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0a9a07e87f522ff1a5629b2c03c2e8a4 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11 | Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \[(10a+b)+(10b+a)=132\] \[11a+11b=132\] \[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\boxed{7}$ ordered pairs. | B | 7 |
0a9a07e87f522ff1a5629b2c03c2e8a4 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11 | Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{7}$ total possibilities. | B | 7 |
af0f1b09251d3dbde61ac767a7ab36e1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_14 | Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?
$\textbf{(A) }525\qquad\textbf{(B) }560\qquad\textbf{(C) }595\qquad\textbf{(D) }665\qquad \textbf{(E) }735$ | Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \boxed{525}$ miles. | A | 525 |
2882957833b02c803b55586eea349ac5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{32}$ | C | 32 |
2882957833b02c803b55586eea349ac5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{32}.$ | C | 32 |
2882957833b02c803b55586eea349ac5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$
Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$
By the Lifting the Exponent Lemma on $n$
\[v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1\] \[=v_2(2)+v_2(4)+v_2(24)-1\] \[=1+2+3-1=5.\]
Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{32}$ | C | 32 |
2882957833b02c803b55586eea349ac5 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.
To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,
and 32 works. So, our answer is $\boxed{32}.$ | C | 32 |
3109fefdad39c0a8f05201810e713677 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$ | Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{5}$ laps. | D | 5 |
3109fefdad39c0a8f05201810e713677 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$ | Call $x$ the distance Annie runs. If Annie is $25\%$ faster than Bonnie, then Bonnie will run a distance of $\frac{4}{5}x$ . For Annie to meet Bonnie, she must run an extra $400$ meters, the length of the track. So $x-\left(\frac{4}{5}\right)x=400 \implies x=2000$ , which is $\boxed{5}$ laps. | D | 5 |
dc1afd732dd393ab5bab2c2e944412f9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17 | An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$ | For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{9990}$ | D | 9990 |
dc1afd732dd393ab5bab2c2e944412f9 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17 | An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$ | Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\boxed{9990}.$ | D | 9990 |
baf017dc3c030807adeb577effbe8617 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$ | From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.
Starting with the first race: \[\frac{216}{6}=36\] \[\frac{36}{6}=6\] \[\frac{6}{6}=1\] Adding all of the numbers in the second column yields $\boxed{43}$ | C | 43 |
baf017dc3c030807adeb577effbe8617 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$ | Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ $5x$ $1$ . Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{43}$ | C | 43 |
baf017dc3c030807adeb577effbe8617 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$ | Since $216$ is a power of $6,$ the answer will be in the form of $6x+1.$ We can see that this is odd and the only option is $\boxed{43}$ ~sanaops9 | C | 43 |
666a73c164a4d2db7a14a01c66ab44fa | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$ . Now, $25n=10000 \rightarrow n=400$ . Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{424}$ | E | 424 |
666a73c164a4d2db7a14a01c66ab44fa | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $x$ be the largest number. Then, $x+(x-2)+(x-4)+\cdots +(x-48)=10000$ . Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 24\right)\right)=10,000$ . Grouping like terms gives $25\left(\frac{x}{2}\right) - 300=5000$ , and continuing down the line, we find $x=\boxed{424}$ | E | 424 |
666a73c164a4d2db7a14a01c66ab44fa | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\cdots +(x+48)=10,000$ . After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$ $25x=9400$ , so $x=376$ . Then, you add $376+48 = \boxed{424}$ | E | 424 |
666a73c164a4d2db7a14a01c66ab44fa | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Dividing the series by $2$ , we get that the sum of $25$ consecutive integers is $5000$ . Let the middle number be $k$ we know that the sum is $25k$ , so $25k=5000$ . Solving, $k=200$ $2k=400$ is the middle term of the original sequence, so the original last term is $400+\frac{25-1}{2}\cdot 2=424$ . So the answer is $\boxed{424}$ | E | 424 |
8508aef2c6cf708960b74b7725150b0e | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_20 | The least common multiple of $a$ and $b$ is $12$ , and the least common multiple of $b$ and $c$ is $15$ . What is the least possible value of the least common multiple of $a$ and $c$
$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$ | We wish to find possible values of $a$ $b$ , and $c$ . By finding the greatest common factor of $12$ and $15$ , we can find that $b$ is 3. Moving on to $a$ and $c$ , in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$ $\rightarrow 4$ . Similarly, with $3$ and $c$ , we obtain $5$ . The least common multiple of $4$ and $5$ is $20 \rightarrow \boxed{20}$ | A | 20 |
8bc20d8ff734f1794c3aa5fdc752cb30 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Let G be the midpoint B and C
Draw H, J, K beneath C, G, B, respectively.
