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901
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_14
| 1
|
A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season?
$\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$
|
Noting that 70% is the same as $\frac{70}{100}=\frac{7}{10}$ , and that, when x is the amount of wins in the last 40 games, the fraction of games won is $\frac{40+x}{50+40}=\frac{40+x}{90}$ , all we have to do is set them equal: \[\frac{40+x}{90}=\frac{7}{10}\] \[40+x=63\] \[x=\boxed{23}\]
| 23
|
902
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_14
| 2
|
A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season?
$\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$
|
Alternatively we can note that they will play a total of $40+50=90$ games and must win $0.7(90)=63$ games. Since they won $40$ games already they need $63-40=\boxed{23}$ more games.
| 23
|
903
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_15
| 1
|
What is the $100^\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$
|
$\frac{4}{37}=\frac{12}{111}=\frac{108}{999}=0.108108108...$
Since this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = $\boxed{1}$
| 1
|
904
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_17
| 1
|
The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\%\\ \textbf{Cleona:}& 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\%\end{tabular}\]
Annville has 100 students and Cleona has 200 students. In the two schools combined, what percent of the students are in grade 6?
$\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%$
|
By the tables, Annville has $11$ 6th graders and Cleona has $34$ . Together they have $45$ 6th graders and $300$ total students, so the percent is $\frac{45}{300}=\frac{15}{100}= \boxed{15}$
| 15
|
905
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_22
| 1
|
The number $6545$ can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
$\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222$
|
The prime factorization of $6545$ is $5\cdot7\cdot11\cdot17 =385\cdot17$ , which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \[5\cdot7=35 \text{, } 11\cdot17=187 \text{ X}\] \[5\cdot11=55 \text{, } 7\cdot17=119 \text{ X}\] \[5\cdot17=85 \text{, } 7\cdot11=77 \text{ }\surd\] \[85+77= \boxed{162}\]
| 162
|
906
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_23
| 1
|
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
$\text{(A)}\ 1120 \qquad \text{(B)}\ 1400 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 2025 \qquad \text{(E)}\ 2500$
|
Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are $5*5*8*7= \boxed{1400}$ numbers.
| 400
|
907
|
https://artofproblemsolving.com/wiki/index.php/1995_AJHSME_Problems/Problem_25
| 1
|
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11$
|
Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - $a:b$ $a$ is for hrs. and $b$ is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed.
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the bus most recently has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all valid times that are 30 minute intervals from 12:45. They are
12:45, 1:15, 1:45, 2:15, 2:45, 3:15, 3:45, 4:15, 4:45, 5:15, and 5:30, your arrival time.
Hence, you will pass $\boxed{10}$ Dallas-bound buses.
| 10
|
908
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_2
| 1
|
$\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$
$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
|
$1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45$
$\frac{45+55}{10} = \dfrac{100}{10} = \boxed{10}$
| 10
|
909
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_5
| 1
|
Given that $\text{1 mile} = \text{8 furlongs}$ and $\text{1 furlong} = \text{40 rods}$ , the number of rods in one mile is
$\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280$
|
\[(1\ \text{mile}) \left( \frac{8\ \text{furlongs}}{1\ \text{mile}} \right) \left( \frac{40\ \text{rods}}{1\ \text{furlong}} \right) = \boxed{320}\]
| 320
|
910
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_6
| 1
|
The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$
|
Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$ . Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{0}$
| 0
|
911
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_6
| 2
|
The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$
|
We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\boxed{0}$
| 0
|
912
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_7
| 1
|
If $\angle A = 60^\circ$ $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$
[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]
$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$
|
The sum of the angles in a triangle is $180^\circ$ . We can find $\angle ABE = 80^\circ$ , so $\angle CBD = 180-80=100^\circ$
\[\angle BDC = 180-100-30=\boxed{50}\]
| 50
|
913
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_8
| 1
|
For how many three-digit whole numbers does the sum of the digits equal $25$
$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$
|
Because $8+8+8=24$ , it follows that one of the digits must be a $9$ . The other two digits them have a sum of $25-9=16$ . The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows
\[799,979,997,889,898,988\]
The number of configurations is $\boxed{6}$
| 6
|
914
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_10
| 1
|
For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$
|
We should list all the positive divisors of $36$ and count them. By trial and error, the divisors of $36$ are found to be $1,2,3,4,6,9,12,18,36$ , for a total of $9$ . However, $1$ and $2$ can't be equal to $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{7}$
| 7
|
915
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_10
| 2
|
For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$
|
To find the number of positive divisors of $36$ , first prime factorize $36$ to get $2^2*3^2$ . Then add $1$ to the power of both $2$ and $3$ to get $3$ . Multiply $3*3$ to get $9$ . Since the problem is asking only for positive integer values of N, subtract $2$ from $9$ (since $2-2$ and $1-2$ result in integers that are not positive) to get $\boxed{7}$
| 7
|
916
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_11
| 1
|
Last summer $100$ students attended basketball camp. Of those attending, $52$ were boys and $48$ were girls. Also, $40$ students were from Jonas Middle School and $60$ were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?
