https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Abhinavg0627&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-21T13:39:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=156254 2021 AMC 10B Problems/Problem 18 2021-06-17T23:48:13Z <p>Abhinavg0627: /* Solution 3 (Quicksolve) */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> There is a &lt;math&gt;\frac36&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}{5}&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{1}{20}}.&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> ==Solution 2==<br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> ==Solution 3 ==<br /> <br /> Note that the problem is basically asking us to find the probability that in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;P_n&lt;/math&gt; denote the probability that the first odd number appears on roll &lt;math&gt;n&lt;/math&gt; and all our conditions are met. We now proceed with complementary counting. <br /> <br /> For &lt;math&gt;n \le 3&lt;/math&gt;, it's impossible to have all &lt;math&gt;3&lt;/math&gt; evens appear before an odd. Note that for &lt;math&gt;n \ge 4,&lt;/math&gt; &lt;cmath&gt;P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{1}}{3^{n-1}}\right)&lt;/cmath&gt; since there's a &lt;math&gt;\frac {1}{2^{n}}&lt;/math&gt; chance that the first odd appears on roll &lt;math&gt;n&lt;/math&gt; (disregarding the other conditions) and the other term is subtracting the probability that less than &lt;math&gt;3&lt;/math&gt; of the evens show up before the first odd roll. Simplifying, we arrive at &lt;cmath&gt;P_n= \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \frac {1}{2 \cdot 3^{n-2}} + \frac{1}{2^{n} \cdot 3^{n-2}}.&lt;/cmath&gt;<br /> <br /> Summing for all &lt;math&gt;n&lt;/math&gt;, we get our answer of &lt;cmath&gt;\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \frac {1}{8} - \frac {1}{12} + \frac{1}{120} = \boxed{\textbf{(C) }\frac{1}{20}.}&lt;/cmath&gt;<br /> <br /> ~ike.chen<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt; E_n &lt;/math&gt; be that probability that the condition in the problem is satisfied given that we need &lt;math&gt; n &lt;/math&gt; more distinct even numbers. Then, <br /> &lt;cmath&gt; E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, &lt;/cmath&gt;<br /> since there is a &lt;math&gt; \frac{1}{3} &lt;/math&gt; probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that &lt;math&gt; E_1=\frac{1}{4} &lt;/math&gt;. <br /> <br /> We can apply the same concept for &lt;math&gt; E_2 &lt;/math&gt; and &lt;math&gt; E_3 &lt;/math&gt;. We find that &lt;cmath&gt; E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; and so &lt;math&gt; E_2=\frac{1}{10} &lt;/math&gt;. Also, &lt;cmath&gt; E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; so &lt;math&gt; E_3=\frac{1}{20} &lt;/math&gt;. Since the problem is asking for &lt;math&gt; E_3 &lt;/math&gt;, our answer is &lt;math&gt; \boxed{\textbf{(C) }\frac{1}{20}} &lt;/math&gt;. -BorealBear<br /> <br /> ==Solution 6 (same as solution 1 but with a little more of explanation)==<br /> The probability of choosing an even number on the first turn is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, now since you already chose that number, it is irrelevant to the problem now, so, if you chose the number again, it doesn't really matter to our problem anymore. Now, with our remaining &lt;math&gt;5&lt;/math&gt; numbers, the probability of choosing another even number is &lt;math&gt;\frac{2}{5}&lt;/math&gt;, and again, after you have chosen that number, it is out of our problem. Now, you just have &lt;math&gt;4&lt;/math&gt; numbers left and the probability of choosing the last even number is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, so the answer is &lt;math&gt;\frac{1}{2} \times \frac{2}{5} \times \frac{1}{4}&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\frac{1}{20}&lt;/math&gt;.