2010 年 AMC 8 真题（P版）

1 答案 C

学生总数为 \(11+8+9 = 28\)。

所以正确答案是 C。

English

There are \(11+8+9 = 28\) students.

Therefore, the answer is C .

2 答案 D

由 \(a\ast b\) 的定义，\[5 \ast 10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}\text{。}\]

所以正确答案是 D。

English

Given our definition of \(a\ast b,\) we have \[5 \ast 10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}.\]

Thus, the answer is D .

3 答案 C

最高价格为 \(17\)，最低价格为 \(10\)。因此百分数为 \[ (\dfrac{17}{10}-1)\cdot 100 = 70\text{。}\]

所以正确答案是 C。

English

The highest price is \(17\) and the lowest price is \(10.\) This means the percent is \[ (\dfrac{17}{10}-1)\cdot 100 = 70.\]

Thus, the answer is C .

4 答案 C

重新排序后的列表是 \(0,0,1,2,3,3,3,4\)。

中位数是中间两个数的平均数，即 \(\dfrac{2+3}2 = 2.5\)。众数是 \(3\)，因为 \(3\) 出现次数最多。平均数为 \[ \dfrac{0+0+1+2+3+3+3+4}{8}\] \[ = \dfrac{16}{8} = 2\text{。}\]

它们的和为 \(2.5+3+2 = 7.5\)。

所以正确答案是 C。

English

The list reordered is \(0,0,1,2,3,3,3,4.\)

The median is the mean of the middle two numbers, which would be \(\dfrac{2+3}2 = 2.5.\) The mode is \(3\) since \(3\) appears the most. The mean is \[ \dfrac{0+0+1+2+3+3+3+4}{8}\]\[ = \dfrac{16}{8} = 2.\]

Their sum is \(2.5+3+2 = 7.5.\)

Thus, the answer is C .

5 答案 B

天花板高度为 \(2.4 \cdot 100 = 240\) 厘米。减去已知的各段高度，得到 \[ 240 - 150 - 46 - 10 = 34\text{。}\]因此凳子的高度是 \(34\) 厘米。所以正确答案是 B。

English

We know the height of the ceiling is \(2.4 \cdot 100 = 240\) cm. Subtracting out all the given values, we get \[ 240 - 150 - 46 - 10 = 34. \] Therefore, the height of the stool is \(34\) cm. Thus, B is the correct answer.

6 答案 E

每条对称轴都必须经过图形的中心，否则中心会只在一侧，反射后无法重合。

如果对称轴经过某条边，它必须经过这条边的中点，并且与这条边垂直，这样反射后该边两侧长度相同且仍落在自身上。

如果对称轴经过某个顶点，它必须经过该角的角平分线，并且两侧边长相同，反射后角才会重合。

现在逐个考察这些图形。等边三角形只有 \(3\) 条经过顶点或边中点的直线，所以至多有 \(3\) 条对称轴。非正方形的菱形只有两条对称轴，因为只有经过相对顶点的直线可行，而经过边中点的直线与该边不垂直。非正方形的长方形也只有两条对称轴，它们经过对边的中点；经过顶点的直线不平分该角，所以不可行。等腰梯形只有 \(1\) 条对称轴，它经过一组对边的中点。正方形有 \(4\) 条对称轴，分别经过相对顶点和相对边的中点。

因此正方形有最多条对称轴。

所以正确答案是 E。

English

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn’t on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, let’s look at each of the shapes. An equilateral triangle has \(3\) lines that intersect a corner or a midpoint, so it has at most \(3\) symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangle only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn’t bisect the angle. An isosceles trapezoid only has \(1\) symmetry line, that goes through the midpoints of opposite sides. A square has \(4\) symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E .

