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Difference between revisions of "2001 AMC 10 Problems"
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is 10. What is the mean?
 
is 10. What is the mean?
  
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(A) 4 (B) 6 (C) 7 (D) 10 (E) 11
+
<math>\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 11</math>
  
−
2. A number x is 2 more than the product of its reciprocal and its additive inverse. In which interval does the number lie?
+
2. A number <math>x</math> is <math>2</math> more than the product of its reciprocal and its additive inverse. In which interval does the number lie?
−
(A) ¡4 · x · ¡2 (B) ¡2 < x · 0 (C) 0 < x · 2
 
−
(D) 2 < x · 4 (E) 4 < x · 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?
+
<math>\mathrm{(A)}\ -4\leq x\leq -2 \qquad\mathrm{(B)}\ -2<x\leq 0 \qquad\mathrm{(C)}\ 0<x\leq 2</math>
−
<math>(A) 2S + 3 (B) 3S + 2 (C) 3S + 6 (D) 2S + 6 (E) 2S + 12</math>
+
 
 +
<math>\mathrm{(D)}\ 2<x\leq 4 \qquad\mathrm{(E)}\ 4<x\leq 6</math>
 +
 
 +
3. The sum of two numbers is <math>S</math>. Suppose <math>3</math> is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
 +
 
 +
<math>\mathrm{(A)}\ 2S+3 \qquad\mathrm{(B)}\ 3S+2 \qquad\mathrm{(C)}\ 3S+6 \qquad\mathrm{(D)}\ 2S+6 \qquad\mathrm{(E)}\ 2S+12</math>
  
 
4. What is the maximum number for the possible points of intersection of a circle and a triangle?
 
4. What is the maximum number for the possible points of intersection of a circle and a triangle?
−
<math>(A) 2 (B) 3 (C) 4 (D) 5 (E) 6</math>
+
 
 +
<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 3 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 5 \qquad\mathrm{(E)}\ 6</math>
 +
 
 +
=== Solutions ===
 +
 
 +
1. The median is <math>n+6</math>, therefore <math>n=4</math>. Computation shows that the sum of all numbers is <math>63</math> and thus the mean is <math>63/9=7</math>.
 +
 
 +
2. The reciprocal of <math>x</math> is <math>\frac 1x</math> and the additive inverse is <math>-x</math>. (Note that <math>x</math> must be non-zero to have a reciprocal.)
 +
The product of these two is <math>\frac 1x \cdot (-x) = -1</math>. Thus <math>x</math> is <math>2</math> more than <math>-1</math>. Therefore <math>x=1</math>.
 +
 
 +
3. The original two numbers are <math>x</math> and <math>y</math>, with <math>x+y=S</math>. The new two numbers are <math>2(x+3)</math> and <math>2(y+3)</math>. Their sum is
 +
<math>2(x+3)+2(y+3)=2x+2y+12=2(x+y)+12 = 2S+12</math>.
 +
 
 +
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>

Revision as of 11:10, 24 January 2009

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 $63$ and thus the mean is $63/9=7$.

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]

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