$\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52$
|
Make a table with the information given.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & & & 52 \\ \text{girls} & 20 & & 48 \\ \text{total} & 40 & 60 & 100 \end{tabular}\]
Because the first two columns must add up to the third column, and the same with the rows, the rest of the empty boxes can be filled in.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & 20 & 32 & 52 \\ \text{girls} & 20 & 28 & 48 \\ \text{total} & 40 & 60 & 100 \end{tabular}\]
The number of boys from Clay is $\boxed{32}$
| 32
|
917
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_14
| 1
|
Two children at a time can play pairball. For $90$ minutes, with only two children playing at time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36$
|
There are $2 \times 90 = 180$ minutes of total playing time. Divided equally among the five children, each child gets $180/5 = \boxed{36}$ minutes.
| 36
|
918
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_16
| 1
|
The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$
|
Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$ . If the perimeter of one is $3$ times the other's, then $a=3b$ . The area of the larger square over the area of the smaller square is
\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{9}\]
| 9
|
919
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_17
| 1
|
Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12$
|
Her driveway has $(4)(10)(3)=120$ cubic yards of snow. After the first hour she would have $120-20=100$ cubic yards, then $100-19=81$ $81-18=63$ $63-17=46$ $46-16=30$ $30-15=15$ , and $15-14=1$ cubic yard after the seventh hour. It will take her a little more than seven hours to shovel it clean, which is closest to $\boxed{7}$
| 7
|
920
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_19
| 1
|
Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$ , has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is
[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); [/asy]
$\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64$
|
The radius of each semicircle is $2$ , half the sidelength of the square. The line straight down the middle of square $ABCD$ is the sum of two radii and the length of the smaller square, which is equivalent to its side length. The area of $ABCD$ is $(4+2+2)^2 = \boxed{64}$
| 64
|
921
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_21
| 1
|
A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$
|
If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{\text{th}}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\boxed{10}$ gumballs.
| 10
|
922
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_24
| 1
|
$2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.
$\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$
|
If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.
Case 1: There are no green squares. This can be done in $1$ way.
Case 2: There is one green square and three red squares. This can only be done when the green square's top and right edges are against the edge, so there is $1$ way.
Case 3: There are two green squares and two red squares. This happens when the two green squares are in the two top squares or two right squares, so there are $2$ ways.
Case 4: There are three green squares and one red square. Similar to case 2, this happens when the red square's left and bottom edges are against the edge, so there is $1$ way.
Case 5: There are four green squares and zero red squares. $1$ way.
\[1+1+2+1+1 = \boxed{6}\]
| 6
|
923
|
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_25
| 2
|
Find the sum of the digits in the answer to
$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$
where a string of $94$ nines is multiplied by a string of $94$ fours.
$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$
|
\[\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6\]
\[4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{846}\]
| 846
|
924
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_2
| 1
|
When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be
$\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133$
|
\begin{align*} \dfrac{49}{84} &= \dfrac{7^2}{2^2\cdot 3\cdot 7} \\ &= \dfrac{7}{2^2\cdot 3} \\ &= \dfrac{7}{12}. \end{align*}
The sum of the numerator and denominator is $7+12=19\rightarrow \boxed{19}$
| 19
|
925
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_6
| 1
|
A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$
|
A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 \times 2 =\boxed{6}$ adults are fed.
| 6
|
926
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_8
| 1
|
To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately
$\text{(A)}\ 1\text{ month} \qquad \text{(B)}\ 4\text{ months} \qquad \text{(C)}\ 6\text{ months} \qquad \text{(D)}\ 8\text{ months} \qquad \text{(E)}\ 1\text{ year}$
|
If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \times 4=240$ days which is about $\boxed{8}$
| 8
|
927
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_9
| 1
|
Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$ . Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$
|
Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=\boxed{4}$
| 4
|
928
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_9
| 2
|
Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$ . Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$
|
By the chart, we can see that the " $*$ " operation is actually multiplication modulo $5$ . Thus, we can do $(2*4)*(1*3)\rightarrow(2\cdot4)\cdot(1\cdot3)=8\cdot3=24\rightarrow\boxed{4}$
| 4
|
929
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_11
| 1
|
Consider this histogram of the scores for $81$ students taking a test:
[asy] unitsize(12); draw((0,0)--(26,0)); draw((1,1)--(25,1)); draw((3,2)--(25,2)); draw((5,3)--(23,3)); draw((5,4)--(21,4)); draw((7,5)--(21,5)); draw((9,6)--(21,6)); draw((11,7)--(19,7)); draw((11,8)--(19,8)); draw((11,9)--(19,9)); draw((11,10)--(19,10)); draw((13,11)--(19,11)); draw((13,12)--(19,12)); draw((13,13)--(17,13)); draw((13,14)--(17,14)); draw((15,15)--(17,15)); draw((15,16)--(17,16)); draw((1,0)--(1,1)); draw((3,0)--(3,2)); draw((5,0)--(5,4)); draw((7,0)--(7,5)); draw((9,0)--(9,6)); draw((11,0)--(11,10)); draw((13,0)--(13,14)); draw((15,0)--(15,16)); draw((17,0)--(17,16)); draw((19,0)--(19,12)); draw((21,0)--(21,6)); draw((23,0)--(23,3)); draw((25,0)--(25,2)); for (int a = 1; a < 13; ++a) { draw((2*a,-.25)--(2*a,.25)); } label("$40$",(2,-.25),S); label("$45$",(4,-.25),S); label("$50$",(6,-.25),S); label("$55$",(8,-.25),S); label("$60$",(10,-.25),S); label("$65$",(12,-.25),S); label("$70$",(14,-.25),S); label("$75$",(16,-.25),S); label("$80$",(18,-.25),S); label("$85$",(20,-.25),S); label("$90$",(22,-.25),S); label("$95$",(24,-.25),S); label("$1$",(2,1),N); label("$2$",(4,2),N); label("$4$",(6,4),N); label("$5$",(8,5),N); label("$6$",(10,6),N); label("$10$",(12,10),N); label("$14$",(14,14),N); label("$16$",(16,16),N); label("$12$",(18,12),N); label("$6$",(20,6),N); label("$3$",(22,3),N); label("$2$",(24,2),N); label("Number",(4,8),N); label("of Students",(4,7),N); label("$\textbf{STUDENT TEST SCORES}$",(14,18),N); [/asy]
The median is in the interval labeled
$\text{(A)}\ 60 \qquad \text{(B)}\ 65 \qquad \text{(C)}\ 70 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 80$
|
Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $\boxed{70}$
| 70
|
930
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_12
| 1
|
If each of the three operation signs, $+$ $\text{--}$ $\times$ , is used exactly ONCE in one of the blanks in the expression
\[5\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}4\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}6\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}3\]
then the value of the result could equal
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 19$
|
There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \times 3 = \boxed{19}$
| 19
|
931
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_13
| 1
|
The word " HELP " in block letters is painted in black with strokes $1$ unit wide on a $5$ by $15$ rectangular white sign with dimensions as shown. The area of the white portion of the sign, in square units, is
[asy] unitsize(12); fill((0,0)--(0,5)--(1,5)--(1,3)--(2,3)--(2,5)--(3,5)--(3,0)--(2,0)--(2,2)--(1,2)--(1,0)--cycle,black); fill((4,0)--(4,5)--(7,5)--(7,4)--(5,4)--(5,3)--(7,3)--(7,2)--(5,2)--(5,1)--(7,1)--(7,0)--cycle,black); fill((8,0)--(8,5)--(9,5)--(9,1)--(11,1)--(11,0)--cycle,black); fill((12,0)--(12,5)--(15,5)--(15,2)--(13,2)--(13,0)--cycle,black); fill((13,3)--(14,3)--(14,4)--(13,4)--cycle,white); draw((0,0)--(15,0)--(15,5)--(0,5)--cycle); [/asy]
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 34 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 38$
|
Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=\boxed{36}$
| 36
|
932
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_14
| 1
|
The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$ . Then $A+B=$
\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
|
The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The last square in the top row cannot be either 1 or 3, so it must be 2.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\boxed{4}$
| 4
|
933
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_15
| 1
|
The arithmetic mean (average) of four numbers is $85$ . If the largest of these numbers is $97$ , then the mean of the remaining three numbers is
$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$
|
Say that the four numbers are $a, b, c,$ $97$ . Then $\frac{a+b+c+97}{4} = 85$ . What we are trying to find is $\frac{a+b+c}{3}$ . Solving, \[\frac{a+b+c+97}{4} = 85\] \[a+b+c+97 = 340\] \[a+b+c = 243\] \[\frac{a+b+c}{3} = \boxed{81}\]
| 81
|
934
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_17
| 1
|
Square corners, 5 units on a side, are removed from a $20$ unit by $30$ unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is
[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,0)--(20,0)--(20,5)--(0,5)--cycle); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)); draw((20+5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5*sqrt(2))--(20,0)); draw((5*sqrt(2),5+5*sqrt(2))--(5*sqrt(2),5*sqrt(2))--(5,5),dashed); draw((5*sqrt(2),5*sqrt(2))--(15+5*sqrt(2),5*sqrt(2)),dashed); [/asy]
$\text{(A)}\ 300 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000$
|
If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $((20)(30)-4(5)(5)) = 600-100 = \boxed{500}$
| 500
|
935
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_18
| 1
|
The rectangle shown has length $AC=32$ , width $AE=20$ , and $B$ and $F$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. The area of quadrilateral $ABDF$ is
[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); [/asy]
$\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340$
|
The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $\triangle BCD$ and $\triangle EFD$ subtracted from the area of the rectangle $ABCD$ . Because $B$ and $F$ are midpoints, we know the dimensions of the two right triangles.
[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); label("$16$",A--B,N); label("$16$",B--C,N); label("$32$",E--D,S); label("$10$",E--F,W); label("$10$",A--F,W); label("$20$",C--D,E); [/asy]
\[(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{320}\]
| 320
|
936
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_20
| 1
|
When $10^{93}-93$ is expressed as a single whole number, the sum of the digits is
$\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833$
|
\begin{align*} 10^2-93&=7\\ 10^3-93&=907\\ 10^4-93&=9907\\ \end{align*}
This can be generalized into $10^n-93$ is equal is $n-2$ nines followed by the digits $07$ . Then $10^{93}-93$ is equal to $91$ nines followed by $07$ . The sum of the digits is equal to $9(91)+7=819+7=\boxed{826}$
| 826
|
937
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_21
| 1
|
If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$
|
If a rectangle had dimensions $10 \times 10$ and area $100$ , then its new dimensions would be $12 \times 15$ and area $180$ . The area is increased by $180-100=80$ or $80/100 = \boxed{80}$
| 80
|
938
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_21
| 2
|
If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$
|
Let the dimensions of the rectangle be $x \times y$ . This rectangle has area $xy$ . The new dimensions would be $1.2x \times 1.5y$ , so the area is $=1.8xy$ , which is $\boxed{80}$ more than the original area.
| 80
|
939
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_22
| 1
|
Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?
