MHB Solving the Lawn Mowing Puzzle: Explained!

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Person A mows a lawn in one hour less than Person B, while Person C takes twice as long as Person B. Their combined mowing rate equals one lawn per hour, leading to the equation 1/t + 1/(t-1) + 1/(2t) = 1. By solving for t, the time it takes Person B to mow the lawn, the individual times for A and C can also be determined. The discussion emphasizes the importance of expressing their rates in terms of a single variable to simplify the problem. The solution reveals the specific times each person takes to mow the lawn alone.
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So person A takes 1 less hour than person B to mow a lawn, Person C takes twice as long as Person B to mow a lawn. All together they mow a lawn in one hour. How long does it take each of them to mow a lawn working alone? Can someone explain this to me please, thanks.
 
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Hi and welcome to the forum.

Let the productivity of A, B, C, measured in lawns per hour, be $a$, $b$ and $c$, respectively. Then A mows the lawn in $1/a$ hours, B does it in $1/b$ hours and C takes $1/c$ hours. Their combined productivity is $a+b+c$, and by assumption $1/(a+b+c)=1$, i.e.,
\[
a+b+c=1.\qquad(*)
\]
Now write the conditions
  1. Person A takes 1 less hour than person B to mow a lawn.
  2. Person C takes twice as long as Person B to mow a lawn
in terms of $a$, $b$ and $c$ and express $a$ and $c$ through $b$. Substituting these expressions into (*) will give you one equation in $b$. It has two solutions, but only one of them has physical sense.
 
Equivalently: let t be the time, in hours, it takes person B to mow the lawn.
so B has rate 1/t "lawns per hour".

person A takes 1 less hour than person B to mow a lawn
So A's time is t- 1 and rate is 1/(t- 1)

Person C takes twice as long as Person B to mow a lawn
So C's time is 2t and rate is 1/(2t)

All together they mow a lawn in one hour.
This last sentence tells us that, together, their rate of work is "1 hour per job".
When people, machines, etc. "work together", their rates of work add.
1/t+ 1/(t-1)+ 1/2t= 1. Solve that for t, the time It takes B to mow a lawn then find the times for A and B.

(Multiplying both sides of the equation by 2t(t-1) will eliminate the fractions.)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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