7 答案 B

要支付 \(4\) 美分，至少需要四枚一美分硬币。有了这四枚后，为了能支付 \(5\) 美分，下一枚硬币的面值至多为 \(5\) 美分，最好选一枚五美分硬币。这五枚硬币总值只有 \(9\) 美分。下一枚硬币的面值至多为 \(10\) 美分；即使选一枚十美分硬币，总值也只有 \(19\) 美分。因此还需要一枚面值至多为 \(20\) 美分的硬币，所以在不产生金额缺口的情况下使用二十五美分硬币之前，至少需要七枚硬币。

若总共只有九枚硬币，在必需的七枚之后最多只剩两枚。前七枚的总值至多为 \(4+5+10+10=29\) 美分，再加两枚二十五美分硬币也只有 \(79\) 美分，所以九枚硬币不能支付从一美分到 \(99\) 美分的每个金额。

十枚硬币确实可以做到：四枚一美分硬币、一枚五美分硬币、两枚十美分硬币和三枚二十五美分硬币，可以支付从 \(1\) 到 \(99\) 美分的每个金额。因此最少需要 \(10\) 枚。

所以正确答案是 B。

English

At least four pennies are necessary to pay \(4\) cents. After those four pennies, the next coin must be worth at most \(5\) cents so that \(5\) cents can be paid; a nickel is the best choice. Those five coins total only \(9\) cents. The next coin must be worth at most \(10\) cents, and even a dime brings the total to only \(19\) cents. One more coin worth at most \(20\) cents is therefore necessary, so at least seven coins are needed before quarters can be used without leaving a gap.

With only nine coins total, at most two coins could remain after those required seven. The first seven can total at most \(4+5+10+10=29\) cents, and two more quarters would bring the total to only \(79\) cents, so nine coins cannot pay every amount through \(99\) cents.

Ten coins do work: four pennies, one nickel, two dimes, and three quarters can make every amount from \(1\) through \(99\) cents. Therefore the minimum is \(10\).

Therefore, the answer is B .

8 答案 D

设 \(d\) 为艾米丽领先埃默森的距离。艾米丽能看到埃默森当且仅当 \(-\dfrac 12 \leq d \leq \frac 12\)。其中 \(t\) 以小时为单位；令 \(t= 0\) 时 \(d = -\frac 12\)。根据两人的相对速度，\(d = (12-8)t - \frac 12 \)。当 \(d = \frac 12\) 时，\[\frac 12 = 4t - \frac 12 \implies 1 = 4t \] \[\implies t = 0.25\text{。}\]因此经过 \(0.25\) 小时，也就是 \(60\cdot0.25 = 15\) 分钟。

所以正确答案是 D。

English

Let \(d\) be how far Emily is ahead of Emerson. Emily sees Emerson if \(-\dfrac 12 \leq d \leq \frac 12.\) Suppose at \(t= 0,\) where \(t\) is in hours, that \(d = -\frac 12.\) Then, \(d = (12-8)t - \frac 12 .\) Since we must find where \(d = \frac 12,\) we find the time where \[\frac 12 = 4t - \frac 12 \implies 1 = 4t \]\[\implies t = 0.25.\] Since \(0.25\) hours passed, we know that \(60\cdot0.25 = 15\) minutes passed.

Therefore, the answer is D .

9 答案 D

三份测试共有 \(25+40+10 = 75\) 题。

第一份测试答对 \(0.8\cdot25 = 20\) 题。

第二份测试答对 \(0.9\cdot40 = 36\) 题。

第三份测试答对 \(0.7\cdot10 = 7\) 题。

因此他总共答对 \(20+36+7=63\) 题，答对比例为 \(\dfrac{63}{75} = 84\%\)。

所以正确答案是 D。

English

There were a total of \(25+40+10 = 75\) problems.

On the first test, he solved \(0.8\cdot25 = 20\) problems.

On the second test, he solved \(0.9\cdot40 = 36\) problems.

On the third test, he solved \(0.7\cdot10 = 7\) problems.

Therefore, he solved a total of \(20+36+7=63.\) This means the fraction he solved is \( \dfrac{63}{75} = 84\%.\)

Therefore, the answer is D .