<br /> <br /> ~math31415926535<br /> <br /> ==Solution 6 (Infinite Geometric Sequence Method)==<br /> Let's say that our even integers are found in the first &lt;math&gt;n&lt;/math&gt; numbers where n must be greater than or equal to &lt;math&gt;3&lt;/math&gt;. Then, we can form an argument based on this. There are &lt;math&gt;3^n&lt;/math&gt; total ways to assign even numbers being &lt;math&gt;(&lt;/math&gt;2&lt;math&gt;, &lt;/math&gt;4&lt;math&gt;, &lt;/math&gt;6&lt;math&gt;)&lt;/math&gt; to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total &lt;math&gt;n&lt;/math&gt; spaces and the case where there are only &lt;math&gt;2&lt;/math&gt; distinct even integers present. There is &lt;math&gt;1&lt;/math&gt; way we can have a single even integer in the entire &lt;math&gt;n&lt;/math&gt; spaces, therefore giving us just &lt;math&gt;1&lt;/math&gt; option for each of the three integers, so we have &lt;math&gt;3&lt;/math&gt; total cases for this. Moreover, the amount of cases with just &lt;math&gt;2&lt;/math&gt; distinct even integers is &lt;math&gt;2^n&lt;/math&gt; but subtract the cases where all of the n spaces is either a single integer giving us &lt;math&gt;2^n-2&lt;/math&gt;, but we multiply by &lt;math&gt;3C2&lt;/math&gt; because of the ways to choose &lt;math&gt;2&lt;/math&gt; distinct even integers that are used in the sequence of &lt;math&gt;n&lt;/math&gt;. Finally, we have &lt;math&gt;\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}&lt;/math&gt; note: we must divide all of this by &lt;math&gt;6^n&lt;/math&gt; for probability. Additionally, over the entire summation, we multiply by &lt;math&gt;1/2&lt;/math&gt; because of the &lt;math&gt;1/2&lt;/math&gt; probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get &lt;math&gt;1/20&lt;/math&gt;. &lt;math&gt;\boxed{\textbf{(C) }\frac{1}{20}.}&lt;/math&gt;<br /> <br /> ~Jske25<br /> <br /> https://www.youtube.com/watch?v=T-kdDFeHzns&amp;t=17s<br /> <br /> ==Solution 7==<br /> The question basically asks for the probability that &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; are all rolled before &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. This probability is equal to the probability that any 3 number combination is rolled before the remaining numbers are rolled (for example, rolling &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; before &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;). There are &lt;math&gt;\binom{6}{3}&lt;/math&gt; combinations possible and summing them all up must result in 1 so the probability of the specific combination &lt;math&gt;2&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; &lt;math&gt;6&lt;/math&gt; being chosen is&lt;math&gt;\boxed{\textbf{(C) }\frac{1}{20}.}&lt;/math&gt;<br /> <br /> ~AwesomeK<br /> <br /> == Video Solution by hurdler (complementary probability) ==<br /> https://www.youtube.com/watch?v=k2Jy4ni9tK8<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/FV9AnyERgJQ?t=480<br /> <br /> ~IceMatrix<br /> ==Video Solution by Interstigation (Simple Bash With PIE)==<br /> (which stands for Principle of Inclusion and Exclusion)<br /> https://youtu.be/2SGmSYZ5bqU<br /> <br /> ~Interstigation<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=146464 2021 AMC 10B Problems/Problem 18 2021-02-13T18:01:19Z <p>Abhinavg0627: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> <br /> There is a &lt;math&gt;\frac{3}{6}&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}{5}&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}{6} * \frac{2}{5} * \frac{1}{4} = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }\frac{1}{20}}.&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> ==Solution 2==<br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> ==Solution 3 (Quicksolve) ==<br /> <br /> Note that the problem is basically asking us to find the probability that in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;P_n&lt;/math&gt; denote the probability that the first odd number appears on roll &lt;math&gt;n&lt;/math&gt; and all our conditions are met. We now proceed with complementary counting. <br /> <br /> For &lt;math&gt;n \le 3&lt;/math&gt;, it's impossible to have all &lt;math&gt;3&lt;/math&gt; evens appear before an odd. Note that for &lt;math&gt;n \ge 4,&lt;/math&gt; &lt;cmath&gt;P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{2}}{3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \left(\frac {1}{2 \cdot 3^{n-2}} - \frac{1}{2^{n} \cdot 3^{n-2}} \right).&lt;/cmath&gt;<br /> <br /> Summing for all &lt;math&gt;n&lt;/math&gt;, we get our answer of &lt;cmath&gt;\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \left (\frac {1}{8} \right) - \left(\frac {1}{12} \right) + \left (\frac{1}{120} \right) = \boxed{\textbf{(C) }\frac{1}{20}.