2001 AMC 10 Problems

1. The median of the list $n; n + 3; n + 4; n + 5; n + 6; n + 8; n + 10; n + 12; n + 15$ is 10. What is the mean?

$\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 11$

2. A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$\mathrm{(A)}\ -4\leq x\leq -2 \qquad\mathrm{(B)}\ -2<x\leq 0 \qquad\mathrm{(C)}\ 0<x\leq 2$

$\mathrm{(D)}\ 2<x\leq 4 \qquad\mathrm{(E)}\ 4<x\leq 6$

3. The sum of two numbers is $S$ . Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$\mathrm{(A)}\ 2S+3 \qquad\mathrm{(B)}\ 3S+2 \qquad\mathrm{(C)}\ 3S+6 \qquad\mathrm{(D)}\ 2S+6 \qquad\mathrm{(E)}\ 2S+12$

4. What is the maximum number for the possible points of intersection of a circle and a triangle?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 5 \qquad\mathrm{(E)}\ 6$

Solutions

1. The median is $n+6$ , therefore $n=4$ . Computation shows that the sum of all numbers is $99$ and thus the mean is $99/9=11$ .

2. The reciprocal of $x$ is $\frac 1x$ and the additive inverse is $-x$ . (Note that $x$ must be non-zero to have a reciprocal.) The product of these two is $\frac 1x \cdot (-x) = -1$ . Thus $x$ is $2$ more than $-1$ . Therefore $x=1$ .

3. The original two numbers are $x$ and $y$ , with $x+y=S$ . The new two numbers are $2(x+3)$ and $2(y+3)$ . Their sum is $2(x+3)+2(y+3)=2x+2y+12=2(x+y)+12 = 2S+12$ .

4. Each side of the triangle can only intersect the circle twice, so the maximum is at most 6. This can be achieved: $[asy] unitsize(0.3cm); draw( circle((0,-0.2),2.2) ); draw( (-2,-2)--(2,-2)--(0,3)--cycle ); [/asy]$

Page

Toolbox

Search

2001 AMC 10 Problems

Solutions