id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
701 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_1 | 1 | circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)}\ 5 \qquad {(E)}\ 6$ | The two lines can both intersect the circle twice, and can intersect each other once, so $2+2+1= \boxed{5}.$ | 5 |
702 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_2 | 1 | How many different combinations of $5 bills and $2 bills can be used to make a total of $17? Order does not matter in this problem.
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 4 \qquad \text {(D)}\ 5 \qquad \text {(E)}\ 6$ | You cannot use more than 4 $<dollar>5$ bills, but if you use 3 $<dollar>5$ bills, you can add another $<dollar>2$ bill to make a combination. You cannot use 2 $<dollar>5$ bills since you have an odd number of dollars that need to be paid with $<dollar>2$ bills. You can also use 1 $<dollar>5$ bill and 6 $<dollar>2$ bill... | 2 |
703 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_3 | 1 | What is the smallest possible average of four distinct positive even integers?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$ | In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\frac{2+4+6+8}{4}=\boxed{5}$ | 5 |
704 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4 | 1 | The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$ | The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{4}$ | 4 |
705 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_4 | 2 | The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$ | The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is $\boxed{4}$ | 4 |
706 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_7 | 1 | The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2... | From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\frac{5}{25} \cdot 100 = \boxed{20}$ | 20 |
707 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_8 | 1 | Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
[asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y <= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(1... | France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$ . The total number of stamps is $15+9=\boxed{24}$ | 24 |
708 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11 | 1 | A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0... | The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{13}$ more tiles than the sixth square. | 13 |
709 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11 | 2 | A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0... | The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh ... | 13 |
710 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_11 | 3 | A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0... | We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that $7^2 - 6^2$ . Now we can see that the seventh square has $(7-6)(7+6) =$ $\boxed{13}$ more tiles than the sixth square. | 13 |
711 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_12 | 1 | A board game spinner is divided into three regions labeled $A$ $B$ and $C$ . The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$ . The probability of the arrow stopping on region $C$ is:
$\text{(A)}\ \frac{1}{12}\qquad\text{(B)}\ \frac{1}{6}\qquad\text{(C)}\ \frac{1}{... | Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\frac{1}{2}-\frac{1}{3}=\boxed{16}$ | 16 |
712 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_13 | 1 | For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
$\text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C... | Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed.
$2^3=8$
Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8.
$8\cdot125= \boxed{1000}$ | 0 |
713 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_14 | 1 | A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is
$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$ | Let's assume that each item is $100$ dollars. First we take off $30\%$ off of $100$ dollars. $100\cdot0.7=70$
Next, we take off the extra $20\%$ as asked by the problem. $70\cdot0.80=56$
So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. So the fina... | 44 |
714 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17 | 1 | In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$ . If she got six right her score would be $(6*5)-(2*4)=22$ . That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$ . Thus, our answer is $\boxed{7}$ . ~avamarora | 7 |
715 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17 | 2 | In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | We can start with the full score, 50, and subtract not only 2 points for each incorrect answer but also the 5 points we gave her credit for. This expression is equivalent to her score, 29.
Let $x$ be the number of questions she answers correctly. Then, we will represent the number incorrect by $10-x$
\begin{align*} 50... | 7 |
716 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_17 | 3 | In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | Suppose she got $x$ questions right. Then she got $10 - x$ questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve $5x - 2(10 - x) = 29$ to get that $x = \boxed{7}$ | 7 |
717 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_18 | 1 | Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?
$\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text... | Converting into minutes and adding, we get that she skated $75*5+90*3+x = 375+270+x = 645+x$ minutes total, where $x$ is the amount she skated on day $9$ . Dividing by $9$ to get the average, we get $\frac{645+x}{9}=85$ . Solving for $x$ \[645+x=765\] \[x=120\] Now we convert back into hours and minutes to get $\boxed{... | 2 |
718 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_18 | 2 | Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?
$\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text... | For the first five days, each day you are $10$ minutes short of $85$ minutes. And for the next three days, you are $5$ minutes above $85$ minutes. So in total you are missing $3*5-5*10$ , which equals to negative $35$ . So on the ninth day, to have an average of $85$ minutes, Gage need to skate for $85+35$ minutes, whi... | 2 |
719 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_19 | 1 | How many whole numbers between 99 and 999 contain exactly one 0?