10 答案 B

每片圆形辣香肠的直径是大圆直径的 \(\dfrac{1}{6}\)，所以它的面积是总面积的 \((\frac 16)^2 = \frac 1{36}\)。

因为有 \(24\) 片辣香肠，所以覆盖面积为 \(24 \cdot \dfrac 1{36} = \frac 23\)。

所以正确答案是 B。

English

Each circle has \(\dfrac{1}{6} \) the diameter of the large circle, so it has \((\frac 16)^2 = \frac 1{36} \) of the total area.

Since there are \(24\) pepperoni, they take up \(24 \cdot \dfrac 1{36} = \frac 23 \) of the area.

Therefore, the answer is B .

11 答案 B

设较高和较矮的树高分别为 \(b\)、\(s\)。则 \(b= s+16\)，且 \(s = 0.75b\)。代入得 \[b = 0.75b + 16 \implies 0.25b = 16 \] \[\implies b = 64\text{。}\]

所以正确答案是 B。

English

Let \(b,\) \(s\) be the heights of bigger and smaller trees respectively. Then, \(b= s+16\) and \( s = 0.75b.\) If we substitute, we get \[b = 0.75b + 16 \implies 0.25b = 16 \]\[\implies b = 64.\]

Thus, the answer is B .

12 答案 D

若 \(r\) 是红球数，则 \[r = 0.8\cdot500 = 400\text{。}\]因此若 \(b\) 是蓝球数，则 \[b = 500-400 = 100\text{。}\]

拿走一些球后，如果红球占 \(75\%\)，那么蓝球占 \(25\%\)。蓝球数量不变，所以剩余球总数为 \(\dfrac{100}{0.25} = 400\)。这表示总球数减少了 \(500-400 = 100\)，也就是拿走了一百个红球。

所以正确答案是 D。

English

If \(r\) is the number of red balls, then \[r = 0.8\cdot500 = 400.\] Therefore, if \(b\) is the number of blue balls, then \[b = 500-400 = 100.\]

If there are \(75\%\) red balls after removing balls, then there are \(25\%\) blue balls. This means the total number of balls is \(\dfrac{100}{0.25} = 400.\) This means the total number of balls decreased by \(500-400 = 100.\)

Thus, the answer is D .

13 答案 E

设最短边长为 \(s\)。三边长度为 \(s\)、\(s+1\)、\(s+2\)，周长为 \(3s+3\)。因为 \(s = 0.3(3s+3)\)，所以 \(s = 0.9s + 0.9 \)。因此 \(0.1s = 0.9\)，得到 \(s = 9\)，最长边为 \(s + 2 = 11\)。

所以正确答案是 E。

English

Let \(s\) be the smallest length. Then, all the side lengths are \(s,\) \(s+1,\) \(s+2.\) This would make the perimeter equal to \(3s+3.\) Since \(s = 0.3(3s+3),\) then \(s = 0.9s + 0.9 .\) This makes \(0.1s = 0.9,\) so \(s = 9\) which makes the longest side length \(s + 2 = 11.\)

Thus, the answer is E .

14 答案 C

首先，质数 \(2\)、\(5\) 是 \(2010\) 的因数，因为它是 \(10\) 的倍数。\(2010\) 除以 \(10\) 得 \(201\)。接着，\(3\) 是 \(201\) 的因数，因为 \(201\) 的数位和是 \(3\) 的倍数。再除以 \(3\) 得到质数 \(67\)。因此质因数为 \(2\)、\(3\)、\(5\)、\(67\)，它们的和为 \(77\)。

所以正确答案是 C。

English

First, the primes \(2,\) \(5\) are factors of \(2010\) since it is a multiple of \(10.\) Dividing \(2010\) by \(10\) is \(201.\) Then, \(3\) is a factor of \(201\) since the digit sum of \(201\) is a multiple of \(3.\) Dividing this by \(3\) yields the prime number \(67.\) This means the prime factors are \(2,\) \(3,\) \(5,\) \(67,\) which makes their sum \(77.\)

Thus, the correct answer is C .

