id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
3,201 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | 1 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$ | 23 |
3,202 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | 2 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$ | 23 |
3,203 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | 3 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this v... | 23 |
3,204 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_5 | 1 | Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. ... | Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{4}$ more dollars than Doug. | 4 |
3,205 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_6 | 1 | What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$
$\textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14$ | Taking the seventh root of both sides, we get $(7x)^2=14x$
Simplifying the LHS gives $49x^2=14x$ , which then simplifies to $7x=2$
Thus, $x=\boxed{27}$ | 27 |
3,206 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | 1 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,... | 6 |
3,207 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | 2 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$... | 6 |
3,208 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | 3 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,... | Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$ | 6 |
3,209 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | 1 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$
Then we get a system of two equations:
$3=4+2b+c$
$3=16+4b+c$
Subtracting the first equation from the second we have:
$0=12+2b$
$b=-6$
Then using $b=-6$ in the first equation:
$0=1+-12+c$
$c=\boxed{11}$ | 11 |
3,210 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | 2 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely $(3,2)$ . Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$ . Expanding this out, we find that $c=\boxed{11}$ | 11 |
3,211 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | 3 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | The points given have the same $y$ -value, so the vertex lies on the line $x=\frac{2+4}{2}=3$
The $x$ -coordinate of the vertex is also equal to $\frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$
$x=\frac{-b}{2a}$
$3=\frac{-b}{2}$
$6=-b$
$b=-6$
Now the equation is of the form $y=x^2-6x+c$ .... | 11 |
3,212 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | 4 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Substituting y into the two equations, we get:
$3=x^2+bx+c$
Which can be written as:
$x^2+bx+c-3=0$
$4$ and $2$ are the solutions to the quadratic. Thus:
$c-3=4\times2$
$c-3=8$
$c=\boxed{11}$ | 11 |
3,213 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_9 | 1 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the n... | 3 |
3,214 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_9 | 2 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] ... | 3 |
3,215 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_10 | 1 | For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
$\textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{... | 11 |
3,216 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_12 | 1 | Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.
[asy] size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0));... | [asy] size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10)); filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,... | 4 |
3,217 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_13 | 1 | A player pays $\textdollar 5$ to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a ... | The probability of rolling an even number on the first turn is $\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\frac{1}{6}$ . The probability of winning is $\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}$ . If the game is to be fair, the amount paid, $5$ dollars, must be $\frac{1}{12}$ the... | 60 |
3,218 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_14 | 1 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring ... | The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\frac{18 \cdot 19}{2} + 2 = \boxed{173}$ | 173 |
3,219 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_14 | 2 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring ... | Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{173}$ | 173 |
3,220 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | 1 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the sta... | Since $d = rt$ , we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twic... | 47 |
3,221 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | 2 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the sta... | We first find the amount of minutes, $k$ , until Odell and Kershaw's next meeting. Let $a$ be the angle in radians between their starting point and the point where they first meet, measured counterclockwise.
Since Kershaw has traveled $300k$ meters at this point and the circumference of his track is $120\pi$ $a=\frac{3... | 47 |
3,222 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | 3 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the sta... | Since Odell's rate is $5/6$ that of Kershaw, but Kershaw's lap distance is $6/5$ that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs \[(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)=... | 47 |
3,223 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_17 | 1 | In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E... | By symmetry $WXYZ$ is a square.
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,... | 12 |
3,224 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_17 | 2 | In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E... | [asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); ... | 12 |
3,225 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | 1 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | The sum of the angles of a triangle is $180$ degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it $\frac{180}{3} = 60$ degrees. The minimum possible value for the smallest angle is $1$ and the highest possible is $59$ (sinc... | 59 |
3,226 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | 2 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | Let the first angle be $x$ , and the common difference be $d$ . The arithmetic progression can now be expressed as $x + (x + d) + (x + 2d) = 180$ . Simplifiying, $x + d = 60$ . Now, using stars and bars, we have $\binom{61}{1} = 61$ .