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, grey); fill((3,0)--(2,4)--(1.5,3)--cycle, grey); draw((1,0)--(1,4)); draw((1.5,0)--(1.5,4)); draw((2,0)--(2,4)); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$G$", (1.5, 4.2)); label("$H$", (1, -0.2)); label("$J$", (1.5, -0.2)); label("$K$", (2, -0.2)); label("$1$", (0.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
Let us take a look at rectangle CDEH. I have labeled E' for convenience. First of all, we can see that EE'H and CE'B are similar triangles because all their three angles are the same. Furthermore, since EH=CB, we can confirm that EE'H and CE'B are identical triangles. Thus, CE'=E'H, which is half of CH.
[asy] fill((0,0)--(1,4)--(1,2)--cycle, grey); draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)); draw((0,0)--(1,4)--(1,2)--(0,0)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$H$", (1, -0.2)); label("$E'$", (1.2, 2)); [/asy]
Then we can see that CEE' has $\frac{1}{4}$ the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have $\frac{1}{4}$ the area of their rectangle. So, the total shaded region is just $\frac{1}{4}$ the area of the total region, or $\frac{1}{4} \times 3 \times 4$ , or $\boxed{3}$ | C | 3 |
8bc20d8ff734f1794c3aa5fdc752cb30 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$ . Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$ . Subtracting these areas from the trapezoid, we get $8-\frac12-\frac92 =\boxed3$ . Therefore, the answer to this problem is $\boxed{3}$ | C | 3 |
8bc20d8ff734f1794c3aa5fdc752cb30 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Set coordinates to the points:
Let $E=(0,0)$ $F=(3,0)$
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); label(scale(0.7)*"$E(0,0)$", (0,-0.2)); label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); label(scale(0.7)*"$F(3,0)$", (3,-0.2)); label(scale(0.7)*"$1$", (0.3, 4), N); label(scale(0.7)*"$1$", (1.5, 4), N); label(scale(0.7)*"$1$", (2.7, 4), N); label(scale(0.7)*"$4$", (3.2, 2), E); [/asy]
Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$ . Hence, the slope of line $CF=-2$
Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$
Doing the same process to line $BE$ , we find that line $BE=2x$
Hence, setting them equal to find the intersection point...
$y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3$
Hence, we find that the intersection point is $(\frac{3}{2},3)$ . Call it Z.
Now, we can see that
$E=(0,0)$
$Z=(\dfrac{3}{2},3)$
$C=(1,4)$
Now use the Shoelace Theorem
$\frac{(0\cdot 3 + \dfrac{3}{2}\cdot 4 + 1\cdot 0)-(\dfrac{3}{2}\cdot 0 + 1\cdot 3 + 4\cdot 0)}{2} = \frac{6-3}{2} = \frac{3}{2}$
Using the Shoelace Theorem , we find that the area of one of those small shaded triangles is $\frac{3}{2}$
Now because there are two of them, we multiple that area by $2$ to get $\boxed{3}$ | C | 3 |
8bc20d8ff734f1794c3aa5fdc752cb30 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | [asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$G$", (1.5, 3.2), N); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
First, it is easy to see that $\triangle CGB \sim \triangle EGF$ . Therefore, the ratio of the height of $\triangle CBG$ to the height of $\triangle EFG$ is $\frac{1}{3}$ . Thus, the area of $\triangle CBG$ is $\frac{1\cdot1}{2} = \frac{1}{2}$ , and the area of $\triangle CBE$ is $\frac{1\cdot4}{2} = 2$ . So, the area of $\triangle CGE$ is $2-\frac{1}{2}$ . Besides, since trapezoid $CBEF$ is isosceles, $\triangle CGE \cong \triangle BGF$ . Hence, the area of the "bat wings" is $2\cdot(2-\frac{1}{2})= \boxed{3}$ | C | 3 |
8bc20d8ff734f1794c3aa5fdc752cb30 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Solution 2/4 are easily better, but if you really wanted to you could use Pick's Theorem for each half of the "bat wings". Unfortunately it isn't immediately applicable since the point common to each bat wing does not lie on a lattice point. We can remedy this by pretending the figure is twice as big and at the end divide the area by 4 (since the area of similar shapes scales quadratically with the scaling factor).
