UND MATHEMATICS TRACK MEET Team Test 2

University of North Dakota Grades 9/10
January 12, 2026
School Team Name
Calculators are NOT allowed. Solutions

• 1.

  A sequence is defined by $a_1=4$ and $a_{n+1}=a_n+3$. What is $a_{20}$?

    (20 pts) 1. $61$

  Solution: We are given $a_1=4$ and $a_{n+1}=a_n+3$. This is an arithmetic sequence with first term $4$ and common difference $3$.

  The $n$-th term of an arithmetic sequence is

  $$a_n=a_1+(n-1)d.$$

  Here $a_1=4$, $d=3$, and $n=20$:

  $$a_{20}=4+(20-1)\cdot 3=4+19\cdot 3=4+57=61.$$

• 2.

  What is the remainder when $7^{2025}$ is divided by $6$?

    (20 pts) 2. $1$

  Solution: We want the remainder when $7^{2025}$ is divided by $6$.

  Notice that

  $$7\equiv 1(mod6),$$

  so

  $$7^{2025}\equiv 1^{2025}\equiv 1(mod6).$$

  Therefore, the remainder is $1$.

• 3.

  The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many more children than adults attended?

    (20 pts) 3. $800$

  Solution: Let $x$ denote the number of adults and let $y$ denote the number of children. Then, we have the system

  So, we solve the system for $x$ and $y$ using whatever method seems easiest. Here, we opt for substitution and solve the first equation for $x$:

  	$x+y$	$=2200$
  	$x$	$=2200-y$

  Now, we substitute our formula for $x$ in to the second equation

  	$4x+1.5y$	$=5050$
  	$4(2200-y)+1.5y$	$=5050$
  	$8800-2.5y$	$=5050$
  	$y$	$={\displaystyle \frac{5050-8800}{-2.5}}=1500.$

  Using our equation from the first step, this gives $x=2200-1500=700$. So, we have our answer: $700$ adults and $1500$ children attended the fair, and $1500-700=800$ more children attended.

• 4.

  A right triangle has legs $6$ and $8$. A square is drawn with the hypotenuse of the right triangle as a side. What is the area of the square?

    (20 pts) 4. $100$

  Solution: theorem:

  $$c^2=6^2+8^2=36+64=100,$$

  so

  $$c=\sqrt{100}=10.$$

  A square on the hypotenuse has side length $10$, so its area is

  $${10}^2=100.$$

• 5.

  A rectangle’s diagonal is $25$ and one side is $7$. What is the perimeter of the rectangle?

    (20 pts) 5. $62$

  Solution: Let the sides of the rectangle be $7$ and $b$. The diagonal is $25$. By the Pythagorean theorem,

  $$7^2+b^2={25}^2.$$

  So

  $$49+b^2=625\Rightarrow b^2=625-49=576,$$

  hence

  $$b=\sqrt{576}=24.$$

  The perimeter is

  $$2(7+24)=2\cdot 31=62.$$

• 6.

  How many distinct ways can the letters in the word “LEVEL” be arranged?

    (20 pts) 6. $30$

  Solution: The word LEVEL has $5$ letters total. The letters are:

  $$\text{L, E, V, E, L}.$$

  We have:

  $$\text{L appears 2 times},\text{E appears 2 times},\text{V appears 1 time}.$$

  The total number of distinct permutations of a multiset is

  $$\frac{5!}{2!\mathrm{\hspace{0.17em}2}!}=\frac{120}{4}=30.$$

• 7.

  A school has $5$ math teams and each team has $4$ students. How many students must you choose to guarantee that at least two are from the same team?

    (20 pts) 7. $6$

  Solution: There are $5$ teams, each with $4$ students.

  To avoid having two students from the same team, you could choose at most one student from each of the $5$ teams, for a total of $5$ students.

  As soon as you choose one more student (the $6$th student), by the pigeonhole principle at least two of your chosen students must come from the same team.

  Therefore, you must choose $6$ students to guarantee that at least two are from the same team.

• 8.

  If the cost of a bat and a baseball combined is $1.10 and the ball costs $1.00 less than the bat, how much is the ball?

    (20 pts) 8. $\mathrm{\$}0.05$

  Solution: Let $x$ denote the cost of the bat and $y$ denote the cost of the ball. We know that $x+y=\mathrm{\$}1.10$ and $x=y+\mathrm{\$}1.00$. Combining the second equation with the first, we produce

  	$x+y$	$=\mathrm{\$}1.10$
  	$(y+\mathrm{\$}1.00)+y$	$=\mathrm{\$}1.10$
  	$2y+\mathrm{\$}1.00$	$=\mathrm{\$}1.10$
  	$2y$	$=\mathrm{\$}0.10$
  	$y$	$=\mathrm{\$}0.05.$

  Therefore, the ball costs $0.05.

• 9.

  Solve for $x$: $\sqrt{x+9}-\sqrt{x-7}=2$.

    (20 pts) 9. $x=16$

  Solution: Solve

  $$\sqrt{x+9}-\sqrt{x-7}=2.$$

  First note the domain: we need $x-7\ge 0$, so $x\ge 7$.

  Set

  $$\sqrt{x+9}=\sqrt{x-7}+2.$$

  Now square both sides:

  $$x+9={(\sqrt{x-7}+2)}^2=(x-7)+4\sqrt{x-7}+4.$$

  Simplify the right-hand side:

  $$x+9=x-3+4\sqrt{x-7}.$$

  Subtract $x$ from both sides:

  $$9=-3+4\sqrt{x-7}.$$

  Add 3 to both sides:

  $$12=4\sqrt{x-7}.$$

  Divide by 4:

  $$3=\sqrt{x-7}.$$

  Square again:

  $$9=x-7\Rightarrow x=16.$$

  Check in the original equation:

  $$\sqrt{16+9}-\sqrt{16-7}=\sqrt{25}-\sqrt{9}=5-3=2,$$

  which works.

• 10.

  In how many ways can $6$ students sit in a row if two particular students insist on sitting next to each other?

    (20 pts) 10. $240$

  Solution: Let the two particular students be $A$ and $B$. They must sit next to each other.

  Think of $A$ and $B$ as a single “block.” Then we have this block plus the other $4$ students, for a total of $5$ objects to arrange.

  The number of ways to arrange $5$ distinct objects in a row is $5!$.

  Inside the block, the order of $A$ and $B$ can be either $AB$ or $BA$, so there are $2!$ ways to arrange them internally.

  Therefore, the total number of seatings with $A$ and $B$ together is

  $$5!\cdot 2!=120\cdot 2=240.$$