However, we must subtract the two cases in which either $x$ or $d$ equal $0$ , so we... | 59 |
3,227 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | 3 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | Consider that we have $(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n$ , where $n \geq 0$ and $n$ is an integer. Since $a \neq 0$ $n=0,1,2,3,\cdots, 58$ which is $\boxed{59}$ solutions. | 59 |
3,228 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_20 | 1 | Six distinct positive integers are randomly chosen between $1$ and $2006$ , inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{3}{5}\qquad\textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{... | For two numbers to have a difference that is a multiple of $5$ , the numbers must be congruent $\bmod{5}$ (their remainders after division by $5$ must be the same).
$0, 1, 2, 3, 4$ are the possible values of numbers in $\bmod{5}$ . Since there are only $5$ possible values in $\bmod{5}$ and we are picking $6$ numbers, b... | 1 |
3,229 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_22 | 1 | problem_id
b36e53894fd42003fe068f1aec8a38d0 Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0 Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us tha... | 30 |
3,230 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_22 | 2 | problem_id
b36e53894fd42003fe068f1aec8a38d0 Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0 Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. S... | 30 |
3,231 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_1 | 1 | What is $(-1)^{1} + (-1)^{2} + ... + (-1)^{2006}$
$\textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006$ | Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$
$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{0}.$ | 0 |
3,232 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_2 | 1 | For real numbers $x$ and $y$ , define $x \spadesuit y = (x+y)(x-y)$ . What is $3 \spadesuit (4 \spadesuit 5)$
$\mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72$ | Since $x \spadesuit y = (x+y)(x-y)$
$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{72}$ | 72 |
3,233 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_3 | 1 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?
$\textbf{(A) } 10\qquad \textbf{(B) } 14\qquad \textbf{(C) } 17\qquad \textbf{(D) } 20\qquad \textbf{(E) } 24$ | Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$ \begin{align*} x+y &= 34 \\ x-y &= 14 \\ 2x &= 48 \\ x &= 24 \\ y &= \boxed{10} | 10 |
3,234 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_4 | 1 | Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
2006amc10b04.gif
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad ... | The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
So we have:
\begin{align*} A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\ A_{blue}&=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)^... | 8 |
3,235 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_5 | 1 | $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
$\textbf{(A) } 16\qquad \textbf{(B) } 25\qquad \textbf{(C) } 36\qquad \... | By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$
Since placing the two re... | 25 |
3,236 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_6 | 1 | A region is bounded by semicircular arcs constructed on the side of a square whose sides measure $\frac{2}{\pi}$ , as shown. What is the perimeter of this region?
[asy] size(90); defaultpen(linewidth(0.7)); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.5)); filldraw(arc((1,0),1,180,0, CCW)--cycle,gray(0.7)); filldr... | Since the side of the square is the diameter of the semicircle, the radius of the semicircle is $\frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi}$
Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is $\frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1$
S... | 4 |
3,237 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_8 | 1 | A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
[asy] defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); [/asy]
$\textbf{(A) } 20\pi\qquad \textbf{(B)... | Since the area of the square is $40$ , the length of a side is $\sqrt{40}=2\sqrt{10}$ . The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\sqrt{10}$
Using the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\sq... | 25 |
3,238 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_9 | 1 | Francesca uses $100$ grams of lemon juice, $100$ grams of sugar, and $400$ grams of water to make lemonade. There are $25$ calories in $100$ grams of lemon juice and $386$ calories in $100$ grams of sugar. Water contains no calories. How many calories are in $200$ grams of her lemonade?
$\textbf{(A) } 129\qquad \textbf... | The calorie to gram ratio of Francesca's lemonade is $\frac{25+386+0}{100+100+400}=\frac{411\textrm{ calories}}{600\textrm{ grams}}=\frac{137\textrm{ calories}}{200\textrm{ grams}}$
So in $200\textrm{ grams}$ of Francesca's lemonade there are $200\textrm{ grams}\cdot\frac{137\textrm{ calories}}{200\textrm{ grams}}=\box... | 137 |
3,239 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_10 | 1 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$ . What is the greatest possible perimeter of the triangle?