[asy] // Original drawing code draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$2$", (0.5, 4), N); label("$2$", (1.5, 4), N); label("$2$", (2.5, 4), N); label("$8$", (3.2, 2), E); // Draw the grid lines for (real i=0.5; i<3; i+=0.5) { draw((i,0)--(i,4), gray+linewidth(0.5)); // Vertical grid lines } for (real j=0.5; j<4; j+=0.5) { draw((0,j)--(3,j), gray+linewidth(0.5)); // Horizontal grid lines } // Boundary points with green dots and black border filldraw(circle((0,0), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,1), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,2), 0.05), green, black+linewidth(0.5)); filldraw(circle((1.5,3), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,4), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,2), 0.05), green, black+linewidth(0.5)); // Interior points with red dots and black border filldraw(circle((.5,1.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,2.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3.5), 0.05), red, black+linewidth(0.5)); [/asy]
Now we can safely use Pick's Theorem on the scaled-up wings:
\[A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12\]
And finally we scale this down to get the original area:
\[A=\frac14A'=\frac14 12=\boxed{3}\] | C | 3 |
17ee4984e12859589162d8bfc1d44e92 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_23 | Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$
$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$ | Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$ . Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$ . Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$
Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$ . Therefore, the answer is $\boxed{120}$ | C | 120 |
17ee4984e12859589162d8bfc1d44e92 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_23 | Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$
$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$ | We know that $\triangle{EAB}$ is equilateral, because all of its sides are congruent radii. Because point $A$ is the center of a circle, $C$ is at the border of a circle, and $E$ and $B$ are points on the edge of that circle, $m\angle{ECB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}$ . Since $\triangle{CED}$ is isosceles, angle $\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{120}$ degrees -SweetMango77. | C | 120 |
1ceac1a1da18955c253e99bbb45a2443 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We see that since $QRS$ is divisible by $5$ $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \equiv 2 \pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R=4$ . This gives $T \equiv 3 \pmod{4}$ , meaning $T=3$ . Now, we see that $Q$ could be either $1$ or $2$ , but $14$ is not divisible by $4$ , but $24$ is. This means that $Q=2$ and $P=\boxed{1}$ | A | 1 |
1ceac1a1da18955c253e99bbb45a2443 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We know that out of $PQRST,$ $QRS$ is divisible by $5$ . Therefore $S$ is obviously 5 because $QRS$ is divisible by 5. So we now have $PQR5T$ as our number. Next, let's move on to the second piece of information that was given to us. $RST$ is divisible by 3. So, according to the divisibility by 3 rule, the sum of $RST$ has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of $RT$ are 4 and 7. So, the possible values for $R$ are 1,3,4,3 and the possible values of $T$ are 3,1,3,4. So, using this we can move on to the fact that $PQR$ is divisible by 4. So, using that we know that $R$ has to be even so 4 is the only possible value for $R$ . Using that we also know that 3 is the only possible value for 3. So, we have $PQRST$ $PQ453$ so the possible values are 1 and 2 for $P$ and $Q$ . Using the divisibility rule of 4 we know that $QR$ has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for $P$ is 1. $P=\boxed{1}$ | A | 1 |
1ceac1a1da18955c253e99bbb45a2443 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We know that $QRS$ is divisible by $5$ , so $S$ would be either $5$ or $0$ . However, $0$ is not a choice, so $S=5$ . Also, $PQR$ is divisible by $4$ , so this means that $QR$ is $12$ $32$ $24$ , or $52$ . If $R=2$ , then $T$ has to be $2$ or $5$ $RST$ is divisible by $3$ ), but both are taken. So, $R=4 \Rightarrow QR=24$ $R+S+T$ must equal $9$ or $12$ , but because $4+5=9$ $R+S+T=12 \Rightarrow T=3$ . This leaves $P=\boxed{1}$ | A | 1 |
1ceac1a1da18955c253e99bbb45a2443 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We can simply try each of the answer choice, and we will see which one works. Trying $P=\boxed{1}$ | A | 1 |
3c8232032694b6d42ee1c7f347b985b4 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$ | First, we multiply $12\cdot9$ . To get that, we need $108$ square feet of carpet to cover the room's floor. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{12}$ | A | 12 |
3c8232032694b6d42ee1c7f347b985b4 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$ | Since there are $3$ feet in a yard, we divide $9$ by $3$ to get $3$ , and $12$ by $3$ to get $4$ . To find the area of the carpet, we then multiply these two values together to get $\boxed{12}$ | A | 12 |
221f776c1bcf236c1af014a0e07b68e5 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_3 | Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$ | Using $d=rt$ , we can set up an equation for when Jill arrives at the swimming pool:
$1=10t$
Solving for $t$ , we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that
Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. Thus, Jill has to wait $15-6=\boxed{9}$ | D | 9 |
6742a5d76df66846c00aba92af7755d1 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_4 | The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$ | There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{12}$ | E | 12 |
56d8dd9b6af035ae9c4ab4eee90c7666 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | We know the semi-perimeter of $\triangle ABC$ is $\frac{29+29+42}{2}=50$ . Next, we use Heron's Formula to find that the area of the triangle is just $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{420}$ | B | 420 |
56d8dd9b6af035ae9c4ab4eee90c7666 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | Splitting the isosceles triangle in half, we get a right triangle with hypotenuse $29$ and leg $21$ . Using the Pythagorean Theorem , we know the height is $\sqrt{29^2-21^2}=20$ . Now that we know the height, the area is $\dfrac{(20)(42)}{2} = \boxed{420}$ | B | 420 |
dd6de2492b141430ef9c7b7ca0d0073d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_8 | What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$ | We know from the Triangle Inequality that the last side, $s$ , fulfills $s<5+19=24$ . Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$ , and because $s+5+19$ is the perimeter of our triangle, $\boxed{48}$ is our answer. | D | 48 |
4b6bcda048abc7f2713aa52eaac555bb | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | First, we have to find how many widgets she makes on Day $20$ . We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$ $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ . The sum of $1,3,5, ... 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{400}$ | D | 400 |
4b6bcda048abc7f2713aa52eaac555bb | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | The sum is just the first $20$ odd counting/natural numbers, which is $20^2=\boxed{400}$ | D | 400 |
4b6bcda048abc7f2713aa52eaac555bb | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | We can easily find out she makes $2\cdot20-1 = 39$ widgets on Day $20$ . Then, we make the sum of $1,3, 5, ... ,35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+...+(19+21)$ , which include $10$ pairs of $40$ . So, the sum of $1,3,5, ...~39$ is $(40\cdot10)=\boxed{400}$ | D | 400 |
ad9959ee3bf4bd73914c282efd0919d4 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_10 | How many integers between $1000$ and $9999$ have four distinct digits?
$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$ | There are $9$ choices for the first number, since it cannot be $0$ , there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three. This means there are $9 \times 9 \times 8 \times 7=\boxed{4536}$ integers between $1000$ and $9999$ with four distinct digits. | B | 4536 |
c4486f04d18c8f3c67dafcbf8f48ce8e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$ . The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$ $\overline{CD}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{CD}$ $\overline{EF}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{GH}$ , and $\overline{CD}\text{ and }\overline{EF}$ . These are $6$ pairs total. We can do the same for the lines in the same direction as $\overline{AE}$ and $\overline{AD}$ . This means there are $6\cdot 3=\boxed{18}$ total pairs of parallel lines. | C | 18 |
c4486f04d18c8f3c67dafcbf8f48ce8e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | Look at any edge, let's say $\overline{AB}$ . There are three ways we can pair $\overline{AB}$ with another edge. $\overline{AB}\text{ and }\overline{EF}$ $\overline{AB}\text{ and }\overline{HG}$ , and $\overline{AB}\text{ and }\overline{DC}$ . There are 12 edges on a cube. 3 times 12 is 36. We have to divide by 2 because every pair is counted twice, so $\frac{36}{2}$ is $\boxed{18}$ total pairs of parallel lines. | C | 18 |
c4486f04d18c8f3c67dafcbf8f48ce8e | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | We can use the feature of 3-Dimension in a cube to solve the problem systematically. For example, in the 3-D of the cube, $\overline{AB}$ $\overline{BC}$ , and $\overline{BF}$ have $4$ different parallel edges respectively. So it gives us the total pairs of parallel lines are $\binom{4}{2}\cdot3 =\boxed{18}$ | C | 18 |
ad6fe8b28947412902a591f1c589fb9d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | Since there will be $9$ elements after removal, and their mean is $6$ , we know their sum is $54$ . We also know that the sum of the set pre-removal is $66$ . Thus, the sum of the $2$ elements removed is $66-54=12$ . There are only $\boxed{5}$ | D | 5 |
ad6fe8b28947412902a591f1c589fb9d | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | We can simply remove $5$ subsets of $2$ numbers while leaving only $6$ behind. The average of this one-number set is still $6$ , so the answer is $\boxed{5}$ | D | 5 |
704160763377cab27e5d71d500e06826 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | Let our $4$ numbers be $n, n+2, n+4, n+6$ , where $n$ is odd. Then, our sum is $4n+12$ . The only answer choice that cannot be written as $4n+12$ , where $n$ is odd, is $\boxed{100}$ | D | 100 |