$\text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180$ | Numbers with exactly one zero have the form $\overline{a0b}$ or $\overline{ab0}$ , where the $a,b \neq 0$ . There are $(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}$ | 162 |
720 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | 1 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.... | The shaded region is a right trapezoid. Assume WLOG that $YZ=8$ . Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$ . Because the line $AB$ is a midsegment, the top base of the trapezoid is $\frac12 AB = \frac14 YZ = 2$ . Also, $AB$ divides $XC$ in two, so the height of the trape... | 3 |
721 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | 2 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.... | Since $A$ and $B$ are the midpoints of $XY$ and $XZ$ , respectively, $AY=AX=BX=BZ$ .
Draw segments $AC$ and $BC$ .
Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that $YC=CZ$ $AB$ is the line that connects the midpoints of two sides of a triangle together, ... | 3 |
722 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | 3 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.... | We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$ . By solving, we have \[\frac{XC\cdot YZ}{2} = 8,\] \[XC\cdot YZ = 16.\]
With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY ... | 3 |
723 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_20 | 4 | The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$ . Altitude $\overline{XC}$ bisects $\overline{YZ}$ . The area (in square inches) of the shaded region is
[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.... | $\usepackage[dvipsnames]{xcolor} \textcolor{BlueViolet}{\text{Super fast after convincing yourself they are congruent}}$ [asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), ... | 3 |
724 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22 | 1 | Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.
[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(... | Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \boxed{26}$ | 26 |
725 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_22 | 2 | Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.
[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(... | We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for $2($ front surface area $+$ side surface area $+$ top surface area $)$ . We find that th... | 26 |
726 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_23 | 1 | A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?
[asy] /* AMC8 2002 #23 Problem */ fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--... | The same pattern is repeated for every $6 \times 6$ tile. Looking closer, there is also symmetry of the top $3 \times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\boxed{49... | 49 |
727 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_24 | 1 | Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
$\text{(A... | A pear gives $8/3$ ounces of juice per pear. An orange gives $8/2=4$ ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be
\[\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{40}\] | 40 |
728 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_24 | 2 | Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
$\text{(A... | Since it doesn't matter how many pears and oranges there are, you can make the number of them whatever you like. In this case, we could use $6$ , because it's the LCM of $2$ and $3$ . Then for the $6$ pears, there are $6/3*8=16$ ounces of pear juice. For the 6 oranges, there are $6/2*8=24$ ounces of orange juice. Since... | 40 |
729 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_25 | 1 | Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
$\text{(A)}\ \frac{1}{... | Since Ott gets equal amounts of money from each friend, we can say that he gets $x$ dollars from each friend. This means that Moe has $5x$ dollars, Loki has $4x$ dollars, and Nick has $3x$ dollars. The total amount is $12x$ dollars, and since Ott gets $3x$ dollars total, $\frac{3x}{12x}= \frac{3}{12} = \boxed{14}$ | 14 |
730 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_25 | 2 | Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
$\text{(A)}\ \frac{1}{... | Assume Moe, Loki, and Nick each give Ott $$ 1$ . Therefore, Moe has $$ 5$ , Loki has $$ 4$ , and Nick has $$ 3$ . After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original; which is $$ 12$ . Ott gets $$ 1$ $+$ $$ 1$ $+$ $$ 1$ $=$ $$ 3$ . Thus, the answe... | 14 |
731 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_1 | 1 | Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$ | It will take him $300\cdot2=600$ seconds to paint all the dimples.
This is equivalent to $\frac{600}{60}=10$ minutes $\Rightarrow \boxed{10}$ | 10 |
732 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2 | 1 | I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$ | Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the corre... | 8 |
733 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_2 | 2 | I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$ | Use the answers to attempt to "reverse engineer" an appropriate pair of numbers. Looking at option $A$ , guess that one of the numbers is $3$ . If the sum of two numbers is $11$ and one is $3$ , then other must be $11 - 3 = 8$ . The product of those numbers is $3\cdot 8 = 24$ , which is the second condition of the p... | 8 |
734 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_3 | 1 | Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?
