Questions on Word Problems: Rate of work, PAINTING, Pool Filling answered by real tutors!

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One crew can detassel a field of corn in 6 days. Another crew can do the same
job in 4 days. If the slower crew works alone for the first two days of
detasseling and then is joined by the faster crew, how long will it take
the two crews to finish?
:
Let t = time required for two crews to finish
:
Let the completed job = 1
:
Crew one time = t+2
Crew two time = t
:
+ = 1
:
Multiply equation by 12 to get rid of the denominators, results:
2(t+2) + 3t = 12
:
2t + 4 + 3t = 12
:
5t = 12 - 4
:
5t = 8
:
t =
t = 1.6 days to complete the job
:
:
Check solution in the original equation
+ =
+ =
.6 + .4 = 1

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sooo...we're now in calculus.com...

r=h/2 __ V=π(r^2)h/3 __ substituting __ V=π(h^3)/12

differentiating __ dV/dT=(π/12)(3h^2)dh/dT __ dh/dT=(dV/dT)/[(π/12)(3h^2)]

dh/dT=8/(π/4)(6^2)=32/(36π)=.283ft/min (approx)

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The intake pipe can fill a certain tank in 6 h when the outlet pipe is closed,
but with the outlet pipe open it takes 9 h. How long would it take to the
outlet pipe to empty a full tank?
:
Let t = time for the outlet pipe to empty the tank.
Let the full tank = 1
:
Filling is positive, draining is negative
:
- = 1
:
Multiply equation by 6t, to get rid of the denominators:
9t - 6(9) = 6t
:
9t - 54 = 6t
:
9t - 6t = 54
:
3t = 54
t =
t = 18 hrs to empty the full tank
;
:
Check solution:
- = 1
- = 1

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Let = the plane usual speed in km/hr
Let = the usual time to make the trip in hours

It is given that
km


Also given is





and



The plane usually flies at 800 km/hr
check:




OK

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a 5 horsepower (hp) pump can empty a pool in 5 hours. a smaller, 2 hp pump empties the same pool in 8 hours. the pumps are used together to begin emptying this pool. after 2 hours, the 2 hp pump breaks down. how long will it take the larger pump to finish emptying the pool?
---------------
5 hp pump DATA:
time = 5 hrs/job ; rate = 1/5 job/hr
----------------------
2 hp pump DATA:
time = 8 hrs/job ; rate = 1/8 job/hr
----------------------
In two hours working together,
2(1/5 + 1/8) = 2(13/40) = 13/20 of the job is done.
That leaves 7/20 of the job to be done by the 5 horsepower pump.
-------------
Let time for the 5 hp to finish the job be "x" hrs.
x(1/5) = 7/20
x = 7/4 hr = 1 3/4 hr. = 1 hr. 45 minutes
=============================
Cheers,
Stan H.

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A cyclist travels 60 km. If he reduces his speed by 2 km/h, he will take one hour longer to complete the trip. What was the original speed of the cyclist?
.
Apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
In our case, we will shift the formula thus:
t = d/r
.
Let x = original speed of cyclist
.
60/(x-2) = 60/x + 1
Multiplying both sides by a common denominator x(x-2)
60x = 60(x-2) + (x-2)x
60x = 60x - 120 + x^2 - 2x
0 = -120 + x^2 - 2x
0 = x^2 - 2x - 120
0 = (x-12)(x+10)
x= {12, -10}
.
We can toss out the negative solution, leaving us with:
x = 12 mph

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A&B work at the rate of 1/42 job per day
B&C work at the rate of 1/31 job per day
A&C work at the rate of 1/20 job per day
Let t=time it takes all workers, working together, to do the job
Let 1/A=rate at which A works
1/B=rate that B works
1/C=rate that C works
Now we know that:
1/A + 1/B=1/42-------------------eq1
1/B + 1/C=1/31---------------------eq2
1/A + 1/C=1/20---------------------eq3
subtract eq3 from eq1:
1/B - 1/C=1/42 - 1/20------------------eq3a
add eq3a and eq2:
2/B=1/42 - 1/20 + 1/31 (13020 is LCM)
2/B=(310-651+420)/13020=79/13020
1/B=79/26040 job per day-------------------------rate at which B works
substitute the value for 1/B into eq1:
1/A + 79/26040=1/42
1/A=1/42 - 79/26040=(620-79)/26040=541/26040 job per day----rate at which A works
substitute the value for 1/A into eq3:
541/26040 + 1/C=1/20
1/C = 1/20 - 541/26040=(1302-541)/26040=761/26040 job per day----rate at which C works
Together, A, B, & C work at the the rate of:
541/26040 +79/26040 + 761/26040= 1381/26040 job per day
Now, our final equation to solve is:
(1381/26040)t=1 (1 job that is) multiply each side by 26040
(rate at which they work * time it take to complete the job=1 job)
1381t=26040 divide by 1381
t= 18.86 days----------------time it takes all the workers working together
CK
A: (541/26040)*18.86=0.3917 of the job
B: (79/26040)*18.86=0.0572 of the job
C: (761/26040)*18.86=0.5512 of the job
0.3917 + 0.0572 + 0.5512=1.000069
1~~~1
Hope this helps---sorry about the first effort----ptaylor