15 答案 C

绿色所占百分比是 \(100\%\) 减去其他颜色百分比之和：\[100\%-30\%-20\%-\] \[15\%-10\% = 25\%\]

因为 \(30\) 颗绿色软糖是总数的 \(25\%\)，所以软糖总数为 \(\dfrac { 30}{0.25} = 120\)。

一开始棕色软糖有 \(0.2\cdot120 = 24\) 颗，蓝色软糖有 \(0.3\cdot120 = 36\) 颗。把一半蓝色换成棕色会增加 \(\dfrac{36}{2} = 18\) 颗棕色软糖。因此棕色软糖共有 \(24+18= 42\) 颗。

所以正确答案是 C。

English

Since we have percentages for every color except green, the percent of green is \(100\%\) minus the sum of the other colors. This would make the percent of green equal to \[100\%-30\%-20\%-\]\[15\%-10\% = 25\%\]

Since we know \(30\) gum drops is \(25\%,\) we know that the total number of gum drops is \(\dfrac { 30}{0.25} = 120.\)

This means there are \(0.2\cdot120 = 24\) brown gum drops to start and \(0.3\cdot120 = 36 \) blue gum drops. If half of the blue gum drops are turned to brown, then \( \dfrac{36}{2} = 18\) more brown gum drops are added. Therefore, we have \(24+18= 42\) brown gum drops.

Thus, the answer is C .

16 答案 B

设正方形边长为 \(s\)，圆半径为 \(r\)。由于面积相同，\(s^2 = \pi r^2 \)。因此 \((\dfrac sr)^2 = \pi\)，所以 \(\frac sr = \sqrt{ \pi} \)。

所以正确答案是 B。

English

Let \(s\) be the side length of the square and let \(r\) be the radius of the circle. Then, since they have the same areas, \(s^2 = \pi r^2 .\) This means \((\dfrac sr)^2 = \pi,\) so \(\frac sr = \sqrt{ \pi} .\)

Thus, the answer is B .

17 答案 D

因为 \(PQ\) 平分面积，直线下方的面积为 \(5\)。去掉右侧的单位正方形后，底部是一个底为 \(5\)、面积为 \(4\) 的三角形。设这个三角形的底为 \(PZ\)。

面积为 \(4\)，所以 \[\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4 \] \[\implies ZQ = 1.6\text{。}\]

因此 \(QY = QZ-1 = 0.6\)，且 \(XQ = 2-QZ = 0.4\)。于是 \(\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23\)。

所以正确答案是 D。

English

Since \(PQ\) bisects the area, the area under the line is \(5.\) Removing the square on the right makes the bottom a triangle of base \(5\) with area \(4.\) Let the base of this triangle be \(PZ.\)

The area being \(4\) means \[\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4 \]\[\implies ZQ = 1.6.\]

Therefore, \(QY = QZ-1 = 0.6\) and \( XQ = 2-QZ = 0.4.\) This would make \(\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.\)

Thus, the answer is D .

18 答案 C

两个半圆合起来是一个直径为 \(d=30\) 的圆，其半径为 \(\dfrac{d}{2}\)。因此两个半圆的总面积为 \[(\dfrac{d}{2})^2 \cdot \pi = \dfrac{\pi d^2}{4}\text{。}\]

因为 \(AD:AB=3:2\) 且 \(AB=d\)，所以 \(AD = \dfrac 32 d\)。长方形面积为 \(\dfrac 32 d^2\)。所求比值为 \(\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}\)。

所以正确答案是 C。

English

Combining the semicircles would make a circle of diameter \(d=30.\) This would make the radius equal to \( \dfrac{d}{2}.\) Therefore, the combined area of the semicircles is \[(\dfrac{d}{2})^2 \cdot \pi = \dfrac{\pi d^2}{4}.\]

Side \(AD = \dfrac 32 d\) because \(AD:AB=3:2\) and \(AB=d.\) The area of the rectangle is therefore \( \dfrac 32 d^2.\) The ratio of the area of the rectangle to the area of the semicircles is \(\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}.\)

Thus, the answer is C .

