https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=T90bag&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:09:31ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems&diff=365892005 AMC 10B Problems2011-02-06T05:15:49Z<p>T90bag: /* Problem 20 */</p>
<hr />
<div>== Problem 1 ==<br />
A scout troop buys <math>1000</math> candy bars at a price of five for <math>\</math><math>2</math>. They sell all the candy bars at a price of two for <math>\</math><math>1</math>. What was the profit, in dollars?<br />
<br />
<math>\mathrm{(A)} 100 \qquad \mathrm{(B)} 200 \qquad \mathrm{(C)} 300 \qquad \mathrm{(D)} 400 \qquad \mathrm{(E)} 500</math><br />
<br />
[[2005 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
A positive number <math>x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>?<br />
<br />
<math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 4 \qquad \mathrm{(C)} 10 \qquad \mathrm{(D)} 20 \qquad \mathrm{(E)} 40</math><br />
<br />
[[2005 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?<br />
<br />
<math>\mathrm{(A)} \frac{1}{10} \qquad \mathrm{(B)} \frac{1}{9} \qquad \mathrm{(C)} \frac{1}{3} \qquad \mathrm{(D)} \frac{4}{9} \qquad \mathrm{(E)} \frac{5}{9} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
For real numbers <math>a</math> and <math>b</math>, define <math>a \diamond b = \sqrt{a^2 + b^2}</math>. What is the value of<br />
<br />
<math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>?<br />
<br />
<math>\mathrm{(A)} 0 \qquad \mathrm{(B)} \frac{17}{2} \qquad \mathrm{(C)} 13 \qquad \mathrm{(D)} 13\sqrt{2} \qquad \mathrm{(E)} 26</math><br />
<br />
[[2005 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?<br />
<br />
<math>\mathrm{(A)} \frac{1}{5} \qquad \mathrm{(B)} \frac{1}{3} \qquad \mathrm{(C)} \frac{2}{5} \qquad \mathrm{(D)} \frac{2}{3} \qquad \mathrm{(E)} \frac{4}{5} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
At the beginning of the school year, Lisa's goal was to earn an A on at least <math>80\%</math> of her <math>50</math> quizzes for the year. She earned an A on <math>22</math> of the first <math>30</math> quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?<br />
<br />
<math>\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 3 \qquad \mathrm{(D)} 4 \qquad \mathrm{(E)} 5</math><br />
<br />
[[2005 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?<br />
<br />
<math>\mathrm{(A)} \frac{\pi}{16} \qquad \mathrm{(B)} \frac{\pi}{8} \qquad \mathrm{(C)} \frac{3\pi}{16} \qquad \mathrm{(D)} \frac{\pi}{4} \qquad \mathrm{(E)} \frac{\pi}{2} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
[[2005 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
One fair die has faces <math>1</math>, <math>1</math>, <math>2</math>, <math>2</math>, <math>3</math>, <math>3</math> and another has faces <math>4</math>, <math>4</math>, <math>5</math>, <math>5</math>, <math>6</math>, <math>6</math>. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?<br />
<br />
<math>\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
In <math>\triangle ABC</math>, we have <math>AC = BC = 7</math> and <math>AB = 2</math>. Suppose that <math>D</math> is a point on line <math>AB</math> such that <math>B</math> lies between <math>A</math> and <math>D</math> and <math>CD = 8</math>. What is <math>BD</math>?<br />
<br />
<math>\mathrm{(A)} 3 \qquad \mathrm{(B)} 2\sqrt{3} \qquad \mathrm{(C)} 4 \qquad \mathrm{(D)} 5 \qquad \mathrm{(E)} 4\sqrt{2} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
The first term of a sequence is <math>2005</math>. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the <math>{2005}^{th}</math> term of the sequence?<br />
<br />
<math>\mathrm{(A)} 29 \qquad \mathrm{(B)} 55 \qquad \mathrm{(C)} 85 \qquad \mathrm{(D)} 133 \qquad \mathrm{(E)} 250 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?<br />
<br />
<math>\mathrm{(A)} \left(\frac{1}{12}\right)^{12} \qquad \mathrm{(B)} \left(\frac{1}{6}\right)^{12} \qquad \mathrm{(C)} 2\left(\frac{1}{6}\right)^{11} \qquad \mathrm{(D)} \frac{5}{2}\left(\frac{1}{6}\right)^{11} \qquad \mathrm{(E)} \left(\frac{1}{6}\right)^{10} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
How many numbers between <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>?<br />
<br />
<math>\mathrm{(A)} 501 \qquad \mathrm{(B)} 668 \qquad \mathrm{(C)} 835 \qquad \mathrm{(D)} 1002 \qquad \mathrm{(E)} 1169 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
[[2005 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
An envelope contains eight bills: <math>2</math> ones, <math>2</math> fives, <math>2</math> tens, and <math>2</math> twenties. Two bills are drawn at random without replacement. What is the probability that their sum is <math>\</math><math>20</math> or more?<br />
<br />
<math>\mathrm{(A)} \frac{1}{4} \qquad \mathrm{(B)} \frac{2}{5} \qquad \mathrm{(C)} \frac{3}{7} \qquad \mathrm{(D)} \frac{1}{2} \qquad \mathrm{(E)} \frac{2}{3} </math><br />
<br />
[[2005 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
