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653ed0ede390cd6e9b7285b1559e0183 | https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_20 | In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then what is the value of $C$
[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N); label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N); label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$ | From this we have \[111A+11B+C=300.\] Clearly, $A<3$ . Since $B,C\leq 9$ \[111A > 201 \Rightarrow A\geq 2.\] Thus, $A=2$ and $11B+C=78$ . From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{1}$ | A | 1 |
653ed0ede390cd6e9b7285b1559e0183 | https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_20 | In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then what is the value of $C$
[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N); label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N); label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$ | Using logic, $a+b+c= 10$ , therefore $b+a+1$ (from the carry over) $= 10$ .
So $b+a=9$ $A+1=3$
Thus, $A=2$ and $11B+C=78$ . From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{1}$ | A | 1 |
a773006f72883bc9636312d7b984b823 | https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_21 | For every $3^\circ$ rise in temperature, the volume of a certain gas expands by $4$ cubic centimeters. If the volume of the gas is $24$ cubic centimeters when the temperature is $32^\circ$ , what was the volume of the gas in cubic centimeters when the temperature was $20^\circ$
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 40$ | We know that \[T=32-3k\Rightarrow V=24-4k.\] Setting $k=4$ , we get the volume when the temperature is $20^\circ$ , which is $24-4(4)=8\rightarrow \boxed{8}$ | A | 8 |
25a80f069c36643742e7ab9e8da631a8 | https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_23 | The Pythagoras High School band has $100$ female and $80$ male members. The Pythagoras High School orchestra has $80$ female and $100$ male members. There are $60$ females who are members in both band and orchestra. Altogether, there are $230$ students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 70$ | There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra. Suppose $x$ males are in both band and orchestra. \[80+100-x=110\Rightarrow x=70.\] Thus, the number of males in band but not orchestra is $80-70=10\rightarrow \boxed{10}$ | A | 10 |
430616845f276646aa05926805eae2a3 | https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_24 | A cube of edge $3$ cm is cut into $N$ smaller cubes, not all the same size. If the edge of each of the smaller cubes is a whole number of centimeters, then $N=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$ | If none of the cubes have edge length $2$ , then all of the cubes have edge length $1$ , meaning they all are the same size, a contradiction.
It is clearly impossible to split a cube of edge $3$ into two or more cubes of edge $2$ with extra unit cubes, so there is one $2\times 2\times 2$ cube and $3^3-2^3=19$ unit cubes.
The total number of cubes, $N$ , is $1+19=20\rightarrow \boxed{20}$ | E | 20 |
30e326008397a5e4f3bd88f738d85e79 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_1 | What is the smallest sum of two $3$ -digit numbers that can be obtained by placing each of the six digits $4,5,6,7,8,9$ in one of the six boxes in this addition problem?
[asy] unitsize(12); draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle); draw((4,1)--(6,1)--(6,3)--(4,3)--cycle); draw((4,4)--(6,4)--(6,6)--(4,6)--cycle); draw((7,1)--(9,1)--(9,3)--(7,3)--cycle); draw((7,4)--(9,4)--(9,6)--(7,6)--cycle); [/asy]
$\text{(A)}\ 947 \qquad \text{(B)}\ 1037 \qquad \text{(C)}\ 1047 \qquad \text{(D)}\ 1056 \qquad \text{(E)}\ 1245$ | Let the two three-digit numbers be $\overline{abc}$ and $\overline{def}$ . Their sum is equal to $100(a+d)+10(b+e)+(c+f)$
To minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ . Similarly, we let $b=6$ $e=7$ , and then $c=8$ and $f=9$ . The sum is \[100(9)+10(13)+(17)=1047 \rightarrow \boxed{1047}\] | C | 1047 |
cffb7a2f091bf2623a79e7c387e5793e | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_2 | Which digit of $.12345$ , when changed to $9$ , gives the largest number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | When dealing with positive decimals, the leftmost digits affect the change in value more. Thus, to get the largest number, we change the $1$ to a $9 \rightarrow \boxed{1}$ | A | 1 |
cb80c78d4ee6a309a22214832b45e401 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_4 | Which of the following could not be the units digit [ones digit] of the square of a whole number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | We see that $1^2=1$ $2^2=4$ $5^2=25$ , and $4^2=16$ , so already we know that either $\text{E}$ is the answer or the problem has some issues.
For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from $0$ through $9$ inclusive. Testing shows that $8$ is unachievable, so the answer is $\boxed{8}$ | E | 8 |
aa4ae4aaba451e134edd444f9dda9363 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_5 | Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$ | Clearly, \[.4<.48017<.5\] Since the function $f(x)=x^3$ is strictly increasing, we can say that \[.4^3<.48017^3<.5^3\] from which it follows that $\text{A}$ is much too small and $\text{C}$ is much too large, so $\boxed{0.110}$ is the answer. | B | 0.110 |
aa4ae4aaba451e134edd444f9dda9363 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_5 | Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$ | Since $0.48017$ is quite close to $0.5$ , or $\dfrac{1}{2}$ , we can look for the answer choice that is just below $(\dfrac{1}{2})^3=\dfrac{1}{8}=0.125$ , which would be $\boxed{0.110}$ | B | 0.110 |
28628d7625d679eb0d59653bdb640573 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_7 | When three different numbers from the set $\{ -3, -2, -1, 4, 5 \}$ are multiplied, the largest possible product is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$ | First we try for a positive product, meaning we either pick three positive numbers or one positive number and two negative numbers.
