UND MATHEMATICS TRACK MEET Individual Test 1

University of North Dakota Grades 11/12
January 12, 2026
School Team Name
Calculators are allowed. Solutions Student Name

• 1.

  Find all real solutions to the equation $2^{x+1}+2^{1-x}=5$.

    (2 pts) 1. $-1,1$

  Solution: Multiply both sides by $2^x$:

  $$2^{2x+1}+2=5\cdot 2^x.$$

  Let $t=2^x$, $t>0$. Then $2t^2-5t+2=0\Rightarrow (2t-1)(t-2)=0$. So $t=\frac{1}{2}$ or $2$. Hence $x=-1$ or $1$.

  $$\boxed{x=-1,\mathrm{\hspace{0.17em}1}.}$$

• 2.

  The expression ${\displaystyle \frac{a^2-b^2}{a-b}}$ simplifies to $a+b$ for all $a\ne b$. If $a=3+\sqrt{5}$ and $b=3-\sqrt{5}$, find the exact value of ${\displaystyle \frac{a^3-b^3}{a-b}}$.

    (3 pts) 2. 32

  Solution: We know ${\displaystyle \frac{a^3-b^3}{a-b}}=a^2+ab+b^2$. Compute:

  $$ab=(3+\sqrt{5})(3-\sqrt{5})=9-5=4,a^2+b^2={(3+\sqrt{5})}^2+{(3-\sqrt{5})}^2=18+2\times 5=28.$$

  Hence $\boxed{a^2+ab+b^2=28+4=32.}$

• 3.

  A rectangle has a diagonal of length 10 cm and one side that is 2 cm longer than the other. Find the dimensions of the rectangle.

    (3 pts) 3. $6$cm$\times 8$cm

  Solution: Let shorter side $x$. Then longer side $x+2$. By Pythagoras: $x^2+{(x+2)}^2={10}^2\Rightarrow 2x^2+4x+4=100$. $2x^2+4x-96=0\Rightarrow x^2+2x-48=0$. $x=6cm$ (positive root). Dimensions: $6$ cm,$8$ cm.

• 4.

  A surveyor stands on level ground and observes the top of a building. The line of sight to the top of the building makes an angle of elevation of ${35}^{\circ }$. The surveyor then walks 50 meters directly toward the building, where the angle of elevation increases to ${50}^{\circ }$. If the surveyor’s eyes are 1.6 meters above the ground, find the horizontal distance from the first observation point to the building to the nearest meter.

    (3 pts) 4. $121m$

  Solution: Let $H=$ height of the building and $x=$ initial distance.

  $$\tan ({35}^{\circ })=\frac{H-1.6}{x},\tan ({50}^{\circ })=\frac{H-1.6}{x-50}.$$

  From the first, $H=x\tan ({35}^{\circ })+1.6$. Substituting in the second:

  $$x=\frac{50\tan ({50}^{\circ })}{\tan ({50}^{\circ })-\tan ({35}^{\circ })}\approx 121.2\Rightarrow \boxed{x\approx 121\text{m}}.$$

• 5.

  Consider the three lines

  $$L_1:2x+3y=12,L_2:y=x-1,L_3:x=2.$$

  Compute the area of the triangle formed by these intersections (three decimal places).

    (3 pts) 5. 0.833

  Solution:

  $$L_2\cap L_3:(2,1),L_1\cap L_3:(2,\frac{8}{3}),L_1\cap L_2:(3,2).$$

  Using the shoelace formula for $A(2,1),B(2,\frac{8}{3}),C(3,2)$:

  $$\text{Area}=\frac{1}{2}\left(2\left(\frac{8}{3}-2\right)+2(2-1)+3\left(1-\frac{8}{3}\right)\right)=\frac{5}{6}.$$

  Decimal: $\boxed{0.833}$

• 6.

  A student council has 12 members: 5 seniors, 4 juniors, and 3 sophomores. A committee of 4 students is selected at random. What is the probability that the committee contains exactly two seniors? (three decimal places)

    (3 pts) 6. $0.424$

  Solution:

  $$P(\text{exactly 2 seniors})=\frac{\left(\genfrac{}{}{0pt}{}{5}{2}\right)\left(\genfrac{}{}{0pt}{}{7}{2}\right)}{\left(\genfrac{}{}{0pt}{}{12}{4}\right)}=\frac{10\times 21}{495}=\boxed{0.424}.$$

• 7.

  A landscape designer plans a triangular flower garden. Two boundary edges measure $110m$ and $75m$, and the angle between them after redesign will be ${63}^{\circ }$. Find the area of the garden to the nearest square meter.

    (3 pts) 7. $3675{\text{ m}}^2$

  Solution: Using the formula for the area of a triangle:

  $$A=\frac{1}{2}ab\sin (C),$$

  with $a=110$, $b=75$, $C={63}^{\circ }$:

  $$A=\frac{1}{2}(110)(75)\sin ({63}^{\circ })\approx 3675.$$

  $$\boxed{A\approx 3675{\text{ m}}^2.}$$