Planes and wheat

Task

A government buys $x$ fighter planes at $z$ dollars each, and $y$ tons of wheat at $w$ dollars each. It spends a total of $B$ dollars, where $B = xz + yw$. In (a)–(c), write an equation whose solution is the given quantity.

1. The number of tons of wheat the government can afford to buy if it spends a total of $100 million, wheat costs $300 per ton, and it must buy 5 fighter planes at $15 million each.
2. The price of fighter planes if the government bought 3 of them, in addition to $10,\!000$ tons of wheat at $500 a ton, for a total of $50 million.
3. The price of a ton of wheat, given that a fighter plane costs $100,\!000$ times as much as a ton of wheat, and that the government bought 20 fighter planes and $15,\!000$ tons of wheat for a total cost of $90 million.

IM Commentary

This is a simple exercise in creating equations from a situation with many variables. By giving three different scenarios, the problem requires students to keep going back to the definitions of the variables, thus emphasizing the importance of defining variables when you write an equation. In order to reinforce this aspect of the problem, the variables have not been given names that remind the student of what they stand for. The emphasis here is on setting up equations, not solving them.

Solution

1. We want to find the value of $y$. We are given $B = 100,\!000,\!000$, $w = 300$, $x = 5$, and $z = 15,\!000,\!000$. So the equation is $$100,\!000,\!000 = 5 \cdot 15,\!000,\!000 + 300 y,$$ or $$100,\!000,\!000 = 75,\!000,\!000 + 300 y.$$
2. We want to find the value of $z$. We are given that $x = 3$, $y = 10,\!000$, $w = 500$, and $B = 50,\!000,\!000$. So the equation is $$50,\!000,\!000 = 3z + 10,\!000\cdot 500,$$ or $$50,\!000,\!000 = 3z + 5,\!000,\!000.$$
3. We want to find the value of $w$. We are given that $x = 20$ and $y = 15,\!000$, $B = 90,\!000,\!000$, and $z = 100,\!000w$. So the equation is $$90,\!000,\!000 = 20 (100,\!000 w) + 15,\!000 w,$$ which simplifies to $$90,\!000,\!000 = 2,\!015,\!000 w.$$