$\textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47$ | Let $x$ be the length of the first side.
The lengths of the sides are: $x$ $3x$ , and $15$
By the Triangle Inequality
$3x < x + 15$
$2x < 15$
$x < \frac{15}{2}$
The greatest integer satisfying this inequality is $7$
So the greatest possible perimeter is $7 + 3\cdot7 + 15 =\boxed{43}$ | 43 |
3,240 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_11 | 1 | What is the tens digit in the sum $7!+8!+9!+...+2006!$
$\textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9$ | Since $10!$ is divisible by $100$ , any factorial greater than $10!$ is also divisible by $100$ . The last two digits of all factorials greater than $10!$ are $00$ , so the last two digits of $10!+11!+...+2006!$ are $00$ .
(*)
So all that is needed is the tens digit of the sum $7!+8!+9!$
$7!+8!+9!=5040+40320+362880=40... | 4 |
3,241 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_15 | 1 | Rhombus $ABCD$ is similar to rhombus $BFDE$ . The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$ . What is the area of rhombus $BFDE$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)... | Using the property that opposite angles are equal in a rhombus $\angle DAB = \angle DCB = 60 ^\circ$ and $\angle ADC = \angle ABC = 120 ^\circ$ . It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$ . Let the lengths of the sides of rhombus $ABCD$ be $s$
The longer diagonal of rhomb... | 8 |
3,242 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_15 | 2 | Rhombus $ABCD$ is similar to rhombus $BFDE$ . The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$ . What is the area of rhombus $BFDE$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)... | Triangle DAB is equilateral so triangles $DEA$ $AEB$ $BED$ $BFD$ $BFC$ and $CFD$ are all congruent with angles $30^\circ$ $30^\circ$ and $120^\circ$ from which it follows that rhombus $BFDE$ has one third the area of rhombus $ABCD$ i.e. $8 \Longrightarrow \boxed{8}$ | 8 |
3,243 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_18 | 1 | Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ $a_2=3$ , and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$ . What is $a_{2006}$
$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$ | Looking at the first few terms of the sequence:
$a_1=2 , a_2=3 , a_3=\frac{3}{2}, a_4=\frac{1}{2} , a_5=\frac{1}{3} , a_6=\frac{2}{3} , a_7=2 , a_8=3 , ....$
Clearly, the sequence repeats every 6 terms.
Since $2006 \equiv 2\bmod{6}$
$a_{2006} = a_2 = \boxed{3}$ | 3 |
3,244 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_18 | 2 | Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ $a_2=3$ , and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$ . What is $a_{2006}$
$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$ | $a_n = \frac{a_{n-1}}{a_{n-2}} = \frac{\frac{a_{n-2}}{a_{n-3}}}{a_{n-2}} = \frac{1}{a_{n-3}}$ , so $a_n = a_{n-6}$ and because $2006 = 2 + 334 \times 6$ , so $a_{2006} = a_2 = \boxed{3}$ | 3 |
3,245 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 1 | In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$ | This solution is the same as Solution 1 up to the point where we find that $y=-42$
We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse $AB$ has legs $200$ and $2000$ , while the triangle with hypotenuse $AD$ has legs $2$ and $20$ . Aha! The two triangles are similar by SAS, w... | 400 |
3,246 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 2 | In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$ | We do not need to solve for y. We form a right triangle with $AB$ as the hypotenuse and two adjacent sides lengths 200 and 2000, respectively. We form another right triangle with $AD$ as the hypotenuse and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because $AD$ and $AB$ are perpendic... | 400 |
3,247 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 3 | In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$ | In order to find the area of rectangle $ABCD$ , we need to find $AB$ first. Using the distance formula , we can derive:
$AB = \sqrt{(2006-6)^2 + (178-(-22))^2} = \sqrt{(2000)^2 + (200)^2 } = 200\sqrt{101}$
Now we can look at the answer choices. Because all of them are integers, then we know that $AD$ has to also contai... | 400 |
3,248 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 1 | A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B ... | Label the points in the figure as shown below, and draw the segment $CF$ . This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$
[asy] unitsize(2cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--... | 18 |
3,249 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 2 | A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B ... | Connect points $E$ and $D$ . Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ratio $3:7$ . Their bases, $EF$ and $FB$ , must be in the same $3:7$ ratio.
