Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

Solution: We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$ , it is less than or equal to $7542$ , the largest number in the set. This happens to be $2754\times2=5508$ . Therefore, the number would have to be between $4914$ and $5508$ , and also even. The only even numbers in the set and in this range are $5472$ and $5274$ . A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\times4=9828$ , well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\times3=7371$ and the greatest is $2475\times3=7425$ , since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\times3=7425, \boxed{7425}$

Answer: D 7425
Problem: There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

Solution: There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.
$5724/2=2862$ $5724/3=1908$ $7245/3=2415$ $7254/2=3612$ $7254/3=2418$ $7425/3=2475$ . The answer is $\boxed{7425}$ . You can obtain the answer in only 6 calculations.

Answer: D 7425
Problem: Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$

Solution: If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old.
If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$
So, the answer is $\boxed{16}$

Answer: B 16
Problem: Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37$

Solution: Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old.
Now we just find  Caitlin's age by doing $21-5$ . This makes $16$ or $\boxed{16}$

Answer: B 16
Problem: Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$ . The reciprocal of $2$ is $1/2$ , but $2$ is not less than $1/2$ . The reciprocal of $-2$ is $-1/2$ , and $-2$ is less than $-1/2$ , so it is $\boxed{2}$

Answer: A 2
Problem: Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$
Multiplication by $x$ can be done if you do it in three parts: $x>0$ $x=0$ , and $x<0$ .  You have to be careful about the direction of the inequality, as you do not know the sign of $x$
If $x>0$ , the sign of the inequality remains the same.  Thus, we have $x^2 < 1$ when $x > 0$ .  This leads to $0 < x < 1$
If $x=0$ , the inequality $x < \frac{1}{x}$ is undefined.
If $x<0$ , the sign of the inequality must be switched.  Thus, we have $x^2 > 1$ when $x < 0$ .  This leads to $x < -1$
Putting the solutions together, we have $x<-1$ or $0 < x < 1$ , or in interval notation, $(-\infty, -1) \cup(0, 1)$ .  The only answer in that range is $\boxed{2}$

Answer: A 2
Problem: Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: Starting again with $x < \frac{1}{x}$ , we avoid multiplication by $x$ .   Instead, move everything to the left, and find a common denominator:
$x < \frac{1}{x}$
$x - \frac{1}{x} < 0$
$\frac{x^2 - 1}{x} < 0$
$\frac{(x+1)(x-1)}{x} < 0$
Divide this expression at $x=-1$ $x=0$ , and $x=1$ , as those are the three points where the expression on the left will "change sign".
If $x<-1$ , all three of those terms will be negative, and the inequality is true.  Therefore, $(-\infty, -1)$ is part of our solution set.
If $-1 < x < 0$ , the $(x+1)$ term will become positive, but the other two terms remain negative.  Thus, there are no solutions in this region.
If $0 < x < 1$ , then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative.  Thus, the entire region $(0, -1)$ is part of the solution set.
If $1 < x$ , then all three terms are positive, and there are no solutions.
At all three "boundary points", the function is either $0$ or undefined.  Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$ , and the only option in that region is $x=-2$ , leading to $\boxed{2}$

Answer: A 2
Problem: Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: We can find out all of their reciprocals. Now we compare and see that the answer is $\boxed{2}$

Answer: A 2
Problem: How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$

Solution: The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ $3$ $4$ $5$ , and $6$ , so the answer is $\boxed{5}$

Answer: D 5
Problem: Each principal of Lincoln High School serves exactly one $3$ -year term. What is the maximum number of principals this school could have during an $8$ -year period?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$

Solution: If the first year of the $8$ -year period was the final year of a principal's
term, then in the next six years two more principals would serve, and the last year of the
period would be the first year of the fourth principal's term. Therefore, the maximum
number of principals who can serve during an $8$ -year period is $4$ , so the answer is $\boxed{4}$ if the terms are divided $1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8$

Answer: C 4
Problem: Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1));  label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: The side of the large square is $1 + 3 + 1 = 5$ , so the area of the large square is $5^2 = 25$
The area of the middle square is $3^2$ , and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$
Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$ .  This is the area of the two congruent L-shaped regions.
So the area of one L-shaped region is $\frac{14}{2} = 7$ , and the answer is $\boxed{7}$

Answer: A 7
Problem: Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1));  label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: The shaded area can be divided into two regions:  one rectangle that is 1 by 3, and one rectangle that is 4 by 1.  (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.)  Either way, the total area of these two regions is $3 + 4 = 7$ , and the answer is $\boxed{7}$

Answer: A 7
Problem: Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1));  label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper.  When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{7}$ as the answer.

