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ff3d1e839192cb832d0132fe693aad94 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_8 | 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]
$\textbf{(A) } 20\pi\qquad \textbf{(B) } 25\pi\qquad \textbf{(C) } 30\pi\qquad \textbf{(D) } 40\pi\qquad \textbf{(E) } 50\pi$ | 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, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$ . So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = \boxed{25}$ | B | 25 |
b3c974f528ef06448cc3f8772d9150b4 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_9 | 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?
$\textbf{(A) } 129\qquad \textbf{(B) } 137\qquad \textbf{(C) } 174\qquad \textbf{(D) } 233\qquad \textbf{(E) } 411$ | 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 $200\textrm{ grams}\cdot\frac{137\textrm{ calories}}{200\textrm{ grams}}=\boxed{137}.$ | B | 137 |
353deab22c46abd630b8350cb6ed1a47 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_10 | 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 $7 + 3\cdot7 + 15 =\boxed{43}$ | A | 43 |
67699eb290742ec9c7c05dc3b1629de7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_11 | 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 $7!+8!+9!$
$7!+8!+9!=5040+40320+362880=408240$
So the tens digit is $\boxed{4}$ | C | 4 |
116ea3a21ce56b4f33db9e9d61c8f009 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_15 | 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)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]
$\textbf{(A) } 6\qquad \textbf{(B) } 4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3}$ | 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 rhombus $BFDE$ is $BD$ . Since $BD$ is a side of an equilateral triangle with a side length of $s$ $BD = s$ . The longer diagonal of rhombus $ABCD$ is $AC$ . Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$ $AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3}$
The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}$ . Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $\left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3}$
Let $x$ be the area of rhombus $BFDE$ . Then $\frac{x}{24} = \frac{1}{3}$ , so $x = \boxed{8}$ | C | 8 |
116ea3a21ce56b4f33db9e9d61c8f009 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_15 | 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)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]
$\textbf{(A) } 6\qquad \textbf{(B) } 4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3}$ | 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. $8 \Longrightarrow \boxed{8}$ | C | 8 |
dd092064856f1e962458fb585ed19913 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_18 | 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 $2006 \equiv 2\bmod{6}$
$a_{2006} = a_2 = \boxed{3}$ | E | 3 |
dd092064856f1e962458fb585ed19913 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_18 | 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 $a_{2006} = a_2 = \boxed{3}$ | E | 3 |
08a6d2bc1da7d33268ae07c2fa01f729 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 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, with one triangle having side lengths $100$ times the other!
Let $AD=x$ . Then from our reasoning above, we have $AB=100x$ . Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400}$ | E | 40400 |
08a6d2bc1da7d33268ae07c2fa01f729 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 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 perpendicular. $\frac{AB}{AD} = \frac{200}{2}$ , so the area $AB \cdot AD = \frac {AB^2}{100} = \frac {2000^2 + 200^2}{100} = \boxed{40400}$ | E | 40400 |
08a6d2bc1da7d33268ae07c2fa01f729 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_20 | 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 contain $\sqrt{101}$ to ensure that $AB*AD$ is an integer. Once you know that, you can guarantee that the answer must be a multiple of 101. (If you're wondering why there can't be a fraction like $1/101$ , there can't because $y$ is an integer the number under the square root in $AD$ , is an integer.) Having that information narrows down the answer choices to just $(B)$ $4040$ , and $(E)$ $40400$
Looking back at what we found for $AB$ , which is $200\sqrt{101}$ , and what we know about $AD$ , that it has to contain $\sqrt{101}$ . we know that the answer is at least $200\sqrt{101}*\sqrt{101}=20200$ , which is already greater than answer choice $(B)$ , which is $4040$ . From this, we can conclude that the answer is $\boxed{40400}$ | E | 40400 |
3e562274b5a2c8ac61f7a3e5d1c50a62 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 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 + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$ | 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--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("$7$",(1.45,0.15)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); draw( C -- F, dashed ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",Ep,NW); label("$F$",F,S); label("$x$",(1,1)); label("$y$",(1.6,1)); [/asy]
Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$
Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$
Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$ , their bases must be in the same ratio, hence $EF:FB = 3:7$
Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$ , their areas must be in the same ratio, hence $x:(y+7) = 3:7$ , which gives us $7x = 3(y+7)$
Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$ , which solves to $x=\frac{15}{2}$ . Then $y=x+3 = \frac{15}{2}+3 = \frac{21}{2}$ , and the total area of the quadrilateral is $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{18}$ | D | 18 |
3e562274b5a2c8ac61f7a3e5d1c50a62 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 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 + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$ | 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 triangle $EFD$ is $3$
Triangle $CED$ and $DEA$ share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases $CE$ and $EA$ . The ratio is $A:(3+3) \Rightarrow A:6$ where $A$ is the area of triangle $CED$
Triangles $CEB$ and $EAB$ also share an altitude. The ratio of their areas is also equal to the ratio of bases $CE$ and $EA$ . The ratio is $(A+3+7):(3+7) \Rightarrow (A+10):10$
Because the two ratios are equal, we get the equation $\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15$ . We add the area of triangle $EDF$ to get that the total area of the quadrilateral is $\boxed{18}$ | D | 18 |
3e562274b5a2c8ac61f7a3e5d1c50a62 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_23 | 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 + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("$7$",(1.25,0.2)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); [/asy]
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$ | 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 $F$ and $C$ , we let the mass at $F$ be x and the mass at $C$ be y. Then $3x = 7y$ and $y = \frac{3}{7}x$ . Then $\frac{10}{7}x = 2$ since we add the masses for the fulcrum mass, and $x = \frac{7}{5}$ and $y = \frac{3}{5}$
To calculate the mass at a, it is merely $\frac{7}{5} - 1 = \frac{2}{5}$ which means $\frac{[BCF]}{[ACF]} = \frac{2}{5}$ or $[ACF] = 25$ . It is easy to see the answer is $\boxed{18}$ | D | 18 |
3af9ca68b9284849f7addc82162ddfe7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 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 the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | 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 $5 \not\in S$ , so for the sake of contradiction, let us suppose that $5 \in S$ . Then $5\cdot 4 = 20 | m$ so the units digit of $m$ is $0$ . Since the number has two distinct digits, each appearing twice, another digit of $m$ must be $0$ . Since Mr. Jones can't be $00$ years old, the last two digits can't be $00$ . Therefore $m$ must be of the form $d0d0$ , where $d$ is a digit. Since $m$ is divisible by $9$ , the sum of the digits of $m$ must be divisible by $9$ (see Divisibility rules for 9 ). Hence $9 | 2d$ which implies $d = 9$ . But $m = 9090$ is not divisible by $4$ , contradiction. So $5 \not\in S$ and $5$ is not the age of one of Mr. Jones' kids. $\boxed{5}$ | B | 5 |
3af9ca68b9284849f7addc82162ddfe7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 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 the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | 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 either be of the form $XX00$ or $X0X0$ (X being the other digit in the license plate) . The condition $XX00$ is impossible as Mr. Jones can't be $00$ years old. If we separate the second condition, $X0X0$ into its prime factors, we get $X\cdot 10\cdot 101$ $101$ is prime, and therefore can't account for allowing $X0X0$ being evenly divisible by the childrens' ages. The $10$ accounts for the 5 and one factor of 2. This leaves $X$ , but no single digit number contains all the prime factors besides 5 and 2 of the childrens' ages, thus $5$ $\boxed{5}$ can't be one of the childrens' ages. | B | 5 |
3af9ca68b9284849f7addc82162ddfe7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 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 the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | 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 $2^2=4$ means indivisibility by $2^3=8)$ . Since the number is divisible by $9$ , it is not the answer, and the digits have to add up to $9$ . Let $(a,b)$ be the digits: since $2\cdot(a+b) | 9$ $(a+b)|9$ . The only possible choices for $(a,b)$ are $(0,9)$ $(1,8)$ $(2,7)$ $(3,6)$ , and $(4,5)$ . The only number that works is $5544$ , but it is not divisible by $5$ $\boxed{5}$ | B | 5 |
3af9ca68b9284849f7addc82162ddfe7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_25 | 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 the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | Since 5 has the "strictest" divisibility rules of the single digit numbers, we check that case first. To be divisible by 9, the other repeating digit must be 4. Seeing that 5544 is the only arrangement of 2 4's and 2 5's that is divisible by the other single digit numbers and itself is not divisible by 5, the answer is $\boxed{5}$ .