The quadratic equation <math>x^2 + mx + n = 0</math> has roots that are twice those of <math>x^2 + px + m = 0</math>, and none of <math>m</math>, <math>n</math>, and <math>p</math> is zero. What is the value of <math>n/p</math>?<br />
<br />
<math>\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 4 \qquad \mathrm{(D)} 8 \qquad \mathrm{(E)} 16 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
Suppose that <math>4^a = 5</math>, <math>5^b = 6</math>, <math>6^c = 7</math>, and <math>7^d = 8</math>. What is <math>a * b * c * d</math>?<br />
<br />
<math>\mathrm{(A)} 1 \qquad \mathrm{(B)} \frac{3}{2} \qquad \mathrm{(C)} 2 \qquad \mathrm{(D)} \frac{5}{2} \qquad \mathrm{(E)} 3 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
All of David's telephone numbers have the form <math>555-abc-defg</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, <math>f</math>, and <math>g</math> are distinct digits and in increasing order, and none is either <math>0</math> or <math>1</math>. How many different telephone numbers can David have?<br />
<br />
<math>\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 7 \qquad \mathrm{(D)} 8 \qquad \mathrm{(E)} 9 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
On a certain math exam, <math>10\%</math> of the students got <math>70</math> points, <math>25\%</math> got <math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> points. What is the difference between the mean and the median score on this exam?<br />
<br />
<math>\mathrm{(A)} 0 \qquad \mathrm{(B)} 1 \qquad \mathrm{(C)} 2 \qquad \mathrm{(D)} 4 \qquad \mathrm{(E)} 5 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
What is the average (mean) of all <math>5</math>-digit numbers that can be formed by using each of the digits <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, and <math>8</math> exactly once?<br />
<br />
<math>\mathrm{(A)} 48000 \qquad \mathrm{(B)} 49999.5 \qquad \mathrm{(C)} 53332.8 \qquad \mathrm{(D)} 55555 \qquad \mathrm{(E)} 56432.8 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
Forty slips are placed into a hat, each bearing a number <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, or <math>10</math>, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let <math>p</math> be the probability that all four slips bear the same number. Let <math>q</math> be the probability that two of the slips bear a number <math>a</math> and the other two bear a number <math>b \neq a</math>. What is the value of <math>q/p</math>?<br />
<br />
<math>\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
For how many positive integers <math>n</math> less than or equal to <math>24</math> is <math>n!</math> evenly divisible by <math>1 + 2 + \ldots + n</math>?<br />
<br />
<math>\mathrm{(A)} 8 \qquad \mathrm{(B)} 12 \qquad \mathrm{(C)} 16 \qquad \mathrm{(D)} 17 \qquad \mathrm{(E)} 21 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>?<br />
<br />
<math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
Let <math>x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits<br />
of <math>x</math>. The integers <math>x</math> and <math>y</math> satisfy <math>x^2 - y^2 = m^2</math> for some positive integer <math>m</math>.<br />
What is <math>x + y + m</math>?<br />
<br />
<math>\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
A subset <math>B</math> of the set of integers from <math>1</math> to <math>100</math>, inclusive, has the property that no two elements of <math>B</math> sum to <math>125</math>. What is the maximum possible number of elements in <math>B</math>?<br />
<br />
<math>\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 </math><br />
<br />
[[2005 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2005 AMC 10B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=49 2005 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_8&diff=361982010 AMC 10A Problems/Problem 82010-12-20T20:59:47Z<p>T90bag: </p>
<hr />
<div>== Problem 8 ==<br />
Tony works <math>2</math> hours a day and is paid &#36;<math>0.50</math> per hour for each full year of his age. During a six month period Tony worked <math>50</math> days and earned &#36;<math>630</math>. How old was Tony at the end of the six month period?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ 12<br />
\qquad<br />
\mathrm{(D)}\ 13<br />
\qquad<br />
\mathrm{(E)}\ 14<br />
</math><br />
<br />
==Solution==<br />
<br />
Tony worked <math>2</math> hours a day and is paid <math>0.50</math> dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is <math>12</math> years old, he gets <math>12</math> dollars a day. We also know that he worked <math>50</math> days and earned <math>630</math> dollars. If he was <math>12</math> years old at the beginning of his working period, he would have earned <math>12 * 50 = 600</math> dollars. If he was <math>13</math> years old at the beginning of his working period, he would have earned <math>13 * 50 = 650</math> dollars. Because he earned <math>630</math> dollars, we know that he was <math>13</math> for some period of time, but not the whole time, because then the money earned would be greater than or equal to <math>650</math>. This is why he was <math>12</math> when he began, but turned <math>13</math> sometime in the middle and earned <math>630</math> dollars in total. So the answer is <math>13</math>.The answer is <math>\boxed{D}</math>. We could find out for how long he was <math>12</math> and <math>13</math>. <math>12 \cdot x + 13 \cdot (50-x) = 630</math>. Then <math>x</math> is <math>20</math> and we know that he was <math>12</math> for <math>20</math> days, and <math>13</math> for <math>30</math> days. Thus, the answer is <math>13</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_7&diff=361972010 AMC 10A Problems/Problem 72010-12-20T20:54:19Z<p>T90bag: </p>