It is clearly impossible to pick three positive numbers. If we try the second case, we want to pick the numbers with the largest absolute values, so we choose $5$ $-3$ and $-2$ . Their product is $30\rightarrow \boxed{30}$ | C | 30 |
c11a7f33cc9b0d7101498fd3f71f1cc8 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_8 | A dress originally priced at $80$ dollars was put on sale for $25\%$ off. If $10\%$ tax was added to the sale price, then the total selling price (in dollars) of the dress was
$\text{(A)}\ \text{45 dollars} \qquad \text{(B)}\ \text{52 dollars} \qquad \text{(C)}\ \text{54 dollars} \qquad \text{(D)}\ \text{66 dollars} \qquad \text{(E)}\ \text{68 dollars}$ | After the price reduction, the sale price is $80-.25\times 80 = 60$ dollars. The tax makes the final price $60+.1\times 60 = 66$ dollars $\rightarrow \boxed{66}$ | D | 66 |
a2c99eb02c231f549fefcfb6d5206017 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_11 | The numbers on the faces of this cube are consecutive whole numbers. The sum of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("$15$",(1.5,1.2),N); label("$11$",(4,2.3),N); label("$14$",(2.5,3.7),N); [/asy]
$\text{(A)}\ 75 \qquad \text{(B)}\ 76 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 81$ | The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$
In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which
it isn't.
Thus, the only possibility is the first case and the sum of the six numbers is $81\rightarrow \boxed{81}$ | E | 81 |
96b32b5cf787b77cbf48c42c4e1b73ca | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_12 | There are twenty-four $4$ -digit numbers that use each of the four digits $2$ $4$ $5$ , and $7$ exactly once. Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is
$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \text{(E)}\ 7524$ | For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.
We can just list from here: $5247,5274,5427,5472,5724 \rightarrow \boxed{5724}$ | B | 5724 |
0a6bb89285f1209d9d76f1450cc97ee4 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_13 | One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce). The postage for a letter weighing $4.5$ ounces was
$\text{(A)}\ \text{96 cents} \qquad \text{(B)}\ \text{1.07 dollars} \qquad \text{(C)}\ \text{1.18 dollars} \qquad \text{(D)}\ \text{1.20 dollars} \qquad \text{(E)}\ \text{1.40 dollars}$ | After the first ounce, there are $3.5$ ounces left. Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4\times 22$ to the cost for the first ounce.
So, the total price is $30+4\times 22=118$ cents. The answer is choice $\boxed{1.18}$ | C | 1.18 |
2359c270713d96d845c3a266caa2091e | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_14 | A bag contains only blue balls and green balls. There are $6$ blue balls. If the probability of drawing a blue ball at random from this bag is $\frac{1}{4}$ , then the number of green balls in the bag is
$\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$ | The total number of balls in the bag must be $4\times 6=24$ , so there are $24-6=18$ green balls $\rightarrow \boxed{18}$ | B | 18 |
e662e7c1894fb8f1d18dbb4883632c7e | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_15 | The area of this figure is $100\text{ cm}^2$ . Its perimeter is
[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]
$\text{(A)}\ \text{20 cm} \qquad \text{(B)}\ \text{25 cm} \qquad \text{(C)}\ \text{30 cm} \qquad \text{(D)}\ \text{40 cm} \qquad \text{(E)}\ \text{50 cm}$ | Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$
There are $10$ sides of this length, so the perimeter is $10(5)=50\rightarrow \boxed{50}$ | E | 50 |
64cf3c5a3a60cbd25608dbb6ac1432cd | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_16 | $1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$ | In the middle, we have $\cdots + 1010-1000+990 -\cdots$
If we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out. In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \rightarrow \boxed{1000}$ | D | 1000 |
64cf3c5a3a60cbd25608dbb6ac1432cd | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_16 | $1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$ | We can see that there are $199$ terms in total. We can also see that the first $198$ numbers form groups of two that add to $10$ each. Dividing to see how many pairs we have, $198$ $2$ $99$ groups of ten, or $990$ . However, we have to remember to add the 199th term ( $10$ ), so we get $990$ $10$ $1000$ , which gives us $\boxed{1000}$ | D | 1000 |
7297961dac5e36f9de8b310ae62353f8 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_17 | A straight concrete sidewalk is to be $3$ feet wide, $60$ feet long, and $3$ inches thick. How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?
$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ \text{more than 20}$ | This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is \[1\times 20\times \frac{1}{12} = 1.\overline{6}.\] Since concrete must be ordered in a whole number of cubic yards, we need $2\rightarrow \boxed{2}$ | A | 2 |
5b42289eb29deb06b9abfba55fab456d | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_18 | Each corner of a rectangular prism is cut off. Two (of the eight) cuts are shown. How many edges does the new figure have?
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewidth(1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 48$
Assume that the planes cutting the prism do not intersect anywhere in or on the prism. | In addition to the original $12$ edges, each original vertex contributes $3$ new edges.
There are $8$ original vertices, so there are $12+3\times 8=36$ edges in the new figure $\rightarrow \boxed{36}$ | C | 36 |
51e6e5e5e6646483d837586620b373b9 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_19 | There are $120$ seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$ | Let $p$ be a person seated and $o$ is an empty seat
The pattern of seating that results in the fewest occupied seats is $\text{opoopoopoo...po}$ .