-liu4505
~edited by mobius247 | B | 5 |
ca71497780243679a574b111429ccc77 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_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 $m-j=20-10=10 \Rightarrow \boxed{10}$ | D | 10 |
3dabece7f16ca606c09bd90430f1c9a9 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_2 | 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 $((1 \star 2) \star 3)$
$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \textrm{This\, value\, is\, not\, defined.}$ | $((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{0}$ | C | 0 |
7776c209b201910c6f778d6b6748a083 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_3 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$ | B | 4 |
a19182f42940ec02b0b659a81261601f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_5 | 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?
$\textbf{(A) } 100\qquad \textbf{(B) } 200\qquad \textbf{(C) } 300\qquad \textbf{(D) } 400\qquad \textbf{(E) } 500$ | 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 $\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3$ free windows.
So they get $3-2=1$ additional window if they purchase the windows together.
Therefore they save $1\cdot100= \boxed{100}$ | A | 100 |
c8cceb64a3dccbb857b566f88df48b95 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_6 | 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 $\frac{1200}{50}=\boxed{24}$ | B | 24 |
3e60bb51cc940e4cce7dd210909af978 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_7 | 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?
$\textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8$ | 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,
$\frac{8}{5}m + m = 13$
$\frac{13}{5}m = 13$
$m = \boxed{5}$ | B | 5 |
4cff37ff77abfa70f3c8e8382e3d7a04 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_8 | 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:
\[1^2 + (HE+1)^2=\sqrt{50}^2\]
\[1 + (HE+1)^2=50\]
\[(HE+1)^2=49\]
\[HE+1=7\]
\[HE=6\] So, the area of the square is $6^2=\boxed{36}$ | C | 36 |
d07e43fcb9a9b660578bec9e551b03d7 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 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
$m^2 = 4, n^2 = 9$
$m = \pm 2, n = \pm 3$
$a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12$
$a = 4$ or $a = -20$
So the desired sum is $(4)+(-20)=\boxed{16}$ | A | 16 |
d07e43fcb9a9b660578bec9e551b03d7 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 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 due to the $\pm$ sign when added). So we must have \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] $\frac{-32}{2} = \boxed{16}$ | A | 16 |
d07e43fcb9a9b660578bec9e551b03d7 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_10 | 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 $\boxed{16}$ | A | 16 |
bf6f86f916f25fe7d3003faa75aaa3fe | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_11 | 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,
$\frac{6n^2}{6n^3}=\frac{1}{4}$
$\frac{1}{n}=\frac{1}{4}$
$n=\boxed{4}$ | B | 4 |
6bb7097144c9d587d7164699cbedcc6d | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_13 | 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 $4 < n < 130$
Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{125}$ | E | 125 |
6bb7097144c9d587d7164699cbedcc6d | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_13 | 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 $n^{100}>4^{100}$ , which means $n>4$
Therefore $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{125}$ | E | 125 |
7d985ae6ef8552dcbcea4b663520572e | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 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 numbers range from $123$ to $420$ . So there are $4$ numbers in this case.
If the middle digit is $3$ , possible numbers range from $135$ to $630$ . So there are $6$ numbers in this case.
If the middle digit is $4$ , possible numbers range from $147$ to $840$ . So there are $8$ numbers in this case.
If the middle digit is $5$ , possible numbers range from $159$ to $951$ . So there are $9$ numbers in this case.
If the middle digit is $6$ , possible numbers range from $369$ to $963$ . So there are $7$ numbers in this case.