<hr />
<div>== Problem 7 ==<br />
Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ \sqrt{2}<br />
\qquad<br />
\mathrm{(C)}\ \sqrt{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt{2}<br />
</math><br />
<br />
==Solution==<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Crystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is <math>\sqrt{2}</math> miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to <math>\sqrt{((\sqrt{2})^2+1^2)}</math>, which is equal to <math>\sqrt{3}</math>. The answer is <math>\boxed{C}</math></div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_6&diff=361962010 AMC 10A Problems/Problem 62010-12-20T20:50:53Z<p>T90bag: </p>
<hr />
<div>== Problem 6 ==<br />
For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit(x, y)</math> is defined as <br />
<br />
<cmath>\spadesuit(x, y) = x -\dfrac{1}{y}</cmath><br />
<br />
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{2}{3}<br />
\qquad<br />
\mathrm{(B)}\ 1<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(D)}\ \dfrac{5}{3}<br />
\qquad<br />
\mathrm{(E)}\ 2<br />
</math><br />
<br />
==Solution==<br />
<math>\spadesuit(2, 2) = 2 - \frac{1}{2} = \frac{3}{2}</math>. Then, <math>\spadesuit(2, \frac{3}{2})</math> is <math>2-\frac{1}{\frac{3}{2}} = 2- \frac{2}{3} = \frac{4}{3}</math><br />
The answer is <math>\boxed{C}</math></div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_5&diff=361952010 AMC 10A Problems/Problem 52010-12-20T20:46:05Z<p>T90bag: </p>
<hr />
<div>== Problem 5 ==<br />
The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 24<br />
\qquad<br />
\mathrm{(D)}\ 36<br />
\qquad<br />
\mathrm{(E)}\ 144<br />
</math><br />
<br />
==Solution==<br />
<br />
If the circumference of a circle is <math>24\pi</math>, the radius would be <math>12</math>. Since the area of a circle is <math>\pi r^2</math>, the area is <math>144\pi</math>. The answer is <math>\boxed{E}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_5&diff=361942010 AMC 10A Problems/Problem 52010-12-20T20:45:48Z<p>T90bag: </p>
<hr />
<div>== Problem 5 ==<br />
The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 24<br />
\qquad<br />
\mathrm{(D)}\ 36<br />
\qquad<br />
\mathrm{(E)}\ 144<br />
</math><br />
<br />
==Solution==<br />
<br />
If the circumference of a circle is <math>24\pi</math>, the radius would be <math>12</math>. Since the area of a circle is <math>\pi r^2</math>, the area is <math>144\pi</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_4&diff=361932010 AMC 10A Problems/Problem 42010-12-20T20:43:16Z<p>T90bag: </p>
<hr />
<div>== Problem 4 ==<br />
A book that is to be recorded onto compact discs takes <math>412</math> minutes to read aloud. Each disc can hold up to <math>56</math> minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?<br />
<br />
<math><br />
\mathrm{(A)}\ 50.2<br />
\qquad<br />
\mathrm{(B)}\ 51.5<br />
\qquad<br />
\mathrm{(C)}\ 52.4<br />
\qquad<br />
\mathrm{(D)}\ 53.8<br />
\qquad<br />
\mathrm{(E)}\ 55.2<br />
</math><br />
<br />
==Solution==<br />
<br />
Assuming that there were fractions of compact discs, it would take <math>412/56 ~= 7.357</math> CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have <math>412/8 = 51.5</math> minutes on each of the 8 discs. The answer is <math>\boxed{B}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_3&diff=361922010 AMC 10A Problems/Problem 32010-12-20T20:38:41Z<p>T90bag: </p>
<hr />
<div>== Problem 3 ==<br />
Tyrone had <math>97</math> marbles and Eric had <math>11</math> marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 13<br />
\qquad<br />
\mathrm{(C)}\ 18<br />
\qquad<br />
\mathrm{(D)}\ 25<br />
\qquad<br />
\mathrm{(E)}\ 29<br />
</math><br />
<br />
==Solution==<br />
<br />
Let <math>x</math> be the number of marbles Tyrone gave to Eric. Then, <math>97-x = 2\cdot(11+x)</math>. Solving for <math>x</math> yields <math>75=3x</math> and <math>x = 25</math>. The answer is <math>\boxed{D}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_2&diff=361912010 AMC 10A Problems/Problem 22010-12-20T20:35:02Z<p>T90bag: </p>
<hr />
<div>== Problem 2 ==<br />
<br />
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? <br />
<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br />
draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);<br />
draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);<br />
draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);<br />
draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);<br />
<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{3}{2}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
==Solution==<br />
<br />
Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_1&diff=361902010 AMC 10A Problems/Problem 12010-12-20T20:28:00Z<p>T90bag: </p>
<hr />
<div>== Problem 1 ==<br />
Mary’s top book shelf holds five books with the following widths, in centimeters: <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>. What is the average book width, in centimeters?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 2<br />
\qquad<br />
\mathrm{(C)}\ 3<br />
\qquad<br />
\mathrm{(D)}\ 4<br />
\qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
==Solution==<br />
<br />
<br />
<br />
To find the average, we add up the widths <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>, to get a total sum of <math>20</math>. Since there are <math>5</math> books, the average book width is <math>\frac{20}{5}=4</math> The answer is <math>\boxed{D}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_1&diff=361892010 AMC 10A Problems/Problem 12010-12-20T20:27:11Z<p>T90bag: </p>
<hr />
<div>==Solution==<br />
<br />
<br />
<br />
To find the average, we add up the widths <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>, to get a total sum of <math>20</math>. Since there are <math>5</math> books, the average book width is <math>\frac{20}{5}=4</math> The answer is <math>\boxed{D}</math>.</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems&diff=361882010 AMC 10A Problems2010-12-20T20:26:31Z<p>T90bag: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
Mary’s top book shelf holds five books with the following widths, in centimeters: <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>. What is the average book width, in centimeters?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 2<br />
\qquad<br />
\mathrm{(C)}\ 3<br />
\qquad<br />
\mathrm{(D)}\ 4<br />
\qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? <br />
<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br />
draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);<br />
draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);<br />
draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);<br />
draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);<br />
<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{3}{2}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Tyrone had <math>97</math> marbles and Eric had <math>11</math> marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 13<br />
\qquad<br />
\mathrm{(C)}\ 18<br />
\qquad<br />
\mathrm{(D)}\ 25<br />
\qquad<br />
\mathrm{(E)}\ 29<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A book that is to be recorded onto compact discs takes <math>412</math> minutes to read aloud. Each disc can hold up to <math>56</math> minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?<br />
<br />
<math><br />
\mathrm{(A)}\ 50.2<br />
\qquad<br />
\mathrm{(B)}\ 51.5<br />
\qquad<br />
\mathrm{(C)}\ 52.4<br />
\qquad<br />
\mathrm{(D)}\ 53.8<br />
\qquad<br />
\mathrm{(E)}\ 55.2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 24<br />
\qquad<br />
\mathrm{(D)}\ 36<br />
\qquad<br />
\mathrm{(E)}\ 144<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit(x, y)</math> is defined as <br />
<br />
<cmath>\spadesuit(x, y) = x -\dfrac{1}{y}</cmath><br />
<br />
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{2}{3}<br />
\qquad<br />
\mathrm{(B)}\ 1<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(D)}\ \dfrac{5}{3}<br />
\qquad<br />
\mathrm{(E)}\ 2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ \sqrt{2}<br />
\qquad<br />
\mathrm{(C)}\ \sqrt{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt{2}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Tony works <math>2</math> hours a day and is paid &#36;<math>0.50</math> per hour for each full year of his age. During a six month period Tony worked <math>50</math> days and earned &#36;<math>630</math>. How old was Tony at the end of the six month period?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ 12<br />
\qquad<br />
\mathrm{(D)}\ 13<br />
\qquad<br />
\mathrm{(E)}\ 14<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A <i>palindrome</i>, such as <math>83438</math>, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 21<br />
\qquad<br />
\mathrm{(C)}\ 22<br />
\qquad<br />
\mathrm{(D)}\ 23<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Marvin had a birthday on Tuesday, May 27 in the leap year <math>2008</math>. In what year will his birthday next fall on a Saturday?<br />
<br />
<math><br />
\mathrm{(A)}\ 2011<br />
\qquad<br />
\mathrm{(B)}\ 2012<br />
\qquad<br />
\mathrm{(C)}\ 2013<br />
\qquad<br />
\mathrm{(D)}\ 2015<br />
\qquad<br />
\mathrm{(E)}\ 2017<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 20<br />
\qquad<br />
\mathrm{(E)}\ 30<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?<br />
<br />
