SOLUTION: Mr. Russo takes three minutes less than Mr. Lloyd to pack a case when each works alone. One day, after Mr. Russo had spent six minutes in packing the case, the boss called him away

Question 408979: Mr. Russo takes three minutes less than Mr. Lloyd to pack a case when each works alone. One day, after Mr. Russo had spent six minutes in packing the case, the boss called him away, and Mr. Lloyd finished packing the case in four more minutes. How many minutes would it take Mr. Russo alone to pack the case?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Mr. Russo takes three minutes less than Mr. Lloyd to pack a case when each works alone.
One day, after Mr. Russo had spent six minutes in packing the case, the boss called him away, and Mr. Lloyd finished packing the case in four more minutes.
How many minutes would it take Mr. Russo alone to pack the case?
:
Let t = time required by Russo alone
then
(t+3) = time required by Lloyd alone
:
Let completed job = 1; (a packed case)
:
Russo + Lloyd = completed job
+ = 1
:
Multiply by t(t+3)
t(t+3)* + t(t+3)* = 1t(t+3)
:
Cancel the denominators, and you have:
6(t+3) + 4t = t^2 + 3t
6t + 18 + 4t = t^2 + 3t
10t + 18 = t^2 + 3t
:
Arrange as a quadratic equation on the right:
0 = t^2 + 3t - 10t - 18
t^2 - 7t - 18 = 0
Factors to
(t-9)(t+2) = 0
:
The positive solution is what we want here:
t = 9 minutes, time required by Russo alone
:
:
Check solution in original shared work equation (Lloyd requires 12 min)
+ = 1; confirms our equation
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