If the middle digit is $7$ , possible numbers range from $579$ to $975$ . So there are $5$ numbers in this case.
If the middle digit is $8$ , possible numbers range from $789$ to $987$ . So there are $3$ numbers in this case.
If the middle digit is $9$ , the only possible number is $999$ . So there is $1$ number in this case.
So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=\boxed{45}$ | E | 45 |
7d985ae6ef8552dcbcea4b663520572e | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 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 $2$ cases:
If both the first digit and the last digit are odd, then we have $1, 3, 5, 7,$ or $9$ as choices for each of these digits, and there are $5\cdot5=25$ numbers in this case.
If both the first and last digits are even, then we have $2, 4, 6, 8$ as our choices for the first digit and $0, 2, 4, 6, 8$ for the third digit. There are $4\cdot5=20$ numbers here.
The total number, then, is $20+25=\boxed{45}$ | E | 45 |
7d985ae6ef8552dcbcea4b663520572e | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 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 digits. Exactly half of these would have their sum be divisible by $2$ (since $90$ is even), so our answer is $\frac{90}{2}=\boxed{45}$ | E | 45 |
7d985ae6ef8552dcbcea4b663520572e | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_14 | 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 digits.
As there are $5$ different possible values for each starting digit of our number, we can multiply that by the number of possible first digits $(1-9)$ to achieve our desired solution. $5 \cdot 9 = \boxed{45}$ | E | 45 |
2b04cff3b664339839a9f192f0e5ce31 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_15 | 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 nonnegative multiples of $3$ that are less than or equal to $8, 5, 2$ and $1,$ respectively.
So:
$a\in\{0,3,6\}$ $3$ possibilities)
$b\in\{0,3\}$ $2$ possibilities)
$c\in\{0\}$ $1$ possibility)
$d\in\{0\}$ $1$ possibility)
So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \boxed{6}$ | E | 6 |
2b04cff3b664339839a9f192f0e5ce31 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_15 | 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 to make bigger expressions. For example, $(2 \cdot 2) \cdot 4 \cdot 4=4 \cdot 4 \cdot 4 = 4^3$ (one 2 comes from the $3!$ , and the other from the $5!$ ). Using this method, we find:
$(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3$
and
$(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3$
So, we have 6 cubes total: $1^3 ,2^3, 3^3, 4^3, 6^3,$ and $12^3$ for a total of $\boxed{6}$ cubes | E | 6 |
3520f855a63a02e7981dd413d746fceb | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_16 | 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 answer is $\boxed{10}$ | D | 10 |
3520f855a63a02e7981dd413d746fceb | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_16 | 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 a two-digit number equal $10a+b$ , where $a$ and $b$ are the tens and units digits of the number.
From the problem, we have $10a+b-(a+b)=9a$
Now let $9a=10x+y$ , where $x$ and $y$ are the tens and units digits of the number. Then it must be that $y=6$ as stated in the problem.
Note that $10a$ ends in $0$ , but $9a$ ends in $6$ , so $a=4$ . We need not to care about $b$ , since it cancels out in the calculation.
So the answer is $\boxed{10}$ , since there are $10$ numbers that have $a=4$ | D | 10 |
644b2af7283c764777048ca965b438e6 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_17 | 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 the middle term of the arithmetic sequence?
2005amc10a17.gif
$\textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13$ | 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 $\frac{60}{5}=\boxed{12}$ | D | 12 |
644b2af7283c764777048ca965b438e6 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_17 | 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 the middle term of the arithmetic sequence?
2005amc10a17.gif
$\textbf{(A) } 9\qquad \textbf{(B) } 10\qquad \textbf{(C) } 11\qquad \textbf{(D) } 12\qquad \textbf{(E) } 13$ | We know that the smallest number in this sequence must be $3 + 5 = 8$ , and the biggest number must be $7 + 9 = 16$ . Since there are $5$ terms in this sequence, we know that $8 + 4d = 16$ , or that $d = 2$ . Thus, we know that the middle term must be $8 + 2 \cdot 2 = \boxed{12}.$ ~yk2007 (Daniel K.) | null | 12 |
cdfa3f03b782515af36a24994d77b30a | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_20 | 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?