<math>\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Angelina drove at an average rate of <math>80</math> kph and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kph. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 80t + 100(\frac{8}{3} -t) = 250<br />
\qquad<br />
\mathrm{(B)}\ 80t = 250 <br />
\qquad<br />
\mathrm{(C)}\ 100t = 250<br />
</math><br/><math><br />
\mathrm{(D)}\ 90t = 250<br />
\qquad<br />
\mathrm{(E)}\ 80(\frac{8}{3} -t) + 100t = 250<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
[[2010 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br />
<br />
Brian: "Mike and I are different species."<br />
<br />
Chris: "LeRoy is a frog."<br />
<br />
LeRoy: "Chris is a frog."<br />
<br />
Mike: "Of the four of us, at least two are toads."<br />
<br />
How many of these amphibians are frogs?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math><br />
<br />
[[2010 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
A solid cube has side length <math>3</math> inches. A <math>2</math>-inch by <math>2</math>-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
[[2010 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Bernardo randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8,9\}</math> and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8\}</math> and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br />
<br />
<math>\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}</math><br />
<br />
[[2010 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A fly trapped inside a cubical box with side length <math>1</math> meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?<br />
<br />
<math><br />
\mathrm{(A)}\ 4+4\sqrt{2}<br />
\qquad<br />
\mathrm{(B)}\ 2+4\sqrt{2}+2\sqrt{3}<br />
\qquad<br />
\mathrm{(C)}\ 2+3\sqrt{2}+3\sqrt{3}<br />
</math><br/><br />
<math><br />
\mathrm{(D)}\ 4\sqrt{2}+4\sqrt{3}<br />
\qquad<br />
\mathrm{(E)}\ 3\sqrt{2}+5\sqrt{3}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
The polynomial <math>x^3 -ax^2 + bx -2010</math> has three positive integer zeros. What is the smallest possible value of <math>a</math>?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 78<br />
\qquad<br />
\mathrm{(B)}\ 88<br />
\qquad<br />
\mathrm{(C)}\ 98<br />
\qquad<br />
\mathrm{(D)}\ 108<br />
\qquad<br />
\mathrm{(E)}\ 118<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math><br />
\mathrm{(A)}\ 28<br />
\qquad<br />
\mathrm{(B)}\ 56<br />
\qquad<br />
\mathrm{(C)}\ 70<br />
\qquad<br />
\mathrm{(D)}\ 84<br />
\qquad<br />
\mathrm{(E)}\ 140<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
Each of 2010 boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math><br />
<br />
[[2010 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math><br />
<br />
[[2010 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Jim starts with a positive integer <math>n</math> and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with <math>n = 55</math>, then his sequence contains <math>5</math> numbers:<br />
<br />
<br />
<cmath>\begin{array}{ccccc}<br />
{}&{}&{}&{}&55\\<br />
55&-&7^2&=&6\\<br />
6&-&2^2&=&2\\<br />
2&-&1^2&=&1\\<br />
1&-&1^2&=&0\\<br />
\end{array}</cmath><br />
<br />
Let <math>N</math> be the smallest number for which Jim’s sequence has <math>8</math> numbers. What is the units digit of <math>N</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 5<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 25|Solution]]</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems&diff=361872010 AMC 10A Problems2010-12-20T20:25:07Z<p>T90bag: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
Mary’s top book shelf holds five books with the following widths, in centimeters: <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>. What is the average book width, in centimeters?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 2<br />
\qquad<br />
\mathrm{(C)}\ 3<br />
\qquad<br />
\mathrm{(D)}\ 4<br />
\qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Problem 2 ==<br />
<br />
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? <br />
<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br />
draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);<br />
draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);<br />
draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);<br />
draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);<br />
<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{3}{2}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Tyrone had <math>97</math> marbles and Eric had <math>11</math> marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 13<br />
\qquad<br />
\mathrm{(C)}\ 18<br />
\qquad<br />
\mathrm{(D)}\ 25<br />
\qquad<br />
\mathrm{(E)}\ 29<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A book that is to be recorded onto compact discs takes <math>412</math> minutes to read aloud. Each disc can hold up to <math>56</math> minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?<br />
<br />
<math><br />
\mathrm{(A)}\ 50.2<br />
\qquad<br />
\mathrm{(B)}\ 51.5<br />
\qquad<br />
\mathrm{(C)}\ 52.4<br />
