Denyse can paint a house in 5 days and Celesta can paint the same house in 10 days. Denyse paints alone for 2 days, then Celesta joins her. How long in total does it take to finish the house?

Step-by-step solution

1. 1

   Convert 'time to finish' into 'fraction of the job per day'. Rates add; times do not. Denyse finishes $1$ house in $5$ days, so her rate is $\tfrac15$ house/day. Celesta's rate is $\tfrac{1}{10}$ house/day. Averaging the times $5$ and $10$ to get $7.5$ would be wrong — that ignores the fact that working together speeds up the job.

2. 2

   Measure the head start. In $2$ days alone, Denyse completes

   $2\times\frac15=\frac25$

   of the house, leaving

   $1-\frac25=\frac35$

   still to be painted.

3. 3

   Add the rates for the joint phase.

   $\frac15+\frac{1}{10}=\frac{2}{10}+\frac{1}{10}=\frac{3}{10}\ \text{house/day}$

   Together they cover $30\%$ of the house each day.

4. 4

   Divide remaining work by the combined rate.

   $t=\frac{3/5}{3/10}=\frac35\cdot\frac{10}{3}=2\ \text{days}$

   The question asks for the total time, so add the $2$ solo days: $2+2=4$ days.

5. 5

   Verify by accounting for each person's output. Denyse works all $4$ days: $4\times\tfrac15=\tfrac45$. Celesta works only the last $2$ days: $2\times\tfrac{1}{10}=\tfrac15$. Total $\tfrac45+\tfrac15=1$ whole house ✓ — exactly one job done, no more and no less.