$\textbf{(A) } \frac72\qquad \textbf{(B) } \frac{7\sqrt2}{2}\qquad \textbf{(C) } \frac{5+4\sqrt2}{2}\qquad \textbf{(D) } \frac{4+5\sqrt2}{2}\qquad \textbf{(E) } 7$ | 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 $\left(\frac{\sqrt2}2\right)^2=\frac12$ , so the area of $7$ squares is $\boxed{72}$ | A | 72 |
cdfa3f03b782515af36a24994d77b30a | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_20 | 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?
$\textbf{(A) } \frac72\qquad \textbf{(B) } \frac{7\sqrt2}{2}\qquad \textbf{(C) } \frac{5+4\sqrt2}{2}\qquad \textbf{(D) } \frac{4+5\sqrt2}{2}\qquad \textbf{(E) } 7$ | 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 $12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{72}$ | A | 72 |
07d8f9f7da5174c54967b0b422387f38 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_21 | 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 $k(n+1) = 12$ for a positive integer $k$
$\frac{12}{n+1}$ is an integer when $n+1$ is a factor of $12$
The factors of $12$ are $1, 2, 3, 4, 6,$ and $12$ , so the possible values of $n$ are $0, 1, 2, 3, 5,$ and $11$
But since $0$ isn't a positive integer, only $1, 2, 3, 5,$ and $11$ are the possible values of $n$ . Therefore the number of possible values of $n$ is $\boxed{5}$ | B | 5 |
eb80804f76664747b0a20e59537b876c | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_22 | 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 the number of multiples of $12$ that are in $S$
Since $4\cdot3=12$ , every $3$ rd element of $S$ will be a multiple of $12$
Therefore the answer is $\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{668}$ | D | 668 |
56681c426b9e8cb1e4362b1dad3c1e3c | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10A_Problems/Problem_24 | 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 between two consecutive square after $576$ is greater than or equal to $49$ ,
Hence we have to consider only the prime numbers till $23$
Squaring prime numbers below $23$ including $23$ we get the following list.
$4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529$
But adding $48$ to a number ending with $9$ will result in a number ending with $7$ , but we know that a perfect square does not end in $7$ , so we can eliminate those cases to get the new list.
$4 , 25 , 121 , 361$
Adding $48$ , we get $121$ as the only possible solution.
Hence the answer is $\boxed{1}$ | B | 1 |
0497a084b3c6c9285a3680d4c98854b9 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_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. | A | 100 |
0497a084b3c6c9285a3680d4c98854b9 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_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?
$\textbf{(A) }\ 100 \qquad \textbf{(B) }\ 200 \qquad \textbf{(C) }\ 300 \qquad \textbf{(D) }\ 400 \qquad \textbf{(E) }\ 500$ | Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \div 10 = 100$ . So our answer is $\boxed{100}$ .