$\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199$
|
There is $1$ two in the one-digit numbers.
The number of two-digit numbers with a two in the tens place is $10$ and the number with a two in the ones place is $9$ . Thus the digit two is used $10+9=19$ times for the two digit numbers.
Now, Pat Peano only has $22-1-19=2$ remaining twos. You must subtract 1 because 22 is counted twice. The last numbers with a two that he can write are $102$ and $112$ . He can continue numbering the last couple pages without a two until $120$ , with the last number he writes being $\boxed{119}$
| 119
|
940
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_24
| 1
|
What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$
|
Notice that a number in row $k$ is $2k$ less than the number directly below it. For example, $5$ , which is in row $3$ , is $(2)(3)=6$ less than the number below it, $11$
From row 1 to row $k$ , there are $k \left(\frac{1+(-1+2k)}{2} \right) = k^2$ numbers in those $k$ rows. Because there are $12^2=144$ numbers up to the 12th row, $142$ is in the $k^{th}$ row. The number directly above is in the 11th row, and is $22$ less than $142$ . Thus the number directly above $142$ is $142-22=\boxed{120}$
| 120
|
941
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_24
| 2
|
What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$
|
Writing a couple more rows, the last number in each row ends in a perfect square. Thus $142$ is two left from the last number in its row, $144$ . One left and one up from $144$ is the last number of its row, also a perfect square, and is $121$ . This is one right and one up from $142$ , so the number directly above $142$ is one less than $121$ , or $\boxed{120}$
| 120
|
942
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_24
| 3
|
What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$
|
We can notice that even numbers are above even numbers so we can narrow our answers down to 2 solutions, $120$ and $122$ . Notice each number on the right end of each row is a perfect square. The perfect square closest to these is $121$ so $122$ is the first number on the next row. Notice $142$ is the third to last number on its row so it is too far away from 122, so our answer is $\boxed{120}$
| 120
|
943
|
https://artofproblemsolving.com/wiki/index.php/1993_AJHSME_Problems/Problem_25
| 1
|
A checkerboard consists of one-inch squares. A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares. The maximum possible value of $n$ is
$\text{(A)}\ 4\text{ or }5 \qquad \text{(B)}\ 6\text{ or }7\qquad \text{(C)}\ 8\text{ or }9 \qquad \text{(D)}\ 10\text{ or }11 \qquad \text{(E)}\ 12\text{ or more}$
|
Using the Pythagorean Theorem , the diagonal of the square $\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2$ . Because this is longer than $2$ ( length of the sides of two adjacent squares), the card can be placed like so, covering $12$ squares. $\rightarrow \boxed{12}$
| 12
|
944
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_1
| 1
|
$\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$
$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$
|
\begin{align*} \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ &= 1 \rightarrow \boxed{1}
| 1
|
945
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_3
| 1
|
What is the largest difference that can be formed by subtracting two numbers chosen from the set $\{ -16,-4,0,2,4,12 \}$
$\text{(A)}\ 10 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 28 \qquad \text{(E)}\ 48$
|
To maximize anything of the form $a-b$ , we maximize $a$ and minimize $b$ . The maximal element of the set is $12$ and the minimal element is $-16$ , so the maximal difference is \[12-(-16)=28\rightarrow \boxed{28}.\]
| 28
|
946
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_4
| 1
|
During the softball season, Judy had $35$ hits. Among her hits were $1$ home run, $1$ triple and $5$ doubles. The rest of her hits were single. What percent of her hits were single?
$\text{(A)}\ 28\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 75\% \qquad \text{(E)}\ 80\%$
|
From the information given, $35-5-1-1=28$ hits were single. Thus, the percentage was \[\dfrac{28}{35}=80\% \rightarrow \boxed{80}.\]
| 80
|
947
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_5
| 1
|
A circle of diameter $1$ is removed from a $2\times 3$ rectangle, as shown. Which whole number is closest to the area of the shaded region?
[asy] fill((0,0)--(0,2)--(3,2)--(3,0)--cycle,gray); draw((0,0)--(0,2)--(3,2)--(3,0)--cycle,linewidth(1)); fill(circle((1,5/4),1/2),white); draw(circle((1,5/4),1/2),linewidth(1)); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$
|
The area of the shaded region is the area of the circle subtracted from the area of the rectangle.
The diameter of the circle is $1$ , so the radius is $1/2$ and the area is \[(1/2)^2\pi = \pi /4.\]
The rectangle obviously has area $2\times 3= 6$ , so the area of the shaded region is $6-\pi / 4$ .
This is closest to $5\rightarrow \boxed{5}$
| 5
|
948
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_6
| 1
|
Suppose that [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$a$",(1,sqrt(3)-0.2),S); label("$b$",(sqrt(3)/10,0.1),ENE); label("$c$",(2-sqrt(3)/10,0.1),WNW); [/asy] means $a+b-c$ .
For example, [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$5$",(1,sqrt(3)-0.2),S); label("$4$",(sqrt(3)/10,0.1),ENE); label("$6$",(2-sqrt(3)/10,0.1),WNW); [/asy] is $5+4-6 = 3$ .
