首页> 重点归纳 > AMC数学竞赛真题2016年B 13-14

AMC数学竞赛真题2016年B 13-14

2018-11-28 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 13

At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$

Solution

We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets.

$amc10数学竞赛报名$

Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields

$\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}$

Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$ , but rather $4c$ . Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$ , leaving us with the number of babies born as quadruplets equal to $\boxed{\textbf{(D)}\ 100}$ .

Alternative Solution

Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\cdot2=24x$ twins, $4x\cdot3=12x$ triplets, and $x\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\frac{1}{10}(1000)=\boxed{\textbf{(D) }100}$

Problem 14

How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$

$美国数学竞赛amc网站$

Solution 1

The region is a right triangle which contains the following lattice points: $(0,0); (1,0)-(1,3); (2,0)-(2,6); (3,0)-(3,9); (4,0)-(4,12); (5,0)-(5,15)$

$amc数学竞赛报名官网$

Squares $1\times 1$ : Suppose that the top-right corner is $(x,y)$ , with $2\le x\le 5$ . Then to include all other corners, we need $1\le y\le 3(x-1)$ . This produces $3+6+9+12=30$ squares.

Squares $2\times 2$ : Here $3\le x\le 5$ . To include all other corners, we need $2\le y\le 3(x-2)$ . This produces $2+5+8=15$ squares.

Squares $3\times 3$ : Similarly this produces $5$ squares.

No other squares will fit in the region. Therefore the answer is $\boxed{\textbf{(D) }50}$ .

Solution 2

The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ . This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of 1x1, 2x2, 3x3, squares and getting 30, 15, and 5 respectively, and we end up with $\boxed{\textbf{(D)}\ 50}$ .