Dean and his younger brother sam are digging a hole to make imu for the luau. Working together it takes them 1.5 hours to finish the hole. If it would take sat twice as long to finish the hole working alone as it would dean how long would it take dean if he had to dig the hole by himself?
Let's start by finding out how much of the job each person can do in one hour.
Dean and Sam can do the job in 1.5 hours, so their combined hourly rate is:
1 job / 1.5 hours = 0.67 jobs per hour
Let's call Dean's hourly rate "D" and Sam's hourly rate "S". We can use this information to set up two equations:
D + S = 0.67 (the combined hourly rate)
S = 2D (it takes Sam twice as long as Dean to do the job alone)
Now we can substitute the second equation into the first equation to solve for D:
D + 2D = 0.67
3D = 0.67
D = 0.22
So Dean can do 0.22 jobs per hour. To find out how long it would take him to do the job alone, we can use the formula:
time = jobs / rate
time = 1 job / 0.22 jobs per hour = 4.55 hours
So it would take Dean 4.55 hours to dig the hole by himself.
Let's assume that it takes Dean x hours to finish the hole by himself.
From the given information, we know that Dean and Sam can complete the hole together in 1.5 hours. So their combined work rate is 1 hole / 1.5 hours = 2/3 hole per hour.
Now, let's determine Sam's work rate. Since it would take Sam twice as long to finish the hole alone as it would Dean, Sam's work rate is 1 hole / (2x) hours = 1/2x hole per hour.
Using the concept of work rates, we can add their individual work rates to get the combined work rate:
Dean's work rate + Sam's work rate = Combined work rate
1/x + 1/(2x) = 2/3
To solve this equation for x:
Multiply both sides of the equation by 6x:
6 + 3 = 4x
9 = 4x
x = 9/4
Therefore, Dean would take 9/4 hours (or 2.25 hours) to finish the hole by himself.