The Timpson Lyceum

1. (2 pts) A juice blend is made by mixing $x$ liters of a $20\%$ -juice drink with $y$ liters of a $50\%$ -juice drink to obtain $9$ liters of a $40\%$ -juice drink. Set up and solve the linear system for $x$ and $y$ .

Solution

1. The total-volume and total-juice equations are

$\begin{aligned} x + y &= 9 \\ 0.2x + 0.5y &= 0.4(9) = 3.6. \end{aligned}$

Substitute $x = 9 - y$ : $0.2(9 - y) + 0.5y = 3.6 \Rightarrow 1.8 + 0.3y = 3.6 \Rightarrow y = 6$ , so $x = 3$ . Mix $3$ L of the $20\%$ drink with $6$ L of the $50\%$ drink. (Check: juice $= 0.2(3) + 0.5(6) = 0.6 + 3 = 3.6$ L, which is $40\%$ of $9$ L.)

2. (3 pts) Two foods A and B are combined to meet nutrient targets. Per serving, Food A supplies $30$ units of protein and $10$ of calcium; Food B supplies $10$ of protein and $40$ of calcium. Find the numbers of servings $a$ and $b$ that provide exactly $90$ units of protein and $140$ units of calcium.

Solution

2. The constraints give

$\begin{aligned} 30a + 10b &= 90 \\ 10a + 40b &= 140. \end{aligned}$

Divide by $10$ : $3a + b = 9$ and $a + 4b = 14$ . From the first, $b = 9 - 3a$ ; substitute: $a + 4(9 - 3a) = 14 \Rightarrow a + 36 - 12a = 14 \Rightarrow -11a = -22 \Rightarrow a = 2$ , so $b = 3$ . Use $2$ servings of A and $3$ of B. (Check: protein $60 + 30 = 90$ , calcium $20 + 120 = 140$ .)

3. (3 pts) Each year $10\%$ of a city's population moves to its suburbs and $20\%$ of the suburban population moves to the city (no other change). Writing the population as $\mathbf{x} = \begin{bmatrix} \text{city} \\ \text{suburb} \end{bmatrix}$ (in thousands), the one-year update is $\mathbf{x}_{1} = M\mathbf{x}_0$ with

$M = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}.$

If $\mathbf{x}_0 = \begin{bmatrix} 600 \\ 400 \end{bmatrix}$ , compute $\mathbf{x}_1$ and verify the total population is unchanged.

Solution

3. $\mathbf{x}_1 = M\mathbf{x}_0 = \begin{bmatrix} 0.9(600) + 0.2(400) \\ 0.1(600) + 0.8(400) \end{bmatrix} = \begin{bmatrix} 540 + 80 \\ 60 + 320 \end{bmatrix} = \begin{bmatrix} 620 \\ 380 \end{bmatrix}.$ The total is $620 + 380 = 1000$ , equal to $600 + 400$ — the population is conserved (no one enters or leaves the system).

4. (3 pts) In a circuit, three currents satisfy Kirchhoff's laws:

$\begin{aligned} I_1 - I_2 - I_3 &= 0 \\ 2I_1 + 3I_2 &= 12 \\ 3I_2 + 4I_3 &= 10. \end{aligned}$

Find $I_1, I_2, I_3$ .

Solution

4. From the first equation, $I_1 = I_2 + I_3$ . Substitute into the second: $2(I_2 + I_3) + 3I_2 = 12$ , i.e. $5I_2 + 2I_3 = 12$ . Pair this with the third, $3I_2 + 4I_3 = 10$ . Multiply the first of these by $2$ : $10I_2 + 4I_3 = 24$ ; subtract the third: $7I_2 = 14$ , so $I_2 = 2$ . Then $2I_3 = 12 - 10 = 2$ , so $I_3 = 1$ , and $I_1 = 2 + 1 = 3$ . Thus $(I_1, I_2, I_3) = (3, 2, 1)$ .

5. (3 pts) An investor splits $10,000 among three funds A, B, C with annual returns $5\%$ , $7\%$ , $10\%$ . She wants a total return of $680 and decides to place twice as much in A as in C. Letting $x, y, z$ be the amounts in A, B, C, set up and solve the system.