Answer: A 7
Problem: Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$ -shaped region is
[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1));  label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]
$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of $4\cdot4=16$ . In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$ . When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{7}$

Answer: A 7
Problem: What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$
$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$

Solution: The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices
exist: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$ .
The latter is smaller, so $\boxed{280}$

Answer: B 280
Problem: problem_id
f530163b6184f697cd4b5054402e0ccf    Three dice with faces numbered $1$ through $6$...
f530163b6184f697cd4b5054402e0ccf    Three dice with faces numbered 1 through 6 are...
Name: Text, dtype: object

Solution: The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers
on the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .
This leaves $63 - 22 = \boxed{41}$ not seen.

Answer: D 41
Problem: Three-digit powers of $2$ and $5$ are used in this "cross-number" puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}.~ 2^m & & \textbf{1}.~ 5^n \end{array}\]
[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0)); draw((3,0)--(2,0)--(2,1)--(3,1)--cycle,linewidth(2));  label("$1$",(0,2),SE); label("$2$",(0,1),SE); [/asy]
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: The $3$ -digit powers of $5$ are $125$ and $625$ , so space $2$ is filled with a $2$ .
The only $3$ -digit power of $2$ beginning with $2$ is $256$ , so the outlined block is filled with
a $\boxed{6}$

Answer: D 6
Problem: Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
$\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

Solution: Shea has grown $20\%$ , if x was her original height, then $1.2x = 60$ , so she was originally $\frac{60}{1.2}=50$ inches tall which is a $60 - 50 = 10$ inch increase.  Ara also started off at $50$ inches.  Since Ara grew half as much as Shea, Ara grew $\frac{10}{2} = 5$ inches.  Therefore, Ara is now $50+5=55$ inches tall which is choice $\boxed{55}.$

Answer: E 55
Problem: The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Solution: Casework by the units digit $u$ will help organize the answer.
$u=0$ gives no solutions, since no real numbers are divisible by $0$
$u=1$ has $4$ solutions, since all numbers are divisible by $1$
$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).
$u=3$ has $1$ solution: $33$ $\pm 10$ or $\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$ $\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$ , but those numbers are $3$ and $63$ , which are out of the range of the problem.
$u=4$ has $2$ solutions: $24$ and $44$ .  Adding or subtracting $10$ will kill divisibility by $4$ , since $10$ is not divisible by $4$
$u=5$ has $4$ solutions:  every number ending in $5$ is divisible by $5$
$u=6$ has $1$ solution: $36$ $\pm 10$ or $\pm 20$ will kill divisibility by $3$ , and thus kill divisibility by $6$
$u=7$ has no solutions.  The first multiples of $7$ that end in $7$ are $7$ and $77$ , but both are outside of the range of this problem.
$u=8$ has $1$ solution: $48$ $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$
$u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$
Totalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\boxed{17}$ , which is 17.

Answer: C 17
Problem: A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
[asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1));  [/asy]
$\text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356$

Solution: Since the bricks are $1$ foot high, there will be $7$ rows.  To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\frac{100}{2} = 50$ bricks to construct.  Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks.
Four rows of $50$ bricks and three rows of $51$ bricks totals $4\cdot 50 + 3\cdot 51 = 200 + 153 = 353$ bricks, giving the answer $\boxed{353}.$

Answer: D 353
Problem: In triangle $CAT$ , we have $\angle ACT =\angle ATC$ and $\angle CAT = 36^\circ$ . If $\overline{TR}$ bisects $\angle ATC$ , then $\angle CRT =$
[asy] pair A,C,T,R; C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570); draw(C--A--T--cycle); draw(T--R); label("$A$",A,N); label("$T$",T,SE); label("$C$",C,SW); label("$R$",R,NW);[/asy]
$\text{(A)}\ 36^\circ\qquad\text{(B)}\ 54^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 90^\circ\qquad\text{(E)}\ 108^\circ$