~HyperVoid | A | 100 |
784389bee001d2855a058930dee7dc63 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_2 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Since $x\%$ means $0.01x$ , the statement " $x\% \text{ of } x \text{ is 4}$ " can be rewritten as " $0.01x \cdot x = 4$ ":
$0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.$ | D | 20 |
784389bee001d2855a058930dee7dc63 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_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. | D | 20 |
953cced3b7cb3da1b93bc3964a635cd8 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_5 | 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?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qquad\textbf{(D) }\ \frac23 \qquad\textbf{(E) }\ \frac45$ | 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 $\boxed{25}$ | C | 25 |
2c4d690318e6d06c7080bf2ecdd4d375 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_6 | 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 $A$
$\textbf{(A) }\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | 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. | B | 2 |
84a6692ef78542d00c137588481dca0f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_7 | 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?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2}$ | 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 inscribed circle is $\dfrac{x\sqrt{2}}{2}$ . Hence, its radius is $\dfrac{x\sqrt{2}}{4}$ . The area of this circle is $\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}$ , and the area of the largest square is $x^2$ . The ratio of the areas is $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\frac{\cancel{x^2}\pi}{8}\cdot\frac{1}{\cancel{x^2}}=\boxed{8}$ | B | 8 |
84a6692ef78542d00c137588481dca0f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_7 | 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?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2}$ | 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 the area of the largest square is therefore $\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{8}.$ | B | 8 |
8982e2646db68b386eccee41d3c156bf | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 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 $BD=\boxed{3}$ | A | 3 |
8982e2646db68b386eccee41d3c156bf | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 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 more time: $8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0$ . The only positive value for $x$ is $3$ , which gives the length of $\overline{BD}$ . Thus the answer is $\boxed{3}$ | A | 3 |
8982e2646db68b386eccee41d3c156bf | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_10 | 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
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$ | A | 3 |
d7635fa432eb29a6d53461130c1937f0 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_11 | 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 $4^{\text{th}}$ terms will equal $133$ $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$ . It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$ , which leads us to the answer of $\boxed{250}$ | E | 250 |
a7bcd60334d66beab823838e6c981607 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_13 | 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 $\frac{2005}{4} = 501 \text{ }R1$ . The multiples of $12$ are $\frac{2005}{12} = 167\text{ }R1.$ So, the answer is $668+501-167-167 = \boxed{835}$ | C | 835 |
a7bcd60334d66beab823838e6c981607 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_13 | 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 $\frac{5}{12} \cdot 2005$ $\boxed{835}$ | C | 835 |
e8579bc0f2f042a144ffbc39aaa2a0c1 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_16 | 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 $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{8}$ | D | 8 |
e8579bc0f2f042a144ffbc39aaa2a0c1 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_16 | 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 into the third equation gives \[n = 4(m)\] Therefore, $n = 8p$ , so $\frac{n}{p}= \boxed{8}$ | D | 8 |
aad68df8e64319f5b3632af8bbbe6ab9 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_18 | 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?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$ | 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 $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{8}$ | D | 8 |
bdbf89594b3a6fe9f0010e26923d8de3 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_19 | 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?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | 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 $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$ . The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$ . The difference between the mean and median, therefore, is $\boxed{1}$ | B | 1 |
bdbf89594b3a6fe9f0010e26923d8de3 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_19 | 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?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | 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 difference between the mean and the median is $\boxed{1}$ | B | 1 |
90e87be084595067a9561d554324bb80 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_20 | What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$ | We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 \div 120$ to get $\boxed{53332.8}$ | C | 53332.8 |
90e87be084595067a9561d554324bb80 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_20 | What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$ | We can first solve for the mean for the digits $1, 3, 5, 7,$ and $9$ since each is $2$ away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the $8$ with a $9$ . We can start how be calculating the gain when the $8$ was in the ones digit. Since there are $4! = 24$ numbers with the $8$ in the ones digit and $1$ was gain from each of them, $24$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the $9$ instead of the $8$ to get $6666600-266664=6399936$ . Finally, we divide this by $120$ to get the average. $\frac{6399936}{120}= \boxed{53332.8}$ | C | 53332.8 |
90e87be084595067a9561d554324bb80 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_20 | What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$ | The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$ .
Values occur in every place so $4.8 * 11111 = \boxed{53332.8}$ | C | 53332.8 |
8eb3dc22b20e6fc18f7f4d1c5962a7c5 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_21 | 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 of the slips bear a number $a$ and the other two bear a number $b \neq a$ . What is the value of $q/p$
$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$ | 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 pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}$
Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{162}$ | A | 162 |
8eb3dc22b20e6fc18f7f4d1c5962a7c5 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_21 | 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 of the slips bear a number $a$ and the other two bear a number $b \neq a$ . What is the value of $q/p$
$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$ | 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 pick which slips you want out of the $4$ of each.
