. It’s been around so long because it works so simply.

 

To add 3/5 and 4/7 , for example, follow these three steps:

 

Step One: Multiply the denominators together to form the new denominator.

鮑蒂方法

自從公司於1981年在紐約市的客廳裡成立以來,鮑蒂方法就一直是《普林斯頓評論》的主要材料。這種方法之所以存在很久是因為它的工作原理如此簡單。

 

例如,要添加 3/5 和 4/7,請執行以下三個步驟:

 

第一步:將分母相乘以形成新的分母。

Step Two: Multiply the first denominator by the second numerator (5 × 4 = 20) and the second denominator by the first numerator (7 × 3 = 21) and place these numbers above the fractions, as shown below.

第二步:將第一分母乘以第二分子(5×4 = 20),將第二分母乘以第一分子(7×3 = 21),並將這些數字放在小數上方,如下所示。

Step Three: Add the products to form the new numerator.

第三步:添加產品以形成新的分子。

Subtraction works the same way.

減法的工作方式相同。

Note that with subtraction, the order of the numerators is important. The new numerator is 21 – 20, or 1. If you somehow get your numbers reversed and use 20 – 21, your answer will be –1/35 , which is incorrect. One way to keep your subtraction straight is to always multiply up from denominator to numerator when you use the Bowtie.

請注意,使用減法時,分子的順序很重要。 新的分子為21 – 20或1。如果以某種方式使數字取反並使用20 – 21,則答案將是–1/35,這是不正確的。 保持直線減法的一種方法是在使用領結時始終將分母乘以分子。.

Multiplying Fractions

To multiply fractions, just multiply the numerators and put the product over the product of the denominators. For example:

乘法分數

要乘以分數,只需乘以分子,然後將乘積放在分母的乘積上。 例如:

Reducing Fractions

When you add or multiply fractions, you often end up with a big fraction that is hard to work with. You can usually reduce such a fraction. To reduce a fraction, find a factor of the numerator that is also a factor of the denominator. It saves time to find the biggest factor they have in common, but this isn’t critical. You may just have to repeat the process a few times. When you find a common factor, cancel it. For example, let’s take the product we just found when we multiplied the fractions above:

減少分數

當您添加或相乘分數時,通常會遇到很大的分數,難以使用。 您通常可以減少這樣的分數。 要減少分數,請找到分子的因數,這也是分母的因數。 這樣可以節省時間來找到他們共同的最大因素,但這並不重要。 您可能只需要重複幾次該過程。 當您發現一個共同因素時,將其取消。 例如,讓我們乘以上面的分數得到的乘積:

Get used to reducing all fractions (if they can be reduced) before you do any work with them. It saves a lot of time and prevents errors in computation.

 

For example, in that last problem, we had to multiply two fractions together:

在使用它們之前,請習慣於減少所有分數(如果可以減少)。 這樣可以節省大量時間並防止計算錯誤。

 

例如,在最後一個問題中,我們必須將兩個分數相乘:

Before you multiplied 2 × 6 and 3 × 5, you could have reduced

在將2×6和3×5相乘之前,您可能已經減少了

Dividing Fractions

To divide one fraction by another, just invert the second fraction and multiply:

分數除法

要將一個分數除以另一個,只需反轉第二個分數並乘以:

which is the same thing as…

與…相同

You may see this same operation written like this:

您可能會看到以下相同的操作:

Again, just invert and multiply. This next example is handled the same way:

同樣,只需反轉並相乘即可。 下一個示例的處理方式相同:

When you invert a fraction, the new fraction is called a reciprocal. 2/3 is the reciprocal of 3/2. The product of two reciprocals is always 1.

 

Converting to Fractions

An integer can be expressed as a fraction by making the integer the numerator and 1 the denominator: 16 = 16/1.

 

The GMAT sometimes gives you numbers that are mixtures of integers and fractions, for example, 3 1/2. It’s easier to work with these numbers if you convert them into fractions. Simply multiply the denominator by the integer, then add the numerator, and place the resulting number over the original denominator.