Then the sum [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$1$",(1,sqrt(3)-0.2),S); label("$3$",(sqrt(3)/10,0.1),ENE); label("$4$",(2-sqrt(3)/10,0.1),WNW); draw((3,0)--(5,0)--(4,sqrt(3))--cycle); label("$2$",(4,sqrt(3)-0.2),S); label("$5$",(3+sqrt(3)/10,0.1),ENE); label("$6$",(5-sqrt(3)/10,0.1),WNW); label("$+$",(2.5,-0.1),N); [/asy] is
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$
|
The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$
Solving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$
Since we have to add the 2 triangles the final answer is $1$ , which is $\boxed{1}$
| 1
|
949
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_7
| 1
|
The digit-sum of $998$ is $9+9+8=26$ . How many 3-digit whole numbers, whose digit-sum is $26$ , are even?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$
|
The highest digit sum for three-digit numbers is $9+9+9=27$ . Therefore, the only possible digit combination is $9, 9, 8$ . Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\boxed{1}$
| 1
|
950
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_8
| 1
|
A store owner bought $1500$ pencils at $$ 0.10$ each. If he sells them for $$ 0.25$ each, how many of them must he sell to make a profit of exactly $$ 100.00$
$\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900$
|
$1500\times 0.1=150$ , so the store owner is $$150$ below profit. Therefore he needs to sell $150+100= 250$ dollars worth of pencils. Selling them at $$0.25$ each gives $250/0.25= \boxed{1000}$
| 0
|
951
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_9
| 1
|
The population of a small town is $480$ . The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
[asy] draw((0,13)--(0,0)--(20,0)); draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0)); label("$\textbf{POPULATION}$",(10,11),N); label("$\textbf{F}$",(5.5,0),S); label("$\textbf{M}$",(13.5,0),S); [/asy]
$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$
|
The graph show that the ratio of men to total population is $\frac{1}{3}$ , so the total number of men is $\frac{1}{3} \times 480= \boxed{160}$
| 160
|
952
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_9
| 2
|
The population of a small town is $480$ . The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
[asy] draw((0,13)--(0,0)--(20,0)); draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0)); label("$\textbf{POPULATION}$",(10,11),N); label("$\textbf{F}$",(5.5,0),S); label("$\textbf{M}$",(13.5,0),S); [/asy]
$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$
|
The graph shows $3$ equal squares, each with value $x$ . So $3x = 480$ , so $x = \boxed{160}$
| 160
|
953
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_10
| 1
|
An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is
[asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewidth(1)); draw((0,c)--(4-c,c),linewidth(1)); draw((c+1,0)--(0,c+1),linewidth(1)); } label("$8$",(2,0),S); label("$8$",(0,2),W); [/asy]
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$
|
Because the smaller triangles are congruent, the shaded area take $\frac{10}{16}$ of the largest triangles area, which is $\frac{8 \times 8}{2}=32$ , so the shaded area is $\frac{10}{16} \times 32= \boxed{20}$
| 20
|
954
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_10
| 2
|
An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is
[asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewidth(1)); draw((0,c)--(4-c,c),linewidth(1)); draw((c+1,0)--(0,c+1),linewidth(1)); } label("$8$",(2,0),S); label("$8$",(0,2),W); [/asy]
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$
|
Each of the triangle has side length of $\frac{1}{4} \times 8=2$ , so the area is $\frac{1}{2} \times 2 \times 2=2$ . Because there are $10$ triangles is the shaded area, its area is $2 \times 10 =\boxed{20}$
| 20
|
955
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_11
| 1
|
The bar graph shows the results of a survey on color preferences. What percent preferred blue?
[asy] for (int a = 1; a <= 6; ++a) { draw((-1.5,4*a)--(1.5,4*a)); } draw((0,28)--(0,0)--(32,0)); draw((3,0)--(3,20)--(6,20)--(6,0)); draw((9,0)--(9,24)--(12,24)--(12,0)); draw((15,0)--(15,16)--(18,16)--(18,0)); draw((21,0)--(21,24)--(24,24)--(24,0)); draw((27,0)--(27,16)--(30,16)--(30,0)); label("$20$",(-1.5,8),W); label("$40$",(-1.5,16),W); label("$60$",(-1.5,24),W); label("$\textbf{COLOR SURVEY}$",(16,26),N); label("$\textbf{F}$",(-6,25),W); label("$\textbf{r}$",(-6.75,22.4),W); label("$\textbf{e}$",(-6.75,19.8),W); label("$\textbf{q}$",(-6.75,17.2),W); label("$\textbf{u}$",(-6.75,15),W); label("$\textbf{e}$",(-6.75,12.4),W); label("$\textbf{n}$",(-6.75,9.8),W); label("$\textbf{c}$",(-6.75,7.2),W); label("$\textbf{y}$",(-6.75,4.6),W); label("D",(4.5,.2),N); label("E",(4.5,3),N); label("R",(4.5,5.8),N); label("E",(10.5,.2),N); label("U",(10.5,3),N); label("L",(10.5,5.8),N); label("B",(10.5,8.6),N); label("N",(16.5,.2),N); label("W",(16.5,3),N); label("O",(16.5,5.8),N); label("R",(16.5,8.6),N); label("B",(16.5,11.4),N); label("K",(22.5,.2),N); label("N",(22.5,3),N); label("I",(22.5,5.8),N); label("P",(22.5,8.6),N); label("N",(28.5,.2),N); label("E",(28.5,3),N); label("E",(28.5,5.8),N); label("R",(28.5,8.6),N); label("G",(28.5,11.4),N); [/asy]
