2013 AMC 10B — Official Competition Problems (February 2013)
📅 2013 B 年11月📝 25题选择题⏱ 40分钟🎯 满分25分✅ 含解题思路👥 612 人已练习
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1
第 1 题
综合
What is \frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6} ?
💡 解题思路
This expression is equivalent to $\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}$
2
第 2 题
几何·面积
Mr. Green measures his rectangular garden by walking two of the sides and finds that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
💡 解题思路
Since each step is $2$ feet, his garden is $30$ by $40$ feet. Thus, the area of $30(40) = 1200$ square feet. Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoe
3
第 3 题
统计
On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3 . In degrees, what was the low temperature in Lincoln that day?
💡 解题思路
Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{\textbf{(C)} \ -5}$
4
第 4 题
计数
When counting from 3 to 201 , 53 is the 51^{st} number counted. When counting backwards from 201 to 3 , 53 is the n^{th} number counted. What is n ?
💡 解题思路
Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{\textbf{(D)}\ 149}$
5
第 5 题
整数运算
Positive integers a and b are each less than 6 . What is the smallest possible value for 2 · a - a · b ? (A)\ -20 {(B)}\ -15 {(C)}\ -10 {(D)}\ 0 {(E)}\ 2
💡 解题思路
Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximize
6
第 6 题
统计
The average age of 33 fifth-graders is 11 . The average age of 55 of their parents is 33 . What is the average age of all of these parents and fifth-graders?
💡 解题思路
The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ peo
7
第 7 题
几何·面积
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles
9
第 9 题
数论
Three positive integers are each greater than 1 , have a product of 27000 , and are pairwise relatively prime. What is their sum?
💡 解题思路
The prime factorization of $27000$ is $2^3*3^3*5^3$ . These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{\textbf{(D) }160}$
10
第 10 题
综合
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
💡 解题思路
Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is
11
第 11 题
方程
Real numbers x and y satisfy the equation x^2+y^2=10x-6y-34 . What is x+y ? (A)\ 1 (B)\ 2 (C)\ 3 (D)\ 6 (E)\ 8
💡 解题思路
If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and
12
第 12 题
概率
Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length?
💡 解题思路
In a regular pentagon, there are 5 sides with the same length and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with
13
第 13 题
计数
Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saying ``1" , so Blair follows by saying ``1, 2" . Jo then says ``1, 2, 3" , and so on. What is the 53^{rd} number said?
💡 解题思路
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4
14
第 14 题
综合
Define a\clubsuit b=a^2b-ab^2 . Which of the following describes the set of points (x, y) for which x\clubsuit y=y\clubsuit x ?
💡 解题思路
$x\clubsuit y = x^2y-xy^2$ and $y\clubsuit x = y^2x-yx^2$ . Therefore, we have the equation $x^2y-xy^2 = y^2x-yx^2$ Factoring out a $-1$ gives $x^2y-xy^2 = -(x^2y-xy^2)$ Factoring both sides further,
15
第 15 题
几何·面积
A wire is cut into two pieces, one of length a and the other of length b . The piece of length a is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is \frac{a}{b} ?
💡 解题思路
Using the area formulas for an equilateral triangle $\left(\frac{{s}^{2}\sqrt{3}}{4}\right)$ and regular hexagon $\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)$ with side length $s$ , plugging $\frac{a}{3}$
16
第 16 题
几何·面积
In triangle ABC , medians AD and CE intersect at P , PE=1.5 , PD=2 , and DE=2.5 . What is the area of AEDC ? [图]
💡 解题思路
Let us use mass points: Assign $B$ mass $1$ . Thus, because $E$ is the midpoint of $AB$ , $A$ also has a mass of $1$ . Similarly, $C$ has a mass of $1$ . $D$ and $E$ each have a mass of $2$ because th
17
第 17 题
综合
Alex has 75 red tokens and 75 blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
💡 解题思路
If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then g
18
第 18 题
规律与数列
The number 2013 has the property that its units digit is the sum of its other digits, that is 2+0+1=3 . How many integers less than 2013 but greater than 1000 have this property?
💡 解题思路
We take cases on the thousands digit, which must be either $1$ or $2$ : If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$
19
第 19 题
规律与数列
The real numbers c,b,a form an arithmetic sequence with a ≥ b ≥ c ≥ 0 . The quadratic ax^2+bx+c has exactly one root. What is this root?
💡 解题思路
It is given that $ax^2+bx+c=0$ has 1 real root, so the discriminant is zero, or $b^2=4ac$ .
20
第 20 题
整数运算
The number 2013 is expressed in the form where a_1 \ge a_2 \ge ·s \ge a_m and b_1 \ge b_2 \ge ·s \ge b_n are positive integers and a_1 + b_1 is as small as possible. What is |a_1 - b_1| ?
💡 解题思路
The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is
21
第 21 题
规律与数列
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is N . What is the smallest possible value of N ?
💡 解题思路
Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1}
22
第 22 题
计数
The regular octagon ABCDEFGH has its center at J . Each of the vertices and the center are to be associated with one of the digits 1 through 9 , with each digit used once, in such a way that the sums of the numbers on the lines AJE , BJF , CJG , and DJH are all equal. In how many ways can this be done? [图]
💡 解题思路
First of all, note that $J$ must be $1$ , $5$ , or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, w
23
第 23 题
几何·面积
In triangle ABC , AB=13 , BC=14 , and CA=15 . Distinct points D , E , and F lie on segments \overline{BC} , \overline{CA} , and \overline{DE} , respectively, such that \overline{AD}\perp\overline{BC} , \overline{DE}\perp\overline{AC} , and \overline{AF}\perp\overline{BF} . The length of segment \overline{DF} can be written as \frac{m}{n} , where m and n are relatively prime positive integers. What is m+n ? [图]
💡 解题思路
Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE
24
第 24 题
规律与数列
A positive integer n is nice if there is a positive integer m with exactly four positive divisors (including 1 and m ) such that the sum of the four divisors is equal to n . How many numbers in the set \{ 2010,2011,2012,\dotsc,2019 \} are nice?
💡 解题思路
A positive integer with only four positive divisors has its prime factorization in the form of $a \cdot b$ , where $a$ and $b$ are both prime positive integers or $c^3$ where $c$ is a prime. One can e
25
第 25 题
计数
Bernardo chooses a three-digit positive integer N and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer S . For example, if N = 749 , Bernardo writes the numbers 10,\!444 and 3,\!245 , and LeRoy obtains the sum S = 13,\!689 . For how many choices of N are the two rightmost digits of S , in order, the same as those of 2N ?
💡 解题思路
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.