We can group the seats in 3s like this: $\text{opo opo opo ... opo}.$
There are a total of $40=\boxed{40}$ groups | B | 40 |
c3002521ec0e64442eaf0c59370410a2 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_20 | The annual incomes of $1,000$ families range from $8200$ dollars to $98,000$ dollars. In error, the largest income was entered on the computer as $980,000$ dollars. The difference between the mean of the incorrect data and the mean of the actual data is
$\text{(A)}\ \text{882 dollars} \qquad \text{(B)}\ \text{980 dollars} \qquad \text{(C)}\ \text{1078 dollars} \qquad \text{(D)}\ \text{482,000 dollars} \qquad \text{(E)}\ \text{882,000 dollars}$ | Let $S$ be the sum of all the incomes but the largest one. For the actual data, the mean is $\frac{S+98000}{1000}$ , and for the incorrect data the mean is $\frac{S+980000}{1000}$ . The difference is $882, or \rightarrow \boxed{882}$ | A | 882 |
e4b7a3ef681039f882bd4e58e7f7a424 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_22 | Several students are seated at a large circular table. They pass around a bag containing $100$ pieces of candy. Each person receives the bag, takes one piece of candy and then passes the bag to the next person. If Chris takes the first and last piece of candy, then the number of students at the table could be
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$ | If this is the case, then if there were only $99$ pieces of candy, the bag would have gone around the table a whole number of times. Thus, the number of students is a divisor of $99$ . The only choice that satisfies this is choice $\boxed{11}$ | B | 11 |
bd11a9e569fa2e8d93856e85e5e98b55 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_24 | Three $\Delta$ 's and a $\diamondsuit$ will balance nine $\bullet$ 's. One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S); label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S); [/asy]
How many $\bullet$ 's will balance the two $\diamondsuit$ 's in this balance?
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | For simplicity, suppose $\Delta = a$ $\diamondsuit = b$ and $\bullet = c$ . Then, \[3a+b=9c\] \[a=b+c\] and we want to know what $2b$ is in terms of $c$ . Substituting the second equation into the first, we have \[4b=6c\Rightarrow 2b=3c\]
Thus, we need $3$ $\bullet$ 's $\rightarrow \boxed{3}$ | C | 3 |
b3ef54ef77e35162ebf268d7c614d3b9 | https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_25 | How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different.
[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$ | We break this into cases.
Case 1: At least one square is a vertex: WLOG , suppose one of them is in the upper-left corner. Then, consider the diagonal through that square. The two squares on that diagonal could be the second square, or the second square is on one side of the diagonal.
The square is reflectionally symmetric about this diagonal, so we only consider the squares on one side, giving another three possibilities.
In this case, there are $2+3=5$ distinct squares.
Case 2: At least one square is on an edge, but no square is on a vertex: There are clearly two edge-edge combinations and one edge-center combination, so this case has $3$ squares.
In total, there are $3+5=8$ distinct squares $\rightarrow \boxed{8}$ | C | 8 |
1bbbb59c284e159c6230671faf2c7e07 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_1 | $(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$ | We make use of the associative and commutative properties of addition to rearrange the sum as \begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\ &= 250 \Longrightarrow \boxed{250} | E | 250 |
1bbbb59c284e159c6230671faf2c7e07 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_1 | $(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$ | Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\] which gives us \[10+30+50+70+90 = 250 = \boxed{250}.\] | E | 250 |
9b8258207736115ee9ca4efc5ce6b3c5 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_2 | $\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$ | \begin{align*} \frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ &= \frac{246}{1000} \\ &= .246 \rightarrow \boxed{.246} | D | .246 |
46aca95513660ba66a82e130e0f1a4ba | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_3 | Which of the following numbers is the largest?
$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$ | We have $.99>.9099>.909>.9009>.9$ , so choice $\boxed{.99}$ is the largest. | A | .99 |
bf074836ce8265b001eb23284dd4832d | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_4 | Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$
$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$ | $401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{2000}\] | E | 2000 |
9def7f6e19a966b21f1354dcc40e34c7 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_5 | $-15+9\times (6\div 3) =$
$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$ | We use the order of operations here to get
\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{3} | D | 3 |
1169ac4f668484d3096e47073fd63103 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_6 | If the markings on the number line are equally spaced, what is the number $\text{y}$
[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a) { draw((4a,-1)--(4a,1)); } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$ | Five steps are taken to get from $0$ to $20$ . Each step is of equal size, so each step is $4$ . Three steps are taken from $0$ to $y$ , so $y=3\times 4=12\rightarrow \boxed{12}$ | C | 12 |
83a21618ba5e776362da64c300272cdc | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_7 | If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$ | We have \begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\ 600 &= 250+10n \\ 35 &= n \implies \boxed{35} | D | 35 |
9addaa55a634c68bd1b1145fe141879b | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8 | $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$ | We use the distributive property to get \[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{26}\] | E | 26 |
9addaa55a634c68bd1b1145fe141879b | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8 | $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$ | Since $\frac12+\frac13+\frac14 > \frac12+\frac14+\frac14 = 1$ , we have \[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\] The only answer choice greater than $24$ is $\boxed{26}$ | E | 26 |
9addaa55a634c68bd1b1145fe141879b | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_8 | $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$ | We can just bash it out, getting $24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{26}$ -fn106068 | E | 26 |
2548763060b44d7ab93cc80c37b32f71 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_10 | What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$ | The smaller angle makes up $5/12$ of the circle which is the clock. A circle is $360^\circ$ , so the measure of the smaller angle is \[\frac{5}{12}\cdot 360^\circ = 150^\circ \rightarrow \boxed{150}\] | D | 150 |
9491c1a3444054b7bbedb911bebb84c0 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_15 | The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$ | Let $[ABC]$ denote the area of figure $ABC$
Clearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,
Since $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$
Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{64}$ | D | 64 |
b7164408c4e0dbb82f9d6f54f1794320 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_16 | In how many ways can $47$ be written as the sum of two primes
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$ | For $47$ to be written as the sum of two integers , one must be odd and the other must be even. There is only one even prime, namely $2$ , so one of the numbers must be $2$ , making the other $45$
However, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $\rightarrow \boxed{0}$ | A | 0 |
83dcd2b02309c3604f00714d1613cf6a | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_17 | The number $\text{N}$ is between $9$ and $17$ . The average of $6$ $10$ , and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$
From what we know, we can deduce that $25<N+16<33$ , and thus \[8.\overline{3}<\frac{N+16}{3}<11\]
The only answer choice that falls in this range is choice $\boxed{10}$ | B | 10 |
b0b5d8e04b173a831759faa57b2bccfd | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_19 | The graph below shows the total accumulated dollars (in millions) spent by the Surf City government during $1988$ . For example, about $.5$ million had been spent by the beginning of February and approximately $2$ million by the end of April. Approximately how many millions of dollars were spent during the summer months of June, July, and August?
