Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: 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
$\text{(A)}\ 81.0 \qquad \text{(B)}\ 82.7 \qquad \text{(C)}\ 83.0 \qquad \text{(D)}\ 84.0 \qquad \text{(E)}\ 84.3$

Solution: 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, \[\frac{a+b+c+97}{4} = 85\] \[a+b+c+97 = 340\] \[a+b+c = 243\] \[\frac{a+b+c}{3} = \boxed{81}\]

Answer: A 81
Problem: 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]
$\text{(A)}\ 300 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000$

Solution: 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 $((20)(30)-4(5)(5)) = 600-100 = \boxed{500}$

Answer: B 500
Problem: 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]
$\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340$

Solution: 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]
\[(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{320}\]

Answer: A 320
Problem: $(1901+1902+1903+\cdots + 1993) - (101+102+103+\cdots + 193) =$
$\text{(A)}\ 167,400 \qquad \text{(B)}\ 172,050 \qquad \text{(C)}\ 181,071 \qquad \text{(D)}\ 199,300 \qquad \text{(E)}\ 362,142$

Solution: We see that $1901=1800+101$ $1902=1800+102$ , etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:
\[(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ =1800 + 1800 + \cdots + 1800\\ =(1800)(93)\\ =\boxed{167,400}\]

Answer: A 167,400
Problem: When $10^{93}-93$ is expressed as a single whole number, the sum of the digits is
$\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833$

Solution: \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 $9(91)+7=819+7=\boxed{826}$

Answer: D 826
Problem: If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$

Solution: 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 $80/100 = \boxed{80}$

Answer: D 80
Problem: If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$ , then the area is increased by
$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$

Solution: 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.

Answer:  80
Problem: 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?
$\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199$

Solution: 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 $\boxed{119}$

Answer: D 119
Problem: What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$

Solution: 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 $142-22=\boxed{120}$

Answer: C 120
Problem: What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$

Solution: 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 $\boxed{120}$

Answer: C 120
Problem: What number is directly above $142$ in this array of numbers?
\[\begin{array}{cccccc}& & & 1 & &\\ & & 2 & 3 & 4 &\\ & 5 & 6 & 7 & 8 & 9\\ 10 & 11 & 12 &\cdots & &\\ \end{array}\]
$\text{(A)}\ 99 \qquad \text{(B)}\ 119 \qquad \text{(C)}\ 120 \qquad \text{(D)}\ 121 \qquad \text{(E)}\ 122$

Solution: 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 $\boxed{120}$

Answer: C 120
Problem: 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
$\text{(A)}\ 4\text{ or }5 \qquad \text{(B)}\ 6\text{ or }7\qquad \text{(C)}\ 8\text{ or }9 \qquad \text{(D)}\ 10\text{ or }11 \qquad \text{(E)}\ 12\text{ or more}$

Solution: 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. $\rightarrow \boxed{12}$

Answer: E 12
Problem: $\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$
$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution: \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}

Answer: B 1
Problem: What is the largest difference that can be formed by subtracting two numbers chosen from the set $\{ -16,-4,0,2,4,12 \}$
$\text{(A)}\ 10 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 28 \qquad \text{(E)}\ 48$

Solution: 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 \[12-(-16)=28\rightarrow \boxed{28}.\]

Answer: D 28
Problem: 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?
$\text{(A)}\ 28\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 75\% \qquad \text{(E)}\ 80\%$

Solution: From the information given, $35-5-1-1=28$ hits were single.  Thus, the percentage was \[\dfrac{28}{35}=80\% \rightarrow \boxed{80}.\]

Answer: E 80
Problem: 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]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: 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 $5\rightarrow \boxed{5}$

Answer: E 5
Problem: 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
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: 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 $\boxed{1}$

Answer: D 1
Problem: The digit-sum of $998$ is $9+9+8=26$ .  How many 3-digit whole numbers, whose digit-sum is $26$ , are even?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: 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 $\boxed{1}$

Answer: A 1
Problem: 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$
$\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900$

Solution: $1500\times 0.1=150$ , 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 $250/0.25= \boxed{1000}$

