UND MATHEMATICS TRACK MEET Individual Test 3

University of North Dakota Grades 7/8
January 12, 2026
School Team Name
Calculators are allowed. Solutions Student Name

• 1.

  Two fair 6-sided dice are rolled and their product is recorded. Find the probability that their product is a multiple of 3. Round your answer to two decimal places.

    (2 pts) 1. $55.56\%$ or $0.56$

  Solution: In order for the product to be a multiple of three, then at least one of the die must be either a 3 or a 6. Looking at the possibilities, we have: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1),(4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2),(6,3), (6,4),(6,5), (6,6). Of these, 20 of the 36 possible outcomes are multiples of 3. Therefore, the probability is $20/36=5/9$.

• 2.

  An isosceles right triangle is inscribed in a circle as shown below. Given that the area of the circle is $9\pi$, find the area of the shaded region. Round your answer to two decimal places.
  

    (3 pts) 2. $2.57$

  Solution: Notice that the shaded region is the same as the top right fourth of the circle minus the right half of the triangle. Furthermore, since the circle has area $9\pi$, then its radius must be $r=3$. Also, then the base of the triangle is $2r=2\cdot 3=6$ and the height is $r=3$. Thus, the shaded region has area ${\displaystyle \frac{1}{4}}(9\pi )-{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{2}}(6)(3)\right)={\displaystyle \frac{9\pi }{4}}-{\displaystyle \frac{9}{2}}\approx 2.57$

• 3.

  Suppose that the population of a bacterial colony is currently 1248. If the population shrinks to half of its size every hour, how many hours will it be until the population reaches 156?

    (3 pts) 3. 3 hours

  Solution: After the first hour, the colony has population 624. After the second, 312. After the third, 156. Therefore, the answer is 3 hours.

• 4.

  Quinn invested $500 in a fund which is growing at a rate of 3.5% per month and $1000 in a fund which is shrinking at a rate of 2% per month. By how much did Quinn’s investment change after one month?

    (3 pts) 4. $-\mathrm{\$}2.50$ or decrease by $2.50

  Solution: The $500 investment increases by $(500)(0.035)=17.5$ dollars. The $1000 investment decreases by $(1000)(0.02)=20$ dollars. Therefore, the total change is $17.5-20=-2.5$ dollars.

• 5.

  Find the set of all $x$ for which ${(2^x-3)}^2+(2^x-3)=2$.

    (3 pts) 5. $\{0,2\}$

  Solution: Subtracting 2 from both sides we have ${(2^x-3)}^2+(2^x-3)-2=0$. Notice that the left-hand side factors giving us $\left((2^x-3)+2\right)\left((2^x-3)-1\right)=0$ or more simply $(2^x-1)(2^x-4)=0$. Setting each factor equal to zero, we have that either $(2^x-1)=0$ or $(2^x-4)=0$. Adding the constants to the other side, we have $2^x=1$ or $2^x=4$. Applying the logarithm base 2 to each side, we have $x=0$ or $x=2$. Therefore, the set of all $x$ which satisfy the equation is $\{0,2\}$.

• 6.

  Suppose the date of the third Friday of October is two times the date of the second Tuesday of October. What day of the week is October 5th?

    (3 pts) 6. Thursday

  Solution: Let $x$ be the date of the first Friday of October. Then the date of the third Friday can be expressed as $x+14$. The date of the second Tuesday would then be either $(x-3)+7$ or $(x+4)+7$. Thus we have that $x+14=2\left((x-3)+7\right)$ or $x+14=2\left((x+4)+7\right)$. Simplifying we have $x+14=2x+8$ or $x+14=2x+22$. Solving for $x$ we either have $x=6$ or $x=-4$. Since dates do not admit negative values, we have that the date of the first Friday of October is $x=6$. Thus, October 5th would be Thursday.

• 7.

  Starting with the number 5/7, we will begin repeating the following process. If the number is bigger than 1, subtract 1. Otherwise we multiply it by 2. Then repeat this again with the new number. How many repetitions do we make before we arrive back at 5/7?

    (3 pts) 7. 5

  Solution: Starting with 5/7, since it is less than 1, we double it to 10/7. Then, since 10/7 is greater than 1, we subtract 1 to get 3/7. We continue this. And in order our numbers become: ${\displaystyle \frac{5}{7}}\to {\displaystyle \frac{10}{7}}\to {\displaystyle \frac{3}{7}}\to {\displaystyle \frac{6}{7}}\to {\displaystyle \frac{12}{7}}\to {\displaystyle \frac{5}{7}}$. Thus, there were 5 repetitions of this process to return to ${\displaystyle \frac{5}{7}}$.