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Bunuel
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hi ,

can anyone provide a solution using variable approach
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Bunuel
One person alone can cut the grass of a yard in 4 hours. If two people both working at that rate were to work on cutting the grass, but one person is to start and the other is to join him later, how long after the first person starts working should the second one join so that the task is completed in 3 hours?

A. Both should start at the same time
B. 45 minutes later
C. 1 hour later
D. 1 hour 20 minutes later
E. 2 hours later

let t=1st person's time working alone
t*1/4+(3-t)*1/2=1
t=2 hours
2nd person joins 2 hours after 1st person starts
E
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Suppose the second person joins after x hours.
So first person works for 3 hours and second person works for 3-x hours
together they will:
3/4 + (3-x)/4 = 1 {rate for both is same}
x = 2
Therefore, the second person joins after two hours
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abhisheknandy08
hi ,

can anyone provide a solution using variable approach

Time taken * Efficiency = Work done.

Efficiency of \(A = B = 25%.\)

\(25t + 50(3-t) = 100\)

\(25t + 150 - 50t = 100\)

\(25t = 50\)

\(t = 2\). Ans - E.
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Bunuel can you explain this in variable form please..

I get 1/4 = work done(w) / time (t)
=> t/4 = w
remaining work => 1 - t/4 = (4-t)/4

putting this in r=w/t equation:

1/2 = w/3
=> 1/2 = {(4-t)/4}/3
=> 1/2 ={4-t}/12
=> 2 = -t
t = - 2 :(
where am I going wrong?
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TheKingInTheNorth
hi ,

can anyone provide a solution using variable approach

Let x be the amount of grass work,

Speed of a person per hour = x/4

Now since the speed of the whole process when two people working will be the sum of the net speed of the individuals in that work, and that it is also given that the time it took to complete the whole process was 3 hours,

Speed of the process = x/3

and

net Speed of person 1 during that time + net speed of person 2 during that time = Speed of the whole process

x/4 + net speed of person 2 = x/3

net speed of person 2 = x/12

Now let t1 be the duration of time in those 3 hours that he worked and t2 be the duration of time wherein he did not work i.e the lag time.

Average/net speed = Total work done by person 2 / Total time

hence, x/12 = (x/4 * t1)/3

which gives t1 = 1 hour

and since t1 + t2 = 3 hours

t2, or lag time, = 3-1 = 2 hours
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One person completes 1/4 of the job in one hour

Within 3 hours the task is completed if one starts at the beginning and the other person joins him later.
So, the first person who starts the work will work for 3 hours.
The 1st person's work = 3* 1/4 = 3/4
remaining work is = 1-3/4 = 1/4
so, the second person has to work for only one hour

The second person will join the work after 2 hours. :thumbsup:
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Here are two approaches to solving this. Hopefully it helps, I've found some of the answers above a bit muddy to read through.

1 - Intuitive Approach

If a person can cut a yard in 4 hours, we get that:
\(1=4r\)
\(r=\frac{1}{4}\)

In other words, a quarter of a lawn is cut in an hour by one person. If the lawn is to be cut in exactly three hours, we can just think through it hour-by-hour. Person 1 will cut 1/4 of the lawn in the first hour, a 1/4 of the lawn in the second hour, and in the third hour, both will cut a quarter each, which is the remaining half of the lawn to be cut.


2 - Variable Approach
When combining the rates of two machines, workers, or whatever, we can always write it as the sum of the rates and time it takes for each individual to do the work. Person 1 will be cutting 1/4 of the lawn for 3 hours, while person 2 will be cutting 1/4 of the lawn for 3-x hours, where x is the amount of time they waited before starting. Solve for x:

\(1=\frac{1}{4}(3)+\frac{1}{4}(3-x)\)
\(1=\frac{3}{4}+\frac{3-x}{4}\)
\(1=\frac{6-x}{4}\)
\(4=6-x\)
\(x=2\)
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