Solution: In $\triangle ACT$ , the three angles sum to $180^\circ$ , and $\angle C = \angle T$
$\angle CAT + \angle ATC + \angle ACT = 180$
$36 + \angle ATC + \angle ATC = 180$
$2 \angle ATC = 144$
$\angle ATC = 72$
Since $\angle ATC$ is bisected by $\overline{TR}$ $\angle RTC = \frac{72}{2} = 36$
Now focusing on the smaller $\triangle RTC$ , the sum of the angles in that triangle is $180^\circ$ , so:
$\angle RTC + \angle TCR + \angle CRT = 180$
$36 + \angle ACT + \angle CRT = 180$
$36 + \angle ATC + \angle CRT = 180$
$36 + 72 + \angle CRT = 180$
$\angle CRT = 72^\circ$ , giving the answer $\boxed{72}$

Answer: C 72
Problem: Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy]
$\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$

Solution: The large triangle $ABC$ has sides of length $4$ .  The medium triangle has sides of length $2$ .  The small triangle has sides of length $1$ .  There are $3$ segment sizes, and all segments depicted are one of these lengths.
Starting at $A$ and going clockwise, the perimeter is:
$AB + BC + CD + DE + EF + FG + GA$
$4 + 4 + 2 + 2 + 1 + 1 + 1$
$15$ , thus the answer is $\boxed{15}$

Answer: C 15
Problem: Triangles $ABC$ $ADE$ , and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$ , respectively. If $AB = 4$ , what is the perimeter of figure $ABCDEFG$
[asy] pair A,B,C,D,EE,F,G; A = (4,0); B = (0,0); C = (2,2*sqrt(3)); D = (3,sqrt(3)); EE = (5,sqrt(3)); F = (5.5,sqrt(3)/2); G = (4.5,sqrt(3)/2); draw(A--B--C--cycle); draw(D--EE--A); draw(EE--F--G); label("$A$",A,S); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,NE); label("$E$",EE,NE); label("$F$",F,SE); label("$G$",G,SE);[/asy]
$\text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21$

Solution: The perimeter of $ABCDEFG$ is the perimeter of the three triangles, minus segments $AD$ and $EG$ , which are on the interior of the figure.  Because each of these segments is on two triangles, each segment must be subtracted two times.
As in solution 1, the sides of the triangles are $4, 2,$ and $1$ , and the perimeters of the triangles are thus $12, 6,$ and $3$
The perimeter of the three triangles is $12 + 6 + 3 = 21$ .  Subtracting the two segments $AD$ and $EG$ two times, the perimeter of $ABCDEFG$ is $21 - 2 - 1 - 2 - 1 = 15$ , and the answer is $\boxed{15}$

Answer: C 15
Problem: In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$

Solution: The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$ , the area is $40\cdot 10=400$ square meters or $\boxed{400}$

Answer: C 400
Problem: Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?
[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,A,CCW)); label("$A$",A,S); label("$B$",B,W); label("$C$",C,N); label("$D$",D,E);[/asy]
$\text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi$

Solution: Draw two squares:  one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ .  These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle.
These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that must be removed.  Thus, the bottom part of the figure has area
$2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2$
$50 - \frac{25\pi}{2}$
This is the area of the part of the figure underneath $\overline{BD}$ .  The part of the figure over $\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}$
Adding the two areas gives a total area of $\boxed{50}$

Answer:  50
Problem: You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution: Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.
You must have $1$ more penny.  If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters.  But you only have $5$ more coins to assign.
Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes.  If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible.  If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible.  If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible.  And all four remaining coins being dimes will not be enough.
Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels.  We quickly see that $2$ nickels and $2$ quarters work.  Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted.  Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents.
There is only $1$ dime in that combo, so the answer is $\boxed{1}$

Answer: A 1
Problem: You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution: We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\boxed{1}$

Answer: A 1