Question 207777: Three girls, Laura, Tammy, and Jeri can do the chores in 1 hour and 20 minutes. If Jeri did all the chores she would take twice as long as Tammy and 4 hours longer than Laura. How long would each girl take to do the chores alone?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! well i must say you picked a good one.
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here's how i think i solved it.
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the problem:
Three girls, Laura, Tammy, and Jeri can do the chores in 1 hour and 20 minutes. If Jeri did all the chores she would take twice as long as Tammy and 4 hours longer than Laura. How long would each girl take to do the chores alone?
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let J = jeri's rate of work
let T = Tammy's rate of work
let L = Laura's rate of work
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General Formula is RT = U which means that Rate * Time = Units produced
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The Units produced = 1 Set of Chores
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The Time it took for all 3 girls working at the same time = 1 and 1/3 hours = 4/3 hours = 1.3333...
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Your formula for when all 3 girls are working is:
4/3 * (J + T + L) = 1
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You say that Jeri takes twice as long as Tammy to do the chores working alone.
This means that if Jeri takes h hours, Tammy only takes h/2 hours because h is 2 times h/2
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J*h = 1 means that J = 1/h
T*(h/2) = 1 means that T = 2/h
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You say that Jeri takes 4 hours longer than Laura.
This means that if Jeri takes h hours, Laura must take (h-4) hours.
L * (h-4) = 1 means that L = 1/(h-4)
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you can now substitute 1/h for J and 2/h for T and 1/(h-4) for L in the original equation of 4/3 * (J + T + L) = 1
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your original equation becomes:
4/3 * ((1/h) + (2/h + 1/(h-4)) = 1
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if you multiply both sides of this equation by h*(h-4) you will be able to remove all the denominators.
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your equation becomes:
4/3 * ((h-4) + 2*(h-4) + h) = h*(h-4)
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this reduces to:
4/3 * (4h-12) = h^2 - 4h
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when you multiply both sides by 3 this becomes:
3h^2 - 28h + 48 = 0
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when you solve this using the quadratic formula, your answers become:
h = 7.070367517
or
h = 2.262965816
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h = 2.2... is rejected because h-4 would become negative.
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your answer becomes h = 7.070367517
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substituting this value into the equation of 3h^2 - 28h + 48 = 0 makes 0 = 0 showing that this is a legitimate root of that equation.
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substituting this value into the equation of 4/3 * ((1/h) + (2/h + 1/(h-4)) = 1 makes 1 = 1 showing that this is a legitimate solution to the original equation.
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J's rate is 1/h = 1/7.070367517 units per hour.
T's rate is 2/h = 2/7.070367517 units per hour.
L's rate is 1/(h-4) = 1/3.070567517 units per hour.
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it takes J (1/h)*x hours to complete the chores.
this means that (1/h)*x = 1 which means that x = 1/(1/h) = h
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it takes T (2/h)*y hours to complete the chores.
this means that (2/h)*y = 1 which means that y = 1/(2/h) = h/2
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it takes L (1/(h-4)*z hours to complete the chores.
this means that (1/(h-4))*z = 1 which means that z = 1/(1/(h-4)) = (h-4)
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everything checks out.
J takes twice as long as T to complete the chores.
J takes 4 hours more than L to complete the chores.
adding up all their rates when they work together produces 1 set of chores.
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