UND MATHEMATICS TRACK MEET Individual Test 4

University of North Dakota Grades 11/12
January 12, 2026
School Team Name
Calculators are NOT allowed. Solutions Student Name

• 1.

  $x$ and $y$ are non-integers with $3x-7y=10$ and $xy=-1$, find the value of $9x^2+49y^2$.

    (2 pts) 1. 58

  Solution: $3x-7y=10$ and $xy=-1$ means that $(\frac{7y+10}{3})y=-1$.
  
  Simplifying to a standard form quadratic, we get $7y^2+10y+3=0$ which factors as $(7y+3)(y+1)=0$. This yields solutions $y=-\frac{3}{7}$ and $y=-1$. We can eliminate the integer solution.
  That means $x=\frac{7}{3}$ and $y=-\frac{3}{7}$. Substituting into $9x^2+49y^2$ results in $9(\frac{49}{9})+49(\frac{9}{49})=58$.
  

• 2.

  In a room of 20 people, if each each person shakes each other person’s hand exactly one time, how many handshakes occur?

    (3 pts) 2. 190

  Solution: $19+18+17+\mathrm{\cdots }+2+1=\frac{19(20)}{2}=190$.
  

• 3.

  Find all positive integers such that $n^2+n+41$ is divisible by 41.

    (3 pts) 3. $40,41$

  Solution: We need $n^2+n$ to be divisible by 41. $n^2+n=n(n+1)$. Since 41 is prime, only $n=40$ and $n=41$ are solutions.
  

• 4.

  Find all real numbers $x,y,z$ satisfying the system of equations.
  

  	$x^2-4y+7=0$
  	$y^2-6z+14=0$
  	$z^2-2x-7=0$

    (3 pts) 4. $(1,2,3)$

  Solution: Adding all three equations together gives us
  
  $x^2-2x+y^2-4y+z^2-6z+14=0$.
  
  Completing the square with each quadratic leads to
  
  ${(x-1)}^2+{(y-2)}^2+{(z-3)}^2=0$.
  
  As each term is a perfect square, the only solution is $(1,2,3)$.
  

• 5.

  The circles below are tangent to the horizontal line and tangent to each other. They have radii of 7 and 10 respectively. Find the distance between $a$ and $b$.
  

    (3 pts) 5. $\sqrt{280}$

  Solution: Create a right triangle in the following way:
  
  

  The horizontal leg of the triangle is what we need to find. The hypotenuse is $17$, the vertical leg is $3$. By the Pythagorean Theorem, our solution is $\sqrt{289-9}=\sqrt{280}.$

• 6.

  A standard deck of 52 playing cards is shuffled and laid out in a row. What is the probability that the four aces appear in alphabetical order from left to right? (The aces don’t need to be consecutive; there can be other cards between them. We only care that when looking at the row of cards left to right, we encounter the aces in the order: Ace of Clubs, Ace of Diamonds, Ace of Hearts, Ace of Spades)

    (3 pts) 6. 1/24

  Solution: We are looking for one arrangement of the aces out of 4! possible arrangements. So, the probability is $\frac{1}{4!}=\frac{1}{24}$

• 7.

  Find the radius of the semicircle enclosed within the right triangle. The semicircle is tangent to both legs of the right triangle.
  

    (3 pts) 7. 120/23

  Solution: Draw the radii perpendicular to the legs of the triangle.
  
  The area of the 8-15-17 triangle can be split into the area of the two smaller right triangles and the square.
  $0.5(8)(15)=0.5(8-r)r+r^2+0.5(15-r)r$
  This simplifies to $60=4r-0.5r^2+r^2+7.5r-0.5r^2$
  $60=11.5r$ or $60=\frac{23}{2}r$, that means $r=\frac{60}{11.5}\text{ or }\frac{120}{23}$