[asy] unitsize(18); for (int a=1; a<13; ++a) { draw((a,0)--(a,.5)); } for (int b=1; b<6; ++b) { draw((-.5,2b)--(0,2b)); } draw((0,0)--(0,12)); draw((0,0)--(14,0)); draw((0,0)--(1,.9)--(2,1.9)--(3,2.6)--(4,4.3)--(5,4.5)--(6,5.7)--(7,8.2)--(8,9.4)--(9,9.8)--(10,10.1)--(11,10.2)--(12,10.5)); label("J",(.5,0),S); label("F",(1.5,0),S); label("M",(2.5,0),S); label("A",(3.5,0),S); label("M",(4.5,0),S); label("J",(5.5,0),S); label("J",(6.5,0),S); label("A",(7.5,0),S); label("S",(8.5,0),S); label("O",(9.5,0),S); label("N",(10.5,0),S); label("D",(11.5,0),S); label("month F=February",(16,0),S); label("$1$",(-.6,2),W); label("$2$",(-.6,4),W); label("$3$",(-.6,6),W); label("$4$",(-.6,8),W); label("$5$",(-.6,10),W); label("dollars in millions",(0,11.9),N); [/asy]
$\text{(A)}\ 1.5 \qquad \text{(B)}\ 2.5 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4.5 \qquad \text{(E)}\ 5.5$ | Since we want to know how much money is spent in June, July and August, we need the difference between the amount of money spent by the beginning of June and the amount of money spent by the end of August.
We estimate these to be about $2.2$ million and $4.8$ million, respectively. The difference is \[4.8-2.2=2.6\approx 2.5 \rightarrow \boxed{2.5}\] | B | 2.5 |
d05d4d84e63fd9015c431f59572af37a | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_20 | The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?
[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); draw((1,0)--(1,1)--(2,1)--(2,0)--cycle); draw((3,1)--(3,0)); label("$1$",(1.5,1.25),N); label("$2$",(1.5,.25),N); label("$3$",(1.5,-.75),N); label("$4$",(2.5,.25),N); label("$5$",(3.5,.25),N); label("$6$",(.5,.25),N); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$ | It is clear that $6$ $5$ , and $4$ will not come together to get a sum of $15$
The faces $6$ $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\rightarrow \boxed{14}$ | D | 14 |
649b0dad2edf7f241d1c28d13a7fd08c | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_21 | Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?
$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111$ | First he gives $128\times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left. Then he gives $96\times .25 = 24$ apples to June, so he has $96-24=72$ left.
Finally, he gives one to the teacher, leaving $71\rightarrow \boxed{71}$ | D | 71 |
014317efba5c693c4fe7a1e920f3cf7d | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_22 | The letters $\text{A}$ $\text{J}$ $\text{H}$ $\text{S}$ $\text{M}$ $\text{E}$ and the digits $1$ $9$ $8$ $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc} & & AJHSME & 1989 \\ & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\ & & ........ & \end{tabular}\]
What is the number of the line on which $\text{AJHSME 1989}$ will appear for the first time?
$\text{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24$ | Every $4\text{th}$ line has $1989$ as part of it and every $6\text{th}$ line has $\text{AJHSME}$ as part of it. In order for both to be part of line $n$ $n$ must be a multiple of $4$ and $6$ , the least of which is $\text{lcm}(4,6)=12\rightarrow \boxed{12}$ | C | 12 |
0989d41d83f86d0c5ca7d153f77cca19 | https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_23 | An artist has $14$ cubes, each with an edge of $1$ meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint?
$\text{(A)}\ 21 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 33 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 42$
[asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy] | We can consider the contributions of the sides of the three layers and the tops of the layers separately.
Layer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.
The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3\times 3$ square, hence this contributes $9$ for the surface area.
You do not have to count the side underneath, since it is not exposed.
