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Question 113478: can you help me?
1.an older computer unit requires 12min more to print a company's payroll than does a newer computer unit. used together the computers can print the payroll in 8 min. find the time the new computer unit, used alone, will require to print the payroll.
2.working together. Dan and Leah can mow a lawn in 1 h and 12 min. Working alone, Leah can mow the lawn in 1 h less time than it would take Dan. Find the time it takes for each of them, working on his/her own, to mow the lawn.
i hope you can help me..
god bless..
Answer by moou_b(3)   (Show Source):
You can put this solution on YOUR website! ANSWER 1.
let the new comp. can print the whole payroll in 'x' min.
therefore in 1 min. it will do 1/x of printing.
the old comp. takes x+12 min. to do the whole printing.
therefore in 1 min. it will do 1/x+12 of printing.
together they do the printing in 8 min.
therefore in 1 min. they do 1/8 of printing.
now, (1/x) + (1/x+12) = 1/8
x+12+x/x(x+12) = 1/8
2x+12/x(x+12) = 1/8
16x+96 = x(x+12)
16x+96 = x^2+12x ( x to the power 2+12x)
x^2+12x-16x = 96
x^2-4x-96 = 0
x^2-12x+8x-96 = 0
x(x-12)+8(x-12) = 0
(x-12)+(x+8) = 0
when x-12 = 0, x=12 and when x+8 = 0, x=(-8) neglecting the negetive term x=12 is the valid answer. therefore, the new comp. will take 12 min. to print the payroll alone.
ANSWER 2.
let dan takes x hrs. to mow the whole lawn.
in 1 hr he will mow 1/x of the lawn.
leah mows the lawn in (x-1) hrs.
in 1 hr he will mow 1/(x-1) of the lawn
leah & dan together mow the lawn in 1 hr. 12 min = 1 and 1/5 hr = 6/5 hrs.
in 1 hr they mow 1/ 6/5 = 5/6 of the lawn.
now, forming equation, we have, (1/x)+(1/x-1)= 5/6
x-1+x/x(x-1) = 5/6
2x-1/x(x-1) = 5/6
12x-6 = 5x^2-5x
5x^2-5x-12x+6= 0
5x^2-17x+6 = 0
5x^2-15x-2x+6=0
5x(x-3)-2(x-3)=0
(5x-2)(x-3) = 0
when 5x-2=0,x=2/5 and when x-3=0, x=3
as 2/5 hr solution is not valid because 2/5 hr=24 min.
so dan can mow the lawn in 3 hrs and leah will mow the lawn 3-1=2 hrs.
hope the solution is understood.
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