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Here is a semi-algebraic explanation. Suppose $A$ works faster than $B$. Imagine that $A$ and $B$ start working on their separate projects at the same time. $A$ completes his project in time $t < 12$ hours. Suppose $A$ now helps $B$ to complete his project.

Since $A$ and $B$ are working together the whole time, the total time to complete the $2$ projects is $12$ hours, and the total time for the two of them is $12 + 12 = 24$ hours. Since $A$ does not actually help $B$ after $t$ hours, $B$ must do the same work that $A$ did in $12 - t$ hours. Since $B$ works slower than $A$, this will take more than $12 - t$ hours, so the total time for the two of them to complete their projects must be greater than $24$ hours.

How do you express this argument using algebraic equations related to rates and times?

I gave this some thought and the algebra is fairly simple, though I don't see any way to relate it to the explanation I gave. Together A and B can do a project in an hour, so their combined work rate is 1/6 and their individual rates are 1/12+x and 1/12-x. For notational convenience, let c=1/12. The total time to complete the two projects is $t = 1/(c+x) + 1/(c-x) = 2c/(c^2 - x^2)$, which will be minimal when $c^2 - x^2$ is maximal, which will be for x=0, giving $t>=2c/c^2 = 2/c$. For c=1/12, this is t=24.

user1153980
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1 Answers1

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In the twelve hours, $A$ completes $1+d$ projects and $B$ completes $1-d$ projects.

This means that $A$ takes $\frac{12}{1+d}$ hours to complete a project, and $B$ takes $\frac{12}{1-d}$ hours.

Adding these together gives $\frac{24}{1-d^2}$, which is greater than $24$ if $d\ne0$.

JMP
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