\qquad<br />
\mathrm{(D)}\ 53.8<br />
\qquad<br />
\mathrm{(E)}\ 55.2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 24<br />
\qquad<br />
\mathrm{(D)}\ 36<br />
\qquad<br />
\mathrm{(E)}\ 144<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit(x, y)</math> is defined as <br />
<br />
<cmath>\spadesuit(x, y) = x -\dfrac{1}{y}</cmath><br />
<br />
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{2}{3}<br />
\qquad<br />
\mathrm{(B)}\ 1<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(D)}\ \dfrac{5}{3}<br />
\qquad<br />
\mathrm{(E)}\ 2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ \sqrt{2}<br />
\qquad<br />
\mathrm{(C)}\ \sqrt{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt{2}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Tony works <math>2</math> hours a day and is paid &#36;<math>0.50</math> per hour for each full year of his age. During a six month period Tony worked <math>50</math> days and earned &#36;<math>630</math>. How old was Tony at the end of the six month period?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ 12<br />
\qquad<br />
\mathrm{(D)}\ 13<br />
\qquad<br />
\mathrm{(E)}\ 14<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A <i>palindrome</i>, such as <math>83438</math>, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 21<br />
\qquad<br />
\mathrm{(C)}\ 22<br />
\qquad<br />
\mathrm{(D)}\ 23<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Marvin had a birthday on Tuesday, May 27 in the leap year <math>2008</math>. In what year will his birthday next fall on a Saturday?<br />
<br />
<math><br />
\mathrm{(A)}\ 2011<br />
\qquad<br />
\mathrm{(B)}\ 2012<br />
\qquad<br />
\mathrm{(C)}\ 2013<br />
\qquad<br />
\mathrm{(D)}\ 2015<br />
\qquad<br />
\mathrm{(E)}\ 2017<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 20<br />
\qquad<br />
\mathrm{(E)}\ 30<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?<br />
<br />
<math>\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Angelina drove at an average rate of <math>80</math> kph and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kph. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 80t + 100(\frac{8}{3} -t) = 250<br />
\qquad<br />
\mathrm{(B)}\ 80t = 250 <br />
\qquad<br />
\mathrm{(C)}\ 100t = 250<br />
</math><br/><math><br />
\mathrm{(D)}\ 90t = 250<br />
\qquad<br />
\mathrm{(E)}\ 80(\frac{8}{3} -t) + 100t = 250<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
[[2010 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br />
<br />
Brian: "Mike and I are different species."<br />
<br />
Chris: "LeRoy is a frog."<br />
<br />
LeRoy: "Chris is a frog."<br />
<br />
Mike: "Of the four of us, at least two are toads."<br />
<br />
How many of these amphibians are frogs?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math><br />
<br />
[[2010 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
A solid cube has side length <math>3</math> inches. A <math>2</math>-inch by <math>2</math>-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
[[2010 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Bernardo randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8,9\}</math> and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8\}</math> and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br />
<br />
<math>\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}</math><br />
<br />
[[2010 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A fly trapped inside a cubical box with side length <math>1</math> meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?<br />
<br />
<math><br />
\mathrm{(A)}\ 4+4\sqrt{2}<br />
\qquad<br />
\mathrm{(B)}\ 2+4\sqrt{2}+2\sqrt{3}<br />
\qquad<br />
\mathrm{(C)}\ 2+3\sqrt{2}+3\sqrt{3}<br />
</math><br/><br />
<math><br />
\mathrm{(D)}\ 4\sqrt{2}+4\sqrt{3}<br />
\qquad<br />
\mathrm{(E)}\ 3\sqrt{2}+5\sqrt{3}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
The polynomial <math>x^3 -ax^2 + bx -2010</math> has three positive integer zeros. What is the smallest possible value of <math>a</math>?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 78<br />
\qquad<br />
\mathrm{(B)}\ 88<br />
\qquad<br />
\mathrm{(C)}\ 98<br />
\qquad<br />
\mathrm{(D)}\ 108<br />
\qquad<br />
\mathrm{(E)}\ 118<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math><br />
\mathrm{(A)}\ 28<br />
\qquad<br />
\mathrm{(B)}\ 56<br />
\qquad<br />
\mathrm{(C)}\ 70<br />
\qquad<br />
\mathrm{(D)}\ 84<br />
\qquad<br />
\mathrm{(E)}\ 140<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
Each of 2010 boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math><br />
<br />
[[2010 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math><br />
<br />
[[2010 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Jim starts with a positive integer <math>n</math> and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with <math>n = 55</math>, then his sequence contains <math>5</math> numbers:<br />
<br />
<br />
<cmath>\begin{array}{ccccc}<br />
{}&{}&{}&{}&55\\<br />
55&-&7^2&=&6\\<br />
6&-&2^2&=&2\\<br />