19 答案 C

两圆之间的面积等于大圆面积减去小圆面积，即 \[(AC)^2\pi - (CB)^2 \pi \] \[= \pi(AC^2 - CB^2)\text{。}\]由勾股定理，\[AC^2 - CB^2 = AB^2\text{。}\]因此只需求 \(AB^2 \pi\)。

因为 \(AB\) 是 \(AD\) 的一半，所以 \(AB = 8\)。于是 \(AB^2 \pi = 64 \pi\)。

所以正确答案是 C。

English

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be \[(AC)^2\pi - (CB)^2 \pi \]\[= \pi(AC^2 - CB^2).\] By the Pythagorean Theorem, we can get \[AC^2 - CB^2 = AB^2.\] Therefore, we need to find \(AB^2 \pi.\)

Since \(AB\) is half of \(AD,\) we get \(AB = 8.\) This makes \(AB^2 \pi = 64 \pi.\)

Thus, the answer is C .

20 答案 A

因为房间中 \(\dfrac 25\) 的人戴手套，所以总人数必须是 \(5\) 的倍数。因为 \(\dfrac 34\) 的人戴帽子，所以总人数必须是 \(4\) 的倍数。因此总人数必须是 \(20\) 的倍数。

另外，由容斥原理可以写出下面的关系：同时戴两者的比例 = 戴手套的比例 + 戴帽子的比例 − 至少戴其中一种的比例。

所求比例等于 \(\dfrac{2}{5} + \dfrac 34 -\) 至少戴一种物品的比例。要使同时戴两者的人数最少，就让至少戴一种物品的比例尽可能大，最多为 \(1\)。所以同时戴两者的比例至少为 \(\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}\)。

取最小的正总人数 \(20\)，在这 \(20\) 人中，同时戴两者的人数为 \(3\)。

所以正确答案是 A。

English

Since our room has \(\dfrac 25\) of the people wearing gloves, the number of people must be a multiple of \(5.\) Since our room has \(\dfrac 34\) of the people wearing hats, the number of people must be a multiple of \(4.\) Therefore, the people in the room must be a multiple of \(20.\)

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to \( \dfrac{2}{5} + \dfrac 34 -\) Fraction of people who wear either. If we wish to minimize the number who wear both, we maximize the fraction of people who wear either, up to \(1.\) Therefore, the fraction of people that wear both is \(\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.\)

Since our number is a (positive) multiple of \(20,\) we have the number of people wearing both as \(3\) if we choose to have just \(20\) people.

Therefore, A is the correct answer.

21 答案 C

第三天后剩下 \(62\) 页。在读最后额外的 \(18\) 页之前，她还剩 \(80\) 页。这是当时剩余页数的 \(\dfrac 23\)，所以第三天开始前她剩下 \(120\) 页。

在读第二天额外的 \(15\) 页之前，她还剩 \(120+15=135\) 页。这是当时剩余页数的 \(\dfrac 34\)，所以第二天开始前她剩下 \(180\) 页。

在读第一天额外的 \(12\) 页之前，她还剩 \(180+12=192\) 页。这是当时剩余页数的 \(\dfrac 45\)，所以第一天开始前她有 \(240\) 页，也就是全书 \(240\) 页。

所以正确答案是 C。

English

The pages left after the third day is \(62.\) Before reading the last \(18\) pages, she had \(80\) pages left. This is \(\dfrac 23\) of the pages remaining, so she had \(120\) pages left before the third day.

Before reading the \(15\) pages, she had \(120+15=135\) pages left. This is \(\dfrac 34\) of the pages remaining, so she had \(180\) pages left before the second day.

