   SEARCH HOME Math Central Quandaries & Queries  Question from Kenneth: Hello: Here is my question: 4 carpenters can build 8 houses in 10 days. 2 carpenters can build how many houses in 15 days? Here is one solution: The number of houses built is jointly proportional to the number of carpenters and the number of days. So h = kcd where k is a constant. Substituting the given: 8 = k*40, so k = 1/5. Our formula is then h = cd/5. I'm not sure how to determine whether or not one unit is jointly proportional to any other unit. However, if 2 carpenters are unknown as in 4 carpenters can build 8 houses in 10 days. How many carpenters can build 6 houses in 15 days? What is the solution using this method? And likewise, if days are unknown, as in 4 carpenters can build 8 houses in 10 days. 2 carpenters can build 6 houses in how many days? I thank you for your assistance. Hi Kenneth,

You let $c$ be the number of carpenters, $d$ the number of days and $h$ the number of houses and developed the equation

$h = k \times c \times d \mbox{ where k is a constant.}$

You then used the information given in the problem to show that $k = \large \frac15$ so that

$h = \frac { c \times d}{5}.$

The question " How many carpenters can build 6 houses in 15 days?" can then be answered by substituting $h = 6$ and $d = 15$ into the equation above and solving for $c.$

You can similarly answer "2 carpenters can build 6 houses in how many days?" by substituting$c = 2$ and $h = 6$ into the equation and solving for $d.$

I hope this helps,
Penny      Math Central is supported by the University of Regina and the Imperial Oil Foundation.