704160763377cab27e5d71d500e06826 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ ; then, the sum is $8n$ . All the integers are divisible by $8$ except $\boxed{100}$ | D | 100 |
704160763377cab27e5d71d500e06826 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$ , the sum is $4a+12$ , and $4a+12$ divided by $4$ gives $a+3$ . This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$ , so the answer is $\boxed{100}$ | D | 100 |
704160763377cab27e5d71d500e06826 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$ . Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$ . Since $n \equiv 1 \pmod{2}$ , we can express it as the equation $n = 2a + 1$ for some integer $a$ . Multiplying 4 to each side of the equation yields $4n = 8a + 4$ , and taking mod 8 gets us $4n \equiv 4 \pmod{8}$ , so $b = 0$ . All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\boxed{100}$ | D | 100 |
704160763377cab27e5d71d500e06826 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \[16=1+3+5+7\] \[40=7+9+11+13\] \[72=15+17+19+21\] \[200=47+49+51+53\] All of the answer choices can be a sum of consecutive odd numbers except $100$ , so the answer is $\boxed{100}$ | D | 100 |
fe5aada0b5e63fbdab4d9f1b13126983 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_15 | At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?
$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$ | We can see that this is a Venn Diagram Problem.
First, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.
$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.
Solving this without a Venn Diagram, we subtract $29$ away from the total, $198$ . Out of the remaining $169$ , we have $149$ people for A and
$119$ people for B. We add this up to get $268$ . Since that is more than what we need, we subtract $169$ from $268$ to get
$\boxed{99}$ | D | 99 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | For starters, we identify d as distance and v as velocity (speed)
Writing the equation gives us: $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$
This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\boxed{9}$ | D | 9 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | $d = rt$ $d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$
$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$
$10r = 270$ so $r = 27$ , plug into the first one and it's $\boxed{9}$ miles to school. | D | 9 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ , giving \[(5)(x)=(3)(x+18)\] \[5x=3x+54\] \[2x=54\] \[x=27\] Hence, $d=\dfrac{27}{3}=\boxed{9}$ | D | 9 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$ , so we have $\frac{2x}{3} = 18$ miles per hour. Solving for $x$ gives us $27$ miles per hour. Because $20$ minutes is a third of an hour, the distance would then be $9$ miles $\boxed{9}$ | D | 9 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | When driving in rush hour traffic, he drives $20$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $3$ distances ( $3d$ ) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $5$ distances ( $5d$ ) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$ , then $d=\frac{18}{2}=9$ . So the distance to the school would be $\boxed{9}$ miles. ----LarryFlora | D | 9 |
1f91a706953ad2b030972f2f75b3afc6 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | (Ratios) $\textbf{requires edits}$
In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is $d$
$d$ $20$ minutes
In no traffic, we can do the same:
$d$ $12$ minutes
We want the ratio to be the distance to $60$ minutes:
$d$ $20$ minutes = $3$ d : $60$ minutes
$d$ $12$ minutes = $5$ d : $60$ minutes
This shows the speed in miles per hour. This means that the speed in the 2nd ratio has to be 18 faster than the first ratio. We get
$3d+18 = 5d \cdot d=9$ , or $\boxed{9}$ | D | 9 |
7354fe1cabf14d20a7a671438b21e2b5 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy] | We begin filling in the table. The top row has a first term $1$ and a fifth term $25$ , so we have the common difference is $\frac{25-1}4=6$ . This means we can fill in the first row of the table: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
The fifth row has a first term of $17$ and a fifth term of $81$ , so the common difference is $\frac{81-17}4=16$ . We can fill in the fifth row of the table as shown: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((4, 4), 25); draw_num((0, 4), 1); draw_num((1, 0), 33); draw_num((2, 0), 49); draw_num((3, 0), 65); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
We must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$ . The common difference of this sequence is $\frac{49-13}4=9$ , so the third term is $13+2\cdot 9=\boxed{31}$ | B | 31 |
7354fe1cabf14d20a7a671438b21e2b5 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy] | The middle term of the first row is $\frac{25+1}{2}=13$ , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$ . Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{31}$ | B | 31 |
7354fe1cabf14d20a7a671438b21e2b5 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy] | The value of $X$ is simply the average of the average values of both diagonals that contain $X$ . This is $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{31}$ | B | 31 |
2e3c2ab37f9a080cbed5a4b83fae3169 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.