Before reading the \(12\) pages, she had \(180+12=192\) pages left. This is \(\dfrac 45\) of the pages remaining, so she had \(240\) pages left before the first day, making the book \(240\) pages.

Thus, the answer is C .

22 答案 E

设个位数字为 \(u\)，十位数字为 \(t\)。则百位数字为 \(u+2\)。原数为 \[100(u+2) +10t+u\] \[=101u+200+10t\text{，}\]倒序数为 \[100u +10t+(u+2)\] \[=101u+2+10t\text{。}\]两数之差为 \[(101u+10t+200)-\] \[(101u+10t+2)=198\text{。}\]因此结果的个位数字是 \(8\)。

所以正确答案是 E。

English

Let the units digit be \(u,\) and let the tens digit be \(t.\) This makes the hundreds digit be \(u+2.\) This makes the number equal to \[100(u+2) +10t+u\]\[=101u+200+10t\] and the reversed number is \[100u +10t+(u+2)\]\[=101u+2+10t.\] This makes the difference equal to \[(101u+10t+200)-\]\[(101u+10t+2)=198.\] This makes the units digit \(8.\)

Therefore, the answer is E .

23 答案 B

每个半圆的面积是 \(\pi \dfrac {r^2}{2} \)。它们的半径都是 \(1\)，所以两个半圆的总面积为 \(\pi \frac{1}{2} + \pi \frac 12 = \pi\)。

大圆半径等于 \(OQ\) 的长度，即 \(\sqrt{1^2+1^2} = \sqrt 2\)。它的面积为 \(\pi r^2 = \pi(\sqrt{2})^2 = 2\pi\)。

因此比值为 \(\dfrac{\pi}{2\pi} = \frac 12\)。

所以正确答案是 B。

English

The area of each of the semicircles is \(\pi \dfrac {r^2}{2} .\) Each of them has a radius of \(1,\) so their combined area is \(\pi \frac{1}{2} + \pi \frac 12 = \pi.\)

Next, the radius of the larger circle is equal to the length of \(OQ,\) which is equal to \( \sqrt{1^2+1^2} = \sqrt 2.\) Its area is \(\pi r^2 = \pi(\sqrt{2})^2 = 2\pi.\)

This means the ratio is \(\dfrac{\pi}{2\pi} = \frac 12.\)

Thus, the answer is B .

24 答案 A

首先，\[2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8\text{。}\]

接着，\[10^8 = (2^8)(5^8) = \] \[(4^4)(5^8) < (5^4)(5^8) = 5^{12}\text{。}\]因此 \(2^{24} < 10^8 < 5^{12} \)。

所以正确答案是 A。

English

First, we get \[2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.\]

Next, we get \[10^8 = (2^8)(5^8) = \]\[(4^4)(5^8) < (5^4)(5^8) = 5^{12}.\] This means \(2^{24} < 10^8 < 5^{12} .\)

Thus, the answer is A .

25 答案 E

设 \(w_n\) 为爬 \(n\) 级楼梯的方法数。对于 \(n\ge4\)，第一步可以爬 \(1\)、\(2\) 或 \(3\) 级，所以 \(w_n=w_{n-1}+w_{n-2}+w_{n-3}\)。

有 \(w_1=1\)、\(w_2=2\)、\(w_3=4\)。因此 \(w_4=4+2+1=7\)、\(w_5=7+4+2=13\)、\(w_6=13+7+4=24\)。

所以正确答案是 E。

English

Let \(w_n\) be the number of ways to climb \(n\) stairs. For \(n\ge4\), the first step can be \(1\), \(2\), or \(3\) stairs, so \(w_n=w_{n-1}+w_{n-2}+w_{n-3}\).

We have \(w_1=1\), \(w_2=2\), and \(w_3=4\). Therefore \(w_4=4+2+1=7\), \(w_5=7+4+2=13\), and \(w_6=13+7+4=24\).

Thus, the answer is E .