反轉分數時,新的分數稱為倒數。 2/3是3/2的倒數。 兩個倒數的乘積始終為1。

 

轉換為分數

通過將整數設為分子,將1設為分母,可以將整數表示為分數:16 = 16/1。

 

GMAT有時會為您提供整數和分數混合的數字,例如3 1/2。 如果將這些數字轉換為分數,則使用起來更容易。 只需將分母乘以整數,然後加上分子,然後將結果數字放在原始分母上。

Comparing Fractions

In the course of a problem, you may have to compare two or more fractions and determine which is greater. This is easy to do as long as you remember that you can compare fractions directly only if they have the same denominator. Suppose you had to decide which of these three fractions is greatest:

比較分數

在問題過程中,您可能必須比較兩個或多個分數,並確定哪個更大。 只要您記得只有分數相同的分數,才可以直接比較分數,這很容易做到。 假設您必須決定這三個部分中哪個最大:

To compare these fractions directly, you need a common denominator, but finding a common denominator that works for all three fractions would be complicated and time consuming. It makes more sense to compare these fractions two at a time. We showed you the classical way to find common denominators when we talked about adding fractions earlier.

 

Let’s start with 1/2 and 5/9. An easy common denominator for these two fractions is 18 (9 × 2).

要直接比較這些分數,您需要一個公分母,但是要找到一個適用於所有三個分數的公分母將既複雜又耗時。 一次比較這兩個分數更有意義。 在前面討論加分數時,我們向您展示了尋找公分母的經典方法。

 

讓我們從1/2和5/9開始。 這兩個分數的一個簡單的公分母是18(9×2)。

1/2                                             5/9

=9/18                                      =10/18

Because 5/9 is greater, let’s compare it with 7/15. Here the easiest common denominator is 45. But before we do that…

 

Two Shortcuts

Comparing fractions is another situation in which we can use the Bowtie. The idea is that if all you need to know is which fraction is greater, you just have to compare the new numerators. Again, simply multiply the denominator of the first fraction by the numerator of the second and the denominator of the second by the numerator of the first, as shown here.

由於5/9更大,因此我們將其與7/15進行比較。 這裡最簡單的公分母是45。但是在我們這樣做之前……

 

兩條捷徑

比較分數是我們可以使用領結的另一種情況。 這個想法是,如果您只需要知道哪個分數更大,則只需比較新的分子即可。 同樣,只需將第一個分數的分母乘以第二個的分子,然後將第二個分母乘以第一個的分子,如此處所示。

10 > 9, therefore  5/9 > 1/2

You could also have saved yourself some time on the last problem by a little fast estimation. Again, which is greater? 1/2 ,5/9 , or 7/15 ?

 

Let’s think about 5/9 in terms of 1/2 . How many ninths equal a half? To put it another way, what is half of 9 ? 4.5. So 4.5/9=1/2. That means 5/9 is greater than 1/2.

 

Now let’s think about 7/15. Half of 15 is 7.5. , 7.5/15=1/2 , which means that 7/15 is less than 1/2.

您也可以通過快速估算節省一些時間來解決最後一個問題。 同樣,哪個更大? 1/2,5/9或7/15?

 

讓我們以1/2來考慮5/9。 九分之幾等於一半? 換句話說,9的一半是多少? 4.5。 因此4.5 / 9 = 1/2。 這意味著5/9大於1/2。

 

現在考慮一下7/15。 15的一半是7.5。 ,7.5 / 15 = 1/2,這意味著7/15小於1/2。

PROPORTIONS

A fraction can be expressed in many ways.  also equals  or , etc. A proportion is just a different way of expressing a fraction. Here’s an example:

比例

分數可以用多種方式表示。 也等於或,等等。比例只是表示分數的另一種方式。 這是一個例子:

If 2 boxes hold a total of 14 shirts, how many shirts are contained in 3 boxes?

Here’s How to Crack It

 

The number of shirts per box can be expressed as a fraction. What you’re asked to do is express the fraction 2/14 in a different way.

如果2盒裝總共14件襯衫,那麼3盒裝多少個襯衫?

