Another work to paint house problem

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In summary, we can solve the problem by setting two equations: one for the rate of painting in days/house and another for the relationship between Clarissa and Shawna's rates. By using the Least Common Denominator (LCD) and factoring, we can find that Clarissa takes 10 days to paint the house by herself.
  • #1
karush
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Clarissa and Shawna, working together, can paint the exterior of a house in \(\displaystyle 6\) days. Clarissa by herself can complete this job in \(\displaystyle 5\) days less than Shawna. How long will it take Clarissa to complete the job by herself?

well if they work equally then

$\frac{1}{12}+\frac{1}{12}=\frac{1}{6}$

but I didn't know how to change this to match what the problem says.

The answer is "Clarissa 10 days by herself".
 
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  • #2
how about this if $c =$ Clarissa then

$$\frac{1}{c+5}+\frac{1}{c}=\frac{1}{6}$$
 
  • #3
For this problem, we'll start more generally:

Given that it takes them 6 days to paint the house, we can let $c$ and $s$ be the rate of painting in days/house. Then, one way of setting an equation (such that the units cancel) and works with the question:

$$6 \text{ days }\cdot\left( \frac{1}{c \frac{\text{days}}{\text{house}}}+\frac{1}{s \frac{\text{days}}{\text{house}}}\right) = 1 \text{ house}$$

From the second part of the question, we can formulate another equation. What is it?
karush said:
how about this if $c =$ Clarissa then

$$\frac{1}{c+5}+\frac{1}{c}=\frac{1}{6}$$

That is correct because we know that $c=s-5$, then $s=c+5$ and we can plug that back into the first equation. How can we solve for $c$? (Wondering)
 
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  • #4
ok from this I got by LCD $$c^2-7c-30=0$$ factoring
$$\left(c+3\right)\left(c-10\right)=0$$
so $C$ has to positive $C=10$ days I saw some other solutions to this on the internet but MHB is really the best place to be. some solutions elsewhere really got messy with wrong answers
 
  • #5


To solve this problem, we need to use the concept of "work rate". Work rate is the amount of work that can be completed in a certain amount of time. In this case, we can say that Clarissa's work rate is 1 job per 5 days and Shawna's work rate is 1 job per 6 days.

We can use the formula: Work rate = Amount of work / Time to complete the work

Let's say the amount of work is 1 (painting the exterior of the house). We can set up the equation as:

Clarissa's work rate = 1 job / 5 days
Shawna's work rate = 1 job / 6 days
Together, their work rate = 1 job / 6 days

Now, we can use the formula for working together: Work rate together = Sum of individual work rates

1/6 = 1/5 + 1/x

where x is the number of days it takes for Clarissa to complete the job by herself.

Solving for x, we get x = 10 days.

Therefore, it will take Clarissa 10 days to complete the job by herself.
 

FAQ: Another work to paint house problem

What is the "Another work to paint house problem"?

The "Another work to paint house problem" is a mathematical problem that involves finding the most efficient way to paint a house with a given number of workers and time constraints.

Why is the "Another work to paint house problem" important?

The "Another work to paint house problem" is important because it has real-world applications in project management and resource allocation. It helps in optimizing the use of resources and minimizing costs.

How is the "Another work to paint house problem" solved?

The problem can be solved using a mathematical technique called linear programming, which involves creating a mathematical model and using algorithms to find the optimal solution.

What are the factors that affect the solution of the "Another work to paint house problem"?

The factors that affect the solution of the problem include the number of workers, their efficiency, the time constraints, and the cost of painting materials.

Can the "Another work to paint house problem" be applied to other scenarios?

Yes, the same approach can be applied to other similar problems such as scheduling tasks with limited resources or optimizing production processes in a factory.

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