$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$ | Since Anjou has $\frac{1}{3}$ the amount of money as Granny Smith and Granny Smith has $ $63$ , Anjou has $\frac{1}{3}\times63=21$ dollars. Elberta has $ $2$ more than this, so she has $ $23$ , or $\boxed{23}$ | 23 |
735 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_6 | 1 | Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 105 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 140$ | There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $\frac{60}{3}=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\times5=100$ feet, $\boxed{100}$ | 100 |
736 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7 | 1 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (in... | The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ , so the area is $\frac{(6)(7)}{2}=21, \boxed{21}$ | 21 |
737 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7 | 2 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (in... | Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside,... | 21 |
738 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_7 | 3 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (in... | Pick's Theorem states: \[\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1\] as the area of a figure on a grid. Counting, we see there are $4$ boundary points and $20$ interior points. Therefore, we have \[\frac{4}{2}+20-1\implies 20+1\implies 21.\] Hence, the answer is $\boxed{21}$ | 21 |
739 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_8 | 1 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] size(85... | Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ , so the diagonals of the large kite are $18$ and $21$ , and the amount of bracing Genevieve needs is... | 39 |
740 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_9 | 1 | To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
[asy] for (in... | The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\times3(7)$ , or $18\times21$ , so the area is $(18)(21)=378$ . The area of the kite is half of the area of the rectangle as you can see, so the area of the waste material is also half the area of the rectangle. Thus, the area... | 189 |
741 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_10 | 1 | A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars... | $2000\%$ is equivalent to $20\times100\%$ . Therefore, $2000\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\times1={20}$ dollars, $\boxed{20}$ | 20 |
742 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_10 | 2 | A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars... | Since $2000\%$ is just $\frac{2000}{100}$ , we can multiply that by $100$ , because four quarters is a $100$ cents. After the multiplication, we get $2000$ . Since our answer is in cents right now, we need to convert it to dollars, which would be $\boxed{20}$ dollars. | 20 |
743 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11 | 1 | Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) {... | [asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewi... | 18 |
744 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_11 | 2 | Points $A$ $B$ $C$ and $D$ have these coordinates: $A(3,2)$ $B(3,-2)$ $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is
[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) {... | Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.
$A_{trap} = A_{tri} + A_{rect}$
$A_... | 18 |
745 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_12 | 1 | If $a\otimes b = \dfrac{a + b}{a - b}$ , then $(6\otimes 4)\otimes 3 =$
$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$ | $6\otimes4=\frac{6+4}{6-4}=5$ $5\otimes3=\frac{5+3}{5-3}=4, \boxed{4}$ | 4 |
746 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_13 | 1 | Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{... | There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\frac{5}{36}$ of the students prefer cherry pie, so $\fra... | 50 |
747 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_14 | 1 | Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$ | There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, ... | 72 |
748 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_15 | 1 | Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?
$\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 33 \qquad \tex... | After the $4$ minutes of Homer peeling alone, he had peeled $4\times3=12$ potatoes. This means that there are $44-12=32$ potatoes left. Once Christen joins him, the two are peeling potatoes at a rate of $3+5=8$ potatoes per minute. So, they finish peeling after another $\frac{32}{8}=4$ minutes. In these $4$ minutes, Ch... | 20 |
749 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_21 | 1 | The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$ | Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given... | 35 |
750 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22 | 1 | On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$ | The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\boxed{97}$ | 97 |
751 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_22 | 2 | On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$ | We can equivalently construct the following rules: You have 100 point at first, but if you give the wrong answer, you will lose 5 points, if you don't answer a question you will lose 4 points. Obviously, you can lose 10 points, 9 points, 8 points, 5 points or 4 points, but you cannot lose 3 points. The answer is $\boxe... | 97 |
752 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23 | 1 | Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(... | There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to $\triangle RST$ There is obviously only $... | 4 |
753 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23 | 2 | Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(... | We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We c... | 4 |
754 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_23 | 3 | Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(... | Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$ $\triangle RYT$ $\triangle RYZ$ $\triangle RST$ . So the answer is $\boxed{4}$ | 4 |
755 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_24 | 1 | Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
[asy] draw((0,0)--(4,4*sqrt(3))); draw(... | Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue t... | 5 |
756 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_8_Problems/Problem_25 | 1 | There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$ | We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is ... | 425 |
757 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1 | 1 | Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$ | If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old.
If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$
So, the answer is $\boxed{16}$ | 16 |
758 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_1 | 2 | Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$ | Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old.