$\text{(A)}\ 20\% \qquad \text{(B)}\ 24\% \qquad \text{(C)}\ 30\% \qquad \text{(D)}\ 36\% \qquad \text{(E)}\ 42\%$
|
The total frequency is $50+60+40+60+40=250$ , with the blue frequency of $60$ . Therefore, the precentage that preferred blue is $\frac{60}{250}=\boxed{24}$
| 24
|
956
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_13
| 1
|
Five test scores have a mean (average score) of $90$ , a median (middle score) of $91$ and a mode (most frequent score) of $94$ . The sum of the two lowest test scores is
$\text{(A)}\ 170 \qquad \text{(B)}\ 171 \qquad \text{(C)}\ 176 \qquad \text{(D)}\ 177 \qquad \text{(E)}\ \text{not determined by the information given}$
|
Because there was an odd number of scores, $91$ must be the middle score. Since there are two scores above $91$ and $94$ appears the most frequent (so at least twice) and $94>91$ $94$ appears twice. Also, the sum of the five numbers is $90 \times 5 =450$ . Thus, the sum of the lowest two scores is $450-91-94-94= \boxed{171}$
| 171
|
957
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_14
| 1
|
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 48$
|
Four gallons is $\frac{1}{2}-\frac{1}{3}= \frac{1}{6}$ . Thus, capacity of the tank in gallons is $6 \times 4= \boxed{24}$
| 24
|
958
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_17
| 1
|
The sides of a triangle have lengths $6.5$ $10$ , and $s$ , where $s$ is a whole number. What is the smallest possible value of $s$
[asy] pair A,B,C; A=origin; B=(10,0); C=6.5*dir(15); dot(A); dot(B); dot(C); draw(B--A--C); draw(B--C,dashed); label("$6.5$",3.25*dir(15),NNW); label("$10$",(5,0),S); label("$s$",(8,1),NE); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$
|
By Triangle Inequality $6.5 + s >10$ and therefore $s>3.5$ . The smallest whole number that satisfies this is $\boxed{4}$
| 4
|
959
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_18
| 1
|
On a trip, a car traveled $80$ miles in an hour and a half, then was stopped in traffic for $30$ minutes, then traveled $100$ miles during the next $2$ hours. What was the car's average speed in miles per hour for the $4$ -hour trip?
$\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$
|
The average speed is given by the total distance traveled divided by the total time traveled. \[\frac{80+0+100}{4} = \boxed{45}\]
| 45
|
960
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_19
| 1
|
The distance between the $5^\text{th}$ and $26^\text{th}$ exits on an interstate highway is $118$ miles. If any two consecutive exits are at least $5$ miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the $5^\text{th}$ and $26^\text{th}$ exits?
$\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98$
|
There are $21$ pairs of consecutive exits. To find the maximum number of miles of one of these, the other $20$ must be equal to the minimum number yielding a total of $(5)(20)=100$ miles. The longest distance must be $118-100=\boxed{18}$
| 18
|
961
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_22
| 1
|
Eight $1\times 1$ square tiles are arranged as shown so their outside edges form a polygon with a perimeter of $14$ units. Two additional tiles of the same size are added to the figure so that at least one side of each tile is shared with a side of one of the squares in the original figure. Which of the following could be the perimeter of the new figure?
[asy] for (int a=1; a <= 4; ++a) { draw((a,0)--(a,2)); } draw((0,0)--(4,0)); draw((0,1)--(5,1)); draw((1,2)--(5,2)); draw((0,0)--(0,1)); draw((5,1)--(5,2)); [/asy]
$\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$
|
One such figure would be
[asy] for (int a=1; a <= 4; ++a) { draw((a,0)--(a,3)); } draw((0,0)--(4,0)); draw((0,1)--(5,1)); draw((1,2)--(5,2)); draw((0,0)--(0,1)); draw((5,1)--(5,2)); draw((2,3)--(1,3)); draw((4,3)--(3,3)); [/asy]
The perimeter of this figure is $\boxed{18}$
| 18
|
962
|
https://artofproblemsolving.com/wiki/index.php/1992_AJHSME_Problems/Problem_25
| 1
|
One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain?
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$
|
1)Model the amount left in the container as follows:
After the first pour $\frac12$ remains, after the second $\frac12 \times \frac23$ remains, etc.
