Printer A can print a report in 3 hrs. Printer B can print the same report in 4 hrs. How long would it take both printers, working together, to print the report?
what i got was let "t" equal the time it will take for them working together to print the report. so the equation is t=(3+4)/2, so t equals 3.5. Is this correct or is there a different way that you are supposed to do it?
Hi, Nicky,
First of all, it always a good idea to examine our answers and try to figure out if they make sense. If printer A can do the report in 3 hrs, but we add printer B to speed things up, then the time should be less than 3 hours. Correct? This would tell us that our first approach was wrong. Lets try something else.
Let’s call the rate that printer A operates at rA, and the rate that B operates at rB. Let’s call the number of reports produced x. Then
rA = (1 report/3 hrs)
rB = (1 report/4hrs)
The number of reports produced equals the rate times time:
x = (rA)(t) + (rB)(t) = (rA + rB)(t)
Substituting in our values:
1 report = ((1 report/3 hrs) + (1 report/4hrs))(t)
Can you rearrange this equation and solve for t
Problems of this type are easily solvable by the following method.
<< If it takes me 2 hours to paint a room and you 3 hours, ow long will it take to paint it together? >>
1--A can paint the house in 5 hours.
2--B can paint the house in 3 hours.
3--A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.
4--B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.
5--Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.
6--Therefore, the time required for both of them to paint the 1 house is 1 house/(A+B)/AB houses/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.
Note - T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.