Now we just find Caitlin's age by doing $21-5$ . This makes $16$ or $\boxed{16}$ | 16 |
759 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2 | 1 | Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$ . The reciprocal of $2$ is $1/2$ , but $2$ is not less than $1/2$ . The reciprocal of $-2$ is $-1/2$ , and $-2$ is less than $-1/2$ , so it is $\boxed{2}$ | 2 |
760 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2 | 2 | Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$
Multiplication by $x$ can be done if you do it in three parts: $x>0$ $x=0$ , and $x<0$ . You have to be careful about the direction of the inequality, as you do not know the sign of $x$
If $x>0$ , the sign of the inequality rem... | 2 |
761 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2 | 3 | Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | Starting again with $x < \frac{1}{x}$ , we avoid multiplication by $x$ . Instead, move everything to the left, and find a common denominator:
$x < \frac{1}{x}$
$x - \frac{1}{x} < 0$
$\frac{x^2 - 1}{x} < 0$
$\frac{(x+1)(x-1)}{x} < 0$
Divide this expression at $x=-1$ $x=0$ , and $x=1$ , as those are the three points wh... | 2 |
762 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_2 | 4 | Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | We can find out all of their reciprocals. Now we compare and see that the answer is $\boxed{2}$ | 2 |
763 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_3 | 1 | How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$ | The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ $3$ $4$ $5$ , and $6$ , so the answer is $\boxed{5}$ | 5 |
764 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6 | 1 | Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4... | The side of the large square is $1 + 3 + 1 = 5$ , so the area of the large square is $5^2 = 25$
The area of the middle square is $3^2$ , and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$
Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$ . This is the area of the two congruent... | 7 |
765 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6 | 2 | Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4... | The shaded area can be divided into two regions: one rectangle that is 1 by 3, and one rectangle that is 4 by 1. (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.) Either way, the total area of these two regions is $3 + 4 = 7$ , and the answer is $\boxed{7}$ | 7 |
766 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6 | 3 | Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4... | Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper. When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{7}$ as the answer. | 7 |
767 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_6 | 4 | Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4... | In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of $4\cdot4=16$ . In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$ . When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{7}$ | 7 |
768 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_7 | 1 | What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$
$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$ | The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices
exist: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$ .
The latter is smaller, so $\boxed{280}$ | 280 |
769 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_8 | 1 | problem_id
f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered $1$ through $6$...
f530163b6184f697cd4b5054402e0ccf Three dice with faces numbered 1 through 6 are...
Name: Text, dtype: object | The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers
on the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .
This leaves $63 - 22 = \boxed{41}$ not seen. | 41 |
770 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_9 | 1 | Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0... | The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .
The only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with
a $\boxed{6}$ | 6 |
771 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_10 | 1 | Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
$\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$ | Shea has grown $20\%$ , if x was her original height, then $1.2x = 60$ , so she was originally $\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase. Ara also started off at $50$ inches. Since Ara grew half as much as Shea, Ara grew $\frac{10}{2} = 5$ inches. Therefore, Ara is now $50+5=55$ inches ... | 55 |
772 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_11 | 1 | The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$ | Casework by the units digit $u$ will help organize the answer.
$u=0$ gives no solutions, since no real numbers are divisible by $0$
$u=1$ has $4$ solutions, since all numbers are divisible by $1$
$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).
$u=3$ has $1$ solution: $33$ $\pm ... | 17 |
773 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_12 | 1 | A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wa... | Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the... | 353 |
774 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_13 | 1 | In triangle $CAT$ , we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$ . If $\overline{TR}$ bisects $\angle ATC$ , then $\angle CRT =$
[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("$A$",A,N); label("$T$",T,SE); labe... | In $\triangle ACT$ , the three angles sum to $180^\circ$ , and $\angle C = \angle T$
$\angle CAT + \angle ATC + \angle ACT = 180$
$36 + \angle ATC + \angle ATC = 180$
$2 \angle ATC = 144$
$\angle ATC = 72$
Since $\angle ATC$ is bisected by $\overline{TR}$ $\angle RTC = \frac{72}{2} = 36$
Now focusing on the smaller $\t... | 72 |
775 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15 | 1 | Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2... | The large triangle $ABC$ has sides of length $4$ . The medium triangle has sides of length $2$ . The small triangle has sides of length $1$ . There are $3$ segment sizes, and all segments depicted are one of these lengths.