}&lt;/cmath&gt;<br /> <br /> ~ike.chen<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt; E_n &lt;/math&gt; be that probability that the condition in the problem is satisfied given that we need &lt;math&gt; n &lt;/math&gt; more distinct even numbers. Then, <br /> &lt;cmath&gt; E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0, &lt;/cmath&gt;<br /> since there is a &lt;math&gt; \frac{1}{3} &lt;/math&gt; probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that &lt;math&gt; E_1=\frac{1}{4} &lt;/math&gt;. <br /> <br /> We can apply the same concept for &lt;math&gt; E_2 &lt;/math&gt; and &lt;math&gt; E_3 &lt;/math&gt;. We find that &lt;cmath&gt; E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; and so &lt;math&gt; E_2=\frac{1}{10} &lt;/math&gt;. Also, &lt;cmath&gt; E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0, &lt;/cmath&gt; so &lt;math&gt; E_3=\frac{1}{20} &lt;/math&gt;. Since the problem is asking for &lt;math&gt; E_3 &lt;/math&gt;, our answer is &lt;math&gt; \boxed{\textbf{(C) }\frac{1}{20}} &lt;/math&gt;. -BorealBear<br /> <br /> == Video Solution by OmegaLearn (Conditional probability) ==<br /> https://youtu.be/IX-Y38KPxqs<br /> <br /> {{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=146463 2021 AMC 12B Problems/Problem 6 2021-02-13T17:59:17Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12 \mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18 \mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24 \mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> <br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi \cdot r^2 \cdot h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate these two expressions because the water volume stays the same like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=qpvS2PVkI8A&amp;t=509s<br /> <br /> == Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==<br /> https://youtu.be/4JhZLAORb8c<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=VzwxbsuSQ80<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2021|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=146462 2021 AMC 12B Problems/Problem 6 2021-02-13T17:58:58Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12 \mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18 \mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24 \mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> <br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi \cdot r^2 \cdot h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate these two expressions because the water volume stays the same like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=qpvS2PVkI8A&amp;t=509s<br /> <br /> == Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==<br /> https://youtu.be/4JhZLAORb8c<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=VzwxbsuSQ80<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2021|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_17&diff=145834 2021 AMC 12A Problems/Problem 17 2021-02-12T05:34:12Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> Trapezoid &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;\overline{AB}\parallel\overline{CD},BC=CD=43&lt;/math&gt;, and &lt;math&gt;\overline{AD}\perp\overline{BD}&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the intersection of the diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;. Given that &lt;math&gt;OP=11&lt;/math&gt;, the length of &lt;math&gt;AD&lt;/math&gt; can be written in the form &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215&lt;/math&gt;<br /> <br /> == Video Solution (Using Similar Triangles, Pythagorean Theorem) ==<br /> https://youtu.be/gjeSGJy_ld4<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}}<br /> {{AMC10 box|year=2021|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_5&diff=145560 2021 AMC 12B Problems/Problem 5 2021-02-12T01:44:28Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90\deg&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y = -x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b - a ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9&lt;/math&gt;<br /> ==Solution==<br /> <br /> The final image of &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(-6,3)&lt;/math&gt;. We know the reflection rule for reflecting over &lt;math&gt;y=-x&lt;/math&gt; is &lt;math&gt;(x,y) --&gt; (-y, -x)&lt;/math&gt;. So before the reflection and after rotation the point is &lt;math&gt;(-3,6)&lt;/math&gt;.