This becomes the product $\frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{9}{10}$
Note that the terms cancel out leaving $\frac{1}{10}$
Now all that remains is to count the number of terms or pourings, as the numerators form an arithmetic sequence with a common difference of 1 and endpoints (1,9), the number of pourings is $\boxed{9}$
| 9
|
963
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_2
| 1
|
$\frac{16+8}{4-2}=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$
|
\begin{align*} \frac{16+8}{4-2} &= \frac{24}{2} \\ &= 12\rightarrow \boxed{12}
| 12
|
964
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_4
| 1
|
If $991+993+995+997+999=5000-N$ , then $N=$
$\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$
|
\begin{align*} 991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\ &\Rightarrow 5000-25=5000-N \\ &\Rightarrow N=25\rightarrow \boxed{25}
| 25
|
965
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_6
| 1
|
Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.) \[\begin{tabular}[t]{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{tabular}\]
$\text{(A)}\ 1 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 15$
|
The largest numbers in the first, second, third, fourth and fifth columns are $13,7,15,12,9$ respectively. Of these, only $7$ is the smallest in its row $\rightarrow \boxed{7}$
| 7
|
966
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_8
| 1
|
What is the largest quotient that can be formed using two numbers chosen from the set $\{ -24, -3, -2, 1, 2, 8 \}$
$\text{(A)}\ -24 \qquad \text{(B)}\ -3 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$
|
Let the two chosen numbers be $a$ and $b$ . To maximize the quotient, we first have either $a,b>0$ or $a,b<0$ , and from there we maximize $|a|$ and minimize $|b|$
For the case $a,b<0$ , we have $a=-24$ and $b=-2$ , which gives us $(-24)/(-2)=12$ . For the case $a,b>0$ , we have $a=8$ and $b=1$ , which gives us $8/1=8$
Since $12>8$ , our answer is $\boxed{12}$
| 12
|
967
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_9
| 1
|
How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?
$\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27$
|
There are $\left\lfloor \frac{46}{3}\right\rfloor =15$ numbers divisible by $3$ $\left\lfloor\frac{46}{5}\right\rfloor =9$ numbers divisible by $5$ , so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$ , except we counted the multiples of $\text{LCM}(3,5)=15$ twice, once for $3$ and once for $5$
There are $\left\lfloor \frac{46}{15}\right\rfloor =3$ numbers divisible by $15$ , so there are $24-3=21$ numbers divisible by $3$ or $5$ $\rightarrow \boxed{21}$
| 21
|
968
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_10
| 1
|
The area in square units of the region enclosed by parallelogram $ABCD$ is
[asy] unitsize(24); pair A,B,C,D; A=(-1,0); B=(0,2); C=(4,2); D=(3,0); draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0)); draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25)); dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E); label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE); label("$A$",A,SW); label("$B$",B,NW); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$
|
The base is $\overline{BC}=4$ . The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\cdot 2=8\Rightarrow \boxed{8}$
| 8
|
969
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_11
| 1
|
There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{ 1,2,3,4,5,6,7,8,9 \}$ . How many of these sets contain a $5$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$
|
Let the three-element set be $\{ a,b,c \}$ and suppose that $a=5$
We need $b+c=10$ and $b\neq c$ . This gives us four solutions, so there are $4$ sets with a $5$ also with the desired properties $\rightarrow \boxed{4}$
| 4
|
970
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_12
| 1
|
If $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$ , then $N=$
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$
|
Note that for all integers $n\neq 0$ \[\frac{(n-1)+n+(n+1)}{n}=3.\] Thus, we must have $N=1991\rightarrow \boxed{1991}$
| 991
|
971
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_13
| 1
|
How many zeros are at the end of the product \[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$
|
\begin{align*} (5^2)^7\times (2^3)^3 &= 5^{14}\times 2^{9} \\ &= 5^5\times 10^9 \end{align*} Since $5^5$ doesn't end in a $0$ , we can conclude that the product ends in $9$ zeroes $\rightarrow \boxed{9}$
| 9
|
972
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_14
| 1
|
Several students are competing in a series of three races. A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$
|
There are two ways for a student to get $11$ $5+5+1$ and $5+3+3$ . Clearly if someone gets one of these combinations someone else could get the other, so we are not guaranteed the most points with $11$
There is only one way to get $13$ points: $5+5+3$ . In this case, the largest score another person could get is $5+3+3=11$ , so having $13$ points guarantees having more points than any other person $\rightarrow \boxed{13}$
| 13
|
973
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_14
| 2
|
Several students are competing in a series of three races. A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$
|
If someone gets $11$ points, the possible combinations are $5,5,1$ and $5,3,3$ If he gets $5,3,3$ then someone else can be $5,3,3$ which would not guarantee victory.
If we have 13 points, the only way to make this is $5,5,3$ . There is no way to get any number of points higher than this, so the answer is $\boxed{13}$ .---stjwyl
| 13
|
974
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_16
| 1
|
The $16$ squares on a piece of paper are numbered as shown in the diagram. While lying on a table, the paper is folded in half four times in the following sequence:
(1) fold the top half over the bottom half
(2) fold the bottom half over the top half
(3) fold the right half over the left half
(4) fold the left half over the right half.
Which numbered square is on top after step $4$
[asy] unitsize(18); for(int a=0; a<5; ++a) { draw((a,0)--(a,4)); } for(int b=0; b<5; ++b) { draw((0,b)--(4,b)); } label("$1$",(0.5,3.1),N); label("$2$",(1.5,3.1),N); label("$3$",(2.5,3.1),N); label("$4$",(3.5,3.1),N); label("$5$",(0.5,2.1),N); label("$6$",(1.5,2.1),N); label("$7$",(2.5,2.1),N); label("$8$",(3.5,2.1),N); label("$9$",(0.5,1.1),N); label("$10$",(1.5,1.1),N); label("$11$",(2.5,1.1),N); label("$12$",(3.5,1.1),N); label("$13$",(0.5,0.1),N); label("$14$",(1.5,0.1),N); label("$15$",(2.5,0.1),N); label("$16$",(3.5,0.1),N); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$
|
Suppose we undo each of the four folds, considering just the top square until we completely unfold the paper. $x$ will be marked in the square if the face that shows after all the folds is face up, $y$ if that face is facing down.