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How many hours are there between 8:30 am today and 3:15am tomorrow?
---------------------
From 0830 to 0830 is 24 hours.
Subtract the hours from 0315 to 0830, which is (830 - 315) 5:15.
24:00 - 5:15 = 18:45, 18 hours, 45 minutes.
-----------------
An easier way is call 0315 tomorrow 2715 (add it to 24 hours), then subtract:
2715
-830
1845

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Jay can clean the house in 6 hours. After Jay and Jim have both been cleaning for an hour, they are joined by Jen and they completed the cleaning in 2 more hours. If it takes Jim 10 hours to clean the house alone, how long would it take Jen to do the job alone?
:
From above information we know:
Jay & Jim have been working a total of 3 hrs,
Jen worked 2 hrs
:
Let t = time required by Jen working alone
:
Let the completed job = 1
:
+ + = 1
:
Multiply equation by 30t to get rid of the denominators;
30t* + 30t* + 30t* = 30t(1)
:
Cancel out the denominators and you have;
5t(3) + 3t(3) + 30(2) = 30t
:
15t + 9t + 60 = 30t
:
24t + 60 = 30t
:
60 = 30t - 24t
:
60t = 6t
t =
t = 10 hrs Jen working alone
:
:
Check solution in original equation
+ + = 1

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Let x=amount of time it takes if all three are working together
Sam works at the rate of 1/4 job per day
Lisa works at the rate of 1/6 job per day
Tom works at the rate of 1/2 job per day
Together, they work at the rate of 1/4 +1/6 +1/2 =
3/12 +2/12 +6/12=11/12 job per day, so our equation to solve is:
(11/12)*x=1 (1 job, that is) multiply each side by 12
11x=12 divide each term by x
x=12/11 days=1.09 days-----time it takes if all three are working together
CK
Sam: (1/4)(12/11)=3/11
Lisa: (1/6)(12/11)=2/11
Tom: (1/2)(12/11)=6/11
3/11+2/11+6/11=11/11
1=1
Hope this helps---ptaylor

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Let's deal in minutes
Let x=amount of time it takes all three working together to fill the pool
Jim fills at the rate of 1/30 pool per min
Sue fills at the rate of 1/45 pool per min
Tony fills at the rate of 1/90 pool per min
Together they fill at the rate of 1/30 + 1/45 + 1/90=
3/90 + 2/90 + 1/90=6/90 =1/15 pool per min, so our equation to solve is:
(1/15)x=1 (1 pool that is) multiply each term by 15
x=15 min-------------------- time it takes all three working together
CK
Jim: (1/30)*15=1/2
Sue: (1/45)*15=1/3
Tony (1/90)*15=1/6
1/2 +1/3 +1/6 =3/6 + 2/6 + 1/6=6/6
1=1

Hope this helps---ptaylor

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Assume the student expected to make D dollars/hour and work for H hours.
That student expected to earn $24. So,

The problem tells you the student had to work longer for the same money. The student worked hours and earned the same $24. The student ended up getting dollars per hour.