<br /> <br /> By definition of rotation, the slope between &lt;math&gt;(-3,6)&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt; must be perpendicular to the slope between &lt;math&gt;(a,b)&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt;. The first slope is &lt;math&gt;\frac{5-6}{1-(-3)} = \frac{-1}{4}&lt;/math&gt;. This means the slope of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Rotations also preserve distance to the center of rotation, and since we only &quot;travelled&quot; up and down by the slope once to get from &lt;math&gt;(3,-6)&lt;/math&gt; to &lt;math&gt;(1,5)&lt;/math&gt; it follows we shall only use the slope once to travel from &lt;math&gt;(1,5)&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Therefore point &lt;math&gt;P&lt;/math&gt; is located at &lt;math&gt;(1+1, 5+4) = (2,9)&lt;/math&gt;. The answer is &lt;math&gt;9-2 = 7 = \boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> == Video Solution by OmegaLearn (Rotation &amp; Reflection tricks) ==<br /> https://youtu.be/VyRWjgGIsRQ<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_5&diff=145559 2021 AMC 12B Problems/Problem 5 2021-02-12T01:44:14Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90\deg&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y = -x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b - a ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9&lt;/math&gt;<br /> ==Solution==<br /> <br /> The final image of &lt;math&gt;P&lt;/math&gt; is &lt;math&gt;(-6,3)&lt;/math&gt;. We know the reflection rule for reflecting over &lt;math&gt;y=-x&lt;/math&gt; is &lt;math&gt;(x,y) --&gt; (-y, -x)&lt;/math&gt;. So before the reflection and after rotation the point is &lt;math&gt;(-3,6)&lt;/math&gt;.<br /> <br /> By definition of rotation, the slope between &lt;math&gt;(-3,6)&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt; must be perpendicular to the slope between &lt;math&gt;(a,b)&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt;. The first slope is &lt;math&gt;\frac{5-6}{1-(-3)} = \frac{-1}{4}&lt;/math&gt;. This means the slope of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;(1,5)&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Rotations also preserve distance to the center of rotation, and since we only &quot;travelled&quot; up and down by the slope once to get from &lt;math&gt;(3,-6)&lt;/math&gt; to &lt;math&gt;(1,5)&lt;/math&gt; it follows we shall only use the slope once to travel from &lt;math&gt;(1,5)&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Therefore point &lt;math&gt;P&lt;/math&gt; is located at &lt;math&gt;(1+1, 5+4) = (2,9)&lt;/math&gt;. The answer is &lt;math&gt;9-2 = 7 = \boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> == Video Solution by OmegaLearn (Rotation &amp; Reflection tricks) ==<br /> https://youtu.be/VyRWjgGIsRQ<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2021|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=145442 2021 AMC 10B Problems/Problem 18 2021-02-11T23:56:26Z <p>Abhinavg0627: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> There is a &lt;math&gt;\frac{3}6&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}5&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}4&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}6 * \frac{2}5 * \frac{1}4 = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt; \boxed{ C) \frac{1}{20} }&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> <br /> ==Solution 2==<br /> <br /> Note that the problem is basically asking us to find the probability that in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=145438 2021 AMC 10B Problems/Problem 18 2021-02-11T23:55:45Z <p>Abhinavg0627: /* Solution 2 (If you don't have much time) */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> There is a &lt;math&gt;\frac{3}6&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}5&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}4&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}6 * \frac{2}5 * \frac{1}4 = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt; \boxed{ C) \frac{1}{20} }&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> <br /> ==Solution 2==<br /> <br /> Note that the problem is basically asking us to find the probability in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=145433 2021 AMC 10B Problems/Problem 18 2021-02-11T23:51:12Z <p>Abhinavg0627: /* Solution 2 (If you don't have much time) */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> There is a &lt;math&gt;\frac{3}6&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}5&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}4&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}6 * \frac{2}5 * \frac{1}4 = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt; \boxed{ C) \frac{1}{20} }&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> <br /> ==Solution 2 (If you don't have much time)==<br /> <br /> Note that the problem is basically asking us to find the probability in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_18&diff=145432 2021 AMC 10B Problems/Problem 18 2021-02-11T23:50:57Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> There is a &lt;math&gt;\frac{3}6&lt;/math&gt; chance that the first number we choose is even.<br /> <br /> There is a &lt;math&gt;\frac{2}5&lt;/math&gt; chance that the next number that is distinct from the first is even.<br /> <br /> There is a &lt;math&gt;\frac{1}4&lt;/math&gt; chance that the next number distinct from the first two is even.<br /> <br /> &lt;math&gt;\frac{3}6 * \frac{2}5 * \frac{1}4 = \frac{1}{20}&lt;/math&gt;, so the answer is &lt;math&gt; \boxed{ C) \frac{1}{20} }&lt;/math&gt;<br /> <br /> ~Tucker<br /> <br /> <br /> Every set of three numbers chosen from &lt;math&gt;\{1,2,3,4,5,6\}&lt;/math&gt; has an equal chance of being the first 3 distinct numbers rolled.<br /> <br /> Therefore, the probability that the first 3 distinct numbers are &lt;math&gt;\{2,4,6\}&lt;/math&gt; is &lt;math&gt;\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}&lt;/math&gt;<br /> <br /> ~kingofpineapplz<br /> <br /> <br /> ==Solution 2 (If you don't have much time)==<br /> <br /> Note that the problem is basically asking us to find the probability in some permutation of &lt;math&gt;1,2,3,4,5,6&lt;/math&gt; that we get the three even numbers in the first three spots.<br /> <br /> There are &lt;math&gt;6!&lt;/math&gt; ways to order the &lt;math&gt;6&lt;/math&gt; numbers and &lt;math&gt;3!(3!)&lt;/math&gt; ways to order the evens in the first three spots and the odds in the next three spots.<br /> <br /> Therefore the probability is &lt;math&gt;\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}&lt;/math&gt;.</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=145425 2021 AMC 12B Problems/Problem 6 2021-02-11T23:46:59Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12\mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18\mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24\mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi \cdot r^2 \cdot h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate these two expressions because the water volume stays the same like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;<br /> <br /> <br /> --abhinavg0627</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=145421 2021 AMC 12B Problems/Problem 6 2021-02-11T23:46:14Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12\mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18\mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24\mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi \cdot r^2 \cdot h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate this two equations like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;<br /> <br /> <br /> --abhinavg0627</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=145419 2021 AMC 12B Problems/Problem 6 2021-02-11T23:45:55Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12\mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18\mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24\mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi \cdot r^2 \cdot h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate this two equations like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=145418 2021 AMC 12B Problems/Problem 6 2021-02-11T23:45:31Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12\mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18\mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24\mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3} \cdot\pi \cdot r^2 \cdot h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pi\cdotr^2\cdoth&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdot h&lt;/math&gt;.