Triangles $EFD$ and $FBD$ share an altitude and their bases are in a $3:7$ ratio. Therefore, their areas are in a $3:7$ ratio and the area of tr... | 18 |
3,250 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 3 | A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
[asy] unitsize(1.5cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B ... | We use mass points (similar to above). Let the triangle be $ABC$ with cevians (lines to opposite side) from $B$ and $C$ . Let the points opposite $B$ and $C$ be $D$ and $F$ respectively and the intersection as $P$
Assign masses of 1 at $B$ and $D$ since $[BPC] = [DPC]$ . Then the mass at $P$ is 2. To find masses at $F$... | 18 |
3,251 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 1 | Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and th... | Let $S$ be the set of the ages of Mr. Jones' children (in other words $i \in S$ if Mr. Jones has a child who is $i$ years old). Then $|S| = 8$ and $9 \in S$ . Let $m$ be the positive integer seen on the license plate. Since at least one of $4$ or $8$ is contained in $S$ , we have $4 | m$
We would like to prove that $5 ... | 5 |
3,252 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 2 | Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and th... | Alternatively, we can see that if one of Mr. Jones' children is of the age 5, then the license plate will have to end in the digit $0$ . The license plate cannot end in the digit $5$ as $2$ is a factor of the number, so it must be even. This means that the license plate would have to have two $0$ digits, and would eith... | 5 |
3,253 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 3 | Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and th... | Another way to do the problem is by the process of elimination. The only possible correct choices are the highest powers of each prime, $2^3=8$ $3^2=9$ $5^1=5$ , and $7^1=7$ , since indivisibility by any powers lower than these means indivisibility by a higher power of the prime (for example, indivisibility by $2^2=4$ ... | 5 |
3,254 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_1 | 1 | While eating out, Mike and Joe each tipped their server $2$ dollars. Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars between their bills?
$\textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20$ | Let $m$ be Mike's bill and $j$ be Joe's bill.
$\frac{10}{100}m=2$
$m=20$
$\frac{20}{100}j=2$
$j=10$
So the desired difference is $m-j=20-10=10 \Rightarrow \boxed{10}$ | 10 |
3,255 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_2 | 1 | For each pair of real numbers $a \neq b$ , define the operation $\star$ as
$(a \star b) = \frac{a+b}{a-b}$
What is the value of $((1 \star 2) \star 3)$
$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\,... | $((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{0}$ | 0 |
3,256 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_3 | 1 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$ | 4 |
3,257 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_5 | 1 | A store normally sells windows at $$100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?
$\textbf{(A) } 100\qquad \textbf{(B) } 200\qquad ... | The store's offer means that every $5$ th window is free.
Dave would get $\left\lfloor\frac{7}{5}\right\rfloor=1$ free window.
Doug would get $\left\lfloor\frac{8}{5}\right\rfloor=1$ free window.
This is a total of $2$ free windows.
Together, they would get $\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15... | 100 |
3,258 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_6 | 1 | The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?
$\textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27$ | Since the average of the first $20$ numbers is $30$ , their sum is $20\cdot30=600$
Since the average of $30$ other numbers is $20$ , their sum is $30\cdot20=600$
So the sum of all $50$ numbers is $600+600=1200$
Therefore, the average of all $50$ numbers is $\frac{1200}{50}=\boxed{24}$ | 24 |
3,259 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_7 | 1 | Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?
$... | Let $m$ be the distance in miles that Mike rode.
Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.
