UND MATHEMATICS TRACK MEET Team Test 2

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

• 1.

  Simplify ${\log }_{10}6250+{\log }_{10}16$.

    (20 pts) 1. 5

  Solution: We have ${\log }_{10}6250+{\log }_{10}16={\log }_{10}(2\cdot 5^5)+{\log }_{10}{2}^4={\log }_{10}(2^5\cdot 5^5)={\log }_{10}{10}^5=5$.

• 2.

  How many numbers from 1 to 120 (including 1 and 120) are divisible by 4 or 5?

    (20 pts) 2. 48

  Solution: Of the numbers from 1 to 120, there are $\lfloor 120/4\rfloor =30$ divisible by 4, $\lfloor 120/5\rfloor =24$ divisible by 5, and $\lfloor 120/\text{lcm}(4,5)\rfloor =\lfloor 120/20\rfloor =6$ divisible by 4 and 5. By inclusion-exclusion counting, there are $30+24-6=48$ such numbers.

• 3.

  Find all positive values for a diameter of a circle for which the area of the circle is equal numerically to half of its circumference.

    (20 pts) 3. 2

  Solution: Let $d$ be the diameter of a circle. The condition $A=\frac{1}{2}C$ for the area $A$ and circumference $C$ can be expressed in terms of $d$ as $\pi {\left(\frac{d}{2}\right)}^2=\frac{1}{2}(\pi d)$. Rearranging gives $\pi d^2-2\pi d=0$ or equivalently $\pi d(d-2)=0$. The only positive solution to the latter is $d=2$.

• 4.

  Simplify

  $$\frac{1}{\frac{1}{5}-0.\overline{18}}$$

    (20 pts) 4. 55

  Solution:

  $$\frac{1}{\frac{1}{5}-0.\overline{18}}=\frac{1}{\frac{1}{5}-\frac{18}{99}}=\frac{1}{\frac{99}{5(99)}-\frac{5(18)}{5(99)}}=\frac{1}{\frac{9}{5(99)}}=5(11)=55.$$

• 5.

  Solve $2\sqrt{x}=\sqrt{9+\sqrt{72}}+\sqrt{9-\sqrt{72}}$.

    (20 pts) 5. 6

  Solution: Observe squaring $2\sqrt{x}=\sqrt{9+\sqrt{72}}+\sqrt{9-\sqrt{72}}$ gives

  	$4x$	$=$	${\left(\sqrt{9+\sqrt{72}}+\sqrt{9-\sqrt{72}}\right)}^2$
  	$4x$	$=$	$9+\sqrt{72}+2\sqrt{9+\sqrt{72}}\cdot \sqrt{9-\sqrt{72}}+\left(9-\sqrt{72}\right)$
  	$4x$	$=$	$9+\sqrt{72}+2\sqrt{9}+\left(9-\sqrt{72}\right)$
  	$4x$	$=$	$24.$

  The latter has solution $x=6$. A check verifies that $x=6$ is a solution to the given equation.

• 6.

  How many sides does a dodecagon have?

    (20 pts) 6. $12$

  Solution: A dodecagon has 12 sides.

• 7.

  Find the area of a square with perimeter 56.

    (20 pts) 7. $196$

  Solution: Let $s$ be the side length of a square. If the perimeter $P$ of the square is 56, we have $4s=56$, so $s=14$. The area of the square is $A=s^2={14}^2=196$.

• 8.

  In degrees, what acute angle does the hour and minute hands make at 3:30?

    (20 pts) 8. ${75}^{\circ }$

  Solution: At 3:30, the hour hand lies halfway between 3 and 4 hour marks on the 12-hour analog clock. There are $\frac{{360}^{\circ }}{12}={30}^{\circ }$ between each hour marks on the clock. Thus the angle at 3:30 between the minute and hour hand is $2({30}^{\circ })+\frac{1}{2}({30}^{\circ })={75}^{\circ }$.

• 9.

  A set of 5 positive integers has a median of 17 and mean of 13. What is the largest possible value in the set?

    (20 pts) 9. $29$

  Solution: Let $n_1\le n_2\le n_3\le n_4\le n_5$ be the five positive integers. Since the median is 17, we have $n_3=17$. A mean of 13 implies

  $$\frac{n_1+n_2+17+n_4+n_5}{5}=13.$$

  Rearranging gives $n_1+n_2+n_4+n_5=48.$ To maximize $n_5$ subject to the conditions given, take $n_1=n_2=1$ and $n_4=17$. This gives $n_5=29$.

• 10.

  Find the number of ways that a red and a blue die can be rolled so that their product is a multiple of 6. Assume each die is standard with 6-sides.

    (20 pts) 10. $15$

  Solution: Let $(r,b)$ be a roll outcome. Then $rb$ is a multiple of 6 if and only if $(1,6)$, $(2,3)$, $(2,6)$, $(3,2)$, $(3,4)$, $(3,6)$, $(4,3)$, $(4,6)$, $(5,6)$, $(6,1)$, $(6,2)$, $(6,3)$, $(6,4)$, $(6,5)$, $(6,6)$. Thus, the number of ways is $15$.