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Question 22848: It takes Billy 2 hours longer than Reena to do a certain job. They worked together for 2 hours; then Reena left and Billy finished the job in 1 hour. How long would it take each of them to do the job alone ?
(this is from Intermediate Algebra, 7th ed. Jerome Kaufmann/Karen Schwitters, pg 330, problem 42, Quadratic Equations and Inequalities...only one I couldn't get)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! LET TIME TAKEN BY REENA ALONE TO DO 1 JOB =X HRS
BILLY ALONE TAKES 2 HRS. MORE =X+2
SO REENA CAN DO ALONE IN 1 HOUR =1/X JOB
BILLY CAN DO ALONE IN 1 HOUR =1/(X+2) JOB
TOGETHER THEY CAN DO IN 1 HR.=(1/X)+(1/(X+2)) JOB
IN 2 HRS TOGETHER THEY CAN DO =2*{(1/X)+(1/(X+2))} JOB
NOW REENA LEFT AND BILL ONLY WORKED FOR 1 HOUR = BILLY ALONE DOES 1/(X+2) JOB IN 1 HOUR
SO TOTAL JOB DONE =2*{(1/X)+(1/(X+2))}+(1/(X+2)) JOB..BUT THIS COMPLETED THE 1 JOB .SO
2*{(1/X)+(1/(X+2))}+(1/(X+2)) =1
2/(X)+2/(X+2)+1/(X+2)=1..L.C.M IS (X)(X+2)
{2(X+2)+2(X)+1(X)}/(X)(X+2)=1
MULTIPLYING BY LCM (X)(X+2) BOTH SIDES WE GET
2X+4+2X+X=(X)(X+2)=X^2+2X
5X+4=X^2+2X
X^2+2X-5X-4=0
X^2-3X-4=0
X^2-4X+X-4=0
X(X-4)+(X-4)=0
(X+1)(X-4)=0
X-4=0
X=4
SO REENA CAN DO THE JOB IN 4 HRS AND BILL CAN DO IN X+2=4+2=6 HRS THE JOB.
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