2&-&1^2&=&1\\<br />
1&-&1^2&=&0\\<br />
\end{array}</cmath><br />
<br />
Let <math>N</math> be the smallest number for which Jim’s sequence has <math>8</math> numbers. What is the units digit of <math>N</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 5<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 25|Solution]]</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems&diff=361862010 AMC 10A Problems2010-12-20T20:23:45Z<p>T90bag: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
Mary’s top book shelf holds five books with the following widths, in centimeters: <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>. What is the average book width, in centimeters?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 2<br />
\qquad<br />
\mathrm{(C)}\ 3<br />
\qquad<br />
\mathrm{(D)}\ 4<br />
\qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
==Solution==<br />
<br />
To find the average, we add up the widths <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>, to get a total sum of <math>20</math>. Since there are <math>5</math> books, the average book width is <math>\frac{20}{5}=4</math> The answer is <math>\boxed{D}</math>.<br />
<br />
== Problem 2 ==<br />
<br />
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? <br />
<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt));<br />
<br />
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br />
draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);<br />
draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);<br />
draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);<br />
draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);<br />
<br />
</asy></center><br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{5}{4}<br />
\qquad<br />
\mathrm{(B)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{3}{2}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 3<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Tyrone had <math>97</math> marbles and Eric had <math>11</math> marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br />
<br />
<math><br />
\mathrm{(A)}\ 3<br />
\qquad<br />
\mathrm{(B)}\ 13<br />
\qquad<br />
\mathrm{(C)}\ 18<br />
\qquad<br />
\mathrm{(D)}\ 25<br />
\qquad<br />
\mathrm{(E)}\ 29<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
A book that is to be recorded onto compact discs takes <math>412</math> minutes to read aloud. Each disc can hold up to <math>56</math> minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?<br />
<br />
<math><br />
\mathrm{(A)}\ 50.2<br />
\qquad<br />
\mathrm{(B)}\ 51.5<br />
\qquad<br />
\mathrm{(C)}\ 52.4<br />
\qquad<br />
\mathrm{(D)}\ 53.8<br />
\qquad<br />
\mathrm{(E)}\ 55.2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The area of a circle whose circumference is <math>24\pi</math> is <math>k\pi</math>. What is the value of <math>k</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 12<br />
\qquad<br />
\mathrm{(C)}\ 24<br />
\qquad<br />
\mathrm{(D)}\ 36<br />
\qquad<br />
\mathrm{(E)}\ 144<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit(x, y)</math> is defined as <br />
<br />
<cmath>\spadesuit(x, y) = x -\dfrac{1}{y}</cmath><br />
<br />
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ \dfrac{2}{3}<br />
\qquad<br />
\mathrm{(B)}\ 1<br />
\qquad<br />
\mathrm{(C)}\ \dfrac{4}{3}<br />
\qquad<br />
\mathrm{(D)}\ \dfrac{5}{3}<br />
\qquad<br />
\mathrm{(E)}\ 2<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ \sqrt{2}<br />
\qquad<br />
\mathrm{(C)}\ \sqrt{3}<br />
\qquad<br />
\mathrm{(D)}\ 2<br />
\qquad<br />
\mathrm{(E)}\ 2\sqrt{2}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Tony works <math>2</math> hours a day and is paid &#36;<math>0.50</math> per hour for each full year of his age. During a six month period Tony worked <math>50</math> days and earned &#36;<math>630</math>. How old was Tony at the end of the six month period?<br />
<br />
<math><br />
\mathrm{(A)}\ 9<br />
\qquad<br />
\mathrm{(B)}\ 11<br />
\qquad<br />
\mathrm{(C)}\ 12<br />
\qquad<br />
\mathrm{(D)}\ 13<br />
\qquad<br />
\mathrm{(E)}\ 14<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A <i>palindrome</i>, such as <math>83438</math>, is a number that remains the same when its digits are reversed. The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 20<br />
\qquad<br />
\mathrm{(B)}\ 21<br />
\qquad<br />
\mathrm{(C)}\ 22<br />
\qquad<br />
\mathrm{(D)}\ 23<br />
\qquad<br />
\mathrm{(E)}\ 24<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Marvin had a birthday on Tuesday, May 27 in the leap year <math>2008</math>. In what year will his birthday next fall on a Saturday?<br />
<br />
<math><br />
\mathrm{(A)}\ 2011<br />
\qquad<br />
\mathrm{(B)}\ 2012<br />
\qquad<br />
\mathrm{(C)}\ 2013<br />
\qquad<br />
\mathrm{(D)}\ 2015<br />
\qquad<br />
\mathrm{(E)}\ 2017<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 6<br />
\qquad<br />
\mathrm{(B)}\ 10<br />
\qquad<br />
\mathrm{(C)}\ 15<br />
\qquad<br />
\mathrm{(D)}\ 20<br />
\qquad<br />
\mathrm{(E)}\ 30<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?<br />