<br /> <br /> We can equate this two equations like this &lt;math&gt;24\cdot24\cdot\pi\cdot h = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&diff=145415 2021 AMC 12B Problems/Problem 6 2021-02-11T23:44:30Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> An inverted cone with base radius &lt;math&gt;12\mathrm{cm}&lt;/math&gt; and height &lt;math&gt;18\mathrm{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has radius of &lt;math&gt;24\mathrm{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6&lt;/math&gt;<br /> ==Solution==<br /> The volume of a cone is &lt;math&gt;\frac{1}{3}\pir^2h&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the base radius and &lt;math&gt;h&lt;/math&gt; is the height. The water completely fills up the cone so the volume of the water is &lt;math&gt;\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi&lt;/math&gt;.<br /> <br /> The volume of a cylinder is &lt;math&gt;\pir^2h&lt;/math&gt; so the volume of the water in the cylinder would be &lt;math&gt;24\cdot24\cdot\pi\cdoth&lt;/math&gt;.<br /> <br /> We can equate this two equations like this &lt;math&gt;24\cdot24\cdot\pi\cdoth = 6\cdot144\pi&lt;/math&gt;. We get &lt;math&gt;4h = 6&lt;/math&gt; and &lt;math&gt;h=\frac{6}{4}&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;1.5 = \boxed{\textbf{(A)}}.&lt;/math&gt;</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=145401 2021 AMC 10B Problems/Problem 2 2021-02-11T23:39:02Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is actually negative, thus the absolute value is not &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; but &lt;math&gt;2\sqrt{3} - 3&lt;/math&gt;.<br /> So the first term equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;.<br /> <br /> Summed up you get &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~bjc and abhinavg0627<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=145396 2021 AMC 10B Problems/Problem 2 2021-02-11T23:37:59Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is actually negative, thus the absolute value is not &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; but &lt;math&gt;2\sqrt{3} - 3&lt;/math&gt;.<br /> So the first term equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;.<br /> <br /> Summed up you get &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~bjc and abhinavg0627<br /> <br /> &lt;math&gt;\phantom{no problem bjc}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=145394 2021 AMC 10B Problems/Problem 2 2021-02-11T23:37:25Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is actually negative, thus the absolute value is not &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; but &lt;math&gt;2\sqrt{3} - 3&lt;/math&gt;.<br /> So the first term equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;<br /> Summed up you get &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~bjc and abhinavg0627<br /> <br /> &lt;math&gt;\phantom{no problem bjc}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=145393 2021 AMC 10B Problems/Problem 2 2021-02-11T23:36:59Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is actually negative, thus the absolute value is not &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; but &lt;math&gt;2\sqrt{3} - 3&lt;/math&gt;.<br /> So the first term equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;<br /> Summed up you get &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~bjc and abhinavg0627<br /> <br /> \phantom{no problem bjc}<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=145392 2021 AMC 10B Problems/Problem 2 2021-02-11T23:36:22Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of a squared number is the absolute value of the number. We know that &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is actually negative, thus the absolute value is not &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; but &lt;math&gt;2\sqrt{3} - 3&lt;/math&gt;.<br /> So the first term equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;<br /> Summed up you get &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~bjc and abhinavg0627<br /> <br /> \phantom{no problem bjc}<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=145049 2021 AMC 12A Problems/Problem 6 2021-02-11T20:06:36Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> ==Solution==<br /> If the probability of choosing a red card is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, the red and black cards are in ratio &lt;math&gt;1:2&lt;/math&gt;. This means at the beginning there are &lt;math&gt;x&lt;/math&gt; red cards and &lt;math&gt;2x&lt;/math&gt; black cards.