Answer: C 1000
Problem: 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]
$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$

Solution: The graph show that the ratio of men to total population is $\frac{1}{3}$ , so the total number of men is $\frac{1}{3} \times 480= \boxed{160}$

Answer: B 160
Problem: 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]
$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$

Solution: The graph shows $3$ equal squares, each with value $x$ . So $3x = 480$ , so $x = \boxed{160}$

Answer:  160
Problem: 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]
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

Solution: 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 $\frac{10}{16} \times 32= \boxed{20}$

Answer: B 20
Problem: 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]
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

Solution: 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 $2 \times 10 =\boxed{20}$

Answer: B 20
Problem: 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]
$\text{(A)}\ 20\% \qquad \text{(B)}\ 24\% \qquad \text{(C)}\ 30\% \qquad \text{(D)}\ 36\% \qquad \text{(E)}\ 42\%$

Solution: The total frequency is $50+60+40+60+40=250$ , with the blue frequency of $60$ . Therefore, the precentage that preferred blue is $\frac{60}{250}=\boxed{24}$

Answer: B 24
Problem: The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first $30,000$ miles the car traveled.  For how many miles was each tire used?
$\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad \text{(E)}\ 37,500$

Solution: In the $30,000$ miles, four tires were always used at one time, so the amount of miles the five tires were used in total is $30,000 \times 4=120,000$ . Five tires were used and each was used equally, so each tire was used for $\frac{120,000}{5}=\boxed{24,000}$

Answer: C 24,000
Problem: 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
$\text{(A)}\ 170 \qquad \text{(B)}\ 171 \qquad \text{(C)}\ 176 \qquad \text{(D)}\ 177 \qquad \text{(E)}\ \text{not determined by the information given}$

Solution: 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 $450-91-94-94= \boxed{171}$

Answer: B 171
Problem: 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
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 48$

Solution: Four gallons is $\frac{1}{2}-\frac{1}{3}= \frac{1}{6}$ . Thus, capacity of the tank in gallons is $6 \times 4= \boxed{24}$

Answer: D 24
Problem: 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]
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution: By Triangle Inequality $6.5 + s >10$ and therefore $s>3.5$ . The smallest whole number that satisfies this is $\boxed{4}$

Answer: B 4
Problem: 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?
$\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

Solution: The average speed is given by the total distance traveled divided by the total time traveled. \[\frac{80+0+100}{4} = \boxed{45}\]

Answer: A 45
Problem: 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?
$\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98$

Solution: 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 $118-100=\boxed{18}$

Answer: C 18
Problem: 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]
$\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution: 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 $\boxed{18}$

Answer: C 18
Problem: Four circles of radius $3$ are arranged as shown.  Their centers are the vertices of a square.  The area of the shaded region is closest to
[asy] fill((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle,lightgray); fill(arc((3,3),(0,3),(3,0),CCW)--(3,3)--cycle,white); fill(arc((3,-3),(3,0),(0,-3),CCW)--(3,-3)--cycle,white); fill(arc((-3,-3),(0,-3),(-3,0),CCW)--(-3,-3)--cycle,white); fill(arc((-3,3),(-3,0),(0,3),CCW)--(-3,3)--cycle,white);  draw(circle((3,3),3)); draw(circle((3,-3),3)); draw(circle((-3,-3),3)); draw(circle((-3,3),3)); draw((3,3)--(3,-3)--(-3,-3)--(-3,3)--cycle); [/asy]
$\text{(A)}\ 7.7 \qquad \text{(B)}\ 12.1 \qquad \text{(C)}\ 17.2 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 27$

Solution: The side of the square of made up of two radii, and the area of the square is $(3+3)^2=36$ . The unshaded region of the square is composed of four quarter circles, and has the area of one circle, which is $9\pi$ . The area of the shaded region is $36-9\pi$ which less than $36-9(3)=9$ , closest to $\boxed{7.7}$

Answer: A 7.7
Problem: 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?
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution: 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 $\boxed{9}$

Answer: D 9