Subbing in the expected return







So the student expected to work 2 hours or -3 hours. Since one cannot work -3 hours, the student expected to work 2 hours at $12/hour. Instead, the student worked 3 hours at $8/hour. A difference of $4 per hour

As far as "how do I do word problems goes", look for pertinent info. Many problems will give you more data -- some of it may be irrelevant. This problem gave you only good data, and no more or no less than required to solve.
The "trick' in this one is to know that wages= WageRate * HoursWorked

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Barry can do a certain job in 3 hours, whereas Sanchez 5 hours to do the same job. How long would it take them to do the job working together?
--------------------------
Barry DATA:
time = 3 hrs/job ; rate = 1/3 job/hr.
---------------------------
Sanchez DATA:
time = 5 hr/job ; rate= 1/5 job/hr
---------------------------
Together DATA:
time = x hr/job ; rate = 1/x job/hr.
---------------
EQUATION:
rate + rate = together rate
1/3 + 1/5 = 1/x
5x + 3x = 15
8x = 15
x = 15/8 hrs. (time to do the job together)
Comment: 15/8 hrs. = 1 hr. and 52 minutes and 50 seconds
=============================================
Cheers,
Stan H.
=====================

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a.
b

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Let x=amount of time it takes B&C to finish the job
A works at the rate of 1/4 job per hour ((1/4)*4=1 job)
B works at the rate of 1/6 job per hour
A&B work at the rate of 1/4 + 1/6 =3/12 +2/12=5/12 of the job per hour
C works at the rate of 1/8 job per hour
B&C works at the rate of 1/6 + 1/8=4/24 + 3/24=7/24 of the job per hour
In 1 hour, A&B completes (5/12)*1 or 5/12 of the job leaving
12/12 -5/12=7/12 of the job for B&C to finish
So, our equation to solve is:
(7/24)*x=7/12 (7/12 of the job, that is) multiply each term by 24
7x=14 divide each side by 7
x=2 hours------------------amount of time it takes B&C to finish
Total time, then would be 1+2=3 hours
CK
In 1 hour
A&B completes (5/12)*1=5/12 of the job
In 2 hours:
B&C completes (7/24)*2=7/12 of the job
So, 5/12 +7/12=12/12=1 job
Hope this helps-----ptaylor














B&C works at the rate of

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This is a good one.
When it was realized she had been given the wrong parcel, the messanger had travelled a certain distance in that 40 mins. Let's see,

------> distance travlled in that 40 mins w/ speed of 51km/hr
Let speed of the 1st messanger,
Now, let speed of the 2nd messanger.
Equating the speeds of the 2nd messanger in 1 hour to the distance travelled by the 1st messanger :



, same speed of the 2nd messanger to reach it or should be more than this to overtake the 1st messanger.
Thank you,
Jojo

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A messanger left a construction site and traveled by jeep at 51km/h. Forty minutes later it was discovered that she had been given the wrong parcel. How fast must a second messanger travel to overtake her in one hour?
----------------------------------------------------------------------
Messenger DATA:
rate=51 km/h ; time =1+(2/3)=(5/3) hrs. ; distance=rt=(5/3)*51 = 85 km
-------------------------
Pursuer DATA:
time = 1 hr ; distance = 85 km; rate = 85 km/hr
=============================
Cheers,
Stan H.

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Let x=time it takes to type 20 pages when both are working together
So, 20/x=rate at which they work when working together
Jeanne types at the rate of 20/40=1/2 page per min
Jim types at the rate of 20/30=2/3 pages per min
Together they work at the rate of 1/2 + 2/3 =7/6 pages per min
So, our equation to solve is:
(7/6)*x=20 multiply each side by6
7x=120
x=17.143 min
Another way:
7/6=20/x multiply each side by 6x
7x=120-----------same as before
CK
(1/2)*17.143= 8.571 pages
(2/3)*17.143=11.429 pages
8.571 + 11.429=20
20=20

Hope this helps---ptaylor

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jean types at a rate of 1/2 page per minute 20/40
Jim types at a rate of 2/3 page per minute 20/30
together: would be 1/2t+2/3t=20 where t is time.
find a common denominator and you get 7/6t=20 multiply both sides by 6 and you get 7t=120 so t=appx 17.1 minutes

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so the 1st 40 hours is 6(40) to get to total wages of $375 one must add 9T where T is over time hours worked . So lets write this our 6(40)240+9T=375 subtract 240 from each side and you arrive at
9T=135 so T=15 regular hours +T(overtime hours)=Total hours worked
40+15=55 hours worked to make $375

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Let = number of hours he must work to make $375
His overtime hours will then be





Willie must work 55 hours to make $375
check answer:




OK

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Peter mows a lawn in 40 minutes. John moes the lawn in 60 minutes. How long will it take them to mow the lawn together?
Let x=amount of time it takes both working together to mow the lawn
Together, they work at the rate of 1/x lawn per hour
Peter works at the rate of 40/60=2/3 lawn per hour
John works at the rate of 1 lawn per hour
Together, they work at the rate of 2/3 + 1=1 2/3 =5/3 lawns per hour
So, our equation to solve is:
(5/3)*x=1 (1 lawn, that is) multiply each term by 3
5x=3 divide both sides by 5
x=3/5 hour=36 min
Another way:
2/3 +1=1/x
5/3=1/x multiply each term by 3x
5x=3
x=3/5
CK
Peter:
(2/3)*(3/5)=2/5 lawn
John:
1*(3/5)=3/5 lawn
2/5 + 3/5=1
1=1
Hope this helps---ptaylor

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If my sister takes twice as long as I to do her taxes but we can do her taxes together in 10 hours, how long will it take me to do her taxes alone?
------------------------
Your DATA:
time = x hr/job ; rate = 1/x job/hr
----------------
Sister DATA:
time = 2x hr/job ; rate = 1/2x job/hr
------------------
Together DATA:
time = 10 hr/job ; rate = 1/10 job/hr.
------------------------
EQUATION:
rate + rate = together rate
1/x + 1/2x = 1/10
Multiply thru by 10x to get:
10 + 5 = x
---------------
x = 15 hrs. (time it will take you alone)
2x = 30 hrs. (time it will take your sister alone)
======================
Cheers,
Stan H.

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Let x=amount of time required for the second press to complete the job alone
Then the second press works at the rate of 1/x of the job per hour
Together, the two presses work at the rate of 1/2 of the job per hour
And the first press works at the rate of 1/6 of the job per hour
So, our equation to solve is:
1/6+1/x=1/2 multiply each term by 6x
x+6=3x subtract x from each side
x-x+6=3x-x collect like terms
2x=6 divide both sides by 2
x=3 number of hours needed for the second press to complete the job
CK
1/6+1/3=1/2
1/6+2/6=1/2
3/6=1/2
1/2=1/2
also
(1/6)*2=1/3
(1/3)*2=2/3
2/3+1/3=1 (1 job, that is)
1=1
Hope this helps---ptaylor

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James can stuff 1000 envelopes in 5 hours. Susan can stuff 1000 envelopes in 3 hours. How long will it take to stuff 1000 envelopes if they worked together?
The solution does not depend on the number 1000. It would be solved the same way even if 10 letters were stuffed, or 100,000 letters were stuffed.
Think of it as completing a job.
James can complete a job 5 hours. Susan can complete the same job in 3 hours. How long will it take to complete the job if they worked together?
Let x=amt of hours each will work. Since they're working together, each of their times =x
James needs 5 hrs by himself. So each hour he completes 1/5.
Susan needs 3 hours by herself, so each hour she completes 1/3
But each don't work 1 hour they work x hours. So,...
James completes x/5
Susan completes x/3
Together they complete x/5+x/3
The whole job is 1
x/5+x/3=1
Multiply each term by the LCM=15 to clear fractions
15x/5 +15x/3=15
3x+5x=15
8x=15
x=15/8
x= 1 7/8 hours
See more word problems like this
worked out step by step in my FREE
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activate, ALL FREE from
www.math911.com
Tell your friends, (and teacher!)
Professor Martin Weissman

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Let x=number of hours it takes with both working together
James stuffs at the rate of 1000/5=200 envelopes per hour
Susan stuffs at the rate of 1000/3=333.333333 =333 1/3 envelopes per hour
Together they stuff at the rate of 200+333.3333=533.3333=533 1/3 envelopes per hour
So, our equation to solve is:
(533 1/3)*x=1000 divide each side by 533 1/3
x=1.875 hours or 1 7/8 hours--------ans
Now we can stay with fractions:
x=1000/(533 1/3)=1000/(1600/3)=1000*(3/1600)=3000/1600=30/16=15/8=1 7/8 hr
Hope this helps----ptaylor

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An oil tanker can be filled in 10 hours from one pipe. It can be filled by another pipe in 8 hrs. How long will it take to fill the tank if both pipelines are used .

