# money

• May 8th 2009, 08:34 PM
Mr Rayon
money
Maya needs to renovate her house. She has enough money to pay a plumber for 28 days or a carpenter for 21 days. For how many days can she pay the tradesmen if they both work at the same time? If Maya's next pay cheque will come in 2 weeks, can she afford to hire both specialists till then?
• May 8th 2009, 08:50 PM
Chris L T521
Quote:

Originally Posted by Mr Rayon
Maya needs to renovate her house. She has enough money to pay a plumber for 28 days or a carpenter for 21 days. For how many days can she pay the tradesmen if they both work at the same time? If Maya's next pay cheque will come in 2 weeks, can she afford to hire both specialists till then?

Let $P$ be the money Maya would pay the plumber daily, and let $C$ represent the money Maya would pay the carpenter daily.

Then if the total amount of money Maya has is $T_0$ dollars, it follows that the daily pay for the plumber is $28P=T_0\implies P=\tfrac{1}{28}T_0$ and the daily pay for the carpenter is $21C=T_0\implies C=\tfrac{1}{21}T_0$

Now, let us say $D_0$ is the number of days she pays both of them to work.

So we want to solve $D_0\left(P+C\right)=T_0\implies D_0\left(\tfrac{1}{28}T_0+\tfrac{1}{21}T_0\right)= T_0$ for $D_0$

Therefore,

\begin{aligned}D_0\left(\tfrac{1}{28}T_0+\tfrac{1} {21}T_0\right)=T_0 & \implies D_0\left(\tfrac{1}{28}+\tfrac{1}{21}\right)=1\\ & \implies D_0\left(\tfrac{3}{84}+\tfrac{4}{84}\right)=1\\ & \implies D_0\left(\tfrac{7}{84}\right)=1\\ & \implies\tfrac{1}{12}D_0=1\\& \implies D_0=12\end{aligned}

Therefore, Maya has enough money to pay both the plumber and the carpenter for 12 days. Given this, she won't have enough money to last her until her paycheck.

Does this make sense?
• May 9th 2009, 05:31 PM
Mr Rayon
Quote:

Originally Posted by Chris L T521
Does this make sense?

Yes. Thank you very much!