<br /> <br /> After &lt;math&gt;4&lt;/math&gt; black cards are added, there are &lt;math&gt;2x+4&lt;/math&gt; black cards. This time, the probability of choosing a red card is &lt;math&gt;\frac{1}{4}&lt;/math&gt; so the ratio of red to black cards is &lt;math&gt;1:3&lt;/math&gt;. This means in the new deck the number of black cards is also &lt;math&gt;3x&lt;/math&gt; for the same &lt;math&gt;x&lt;/math&gt; red cards.<br /> <br /> So, &lt;math&gt;3x = 2x + 4&lt;/math&gt; and &lt;math&gt;x=4&lt;/math&gt; meaning there are &lt;math&gt;4&lt;/math&gt; red cards in the deck at the start and &lt;math&gt;2(4) = 8&lt;/math&gt; black cards. <br /> <br /> So the answer is &lt;math&gt;8+4 = 12 = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==Note==<br /> See [[2021 AMC 12A Problems/Problem 1|problem 1]].<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=145047 2021 AMC 12A Problems/Problem 6 2021-02-11T20:06:14Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> ==Solution==<br /> If the probability of choosing a red card is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, the red and black cards are in ratio &lt;math&gt;1:2&lt;/math&gt;. This means at the beginning there are &lt;math&gt;x&lt;/math&gt; red cards and &lt;math&gt;2x&lt;/math&gt; black cards.<br /> <br /> After &lt;math&gt;4&lt;/math&gt; black cards are added, there are &lt;math&gt;2x+4&lt;/math&gt; black cards. This time, the probability of choosing a red card is &lt;math&gt;\frac{1}{4}&lt;/math&gt; so the ratio of red to black cards is &lt;math&gt;1:3&lt;/math&gt;. This means in the new deck the number of black cards is also &lt;math&gt;3x&lt;/math&gt; for the same &lt;math&gt;x&lt;/math&gt; red cards.<br /> <br /> So, &lt;math&gt;3x = 2x + 4&lt;/math&gt; and &lt;math&gt;x=4&lt;/math&gt; meaning there are &lt;math&gt;4&lt;/math&gt; red cards in the deck at the start and &lt;math&gt;2(4) = 8&lt;/math&gt; black cards. <br /> <br /> So the answer is &lt;math&gt;8+4 = 12 = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Note==<br /> See [[2021 AMC 12A Problems/Problem 1|problem 1]].<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=145045 2021 AMC 12A Problems/Problem 6 2021-02-11T20:05:57Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> ==Solution==<br /> If the probability of choosing a red card is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, the red and black cards are in ratio &lt;math&gt;1:2&lt;/math&gt;. This means at the beginning there are &lt;math&gt;x&lt;/math&gt; red cards and &lt;math&gt;2x&lt;/math&gt; black cards.<br /> <br /> After &lt;math&gt;4&lt;/math&gt; black cards are added, there are &lt;math&gt;2x+4&lt;/math&gt; black cards. This time, the probability of choosing a red card is &lt;math&gt;\frac{1}{4}&lt;/math&gt; so the ratio of red to black cards is &lt;math&gt;1:3&lt;/math&gt;. This means in the new deck the number of black cards is also &lt;math&gt;3x&lt;/math&gt; for the same &lt;math&gt;x&lt;/math&gt; red cards.<br /> <br /> So, &lt;math&gt;3x = 2x + 4&lt;/math&gt; and &lt;math&gt;x=4&lt;/math&gt; meaning there are &lt;math&gt;4&lt;/math&gt; red cards in the deck at the start and &lt;math&gt;2(4) = 8&lt;/math&gt; black cards. <br /> <br /> So the answer is &lt;math&gt;8+4 = 12 = \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Note==<br /> See [[2021 AMC 12A Problems/Problem 1|problem 1]].<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=145039 2021 AMC 12A Problems/Problem 6 2021-02-11T20:01:12Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> ==Solution==<br /> The solutions will be posted once the problems are posted.<br /> ==Note==<br /> See [[2021 AMC 12A Problems/Problem 1|problem 1]].<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_6&diff=145037 2021 AMC 12A Problems/Problem 6 2021-02-11T20:01:01Z <p>Abhinavg0627: /* Problem */</p> <hr /> <div>==Problem==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally.<br /> <br /> ==Solution==<br /> The solutions will be posted once the problems are posted.<br /> ==Note==<br /> See [[2021 AMC 12A Problems/Problem 1|problem 1]].<br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_4&diff=145022 2021 AMC 12A Problems/Problem 4 2021-02-11T19:54:30Z <p>Abhinavg0627: /* Text Solution */</p> <hr /> <div>==Problem==<br /> Tom has a collection of &lt;math&gt;13&lt;/math&gt; snakes, &lt;math&gt;4&lt;/math&gt; of which are purple and &lt;math&gt;5&lt;/math&gt; of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add. Which of these conclusions can be drawn about Tom's snakes?