這是破解方法

 

每盒襯衫的數量可以表示為分數。 您需要做的是用不同的方式表示分數2/14。

To find the answer, all you need to do is find a value for x such that 2/14=3/x. The easiest way to do this is to cross-multiply.

 

2x = 42, which means that x = 21. There are 21 shirts in 3 boxes.

要找到答案,您要做的就是找到x的值,使得2/14 = 3 / x。 最簡單的方法是相乘。

 

2x = 42,這意味著x =21。3個盒子中有21件襯衫。

DRILL 4

The answers to these questions can be found in Part VI.

 

1. 5 3/4 + 3/8 =

 

2. Reduce 12/60

 

3. Convert 9 2/3 to a fraction

 

4. Solve for x in 9/2 = x/4

 

A relatively easy GMAT fraction problem might look like this:

練習 4

這些問題的答案可以在第六部分中找到。

 

1. 5 3/4 + 3/8 =

 

2.降低12/60

 

3.將9 2/3轉換為分數

 

4.求解x in 9/2 = x / 4

 

一個相對簡單的GMAT分數問題可能看起來像這樣:

gmat-quantity-order-201128-42

3/100

3/16

1/3

1

7/16

Fractions: Advanced Principles

Now that you’ve been reacquainted with the basics of fractions, let’s go a little further. More complicated fraction problems usually involve all of the rules we’ve just mentioned, with the addition of two concepts: part/whole  , and the rest. Here’s a typical medium fraction problem:

分數:高級原理

現在,您已經熟悉了分數的基礎知識,讓我們再進一步。 更為複雜的分數問題通常涉及我們剛才提到的所有規則,另外還包括兩個概念:part / whole和其餘。 這是一個典型的中等分數問題:

A cement mixture is composed of 3 elements. By weight,  1/3 of the mixture is sand,  3/5 of the mixture is water, and the remaining 12 pounds of the mixture is gravel. What is the weight of the entire mixture in pounds?

    4

    8

  36

  60

180

水泥混合物由3種元素組成。 按重量計,混合物的1/3為沙子,混合物的3/5為水,剩餘的12磅混合物為礫石。 整個混合物的重量(以磅為單位)是多少?

     4

     8

   36

   60

180

Easy Eliminations

Before we even start doing serious math, let’s use some common sense. The weight of the gravel alone is 12 pounds. Because we know that sand and water make up the bulk of the mixture—sand 1/3 , water 3/5  (which is a bit more than half)—the entire mixture must weigh a great deal more than 12 pounds. Choices (A) and (B) are out of the question. Eliminate them.

簡單除去

在開始進行認真的數學運算之前,請先使用一些常識。 僅礫石的重量為12磅。 因為我們知道沙子和水構成了混合物的大部分(沙子1/3,水3/5(略大於一半)),所以整個混合物的重量必須超過12磅。 選擇(A)和(B)是不可能的。 消除它們。

Here’s How to Crack It

 

The difficulty in solving this problem is that sand and water are expressed as fractions, while gravel is expressed in pounds. At first there seems to be no way of knowing what fractional part of the mixture the 12 pounds of gravel represent; nor do we know how many pounds of sand and water there are.

 

The first step is to add up the fractional parts that we do have:

這是破解方法

 

解決此問題的困難在於,沙子和水以分數表示,而礫石以磅表示。 首先,似乎沒有辦法知道這12磅礫石所代表的混合物的分數。 我們也不知道那裡有多少磅的沙子和水。

 

第一步是將我們擁有的分數部分相加:

Sand and water make up 14 parts out of the whole of 15. This means that gravel makes up what is left over—the rest: 1 part out of the whole of 15. Now the problem is simple. Set up a proportion between parts and weights.

沙子和水在15的全部中佔14的一部分。這意味著礫石佔了剩餘的部分,其餘部分在15的全部中佔1的部分。現在問題很簡單了。 在零件和重量之間設置比例。

Cross-multiply: x = 180. The answer is (E).

互乘:x =180。答案是(E)。

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