We have $x+y+z=12$ $x+3y+4z=24$ , and $x,y,z \ge 1$
Now, we subtract to find $2y+3z=12$ , and $y,z \ge 1$ .
It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = multiple of 3, so we must have $2y=6$
Therefore, $y=3$ , and it follows that $z=2$ . Now, $x=12-y-z=12-3-2=\boxed{7}$ , as desired. | D | 7 |
2e3c2ab37f9a080cbed5a4b83fae3169 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Since the total cost of the socks was $$24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{$24}{12} = $2$
There are two ways to make packages of socks that average to $$2$ . You can have:
$\bullet$ Two $$1$ pairs and one $$4$ pair (package adds up to $$6$
$\bullet$ One $$1$ pair and one $$3$ pair (package adds up to $$4$
Now, we need to solve \[6a+4b=24,\] where $a$ is the number of $$6$ packages and $b$ is the number of $$4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$ , which yields the answer of $2\times2+3\times1 = \boxed{7}$ | D | 7 |
2e3c2ab37f9a080cbed5a4b83fae3169 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ . This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$ s and $3$ s is $2+2+3$ . If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{7}$ pairs. | D | 7 |
2e3c2ab37f9a080cbed5a4b83fae3169 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | Let the amount of $1$ dollar socks be $a$ $3$ dollar socks be $b$ , and $4$ dollar socks be $c$ . We then know that $a+b+c=12$ and $a+3b+4c=24$ . We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$ . It's not hard to see $b=3$ and $c=2$ . Now that we know $b$ and $c$ , we know that $a=7$ , meaning the number of $1$ dollar socks Ralph bought is $\boxed{7}$ | D | 7 |
2e3c2ab37f9a080cbed5a4b83fae3169 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20 | Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | If Ralph bought one sock of each kind, he already used $$8$ , so there are $$16$ left and 9 socks. If we split the $$16$ into four $$4$ sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $$3$ pair, he would need to buy a $$1$ pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $$1$ pair, he would either need to buy three $$1$ pairs or a $$3$ pair. If Ralph bought a $$4$ pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that $1 + 2 + 2 + 4 = 9$ , or a $$4$ pair, two $$3$ pairs, and six $$1$ pairs. Because we subtracted the necessary one of each kind, there are two $$4$ pairs, three $$3$ pairs, and seven $$1$ pairs. Therefore, the number of $$1$ pairs Ralph bought is $\boxed{7}$ .
~strongstephen | D | 7 |
3a0816a028efddcc524a06386d8f9f08 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21 | In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$ | Clearly, since $\overline{FE}$ is a side of a square with area $32$ $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$ . Now, since $\overline{FE} = \overline{BC}$ , we have $\overline{BC} = 4 \sqrt{2}$
Now, $\overline{JB}$ is a side of a square with area $18$ , so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$ . Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$
Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$ , so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$ . Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{12}$ | C | 12 |
3a0816a028efddcc524a06386d8f9f08 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21 | In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$ | Since $\overline{FE} = \sqrt{32}$ , and $\overline{FE} = \overline{BC}$ $\overline{BC} = 4\sqrt{2}$ . Meanwhile, $\overline{JB} = 3\sqrt{2}$ , and since $\triangle JBK$ is equilateral, $\overline{BK} = 3\sqrt{2}$ . If $ABCDEF$ is equiangular, $\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}$ , where $n$ is the number of sides of the shape. Adding all the angles around $B$ gives $270^{\circ}$ , so $\angle KBC = 360 - 90 = 270^{\circ}$ . Because $\triangle KBC$ is right, the area of $\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12$ . Therefore, the answer is $\boxed{12}$ . ~strongstephen | C | 12 |
0abc92b282031b086768da6717979226 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_22 | On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$ | The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{60}$ | C | 60 |
0abc92b282031b086768da6717979226 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_22 | On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$ | Since we know the number must be a multiple of $15$ , we can eliminate $A$ . We also know that after $12$ days, the students can't find any more arrangements, meaning the number has $12$ factors. Now, we just list the factors of every number, starting with $30$ \[30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6\] \[60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10\] $60$ has $12$ factors, so the answer is $\boxed{60}$ | C | 60 |
20a2f302127a4849ca8aa2038fe6ffb7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24 | A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$ | On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$
Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$
Next, $M=6$ does not work because $52$ is not divisible by $3$
We try $M=7$ does work by giving $N=16$ $~M=7$ and thus $3\times 16=\boxed{48}$ games in their division. | B | 48 |
20a2f302127a4849ca8aa2038fe6ffb7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24 | A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$ | $76=3N+4M > 10M$ , giving $M \le 7$ .