Thus, the artist paints $24+9=33 \rightarrow \boxed{33}$ square meters. | C | 33 |
5dd9547ab64e33b16a36db0690a3ad00 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_1 | The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of
[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2.6,0),E); label("$11$",(2.6,0),W); [/asy]
$\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6$ | Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $\text{C}$ and $\text{D}$ are possible.
Looking, we see that the arrow is closer to $10.3$ , so $\boxed{10.3}$ | D | 10.3 |
01b65ee29df759e53d2b8cb692e124f7 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_2 | The product $8\times .25\times 2\times .125 =$
$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ | Converting the decimals to fractions , this is \begin{align*} 8\times \frac{1}{4} \times 2\times \frac{1}{8} &= \frac{8\times 2}{4\times 8} \\ &= \frac{16}{32} \\ &= \frac{1}{2} \rightarrow \boxed{12} | C | 12 |
b7b6acf5807c2c2e12cbad39614e499b | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_3 | $\frac{1}{10}+\frac{2}{20}+\frac{3}{30} =$
$\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6$ | Each of the fractions simplify to $\frac{1}{10}$ , so this sum is \begin{align*} \frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ &= .3 \rightarrow \boxed{.3} | D | .3 |
f90586c0bf9782cc25af926bb57860cb | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_4 | The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(12,1)--(12,4)--(3,4)--cycle,white); //Black Squares, Gray Border (blends better than white) for(int a=0; a<7; ++a) { filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray); } for(int b=7; b<15; ++b) { filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray); } for(int c=1; c<7; ++c) { filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray); } filldraw((6,4)--(7,4)--(7,5)--(6,5)--cycle,black,gray); filldraw((7,5)--(8,5)--(8,6)--(7,6)--cycle,black,gray); filldraw((8,4)--(9,4)--(9,5)--(8,5)--cycle,black,gray); //White Squares, Black Border filldraw((7,4)--(8,4)--(8,5)--(7,5)--cycle,white,black); for(int a=0; a<7; ++a) { filldraw((2a+1,0)--(2a+2,0)--(2a+2,1)--(2a+1,1)--cycle,white,black); } for(int b=9; b<15; ++b) { filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black); } for(int c=1; c<7; ++c) { filldraw((c+1,c)--(c+2,c)--(c+2,c+1)--(c+1,c+1)--cycle,white,black); } label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy] | It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\rightarrow \boxed{8}$ | B | 8 |
7447a6fbbfc4e5c3a72381c1a662ab4b | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_5 | If $\angle \text{CBD}$ is a right angle , then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately
[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]
$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ$ | We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{50}$ | C | 50 |
bd70525463c4e1a316a0e627eba19204 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_12 | Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is
$\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}$ | We want the cost per person, which is \begin{align*} \frac{20\text{ billion}}{250\text{ million}} &= \frac{20000\text{ million}}{250\text{ million}} \\ &= 80 \rightarrow \boxed{80} | C | 80 |
0fbe2a5a6d481e33b99a9c8593cdd98d | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_13 | If rose bushes are spaced about $1$ foot apart, approximately how many bushes are needed to surround a circular patio whose radius is $12$ feet?
$\text{(A)}\ 12 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 48 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 450$ | The circumference of the patio is $2\cdot 12\cdot \pi =24\pi \approx 75.398$ . Since the bushes are spaced $1$ foot apart, about $75\rightarrow \boxed{75}$ are needed. | D | 75 |
d052be5e8b4e557a4a50941879fc6de9 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_14 | $\diamondsuit$ and $\Delta$ are whole numbers and $\diamondsuit \times \Delta =36$ . The largest possible value of $\diamondsuit + \Delta$ is
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20\ \qquad \text{(E)}\ 37$ | Since it doesn't take too long, we can just make a table with all the possible values of the sum \[\begin{array}{|c|c|c|} \multicolumn{3}{}{} \\ \hline \diamondsuit & \Delta & \diamondsuit + \Delta \\ \hline 36 & 1 & 37 \\ \hline 18 & 2 & 20 \\ \hline 12 & 3 & 15 \\ \hline 9 & 4 & 13 \\ \hline 6 & 6 & 12 \\ \hline \end{array}\]
Clearly $37\rightarrow \boxed{37}$ is the largest. | E | 37 |
08e56c3cab6fd376138bc7bfb661ec06 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_16 | Placing no more than one $\text{X}$ in each small square , what is the greatest number of $\text{X}$ 's that can be put on the grid shown without getting three $\text{X}$ 's in a row vertically, horizontally, or diagonally?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
[asy] for(int a=0; a<4; ++a) { draw((a,0)--(a,3)); } for(int b=0; b<4; ++b) { draw((0,b)--(3,b)); } [/asy] | By the Pigeonhole Principle , if there are at least $7$ $\text{X}$ 's, then there will be some row with $3$ $\text{X}$ 's. We can put in $6$ by leaving out the three boxes in one of the main diagonals.