Since their combined distance was $13$ miles,
$\frac{8}{5}m + m = 13$
$\frac{13}{5}m = 13$
$m = \boxed{5}$ | 5 |
3,260 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_8 | 1 | In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$ . What is the area of the inner square $EFGH$
AMC102005Aq.png
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$ | We see that side $BE$ , which we know is $1$ , is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, $AH = 1$ . Then $HB = HE + BE = HE + 1$ , and $HE$ is one of the sides of the square whose area we want to find. So:
\[1^2 + (HE+1)^2=\sqrt{50}^2\]
\[1 + (HE+1)^2=50\]... | 36 |
3,261 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 1 | There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$ | quadratic equation has exactly one root if and only if it is a perfect square . So set
$4x^2 + ax + 8x + 9 = (mx + n)^2$
$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$
Two polynomials are equal only if their coefficients are equal, so we must have
$m^2 = 4, n^2 = 9$
$m = \pm 2, n = \pm 3$
$a + 8= 2mn = \pm 2\cdot 2\cdot 3... | 16 |
3,262 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 2 | There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$ | Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have \[(a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80.\] We can use the quadratic formula to solve for its roots (we can ignore the expression in the radical sign as it will cancel out... | 16 |
3,263 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 3 | There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$ | There is only one positive value for $k$ such that the quadratic equation would have only one solution. $k-8$ and $-k-8$ are the values of $a$ $-8-8=-16$ , so the answer is $\boxed{16}$ | 16 |
3,264 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_11 | 1 | A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$
$\textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7$ | Since there are $n^2$ little faces on each face of the big wooden cube , there are $6n^2$ little faces painted red.
Since each unit cube has $6$ faces, there are $6n^3$ little faces total.
Since one-fourth of the little faces are painted red,
$\frac{6n^2}{6n^3}=\frac{1}{4}$
$\frac{1}{n}=\frac{1}{4}$
$n=\boxed{4}$ | 4 |
3,265 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_13 | 1 | How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$ | We're given $(130n)^{50} > n^{100} > 2^{200}$ , so
$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus
$130n > n^2 > 2^4$
$130n > n^2 > 16$
Solving each part separately:
$n^2 > 16 \Longrightarrow n > 4$
$130n > n^2 \Longrightarrow 130 > n$
So $4 < n < 130$
Theref... | 125 |
3,266 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_13 | 2 | How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$ | We're given $\left(130n\right)^{50}>n^{100}>2^{200}$
Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$ . Because the exponents are equal, we can ignore them and solve for $n$ $130n>n^{2}$ , or $n<130$
The second half: $n^{100}>2^{200}$ , or $n^{100}>4^{100}$ ,... | 125 |
3,267 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 1 | How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$ | If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework
If the middle digit is $1$ , possible numbers range from $111$ to $210$ . So there are $2$ numbers in this case.
If the middle digit is $2$ , possible number... | 45 |
3,268 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 2 | How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$ | Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into $... | 45 |
3,269 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 3 | How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$ | As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have $9$ possibilities $(1-9)$ , and the third digit can have $10$ possibilities $(0-9)$ . This means there can be $9\cdot10=90$ possible two-digit numbers in which the first digit and the third digit are d... | 45 |
3,270 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 4 | How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$ | If our first digit is odd, the last digit also has to be odd for there to be a whole number middle digit that is the average of the first and the last digits.
If our first digit is even, the last digit has to be even, or $0$ for there to be a whole number middle digit that is the average of the first and the last digit... | 45 |
3,271 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_15 | 1 | How many positive cubes divide $3! \cdot 5! \cdot 7!$
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6$ | $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$
Therefore, a perfect cube that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$ where $a$ $b$ $c$ , and $d$ are nonnegati... | 6 |
3,272 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_15 | 2 | How many positive cubes divide $3! \cdot 5! \cdot 7!$
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6$ | $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$
In the expression, we notice that there are 3 $3's$ , 3 $2's$ , and 3 $1's$ . This gives us our first 3 cubes: $3^3$ $2^3$ , and $1^3$
However, we can also multiply smaller numbers in the expression ... | 6 |
3,273 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_16 | 1 | The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$ . How many two-digit numbers have this property?
$\textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19$ | Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.
So $(10a+b)-(a+b)=9a$ must have a units digit of $6$
This is only possible if $9a=36$ , so $a=4$ is the only way this can be true.