Starting at $A$ and going clockwise, the perimeter is:
$AB + BC + CD + DE + EF + FG + GA$
$4 +... | 15 |
776 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_15 | 2 | Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2... | The perimeter of $ABCDEFG$ is the perimeter of the three triangles, minus segments $AD$ and $EG$ , which are on the interior of the figure. Because each of these segments is on two triangles, each segment must be subtracted two times.
As in solution 1, the sides of the triangles are $4, 2,$ and $1$ , and the perimeter... | 15 |
777 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_16 | 1 | In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$ | The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$ , the area is $40\cdot 10=400$ square meters or $\boxed{400}$ | 400 |
778 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_19 | 1 | Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?
[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,... | Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle.
These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that ... | 50 |
779 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20 | 1 | You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.
You must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and q... | 1 |
780 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20 | 2 | You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\boxed{1}$ | 1 |
781 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_20 | 3 | You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$ . Now we have $ $1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $ $0.75$ with $6$ coins. We take a few educated gues... | 1 |
782 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_21 | 1 | Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$ | Divide it into $2$ cases:
1) Keiko and Ephriam both get $0$ heads:
This means that they both roll all tails, so there is only $1$ way for this to happen.
2) Keiko and Ephriam both get $1$ head:
For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways sin... | 38 |
783 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_22 | 1 | A cube has edge length $2$ . Suppose that we glue a cube of edge length $1$ on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
[asy] draw((0,0)--... | The original cube has $6$ faces, each with an area of $2\cdot 2 = 4$ square units. Thus the original figure had a total surface area of $24$ square units.
The new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of of $6$ additional square units ad... | 17 |
784 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_23 | 1 | There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qqu... | Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$
So if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \cdot 5 = 20$
If the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \cdot 8 = ... | 6 |
785 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_24 | 1 | If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$ , then $\angle B+\angle D =$
[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); ... | As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone. If you see $\triangle BFD$ , then you can see that the problem is solved quickly after determining $\angle... | 80 |
786 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25 | 1 | The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$"... | To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions. Labelling the right midpoint as $M$ , and the bottom midpo... | 27 |
787 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25 | 2 | The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$"... | The above answer is fast, but satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$
Labelling $m$ and $n$ as the right... | 27 |
788 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_8_Problems/Problem_25 | 3 | The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$"... | Let's assume, for simplicity, that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,$ $4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,$ $4\cdot 9\cdot\frac{1}{2} = 18.$ Adding these up, we get $45$ , and subtracting that from $72$ , we get $27$ , so th... | 27 |
789 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_2 | 1 | What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SS... | At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$
This makes them $\frac{1}{6}$ th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{60}$ | 60 |
790 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_2 | 2 | What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SS... | We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$ .
If the time is $10:00$ , then the hour hand will be on $10$ , and the minute hand will be on, $12$ , making them $2$ numbers apart, so they will be $... | 60 |
791 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4 | 1 | The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label... | After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\boxed{15}$ | 15 |
792 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_4 | 2 | The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9)); label... | We see that each dot is $15$ units away from the nearest one above it. So the answer is $\boxed{15}$ | 15 |
793 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_5 | 1 | A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)... | We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \times 60$ which is $1200$ $1600-1200=400$ w... | 400 |
794 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_7 | 1 | The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \te... | There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\boxed{130}$ | 130 |
795 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_9 | 1 | Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N);... | Plants shared by two beds have been counted
twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{1150}$ | 150 |
796 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_9 | 2 | Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N);... | Bed A has $350$ plants it doesn't
share with B or C. Bed B has $400$ plants it doesn't
share with A or C. And C has $250$ it doesn't share
with A or B. The total is $350 + 400 + 250 + 50 + 100 = \boxed{1150}$ plants. | 150 |
797 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | 1 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{27}$ | 27 |
798 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_12 | 2 | The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$ | The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27}.$ -Clara Garza | 27 |
799 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_13 | 1 | The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qqua... | First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is div... | 28 |
800 | https://artofproblemsolving.com/wiki/index.php/1999_AMC_8_Problems/Problem_14 | 1 | In trapezoid $ABCD$ , the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is
[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16... | [asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8)); label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$8$",(8,0),S); label("$3$",(4,1.5),E); label("$4$",(2,0),S... | 34 |
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