$\rightarrow \boxed{6}$ | E | 6 |
7b1c48bd7edb814f1577c48f38a6d779 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_18 | The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds. The average weight of the $10$ children is
$\text{(A)}\ 135\text{ pounds} \qquad \text{(B)}\ 137\text{ pounds} \qquad \text{(C)}\ 138\text{ pounds} \qquad \text{(D)}\ 140\text{ pounds} \qquad \text{(E)}\ 141\text{ pounds}$ | Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ . We are given
\begin{align*} \frac{S_B}{6} &= 150 \\ \frac{S_G}{4} &= 120 \end{align*}
We want the average of the $10$ children, which is \[\frac{S_B+S_G}{10}\] From the first two equations , we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ . Therefore, the average we desire is \[\frac{1380}{10}=138 \rightarrow \boxed{138}\] | C | 138 |
62785e4d822408d8677d545eb451fd51 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_19 | What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$ | To get from the $1^\text{st}$ term of an arithmetic sequence to the $100^\text{th}$ term, we must add the common difference $99$ times. The first term is $1$ and the common difference is $5-1=9-5=13-9=\cdots = 4$ , so the $100^\text{th}$ term is \[1+4(99)=397 \rightarrow \boxed{397}\] | A | 397 |
62785e4d822408d8677d545eb451fd51 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_19 | What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$ | Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$
$a_{100}=4(100)-3=397$ , or $\boxed{397}$ | A | 397 |
4a711e79f44f0067e522bbf882607dce | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_20 | The glass gauge on a cylindrical coffee maker shows that there are $45$ cups left when the coffee maker is $36\%$ full. How many cups of coffee does it hold when it is full?
$\text{(A)}\ 80 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 130 \qquad \text{(E)}\ 262$
[asy] draw((5,0)..(0,-1.3)..(-5,0)); draw((5,0)--(5,10)); draw((-5,0)--(-5,10)); draw(ellipse((0,10),5,1.3)); draw(circle((.3,1.3),.4)); draw((-.1,1.7)--(-.1,7.9)--(.7,7.9)--(.7,1.7)--cycle); fill((-.1,1.7)--(-.1,4)--(.7,4)--(.7,1.7)--cycle,black); draw((-2,11.3)--(2,11.3)..(2.6,11.9)..(2,12.2)--(-2,12.2)..(-2.6,11.9)..cycle); [/asy] | Let the amount of coffee the maker will hold when full be $x$ . Then, \[.36x=45 \Rightarrow x=125 \rightarrow \boxed{125}\] | C | 125 |
17f2ce9442987dc004883a81eef83c26 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_21 | A fifth number, $n$ , is added to the set $\{ 3,6,9,10 \}$ to make the mean of the set of five numbers equal to its median . The number of possible values of $n$ is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than }4$ | The possible medians after $n$ is added are $6$ $n$ , or $9$ . Now we use casework
Case 1: The median is $6$
In this case, $n<6$ and \[\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2\] so this case contributes $1$
Case 2: The median is $n$
We have $6<n<9$ and \[\frac{3+6+n+9+10}{5}=n \Rightarrow n=7\] so this case also contributes $1$
Case 3: The median is $9$
We have $9<n$ and \[\frac{3+6+9+n+10}{5}=9 \Rightarrow 17\] so this case adds $1$
In all there are $3\rightarrow \boxed{3}$ possible values of $n$ | C | 3 |
41fad770328077566b6c27db8cca3d26 | https://artofproblemsolving.com/wiki/index.php/1988_AJHSME_Problems/Problem_23 | Maria buys computer disks at a price of $4$ for $$5$ and sells them at a price of $3$ for $$5$ . How many computer disks must she sell in order to make a profit of $$100$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200$ | This is the equivalent of saying she buys $12$ for $$15$ and sells $12$ for $$20$ , so for every dozen disks she sells, she profits $$5$
She needs to profit $$100$ , so she needs to sell $\frac{100}{5}=20$ dozen disks, which is $240\rightarrow \boxed{240}$ | D | 240 |
54dd0da2f86cabbd4aa34fc3d3fc38c5 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_1 | $.4+.02+.006=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426$ | $.4+.02+.006 = .400 + .020 + .006 = .426\rightarrow \boxed{.426}$ | E | .426 |
e5fd344fcb4ed49d9e393f06f3b3e6f4 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_2 | $\frac{2}{25}=$
$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5$ | $\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{.08}$ | B | .08 |
a6a7d492b82eb3b6c08e6be04f36a89b | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_3 | $2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$ | Find that \[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\] Which gives us \begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{1800} | E | 1800 |
a6a7d492b82eb3b6c08e6be04f36a89b | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_3 | $2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$ | $2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ $83+97$ $85+95$ $87+93$ $89+91$ $180$ . Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ . But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\boxed{1800}$ | E | 1800 |
09609ce08eeeb2d722796c9c5c2c57b6 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_4 | Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250$ | The right angle is $1/4$ of the circle, hence it contains $500/4=125\rightarrow \boxed{125}$ clerts. | C | 125 |
e42bf456ea609104d1c83fbdf15ab82b | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_6 | The smallest product one could obtain by multiplying two numbers in the set $\{ -7,-5,-1,1,3 \}$ is
$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$ | To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $\boxed{21}$ | B | 21 |
7abadfa0a9207198b0db836b624dd00a | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_7 | The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy] | Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.
There are $10$ painted unit squares on the half of the cube shown, so there are $10\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\boxed{20}$ | C | 20 |
7abadfa0a9207198b0db836b624dd00a | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_7 | The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy] | Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus $27 - 7 = 20$ our answer is $\boxed{20}$ | C | 20 |
b74c26716ea70188c67c5a4eb4c88eb8 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_9 | When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$ , the least common denominator used is
$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$ | We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\boxed{420}$ | C | 420 |
bae9b4da6ae4d16e7fa2b502baf5c98a | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_10 | $4(299)+3(299)+2(299)+298=$
$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$ | We can make use of the distributive property as follows: \begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}
$\boxed{2989}$ | B | 2989 |
f76e2cdc836fa34a8069608958b4c7fd | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_14 | A computer can do $10,000$ additions per second. How many additions can it do in one hour?