Since $M>4$ , we have $M=5,6,7$ .
Since $4M$ is $1$ $\pmod{3}$ , we must have $M$ equal to $1$ $\pmod{3}$ , so $M=7$
This gives $3N=48$ , as desired. The answer is $\boxed{48}$ | B | 48 |
20a2f302127a4849ca8aa2038fe6ffb7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24 | A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$ | Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$
Since each team plays $M$ games against each of the four other teams in the other division, that's $4M$
So $3N+4M=76$ , with $M>4, N>2M$
Let's start by solving this Diophantine equation. In other words, $N=\frac{76-4M}{3}$
So $76-4M\equiv0 \pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\pmod{3}$ ), we find that $M\equiv1 \pmod{3}$
Since $M>4$ , so the minimum possible value of $M$ is $7$ . However, remember that $N>2M$ ! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$ ). Plugging $N=2M$ in, $10M=76$ . So $M<7.6$ . Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\equiv1 \pmod{3}$ . After checking $M=7$ , we find that it works.
So $M=7, N=16$ . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$ . Select $\boxed{48}$ | B | 48 |
3e030bf775c0f9a91e5f21ec86c353e7 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_25 | One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]
$\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17$ | Proceed as in solution 1 to get $x^2 - 5x + 5 = 0$
Notice that the side length of the square is $\sqrt{x^2+(5-x)^2}$ by the Pythagorean Theorem. Thus, the area of the square is \[\left(\sqrt{x^2+(5-x)^2}\right)^2\] \[= x^2 + x^2 - 10x + 25\] \[= (2x^2 - 10x + 10) + 15\] \[= 2(x^2 - 5x + 5) + 15\] \[= 2(0) + 15\] \[= \boxed{15}\] | C | 15 |
31ae8961d2b1a41d4e67ffccb4d552df | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_1 | Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry's answer is $H$ and Terry's answer is $T$ , what is $H-T$
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$ | We have $H=8-7=1$ and $T=8-2+5=11$ . Clearly $1-11=-10$ , so our answer is $\boxed{10}$ | A | 10 |
cc749ca8ba96cff82b735668d63408c6 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_2 | Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$ , if he only uses nickels. Therefore we have $7-2=\boxed{5}$ | E | 5 |
a0d07bf57995655f037b7b2f85b92680 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_3 | Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?
$\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280$ | Isabella read $3\cdot 36+3\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\cdot (36+44)=3\cdot 80$ , which gives that she read $240$ pages. On her last day, she read $10$ more pages for a total of $240+10=\boxed{250}$ pages. | B | 250 |
a0891fcd63c4c8b72ff77136a6448941 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_4 | The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$ | Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\cdot2=\boxed{166}$ | E | 166 |
66edcf89e2632e38344afcac62a21521 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_5 | Margie's car can go $32$ miles on a gallon of gas, and gas currently costs $ $4$ per gallon. How many miles can Margie drive on $\textdollar 20$ worth of gas?
$\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad \textbf{(E) }640$ | Margie can afford $20/4=5$ gallons of gas. She can go $32\cdot5=\boxed{160}$ miles on this amount of gas. | C | 160 |
bac44cda675e620cf64645a855a86b34 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_6 | Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$ , and $36$ . What is the sum of the areas of the six rectangles?