Hence, the probability is $\frac{45\cdot6\cdot6}{\binom{40}{4}}$
Hence, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{162}$ | A | 162 |
d4b44cf807589ce17cf239ee81816abb | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_22 | For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \cdots + n?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21$ | Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ , the condition is equivalent to having an integer value for $\frac{n!} {\frac{n(n+1)}{2}}$ . This reduces, when $n\ge 1$ , to having an integer value for $\frac{2(n-1)!}{n+1}$ . This fraction is an
integer unless $n+1$ is an odd prime. There are $8$ odd primes less than or equal to $24$ , so there
are $24 - 8 = \boxed{16}$ numbers less than or equal to $24$ that satisfy the condition. | C | 16 |
8cae1bb2d71c7f982f824f83b40c2d90 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23 | 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 $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$ | 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
$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$ , and
$\frac{1}{2}AB=\frac{5}{2}DC$
$\frac{AB}{DC} = \boxed{5}$ | C | 5 |
8cae1bb2d71c7f982f824f83b40c2d90 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23 | 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 $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$ | Mark $DC=z$ $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ $FECD$ are the same. Mark the height to be $h$
Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$
From this, we get that $x=2z+y$
We also get that $\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)$
Simplifying, we get that $2x=z+3y$
Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$
Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$
Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$
Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=\boxed{5}$ | C | 5 |
8cae1bb2d71c7f982f824f83b40c2d90 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_23 | 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 $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$ | Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$ . The area of the smaller trapezoid is $A$ $h \cdot \frac {b_1 + b_2}{2}$ . The area of the larger trapezoid is $A$ $h \cdot \frac {b_1 + x}{2}$ . Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$
The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$ . We have the equation $\frac {x}{2} = 5$ $x$ = 10, and our answer is $\boxed{5}$ | C | 5 |
b2e56efec5a9e6415ac0c8d212c184ac | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24 | Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$ | Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero.
Then, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$
Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 \cdot 11$ , so $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum value of $a+b$ is $9+8=17$ , so $a+b=11$
Then, we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=\boxed{154}$ | E | 154 |
b2e56efec5a9e6415ac0c8d212c184ac | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24 | Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$ | The first steps are the same as Solution 1. Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than $10$ ).
Like Solution 1, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . This is where the solution diverges.
We know that the left side of the equation is a perfect square because $m$ is an integer. If we factor $99$ into its prime factors, we get $3^2\cdot 11$ . In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals $11$ while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ . The sum of the numbers is $\boxed{154}$ | E | 154 |
b2e56efec5a9e6415ac0c8d212c184ac | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24 | Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$ | Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$
Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$
Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$
A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $\boxed{154}$ | E | 154 |
b2e56efec5a9e6415ac0c8d212c184ac | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_24 | Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$ | Continue the same as Solution $3$ until we get $33$ . Knowing that $33^2 = 1089$ , we have narrowed down our Pythagorean triples. We know that the $2$ other squares should be larger than $33^2$ , so we can start testing.
If we start testing the $40$ s, it is fruitless since the closest to $33^2$ would be $33 - 45 - 54$ which is not a Pythagorean triple. We can start by testing out the $50$ s, and it turns our that $33 - 56 - 65$ is a Pythagorean triple. Therefore, our answer is $33+56+65$ $\boxed{154}$ | E | 154 |
5918d3689858b30072da883c77941fab | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25 | subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$ | The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26+99$ $27+98$ $\ldots$ $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{62}$ | C | 62 |
5918d3689858b30072da883c77941fab | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25 | subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$ | "Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\boxed{62}$ | C | 62 |
5918d3689858b30072da883c77941fab | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25 | subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$ | The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$ . So, we have to find two consecutive numbers, $n$ and $n+1$ , whose sum is $125$ . Setting up our equation, we have $n+(n+1) = 2n+1 = 125$ . When we solve for $n$ , we get $n =\boxed{62}$ | C | 62 |
5918d3689858b30072da883c77941fab | https://artofproblemsolving.com/wiki/index.php/2005_AMC_10B_Problems/Problem_25 | subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$ | We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get $B =\boxed{62}$ | C | 62 |
5b873b3e671c5ab18709adeac9b14e18 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_1 | You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?