$\text{(A)}\ 6\text{ million} \qquad \text{(B)}\ 36\text{ million} \qquad \text{(C)}\ 60\text{ million} \qquad \text{(D)}\ 216\text{ million} \qquad \text{(E)}\ 360\text{ million}$ | There are $3600$ seconds per hour, so we have \begin{align*} \frac{3600\text{ seconds}}{\text{hour}}\cdot \frac{10,000\text{ additions}}{\text{second}} &= \frac{36,000,000\text{ additions}}{\text{hour}} \\ &= 36\text{ million additions per hour} \end{align*}
$\boxed{36}$ | B | 36 |
dde85c5a12d93257c764a885c68b6dfc | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_15 | The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale. What was the regular price of one tire?
$\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}\ 79\text{ dollars} \qquad \text{(E)}\ 80\text{ dollars}$ | Let the regular price of one tire be $x$ . We have \begin{align*} 3x+3=240 &\Rightarrow 3x=237 \\ &\Rightarrow x=79 \end{align*}
$\boxed{79}$ Good Job! | D | 79 |
f34033a0ee10dde03bfbe2f721d16707 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_16 | Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40\%$ . In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$ . How many of these $10$ shots did she make?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | After the fourth game, she took $40$ shots, $50\%$ of which she made, so she made $40\times .5=20$ shots. Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.
$\boxed{8}$ | E | 8 |
a75a17899ed26dd28851f2dd8848715a | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_18 | Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$ | Let the original number of people in the room be $x$ . Half of them left, so $\frac{x}{2}$ of them are left in the room.
After that, one third of this group is dancing, so $\frac{x}{2}-\frac{1}{3}\left( \frac{x}{2}\right) =\frac{x}{3}$ people are not dancing.
This is given to be $12$ , so \[\frac{x}{3}=12\Rightarrow x=36\]
$\boxed{36}$ | C | 36 |
a75a17899ed26dd28851f2dd8848715a | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_18 | Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$ | First note that of the $\frac{1}{2}$ people remaining in the room, $\frac{2}{3}$ are not dancing. Therefore $\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}$ of the original amount of people in the room is $12$ . The answer is $\boxed{36}$ | C | 36 |
afceb5475005b1e83a320f537b30494c | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_20 | "If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$ | To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for $n=9,12,16$ . If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime. However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.
Therefore, the answer is $\boxed{9}$ , 9. | A | 9 |
5097390289bbbd7014a8ca49310c964b | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_21 | Suppose $n^{*}$ means $\frac{1}{n}$ , the reciprocal of $n$ . For example, $5^{*}=\frac{1}{5}$ . How many of the following statements are true?
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$ | We can just test all of these statements: \begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac{1}{5} = 5^* \end{align*}
The last two statements are true and the first two aren't, so $\boxed{2}$ | C | 2 |
4609b6065f7244ec1de7aadd712e05bc | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_23 | Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group. To the nearest percent , what percent of the U.S. Black population lived in the South?
\[\begin{tabular}[t]{c|cccc} & NE & MW & South & West \\ \hline White & 42 & 52 & 57 & 35 \\ Black & 5 & 5 & 15 & 2 \\ Asian & 1 & 1 & 1 & 3 \\ Other & 1 & 1 & 2 & 4 \end{tabular}\]
$\text{(A)}\ 20\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 80\%$ | There are $5+5+15+2=27$ million Blacks living in the U.S. Out of these, $15$ of them live in the South, so the percentage is $\frac{15}{27}\approx \frac{60}{108}\approx 56\%$
$\boxed{56}$ | D | 56 |
1224f072a5b52fa771ede8313175df25 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_24 | A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$ | Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that \begin{align} c+w+b &= 20 \\ 5c-2w &= 48 \end{align}
Adding equation $(2)$ to double equation $(1)$ , we get \[7c+2b=88\]
Since we want to maximize the value of $c$ , we try to find the largest multiple of $7$ less than $88$ . This is $84=7\times 12$ , so let $c=12$ . Then we have \[7(12)+2b=88\Rightarrow b=2\]
Finally, we have $w=20-12-2=6$ . We want $c$ , so the answer is $12$ , or $\boxed{12}$ | D | 12 |
1224f072a5b52fa771ede8313175df25 | https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_24 | A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$ | If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 \cdot 5 - 4 \cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 \cdot 5 - 6 \cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $\boxed{12}$ | D | 12 |
ed79581a751b8c7ab13936224470113e | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_3 | The smallest sum one could get by adding three different numbers from the set $\{ 7,25,-1,12,-3 \}$ is
$\text{(A)}\ -3 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 21$ | To find the smallest sum, we just have to find the smallest 3 numbers and add them together.
Obviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$
$\boxed{3}$ | C | 3 |
0adab5b7da9e31a005190c6179afcb87 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_4 | The product $(1.8)(40.3+.07)$ is closest to
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$ | Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\boxed{74}$ | C | 74 |
0b529da4dcc697199d8c00f6d635398e | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_7 | How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.