Step 0: [asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); label("$x$",(0.5,0.1),N); [/asy] Step 1: [asy] unitsize(18); draw((0,0)--(2,0)--(2,1)--(0,1)--cycle); draw((1,0)--(1,1)); label("$y$",(0.5,0.1),N); [/asy] Step 2: [asy] unitsize(18); draw((0,0)--(4,0)--(4,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((3,0)--(3,1)); label("$y$",(0.5,0.1),N); [/asy] Step 3: [asy] unitsize(18); draw((0,0)--(4,0)--(4,2)--(0,2)--cycle); draw((0,1)--(4,1)); draw((1,0)--(1,2)); draw((2,0)--(2,2)); draw((3,0)--(3,2)); label("$y$",(0.5,1.1),N); [/asy] Step 4: [asy] unitsize(18); for(int a=0; a<5; ++a) { draw((a,0)--(a,4)); } for(int b=0; b<5; ++b) { draw((0,b)--(4,b)); } label("$y$",(0.5,1.1),N); [/asy] The marked square is in the same spot as the number $9\rightarrow \boxed{9}$
| 9
|
975
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_17
| 1
|
An auditorium with $20$ rows of seats has $10$ seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is
$\text{(A)}\ 150 \qquad \text{(B)}\ 180 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 460$
|
We first note that if a row has $n$ seats, then the maximum number of students that can be seated in that row is $\left\lceil \frac{n}{2} \right\rceil$ , where $\lceil x \rceil$ is the smallest integer greater than or equal to $x$ . If a row has $2k$ seats, clearly we can only fit $k$ students in that row. If a row has $2k+1$ seats, we can fit $k+1$ students by putting students at the ends and then alternating between skipping a seat and putting a student in.
For each row with $10+k$ seats, there is a corresponding row with $29-k$ seats. The sum of the maximum number of students for these rows is \[\left\lceil \frac{10+k}{2}\right\rceil +\left\lceil \frac{29-k}{2} \right\rceil = 20.\] There are $20/2=10$ pairs of rows, so the maximum number of students for the exam is $20\times 10=200\rightarrow \boxed{200}$
| 200
|
976
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_19
| 1
|
The average (arithmetic mean) of $10$ different positive whole numbers is $10$ . The largest possible value of any of these numbers is
$\text{(A)}\ 10 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 55 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 91$
|
If the average of the numbers is $10$ , then their sum is $10\times 10=100$
To maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ . Their sum is $45$
The sum of nine of the numbers is $45$ , and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{55}$
| 55
|
977
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_20
| 1
|
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 $C=1\rightarrow \boxed{1}$
| 1
|
978
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_20
| 2
|
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 $C=1\rightarrow \boxed{1}$
| 1
|
979
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_21
| 1
|
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 $24-4(4)=8\rightarrow \boxed{8}$
| 8
|
980
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_23
| 1
|
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 $80-70=10\rightarrow \boxed{10}$
| 10
|
981
|
https://artofproblemsolving.com/wiki/index.php/1991_AJHSME_Problems/Problem_24
| 1
|
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 $1+19=20\rightarrow \boxed{20}$
| 20
|
982
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_1
| 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 \[100(9)+10(13)+(17)=1047 \rightarrow \boxed{1047}\]
| 47
|
983
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_2
| 1
|
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 $9 \rightarrow \boxed{1}$
| 1
|
984
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_4
| 1
|
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 $\boxed{8}$
| 8
|
985
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_7
| 1
|
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 $30\rightarrow \boxed{30}$
| 30
|
986
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_8
| 1
|
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 $\rightarrow \boxed{66}$
| 66
|
987
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_11
| 1
|
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 $81\rightarrow \boxed{81}$
| 81
|
988
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_12
| 1
|
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: $5247,5274,5427,5472,5724 \rightarrow \boxed{5724}$
| 724
|
989
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_14
| 1
|
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 $\rightarrow \boxed{18}$
| 18
|
990
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_15
| 1
|
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 $10(5)=50\rightarrow \boxed{50}$
| 50
|
991
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_16
| 1
|
$1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$
|
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 $1000 \rightarrow \boxed{1000}$
| 0
|
992
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_16
| 2
|
$1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$
|
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 $\boxed{1000}$
| 0
|
993
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_17
| 1
|
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 $2\rightarrow \boxed{2}$
| 2
|
994
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_18
| 1
|
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 $\rightarrow \boxed{36}$
| 36
|
995
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_19
| 1
|
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
| 40
|
996
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_20
| 1
|
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 $882, or \rightarrow \boxed{882}$
| 882
|
997
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_22
| 1
|
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 $\boxed{11}$
| 11
|
998
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_24
| 1
|
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 $\rightarrow \boxed{3}$
| 3
|
999
|
https://artofproblemsolving.com/wiki/index.php/1990_AJHSME_Problems/Problem_25
| 1
|
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 $\rightarrow \boxed{8}$
| 8
|
1,000
|
https://artofproblemsolving.com/wiki/index.php/1989_AJHSME_Problems/Problem_1
| 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}
| 250
|
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