UND MATHEMATICS TRACK MEET Team Test 1

University of North Dakota Grades 9/10
January 12, 2026
School Team Name
Calculators are allowed. Solutions

• 1.

  Determine exactly the value of $x$ for which ${\displaystyle \frac{3x-5}{2}}={\displaystyle \frac{5}{4}}$.

    (20 pts) 1. ${\displaystyle \frac{5}{2}}$

  Solution: Multiply both sides by $8$ to obtain $4(3x-5)=10⟹12x-20=10⟹12x=30⟹x=\frac{30}{12}=\frac{5}{2}$.

• 2.

  If $a$ and $b$ are positive real numbers and $\frac{a^2+b^2}{\frac{1}{a^2}+\frac{1}{b^2}}=10$, determine exactly the value of $\frac{a^3+b^3}{\frac{1}{a^3}+\frac{1}{b^3}}$.

    (20 pts) 2. $10\sqrt{10}$

  Solution: The first expression can be rewritten as $\frac{a^2+b^2}{\frac{a^2+b^2}{a^2{b}^2}}=10$. Therefore, $a^2{b}^2=10⟹ab=\sqrt{10}$. The second expressison can likewise be written as $\frac{a^3+b^3}{\frac{a^3+b^3}{a^3{b}^3}}=a^3{b}^3={(ab)}^3={(\sqrt{10})}^3=10\sqrt{10}$.

• 3.

  $a$, $b$, and $c$ are positive integers such that the sum of $a$ and $b$ is $3$ more than the value of $c$, the value of $a$ is double that of $b$, and the value of $b$ is five less than $c$. Determine exactly the value of $b$.

    (20 pts) 3. 4

  Solution: If the sum of $a$ and $b$ is $3$ more than $c$, then $a+b=c+3$. Since $a$ is double $b$ and $b$ is five less than $c$, then $a=2b$ and $b=c-5$ respectively. Substituting the second and third equations into the first gives us:

  $$2(c-5)+(c-5)=c+3\Rightarrow 3(c-5)=c+3\Rightarrow 2c=18\Rightarrow c=9$$

  Therefore, $b=9-5=4$

• 4.

  The base of a triangular flower bed is $4$ meters longer than its height. If the area of the flower bed is $30$ m², find the height of the triangle.

    (20 pts) 4. $6$ m

  Solution: Let the height of the triangle be $h$. Then, the base is $h+4$. Substitute into the formula for the area of a triangle:

  	$30$	$={\displaystyle \frac{1}{2}}\cdot (h+4)\cdot h$
  	$30$	$={\displaystyle \frac{1}{2}}(h^2+4h)$
  	$60$	$=h^2+4h$
  	$0$	$=h^2+4h-60$
  	$0$	$=(h+10)(h-6)$

  Solving for $h$, we get $h=-10$ and $h=6$, but height here cannot be negative, so our height must be $6$ meters.

• 5.

  Determine exactly the coordinates of the intersection of the lines $x+y=15$ and $5x+8y=87$.

    (20 pts) 5. $(11,4)$

  Solution: Subtracting $5x+5y=75$ from $5x+8y=87$ yields $3y=12$ or $y=4$. Then, $5x=55$ and $x=11$, so the coordinates are $(11,4)$.

• 6.

  Buddy delivers the same number of local newspapers every day during the summer. He gets paid $\mathrm{\$}0.25$ for each paper delivered, except on Sundays, when he gets paid $\mathrm{\$}1.00$ per paper. After three full weeks, Buddy has earned $\mathrm{\$}900$. how many papers does he deliver daily?

    (20 pts) 6. $120$

  Solution: Let $p=\text{the number of papers delivered daily}$. Then, Buddy earns $0.25p$ on each of six days a week, and $1.00p$ on Sundays. His weekly pay is $0.25p(6)+1.00p=2.50p$. Write an equation representing $3$ weeks’ pay: $2.50p(3)=900⟹2.50p=300⟹p=120$.

• 7.

  Five years ago, Maria was three times as old as Alex. Five years from now, Maria will be twice as old as Alex. What is Alex’s current age?

    (20 pts) 7. $15$

  Solution:

  1. 1.

     Five years ago:
     Maria’s age was $y-5$, and Alex’s age was $x-5$. From the problem:

     $$y-5=3(x-5).$$

  2. 2.

     Five years from now:
     Maria’s age will be $y+5$, and Alex’s age will be $x+5$. From the problem:

     $$y+5=2(x+5).$$

  3. 3.

     Solve the system of equations:
     From the first equation:

     $$y-5=3x-15\Rightarrow y=3x-10.$$

     Substitute $y=3x-10$ into the second equation:

     $$(3x-10)+5=2(x+5).$$

     Simplify:

     $$3x-5=2x+10.$$

     Solve for $x$:

     $$x=15.$$

  Hence, Alex is currently $15$ years old.

• 8.

  Find the constants $A$, $B$, and $C$ such that ${\displaystyle \frac{A}{(x-2)(x-4)}}+{\displaystyle \frac{B}{x-2}}+{\displaystyle \frac{C}{x(x-4)}}={\displaystyle \frac{1}{x}}$ for all permissible $x$.

    (20 pts) 8. $A=2$, $B=1$, $C=-4$

  Solution: Multiplying the equation by $x(x-2)(x-4)$ produces: $Ax+Bx(x-4)+C(x-2)=(x-2)(x-4)$. Expanding and simplifying yields: $B{x}^2+(A-4B+C)x-2C=x^2-6x+8$. Therefore, $B=1$, $A-4B+C=-6$, and $-2C=8$, so $C=-4$ and $A=-6+4+4=2$.

• 9.

  $(x,y)$ is the intersection of ${\displaystyle \frac{3x}{y}}-4=11$ and $x-y=44$. Determine the value of $x+y$.

    (20 pts) 9. $66$

  Solution: $3x=15y\Rightarrow x=5y.$ Therefore, $5y-y=44\Rightarrow y=11$. So, $x=55$ and $x+y=66$

• 10.

  How many pairs of positive integers $(x,y)$, where $x+y\le 2026$, satisfy the equation ${\displaystyle \frac{x+y^{-1}}{x^{-1}+y}}=11$?

    (20 pts) 10. $168$

  Solution: Multiply numerator and denominator by $xy$ to obtain $\frac{x^{2}y+x}{y+x{y}^2}=\frac{x(xy+1)}{y(1+xy)}=\frac{x}{y}$. Therefore, $\frac{x}{y}=11$, or $x=11y$. Since $x+y\le 2026$, $12y\le 2026$. Therefore, , and each one of these $168$ values of $y$ produces an ordered pair $(11y,y)$ that is a solution.