SOLUTION: Kay can solve equations at the rate of 26 per hour. Dan can solve them at the rate of 20 per hour. Prior to starting to work their equations, Kay solved 3 equations in study hall w

Question 156569: Kay can solve equations at the rate of 26 per hour. Dan can solve them at the rate of 20 per hour. Prior to starting to work their equations, Kay solved 3 equations in study hall while Dan solved 8.
a. defina the variable for the number of hours they worked in study hall
b. how many equations has each solved after 20 minutes
c. How long will it take until each has solved the same number of equations?
d. There are 27 total equations. How long does it take each to complete?

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
here goes .................
question a: define the variable for the number of hours they worked in study hall.
h(k) = number of hours kay worked.
h(d) = number of hours dan worked.
hours worked = number of units completed / rate per unit.
kay does 26 per hour so h(k) = ( 3 / 26 ) hours = .115384615 hours.
dan does 20 per hour so h(d) = ( 8 / 20 ) hours = .4 hours.
question b: how many equations has each solved after 20 minutes
formula is number of units completed = time worked * units per hour
kay completed 26 * (20 / 60) = 8.666... = 8 units in 20 minutes (working into her ninth but not done yet).
dan completed 20 * (20 / 60) = 6.666... = 6 units in 20 minutes (working into his 7th but not done yet).
if you assume that they have to add the units already completed, then...
kay completed 8 + 3 = 11 total (3 already done plus 8 in additional 20 minutes).
dan completed 6 + 8 = 14 total (8 already done plus 6 in additional 20 minutes).
question c: how long will it take until each has solved the same number of equations?
in order for them to solve the same number of equations kay has to solve 5 more than dan because she started at 3 and he started at 8.
if x = number of equations dan has to solve then x+5 = number of equations kay has to solve.
for the total solved to be equal then (x+5) solved at the rate of 26 per hour has to equal x solved at the rate of 20 per hour.
the equation is then (x+5)/26 = x/20.
solving for x yields x = 16.6667.
dan has to solve 16.6667 and kay has to solve 21.6667.
time to solve 21.6667 at 26 per hour = .8333346... hours.
time to solve 16.6667 at 20 per hour = .833335 hours.
the hours are the same within rounding errors so the equation appears valid.
as a practical matter the answer should be rounded down to x = 16.
after doing that they kay has to solve 21 at 26 per hours and dan has to solve 16 at 20 per hour.
time to solve 21 at 26 per hour = .807692308 hours.
time to solve 16 at 20 per hour = .8 hours
number of hours required is then .807692308 to achieve equality (the longer of the two differing hours since dan will actually complete his share of the equality a little ahead of kay).
testing the answer is then as follows:
in .807692308 hours dan has solved 16.15 + 8 already done = 24.15 = 24 solved and working on his 25th but not done yet.
in the same time kay has solved 21.00000... + 3 already done = 24.00000... = 24 solved and just starting on her 25th looks like.
answer is then .807692308 hours rounded to wherever you want it to be.
question d: there are 27 total equations. how long does it take each to complete?
total time required for kay to complete = 27 / 26 = 1.038461538 hours.
total time required for dan to complete = 27 / 20 = 1.35 hours.
take away ones already done, then..............
total time required for kay to complete = 24 / 26 = .923076923 hours.
total time required for dan to complete = 19 / 20 = .95 hours.

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