Let x=amount of time it takes to fill the tanker when both pipelines are used
First pipe fills at the rate of 1/10 tanker per hour
Another pipe fills at the rate of 1/8 tanker per hour
Together they fill at the rate of 1/10 + 1/8=4/40 +5/40=9/40 tanker per hour.
So our equation to solve is
(9/40)*x=1 ( 1 tanker, that is) multiply each side by 40
9x=40 divide each side by 9
x=4 4/9 hours----------------------------answer
CK
First pipe:
(1/10)*(4 4/9)=(1/10)*(40/9)=4/9 of the tanker
Second pipe:
(1/8)*(40/9)=5/9
4/9 + 5/9=1
9/9=1
1=1
I THINK YOUR BOOK IS WRONG.
Hope this helps---ptaylor

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10*8/(10+8)=80/18=4.444 HOURS TO FILL THE TANK USING BOTH PIPES.
BELIEVE YOUR ANSWER OF 2 11/12 IS INCORRECT.

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10*8/(10+8)=80/18=4.444 HOURS TO FILL THE TANK USING BOTH PIPES.

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Melanie paints at the rate of 1/x barns per day
Melissa paints at the rate of 1/2x barns per day
Together, they paint at the rate of 1/x + 1/2x =2/2x+1/2x=3/2x barns per day, so in 1 day they paint 3/2x of the barn.
If x=5:
3/2x=3/2*5=3/10 of the barn
CK
Melanie Paints at the rate of 1/x barns per day
if x=5, she paints 1/5 of the barn per day
Melissa paints at the rate of 1/2x barns per day
if x=5, she paints 1/10 of the barn per day
So, 1/5 + 1/10=2/10+1/10=3/10 of the barn
Hope this helps-----ptaylor

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An executive flew in the corporate jet to a meeting in a city 1500 kilometers
away. After traveling the same amount of time on the return flight, the pilot
mentioned that they still had 300 kilometers to go. If the air speed of the
plane was 600 kilometers per hour, how fast was the wind blowing?
(Assume that the wind direction was parallel to the flight path and constant all day).
:
Let w = speed of the wind
:
Then we know:
(600+w) = ground speed with the wind
(600-w) = speed against the wind
:
= time to fly the outbound trip
this time is equal to flying: 1500 - 300 = 1200 mi against the wind
:
Write a time equation:
=
Cross multiply:
1200(600+w) = 1500(600-w)
:
720000 + 1200w = 900000 - 1500w
:
1200w + 1500w = 900000 - 720000
:
2700w = 18000
w =
w = 66.67 km/hr is the wind speed
:
:
Check solution find out if the times are equal
1500/666.67 = 2.25 hrs
1200/533.33 = 2.25 hrs also, confirms our solution
:
Actually we can simply state the problem,
The plane flew 1500 with the wind in the same time it flew 1200 mi against the wind.

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you managed to "stumble" onto the correct answer

you defined Z as the rate __ "PIPE B RATE = Z"
__ but then used 1/Z in your calculations

"T*R = w"
"T(1/12) + T-4.5(1/Z)=1"
"9(1/12) + 9-4.5(1/Z)=1"

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A pump fills a pool in 4 hours. The pool can drain in 1 hour. How long will it take to fill the pool if the drain is left open?
-----------------
It will never fill, since it drains 4 times as fast it fills.

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x*3x/(x+3x)=8
3x^2=8*4x
3x^2=32x
3x^2-32x=0
x(3x-32)=0
3x-32=0
3x=32
x=32/3 or 10.667 Andrew's time.
3x=3*10.66=32 John's time.
Proof:
10.667*32/(10.667+32)=8
341.333/42.667=8
8=8

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The radiator of a car holds 16 quarts of liquid of which 8 are antifreeze. The radiator has a leak and loses 2 quarts of liquid each week. Every week, 2 quarts of water are added. After 3 weeks how much antifreeze is left in the radiator?
:
Let x = fractional amt of antifreeze after 2 qts leak out & 2 qt water added
:
Week 1:
(16-2) = 16x (2 qts of water added to make it 16 again)
(14) = 16x
7 = 16x
x = ;
Therefore: *16 = 7 qts, of antifreeze after week 1
:
Week 2;
(16-2) = 16x
14 = 16x
= 16x
x = *
x = ;
Therefore: *16 = 6.125 qts of antifreeze after week 2
:
Week 3:
(16-2) = 16x
(14) = 16x
= 16x
x = *
x = ;
therefore: *16 = 5.34 qts of antifreeze after week 3
:
:
Hope this makes sense.

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1.50x=2(x-5)-18
1.50x=2x-10-18
1.5x-2x=-28
-.5x=-28
x=-28/-.5
x=56 dozen roses were bought.
Proof:
1.50*56=2(56-5)-18
84=2*51-18
84=102-18
84=84

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Let X be the number of dozen he bought.
Then (X-5) is the number of dozen he sold.