$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$ | The sum of the areas is equal to $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$ . This is equal to $2(1+4+9+16+25+36)$ , which is equal to $2\cdot91$ . This is equal to our final answer of $\boxed{182}$ | D | 182 |
9662156b910aaf65dba41e8cd2459bbd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_8 | Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$ | Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{3}$ ~SparklyFlowers | D | 3 |
9662156b910aaf65dba41e8cd2459bbd | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_8 | Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$ | We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$ . Thus, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0 = 0$ . Thus, $A = \boxed{3}$ ~fn106068 | D | 3 |
c5179fa9f4dbee24e8dc213e1fbbc8df | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_10 | The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$ | The seventh AMC 8 would have been given in $1991$ . If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$
Our answer is $\boxed{1979}$ corrections made by DrDominic | A | 1979 |
c5179fa9f4dbee24e8dc213e1fbbc8df | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_10 | The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$ | Since she was 12 when she took the seventh AMC 8, she should be $12-6=6$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-6=\boxed{1979}$ ~SweetMango77
corrections made by DrDominic | A | 1979 |
0fbabfbdfa4aea9e5bf719df3c6391da | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14 | Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$ | The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$ | B | 13 |
0fbabfbdfa4aea9e5bf719df3c6391da | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14 | Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$ | The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a $5-12-13$ triangle. So, the hypotenuse $DE$ has to be $\boxed{13}$ | B | 13 |
0fbabfbdfa4aea9e5bf719df3c6391da | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14 | Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$ | This problem can be solved with the Pythagorean Theorem ( $a^2 + b^2 = c^2$ ). We know $AB = DC$ , so $DC = 5$ $CE$ is twice the length of $AD$ , so $CE = 12$ $5^2 + 12^2 = c^2$ $5^2 = 25$ $12^2 = 144$ $25 + 144 = 169$ $169$ has a square root of $13$ , so the hypotenuse or $DE$ is $13$ . The answer is $\boxed{13}$ | B | 13 |
c1a1311f0a7d0a0590897bbd9fdd6983 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_16 | The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$ | Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.
Therefore, the total number of games is $56+32 = \boxed{88}$ is our answer. | B | 88 |
81cbb4b46f8927477d911d2a4d3c2388 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17 | George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$ | Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6}$ is the answer. | B | 6 |
81cbb4b46f8927477d911d2a4d3c2388 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17 | George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$ | Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half $x$ , we can make the expression:
$\frac{2 * 2x}{2 + x}$
Which simplifies to:
$\frac{4x}{2 + x}$
Then, because we know that since he goes at $3$ mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar:
$\frac{4x}{2 + x} = 3$
Solving that equation gives us:
$\boxed{6}$ | B | 6 |
c37b944e047c68d72f17a349a9d7d86c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_20 | Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy]
$\text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5$ | The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.
The area of the rectangle is $3\cdot5 =15$ . The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$ . Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$ $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{4.0}$ | B | 4.0 |
b398bbdad9ef8a73baab5815377d1f23 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_21 | The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$ | The number $\mod{3}$ is congruent to sum of a number's digits $\mod{3}$ is congruent to the number $\mod{3}$ $74A52B1 \pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \pmod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{1}$ | A | 1 |
b398bbdad9ef8a73baab5815377d1f23 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_21 | The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$ | Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$ , to get $23$ , which shows us that C = $1$ or $4$ or $7$ ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\boxed{1}$ is our answer. | A | 1 |
17cab8f2b951b6238bb36bd9c03b563d | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_1 | Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$ | The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{1}$ more model car.
~avamarora | A | 1 |
17cab8f2b951b6238bb36bd9c03b563d | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_1 | Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$ | 6 x 4 = 24, which is 1 more than 23. So, the answer is $\boxed{1}$ | A | 1 |
422c97304a9b0722546465f890e2d703 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_2 | A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | 50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{12}$ dollars. | D | 12 |
422c97304a9b0722546465f890e2d703 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_2 | A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$ | Suppose a full pound at normal price costs $x$ dollars. Then, with the 50% off deal, the full pound would cost $x/2$ dollars. A half pound of this would cost $x/4$ dollars (including the 50% off). 50% off half a pound is 3, so we form the equation $x/4=3\Rightarrow x=\boxed{12}$ | D | 12 |
0451adb95a0a866e3ca6a9983276a060 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_3 | What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$
$\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$ | We group the addends inside the parentheses two at a time: \begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500. \end{align*} Then the desired answer is $4 \times 500 = \boxed{2000}$ | E | 2000 |
9cc5116396ed37aa668e5a70fb0094a2 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_4 | Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$ | Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\cdot8=\boxed{140}$ to find the total the 8 friends paid. | C | 140 |
9cc5116396ed37aa668e5a70fb0094a2 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_4 | Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$ | Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\frac{m}{8}$ . Since Judi forgot her money, we have to add $$\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5$ . This is equal to $m$ because both sides represent the total bill. Solving for $m$ \[\frac{7m}{8} + 17.5=m\] \[\frac{m}{8}=17.5\] \[m=\boxed{140}.\] | C | 140 |