So the numbers that have this property are $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$
Therefore the an... | 10 |
3,274 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_17 | 1 | In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$ , although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$ $BC$ $CD$ $DE$ , and $EA$ form an arithmetic sequence, although not necessarily in that order. What is... | Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ( $A$ goes to $AE$ and $AB$ $D$ goes to $DC$ and $DE$ ). The sum of every term is equal to $2(3+5+6+7+9)=60$
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=\boxed{12}$ | 12 |
3,275 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_20 | 1 | An equilangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$ , arranged so that no two consecutive sides have the same length. What is the area of the octagon?
$\textbf{(A) } \frac72\qquad \textbf{(B) } \frac{7\sqrt2}{2}\qquad \textbf{(C) } \frac{5+4\sqrt2}{2}\qquad \textbf{(D) ... | The area of the octagon can be divided up into $5$ squares with side $\frac{\sqrt2}2$ and $4$ right triangles, which are half the area of each of the squares.
Therefore, the area of the octagon is equal to the area of $5+4\left(\frac12\right)=7$ squares.
The area of each square is $\left(\frac{\sqrt2}2\right)^2=\frac12... | 72 |
3,276 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_20 | 2 | An equilangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$ , arranged so that no two consecutive sides have the same length. What is the area of the octagon?
$\textbf{(A) } \frac72\qquad \textbf{(B) } \frac{7\sqrt2}{2}\qquad \textbf{(C) } \frac{5+4\sqrt2}{2}\qquad \textbf{(D) ... | Refer to the following diagram:
AMC10 2005A P20.png
(Picture made on Geogebra)
Note that each square has area $\frac14$ , and each triangle has area $\frac18$ . The total area is $12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{72}$ | 72 |
3,277 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_21 | 1 | For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$
$\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11$ | If $1+2+...+n$ evenly divides $6n$ , then $\frac{6n}{1+2+...+n}$ is an integer
Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the RHS in the above fraction . So the problem asks us for how many positive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ f... | 5 |
3,278 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_22 | 1 | Let $S$ be the set of the $2005$ smallest positive multiples of $4$ , and let $T$ be the set of the $2005$ smallest positive multiples of $6$ . How many elements are common to $S$ and $T$
$\textbf{(A) } 166\qquad \textbf{(B) } 333\qquad \textbf{(C) } 500\qquad \textbf{(D) } 668\qquad \textbf{(E) } 1001$ | Since the least common multiple $\mathrm{lcm}(4,6)=12$ , the elements that are common to $S$ and $T$ must be multiples of $12$
Since $4\cdot2005=8020$ and $6\cdot2005=12030$ , several multiples of $12$ that are in $T$ won't be in $S$ , but all multiples of $12$ that are in $S$ will be in $T$ . So we just need to find t... | 668 |
3,279 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_24 | 1 | For each positive integer $n > 1$ , let $P(n)$ denote the greatest prime factor of $n$ . For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$
$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | If $P(n) = \sqrt{n}$ , then $n = p_{1}^{2}$ , where $p_{1}$ is a prime number
If $P(n+48) = \sqrt{n+48}$ , then $n + 48$ is a square, but we know that n is $p_{1}^{2}$
This means we just have to check for squares of primes, add $48$ and look whether the root is a prime number.
We can easily see that the difference betw... | 1 |
3,280 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_1 | 1 | A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?