$\mathrm{(A) \ } 250\qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 1500 \qquad \mathrm{(D) \ } 7500 \qquad \mathrm{(E) \ } 9000$ | There are $6$ people to split the $1500$ dollars among, so each person must raise $\frac{1500}6=250$ dollars. $\Rightarrow\boxed{250}$ | A | 250 |
949ed950e355402ca376b73686dada44 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_3 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | $20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{29}$ | E | 29 |
949ed950e355402ca376b73686dada44 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_3 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{29}$ | E | 29 |
5c1673081291934e4f151e175b303016 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_4 | What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$ | $|x-1|$ is the distance between $x$ and $1$ $|x-2|$ is the distance between $x$ and $2$
Therefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{32}$ | D | 32 |
5c1673081291934e4f151e175b303016 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_4 | What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$ | We know that either $x-1=x-2$ or $x-1=-(x-2)$ . The first equation simplifies to $-1=-2$ , which is false, so $x-1=-(x-2)$ . Solving, we get $x=\frac{3}{2}\Rightarrow\boxed{32}$ | D | 32 |
7b14be6140a36bcf55b3ac72def941ff | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$ | E | 26 |
7b14be6140a36bcf55b3ac72def941ff | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$ | null | 26 |
7b14be6140a36bcf55b3ac72def941ff | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_6 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters. Adding them together gives the answer of $\boxed{26}$ | E | 26 |
b5ed9ad9ad5679b54f466837b74d9e37 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_7 | A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?
$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$ | There are $5\times8=40$ oranges on the $1^{\text{st}}$ layer of the stack. The $2^{\text{nd}}$ layer that is added on top of the first will be a layer of $4\times7=28$ oranges. When the third layer is added on top of the $2^{\text{nd}}$ , it will be a layer of $3\times6=18$ oranges, etc.
Therefore, there are $5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100$ oranges in the stack. $\boxed{100}$ | C | 100 |
ac902253c0c7c91284afb055593e6081 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.
After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{37}$ rounds until the game ends. | B | 37 |
ac902253c0c7c91284afb055593e6081 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.
First round: $15,14,13$ (given)
Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$
We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\boxed{37}$ | B | 37 |
ac902253c0c7c91284afb055593e6081 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\boxed{37}$ | B | 37 |
4fdca0ad68e25e359c10aadb7f68f6d6 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ $[\triangle CDB]$ . So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$ . Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$ | B | 4 |
4fdca0ad68e25e359c10aadb7f68f6d6 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Since $AE \perp AB$ and $BC \perp AB$ $AE \parallel BC$ . By alternate interior angles and $AA\sim$ , we find that $\triangle ADE \sim \triangle CDB$ , with side length ratio $\frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$ . Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$ | B | 4 |
4fdca0ad68e25e359c10aadb7f68f6d6 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Let $[X]$ represent the area of figure $X$ . Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{4}$ | B | 4 |
4fdca0ad68e25e359c10aadb7f68f6d6 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_9 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$ .
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen easily in the diagram.
$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$
$Area(\triangle ABC) = 0.5 * 4 * 16 = 12$
$Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{4}$ | B | 4 |
2393dce2ca11fb3d451c14d27465d347 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_11 | A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$ | When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$ | C | 36 |
3d6ba4a9134cc345e3a668e2330c196f | https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_12 | Henry's Hamburger Haven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two,or three meat patties and any collection of condiments. How many different kinds of hamburgers can be ordered?
$\text{(A) \ } 24 \qquad \text{(B) \ } 256 \qquad \text{(C) \ } 768 \qquad \text{(D) \ } 40,320 \qquad \text{(E) \ } 120,960$ | For each condiment, a customer may either choose to order it or not. There are $8$ total condiments to choose from. Therefore, there are $2^8=256$ ways to order the condiments. There are also $3$ choices for the meat, making a total of $256\times3=768$ possible hamburgers. $\boxed{768}$ | C | 768 |
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