Solution

5. The equations are

$\begin{aligned} x + y + z &= 10000 \\ 0.05x + 0.07y + 0.10z &= 680 \\ x &= 2z. \end{aligned}$

Substitute $x = 2z$ into the first: $y = 10000 - 3z$ . Substitute $x = 2z$ and this $y$ into the second: $0.05(2z) + 0.07(10000 - 3z) + 0.10z = 680 \Rightarrow 0.1z + 700 - 0.21z + 0.1z = 680 \Rightarrow 700 - 0.01z = 680 \Rightarrow z = 2000$ . Then $x = 4000$ and $y = 10000 - 6000 = 4000$ . Invest $4000 in A, $4000 in B, $2000 in C. (Check: return $= 200 + 280 + 200 = 680$ .)

6. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) In a migration (stochastic) matrix like $M$ above, each column sums to $1$ .

(b) Every linear model has a unique solution.

(c) In a diet model, each equation corresponds to one nutrient constraint.

Solution

6.

(a) True. Column $j$ lists the fractions of group $j$ 's population going to each destination; since everyone must go somewhere, those fractions sum to $1$ . Here column 1 is $0.9 + 0.1 = 1$ and column 2 is $0.2 + 0.8 = 1$ .

(b) False. Linear models can be inconsistent (no solution) or underdetermined (infinitely many), as with network-flow models; uniqueness is not guaranteed.

(c) True. Each nutrient (protein, calcium, …) contributes one linear equation equating the amount supplied by the foods to the required amount.

7. (3 pts) Explain what each column of the migration matrix $M = \begin{bmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{bmatrix}$ represents, and explain why each column must sum to $1$ .

Solution

7. Column $1 = \begin{bmatrix} 0.9 \\ 0.1 \end{bmatrix}$ describes what happens to the city population in one year: $0.9$ (i.e. $90\%$ ) stays in the city and $0.1$ ( $10\%$ ) moves to the suburb. Column $2 = \begin{bmatrix} 0.2 \\ 0.8 \end{bmatrix}$ describes the suburb population: $0.2$ moves to the city and $0.8$ stays. Each column must sum to $1$ because the entries account for all of that group's people — every resident either stays or moves, and no one is created or lost.

8. (4 pts ••) Find the population distribution, expressed as fractions of the total, that is unchanged from one year to the next under $M$ above — i.e. the vector $\mathbf{x}$ with $M\mathbf{x} = \mathbf{x}$ and entries summing to $1$ . Verify that $M\mathbf{x} = \mathbf{x}$ for your answer.

Solution

8. Require $M\mathbf{x} = \mathbf{x}$ , i.e. $(M - I)\mathbf{x} = \mathbf{0}$ :

$M - I = \begin{bmatrix} -0.1 & 0.2 \\ 0.1 & -0.2 \end{bmatrix}, \qquad -0.1x_1 + 0.2x_2 = 0 \;\Rightarrow\; x_1 = 2x_2.$

(The second row is the negative of the first, so it gives no new information.) Imposing $x_1 + x_2 = 1$ : $2x_2 + x_2 = 1 \Rightarrow x_2 = \tfrac{1}{3}$ , $x_1 = \tfrac{2}{3}$ . The steady distribution is

$\mathbf{x} = \begin{bmatrix} 2/3 \\ 1/3 \end{bmatrix} \;(\text{city : suburb} = 2:1).$

Verification: $M\mathbf{x} = \begin{bmatrix} 0.9\cdot\tfrac{2}{3} + 0.2\cdot\tfrac{1}{3} \\ 0.1\cdot\tfrac{2}{3} + 0.8\cdot\tfrac{1}{3} \end{bmatrix} = \begin{bmatrix} 0.6 + \tfrac{0.2}{3} \\ \tfrac{0.1\cdot 2}{3} + \tfrac{0.8}{3} \end{bmatrix} = \begin{bmatrix} 2/3 \\ 1/3 \end{bmatrix} = \mathbf{x}.$ $\blacksquare$