$\textbf{(A) }\ 100 \qquad \textbf{(B) }\ 200 \qquad \textbf{(C) }\ 300 \qquad \textbf{(D) }\ 400 \qquad \textbf{(E) }\ 500$ | \begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{100} Note: Revenue is a gain. | 100 |
3,281 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_2 | 1 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Since $x\%$ means $0.01x$ , the statement " $x\% \text{ of } x \text{ is 4}$ " can be rewritten as " $0.01x \cdot x = 4$ ":
$0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.$ | 20 |
3,282 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_2 | 2 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\boxed{20}.$ Very fast. | 20 |
3,283 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_5 | 1 | Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qq... | Let $m =$ Brianna's money. We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ . Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{25}$ | 25 |
3,284 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_6 | 1 | At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$
$\textbf{(A) }\ 1 \qquad... | Lisa's goal was to get an $A$ on $80\% \cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\boxed{2}$ of them. | 2 |
3,285 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_7 | 1 | A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \... | Let the side of the largest square be $x$ . It follows that the diameter of the inscribed circle is also $x$ . Therefore, the diagonal of the square inscribed inscribed in the circle is $x$ . The side length of the smaller square is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$ . Similarly, the diameter of the smaller ins... | 8 |
3,286 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_7 | 2 | A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \... | Let the radius of the smallest circle be $r$ . Then the side length of the smaller square is $2r$ . The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$ . Hence the largest square has sides of length $2\sqrt{2}r$ . The ratio of the area of the smallest circle to... | 8 |
3,287 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 1 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\sqrt{48}$ . Since $CD=8$ $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{3}$ | 3 |
3,288 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 2 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one ... | 3 |
3,289 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 3 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Let $BD=k$ . Then, by Stewart's Theorem
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$ | 3 |
3,290 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_11 | 1 | The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$ | Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ $3^{\text{rd}}$ , and $4^{\text{th}}$ t... | 250 |
3,291 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_13 | 1 | How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$
$\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169$ | To find the multiples of $3$ or $4$ but not $12$ , you need to find the number of multiples of $3$ and $4$ , and then subtract twice the number of multiples of $12$ , because you overcount and do not want to include them. The multiples of $3$ are $\frac{2005}{3} = 668\text{ }R1.$ The multiples of $4$ are $\frac{2005}{4... | 835 |
3,292 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_13 | 2 | How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$
$\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169$ | From $1$ $12$ , the multiples of $3$ or $4$ but not $12$ are $3, 4, 6, 8,$ and $9$ , a total of five numbers. Since $\frac{5}{12}$ of positive integers are multiples of $3$ or $4$ but not $12$ from $1$ $12$ , the answer is approximately $\frac{5}{12} \cdot 2005$ $\boxed{835}$ | 835 |
3,293 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_16 | 1 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$ . Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \... | 8 |
3,294 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_16 | 2 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \[2a + 2b = -m\] \[a + b = -p\] \[2a(2b) = n\] \[a(b) = m\] Therefore, substituting the second equation into the first equation gives \[m = 2(p)\] and substituting the fourth equation ... | 8 |
3,295 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_18 | 1 | All of David's telephone numbers have the form $555-abc-defg$ , where $a$ $b$ $c$ $d$ $e$ $f$ , and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$ . How many different telephone numbers can David have?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 ... | The only digits available to use in the phone number are $2$ $3$ $4$ $5$ $6$ $7$ $8$ , and $9$ . There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$ . The answer is then $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{8}$ | 8 |
3,296 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_19 | 1 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^{\text... | 1 |
3,297 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_19 | 2 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \text... | The remaining $30\%$ of the students got $95$ points.
The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$ .
The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$ , the median is $85$ . The di... | 1 |
3,298 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_21 | 1 | Forty slips are placed into a hat, each bearing a number $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two o... | There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$
There are ${10 \choose 2} = 45$ ways to determine which two numbers to... | 162 |
3,299 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_21 | 2 | Forty slips are placed into a hat, each bearing a number $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two o... | For probability $p$ , there are $\binom{10}{1}=10$ ways to choose the number you want to show up $4$ times.
Hence, the probability is $\frac{10}{\binom{40}{4}}$
For probability $q$ , there are $\binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $\binom{4}{2}\cdot\binom{4}{2}$ ways to ... | 162 |
3,300 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23 | 1 | In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \tex... | Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$
$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$
$\frac{AB+EF}{2}=DC+EF$ , so
$AB+EF=2DC+2EF$
$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\frac{AB+DC}{2}$
$AB+\frac{AB+DC}{2}=2DC+AB+DC$ , so
$\frac{3}{2}AB+\frac... | 5 |
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