He bought 56 dozen and sold 51 of them.
That's a lot of flowers and not a lot of profit :(

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One person can do a job in 8 hours. A second person can do it in 12 hours. If the first person works 2 hours less than the second, how many hours will it take them working together?
:
Let t = time that they worked together
then
(t+2) = total time worked by the 2nd person
:
Let the completed job = 1
;
Each will do a fraction of the work which will add up to 1:
:
1st person + 2nd person = completed job
+ = 1
:
Multiply equation by 24 to get rid of the denominator, rsults:
3t + 2(t+2) = 24
3t + 2t + 4 = 24
3t + 2t = 24 - 4
5t = 20
t =
t = 4 hrs working together
;
:
Check solution:
4/8 + 6/12 = 1

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You're looking for the rate of change of SA with respect to R.
That's the same as the derivative.
Take a small step in R, calculate the SA, subtract the original SA, and divide by the small step.



Assume that is small, then is even smaller (=0)


.
.
.
That's the hard way to find the derivative of SA with respect to R.
You can also differentiate.


So when R=20

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Let x=number of hours machine 1 is in operation (30 components per hr)
And y=number of hours machine 2 is in operation
And we are told that:
x+y=15-------------------------------eq1
Machine 1 operates at the rate of 30 components per hour
Machine 2 operates at the rate of 40 components per hour
Now we know that:
30x+40y=545 divide each term by 5 just to reduce the size of the numbers and we get:
6x+8y=109------------------------------eq2
multiply eq1 by 6(we get 6x+6y=90) and then subtract it from eq2
2y=19 divide each side by 2
y=9.5 hrs------------------number of hours machine 2 is in operation
Substitute y=9.5 into eq1:
x+9.5=15 subtract 9.5 from each side
x=15-9.5=5.5 hrs------------------number of hours machine 1 is in operation
CK
30*5.5+40*9.5=545
165+380=545
545=545
Hope this helps---ptaylor

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A vertical cylindrical storage tank, with diameter 20 feet, is filled with oil
to a depth of 40 feet. Sometime later, the oil is drained, decreasing the
depth a rate of 8 inches per hour. Write an equation for the volume of oil
(ft^3) remaining in the tank t hours later as a function of t.
Draw a geometrically correct sketch, supporting your solution.
:
This is a linear equation so we can find the slope using the time/volume
:
Find the volume at 40 ft, (radius = 10 ft)
V =
V = 12566 cu ft (rounded to nearest cu ft)
:
Therefore: t=0; v=12566
:
Find out how many hours to empty the tank
:
At 8 in/hr how long to lower it 40 ft?
t = = 60 hrs (vol = 0)
Therefore: t=60; v=0
:
Find the slope: m = = is the slope
:
An equation: V = t + 12566
:
You can illustrate this with a graph. V = vertical axis; t = horizontal axis

:
You can prove this:
t = 30 hr
At 30 hrs the depth would be 30*8" = 240" = 20 feet (40-20)
Find the volume at 20ft
V =
V = 6283 cu ft
:
Using the equation/graph
v = (30) + 12566
v = 6283 cu ft
:
Did this sense to you? I am not sure what kind of sketch they have in mind, I'll
leave that up to you.

You can put this solution on YOUR website!

You can put this solution on YOUR website!

You can put this solution on YOUR website!
Let x=amount of time it takes to do the job when they work together
Jerry works at the rate of 1/5 of the job per week
Sam works at the rate of 1/3 of the job per week
Together they work at the rate of:
1/5 + 1/3=3/15 +5/15=8/15 of the job per week
So, our equation to solve is:
(8/15)x=1 (1 job, that is) multiply each side by 15
8x=15 divide each side by 8
x=1.875 weeks---------------------------ans
Another way:
if x=amount of time it takes to do the job when they work together, then, together, they work at the rate of 1/x job per week, so:
1/5 + 1/3 =1/x multiply each term by 15x
3x+5x+15
8x=15----same as before
CK
Amount of work that Jerry does in 1.875 weeks plus the amount of work that Sam does in 1.875 weeks has to equal 1 job completed,so:
(1/5)*1.875 + (1/3)*1.875=1
0.375+0.625=1
1=1
Hope this helps---ptaylor

You can put this solution on YOUR website!
12*8/(12+8)=96/20=4.8 MINUTES TO FILL THE TUB WITH BOTH PIPES OPERATING.