<br />
<math>\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Angelina drove at an average rate of <math>80</math> kph and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kph. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 80t + 100(\frac{8}{3} -t) = 250<br />
\qquad<br />
\mathrm{(B)}\ 80t = 250 <br />
\qquad<br />
\mathrm{(C)}\ 100t = 250<br />
</math><br/><math><br />
\mathrm{(D)}\ 90t = 250<br />
\qquad<br />
\mathrm{(E)}\ 80(\frac{8}{3} -t) + 100t = 250<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?<br />
<br />
<math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math><br />
<br />
[[2010 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br />
<br />
Brian: "Mike and I are different species."<br />
<br />
Chris: "LeRoy is a frog."<br />
<br />
LeRoy: "Chris is a frog."<br />
<br />
Mike: "Of the four of us, at least two are toads."<br />
<br />
How many of these amphibians are frogs?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math><br />
<br />
[[2010 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
Nondegenerate <math>\triangle ABC</math> has integer side lengths, <math>\overline{BD}</math> is an angle bisector, <math>AD = 3</math>, and <math>DC=8</math>. What is the smallest possible value of the perimeter?<br />
<br />
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37</math><br />
<br />
[[2010 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
A solid cube has side length <math>3</math> inches. A <math>2</math>-inch by <math>2</math>-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?<br />
<br />
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15</math><br />
<br />
[[2010 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Bernardo randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8,9\}</math> and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8\}</math> and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br />
<br />
<math>\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}</math><br />
<br />
[[2010 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
[[2010 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
A fly trapped inside a cubical box with side length <math>1</math> meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?<br />
<br />
<math><br />
\mathrm{(A)}\ 4+4\sqrt{2}<br />
\qquad<br />
\mathrm{(B)}\ 2+4\sqrt{2}+2\sqrt{3}<br />
\qquad<br />
\mathrm{(C)}\ 2+3\sqrt{2}+3\sqrt{3}<br />
</math><br/><br />
<math><br />
\mathrm{(D)}\ 4\sqrt{2}+4\sqrt{3}<br />
\qquad<br />
\mathrm{(E)}\ 3\sqrt{2}+5\sqrt{3}<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
The polynomial <math>x^3 -ax^2 + bx -2010</math> has three positive integer zeros. What is the smallest possible value of <math>a</math>?<br />
<br />
<br />
<math><br />
\mathrm{(A)}\ 78<br />
\qquad<br />
\mathrm{(B)}\ 88<br />
\qquad<br />
\mathrm{(C)}\ 98<br />
\qquad<br />
\mathrm{(D)}\ 108<br />
\qquad<br />
\mathrm{(E)}\ 118<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math><br />
\mathrm{(A)}\ 28<br />
\qquad<br />
\mathrm{(B)}\ 56<br />
\qquad<br />
\mathrm{(C)}\ 70<br />
\qquad<br />
\mathrm{(D)}\ 84<br />
\qquad<br />
\mathrm{(E)}\ 140<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
Each of 2010 boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?<br />
<br />
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math><br />
<br />
[[2010 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math><br />
<br />
[[2010 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
Jim starts with a positive integer <math>n</math> and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with <math>n = 55</math>, then his sequence contains <math>5</math> numbers:<br />
<br />
<br />
<cmath>\begin{array}{ccccc}<br />
{}&{}&{}&{}&55\\<br />
55&-&7^2&=&6\\<br />
6&-&2^2&=&2\\<br />
2&-&1^2&=&1\\<br />
1&-&1^2&=&0\\<br />
\end{array}</cmath><br />
<br />
Let <math>N</math> be the smallest number for which Jim’s sequence has <math>8</math> numbers. What is the units digit of <math>N</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 5<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
[[2010 AMC 10A Problems/Problem 25|Solution]]</div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_22&diff=361852010 AMC 10A Problems/Problem 222010-12-20T20:19:59Z<p>T90bag: /* Problem */</p>
<hr />
<div>==Problem==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math><br />
<br />
==Solution==<br />
To choose 3 points on a circle with 8 points, we simply have <math>{{8}\choose{3}}</math> to get the answer <math>\boxed{56}</math></div>T90baghttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_22&diff=361732010 AMC 10A Problems/Problem 222010-12-19T05:47:31Z<p>T90bag: /* Problem */</p>
<hr />
<div>==Problem==<br />
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?<br />
<br />
<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math><br />
<br />
To choose 3 points on a circle with 8 points, we simply have <math>{{8}\choose{3}}</math> to get the answer <math>\boxed{56}</math></div>T90bag