Clearly it must be true that for any positive integers $a$ $b$ , and $c$ with $a>b>c$ \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]
If we let $a=9$ $b=8$ , and $c=4$ , then we get \[\sqrt{9}>\sqrt{8}>\sqrt{4}\] \[3>\sqrt{8}>2\]
Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$
Similarly, if we let $a=81$ $b=80$ , and $c=64$ , we get \[\sqrt{81}>\sqrt{80}>\sqrt{64}\] \[9>\sqrt{80}>8\]
So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$
So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.
$\boxed{6}$ | B | 6 |
c9799e5a84c71256bfc145025255881e | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_8 | In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$ , so $\text{B}$ is either $3$ or $8$ . If $\text{B}=3$ , then the product is $32\times 73$ , which is clearly too small, so $\text{B}=8$
$\boxed{8}$ | E | 8 |
0577b5a0dce39fd0d08c5983dca37fc5 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_9 | Using only the paths and the directions shown, how many different routes are there from $\text{M}$ to $\text{N}$
[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.
Since A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.
Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.
M can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.
6 is $\boxed{6}$ | E | 6 |
8c86177ecec1074724194e7b618a05a3 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_11 | If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$ , then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$ | We just plug in and evaluate: \begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}
$\boxed{6}$ | A | 6 |
712fd756b8872f5622ec0fb712a29665 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_13 | The perimeter of the polygon shown is
$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$
$\text{(E)}\ \text{cannot be determined from the information given}$ | You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!
[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]
So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{28}$ | C | 28 |
3643cfc0cee5fe0851b66d179a049bc2 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_16 | A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar indicating the number sold during the winter is covered by a smudge. If exactly $25\%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?
[asy] size(250); void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth, string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep); for(real y=ymin; y<ymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); } int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;i<n;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;i<n;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); } string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;i<LX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4}; bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4.5$ | What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $\frac{1}{4}$ of the total number of hamburgers sold, so we can say that...
$4 \times \text{Fall} = \text{Spring} + \text{Winter} + \text{Fall} + \text{Summer}$
$4 \times 4 = 4.5 + 4 + x + 5$
$16 = x + 13.5$
$2.5 = x$
The answer is 2.5, or $\boxed{2.5}$ | A | 2.5 |
e6d200e3ec11fa4ca50d467260b3e0c9 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_18 | A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60$ m?
[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$ | Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$ m by $60$ m rectangle, so the $60$ m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$ m. We can start by counting $60\div12+1$ on the $60$ m fence (since we also count the $0$ m post). Next, we have the two $36$ m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$ m and $60$ m fence posts are also part of the $36$ m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$ m by $60$ m grazing area, or answer $\boxed{12}$ | B | 12 |
c141003957705de10b0f959ba89aa18c | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_19 | At the beginning of a trip, the mileage odometer read $56,200$ miles. The driver filled the gas tank with $6$ gallons of gasoline. During the trip, the driver filled his tank again with $12$ gallons of gasoline when the odometer read $56,560$ . At the end of the trip, the driver filled his tank again with $20$ gallons of gasoline. The odometer read $57,060$ . To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?
$\text{(A)}\ 22.5 \qquad \text{(B)}\ 22.6 \qquad \text{(C)}\ 24.0 \qquad \text{(D)}\ 26.9 \qquad \text{(E)}\ 27.5$ | The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. Hence the average MPG ratio is $860 / 32 \approx 26.9 \rightarrow \boxed{26.9}$ | D | 26.9 |
618adc7b40507584a0bf1093e6e54a21 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_20 | The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$ | \[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\] Which is closest to $3\rightarrow\boxed{3}$ | D | 3 |
38a9060e685763518e915f00e3bd5f49 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_21 | Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.
In total, there are $6\rightarrow\boxed{6}$ good possibilities. | E | 6 |
38a9060e685763518e915f00e3bd5f49 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_21 | Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | Fold the four squares into the four sides of a cube. Then, there are six edges "open" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6\rightarrow\boxed{6}$ | E | 6 |
106e2b074943d2322c524b83a771cbb5 | https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_23 | The large circle has diameter $\text{AC}$ . The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$ , the center of the large circle. If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed); label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]
$\text{(A)}\ \text{between }\frac{1}{2}\text{ and 1} \qquad \text{(B)}\ 1 \qquad \text{(C)}\ \text{between 1 and }\frac{3}{2}$
$\text{(D)}\ \text{between }\frac{3}{2}\text{ and 2} \qquad \text{(E)}\ \text{cannot be determined from the information given}$ | The small circle has radius $1$ , thus its area is $\pi$
The large circle has radius $2$ , thus its area is $4\pi$
The area of the semicircle above $AC$ is then $2\pi$
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$ . This means that the area of the shaded part is $2\pi-\pi=\pi$ .
This is equal to the area of a small circle, hence the correct answer is $\boxed{1}$ | B | 1 |
602432064a92be33b3fb5297d7f7b10c | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1 | [katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex] | By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{1}[/katex] | A | 1 |
602432064a92be33b3fb5297d7f7b10c | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1 | [katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex] | Notice that the $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\boxed{1}$ | A | 1 |
5a853380761c42487ee1f6ee5955bafe | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{945}[/mathjax]. | B | 945 |
5a853380761c42487ee1f6ee5955bafe | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{945} | B | 945 |
5a853380761c42487ee1f6ee5955bafe | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | Instead of breaking the sum then rearranging, we can rearrange directly: \begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= \boxed{945} | B | 945 |