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; Purple snakes can add.<br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; Purple snakes are happy.<br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; Snakes that can add are purple.<br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; Happy snakes are not purple.<br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; Happy snakes can't subtract.<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=MUHja8TpKGw&amp;t=259s (Note that there's a slight error in the video I corrected in the description)<br /> <br /> <br /> <br /> ==Solution 1==<br /> <br /> We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=3|num-a=5}}<br /> {{AMC10 box|year=2021|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_4&diff=145021 2021 AMC 12A Problems/Problem 4 2021-02-11T19:54:02Z <p>Abhinavg0627: /* Video Solution by Punxsutawney Phil */</p> <hr /> <div>==Problem==<br /> Tom has a collection of &lt;math&gt;13&lt;/math&gt; snakes, &lt;math&gt;4&lt;/math&gt; of which are purple and &lt;math&gt;5&lt;/math&gt; of which are happy. He observes that all of his happy snakes can add, none of his purple snakes can subtract, and all of his snakes that can't subtract also can't add. Which of these conclusions can be drawn about Tom's snakes?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; Purple snakes can add.<br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; Purple snakes are happy.<br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; Snakes that can add are purple.<br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; Happy snakes are not purple.<br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; Happy snakes can't subtract.<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtube.com/watch?v=MUHja8TpKGw&amp;t=259s (Note that there's a slight error in the video I corrected in the description)<br /> <br /> <br /> <br /> ==Text Solution==<br /> <br /> We know that purple snakes cannot subtract, thus they cannot add either. Since happy snakes must be able to add, the purple snakes cannot be happy. Therefore, we know that the happy snakes are not purple and the answer is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=3|num-a=5}}<br /> {{AMC10 box|year=2021|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144995 2021 AMC 12A Problems/Problem 3 2021-02-11T19:42:06Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238 = \boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144993 2021 AMC 12A Problems/Problem 3 2021-02-11T19:41:16Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238 = \boxed{\text{D}}&lt;/math&gt;.<br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144991 2021 AMC 12A Problems/Problem 3 2021-02-11T19:40:44Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238 = \boxed{\text{\bold{D}}}&lt;/math&gt;.<br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144988 2021 AMC 12A Problems/Problem 3 2021-02-11T19:40:15Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238 = \boxed{\text{[b](D)[/b]}}&lt;/math&gt;.<br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144984 2021 AMC 12A Problems/Problem 3 2021-02-11T19:39:37Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238 = \boxed{D}&lt;/math&gt;.<br /> <br /> --abhinavg0627<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_3&diff=144982 2021 AMC 12A Problems/Problem 3 2021-02-11T19:39:06Z <p>Abhinavg0627: /* Solution */</p> <hr /> <div>==Problem==<br /> The sum of two natural numbers is &lt;math&gt;17{,}402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The units digit of a multiple of &lt;math&gt;10&lt;/math&gt; will always be &lt;math&gt;0&lt;/math&gt;. We add a &lt;math&gt;0&lt;/math&gt; whenever we multiply by &lt;math&gt;10&lt;/math&gt;. So, removing the units digit is equal to dividing by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> Let the smaller number (the one we get after removing the units digit) be &lt;math&gt;a&lt;/math&gt;. This means the bigger number would be &lt;math&gt;10a&lt;/math&gt;.<br /> <br /> We know the sum is &lt;math&gt;10a+a = 11a&lt;/math&gt; so &lt;math&gt;11a=17402&lt;/math&gt;. So &lt;math&gt;a=1582&lt;/math&gt;. The difference is &lt;math&gt;10a-a = 9a&lt;/math&gt;. So, the answer is &lt;math&gt;9(1582) = 14238&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Abhinavg0627