Sara take 3hr longer to paint a floor than it takes kate. When they work together, it takes them 2hrs. how lon

Sara take 3hr longer to paint a floor than it takes Kate. When they work together, it takes them 2hrs. how long would each take to do the job alone.Sara take 3hr longer to paint a floor than it takes kate. When they work together, it takes them 2hrs. how lon
It depends what color they are using.... and how much the paint cost...Sara take 3hr longer to paint a floor than it takes kate. When they work together, it takes them 2hrs. how lon
sara - 6hrs. alone


kate - 3hrs. alone
1/Kate+1/(Kate+3)=1/2


Kate+3+Kate=1/2*Kate*(Kate+3)


2*Kate+3=1/2*Kate^2+3/2*Kate


1/2*Kate^2-1/2*Kate-3=0


Kate=3 hours


Sara=Kate+3=6 hours
Answer: Kate does it in 3hrs and Sarah in 6hrs.





Assume Kate does the job on her own in x hrs.


Therefore Sarah can do it in x + 3 hrs.


In 1 hr, Kate can do a fraction 1/x of the job.


In 1 hr, Sarah can do 1/(x+3) of the job


Together, they finish 1/x + 1/(x+3) of the job in 1 hr.


Given that together they do the job in 2 hrs.


So, together they do 1/2 of the job in 1 hr.


OK?


So, 1/x + 1/(x+3) = 1/2


Solving this equation (it turns out to be a quadratic) gives


x =3
note


the set up for these kinds of word problems is always the same


(1/x)w + (1/y)w = z





Sara takes 1/x to paint in one unit of time


Kate takes 1/y to paint in one unit of time


z = the proportion of floor painted


w = the amount of time it takes for both to paint floor





(1/x)w + (1/y)w = z





1/(y +3) * 2 + (1/y)*2 = 1


multiply each term by the common denominator y(y + 3)





2y + 2(y + 3) = y(y + 3)


4y + 6 = y^2 + 3y


solve for zero


y^2 - y - 6 = 0


Factor


(y - 3)(y + 2) = 0


y = 3 {not y = -2, which is extraneous}





Kate would take 3 hours


Sara would take 6 hours
Let x = number of hrs Sara takes to paint the floor.


and y = number of hrs Kate takes to paint the floor.





Then x = y + 3.





Also,


1/x = portion of floor Sara can paint in 1 hr


1/y = portion of floor Kate can paint in 1 hr





1/x + 1/y = portion of floor they can paint together in 1 hr.





Since they can paint the floor together in 2 hrs, they can paint 1/2 of the floor in 1 hr.


Therefore


1/x + 1/y = 1/2





Substitution x = y + 3, we get


1/(y+3) + 1/y = 1/2





Get a common denominator


y/(y)(y+3) + (y+3)/(y)(y+3) = 1/2


(y + y + 3)/(y)(y+3) = 1/2





Cross multiply


2(y + y + 3) = (y)(y + 3)





4y + 6 = y^2 + 3y


0 = y^2 - y - 6





Factor or use the quadratic formula to solve for y.


You get y = 3 or y = -2





But y = -2 is obviously incorrect. It is what is called an extraneous root -- it's a correct solution to the equation, but it does not fit the original conditions.





So we discard y = -2, and we are left with the correct solution y = 3.


From this, we get x = y+3 = 6.
You have to be clear on the difference between distance (d), speed (v) and time (t).





d = vt





The ';distance'; in this case is painting a given floor. All you're given is times but this floor is always the same one.





Kate: d = v[K]*t[K]


Sara: d = v[S]*t[S]





We are told Sara's time is 3 hrs longer,





t[S] = t[K] + 3





and that if they work together, they take 2 hours





d = (v[S]+v[K]) * 2





Change all variables to time since you are only asked about the time each would take individually. So solve for t[S] and t[K]. Changes speeds to distance/time:








d = (d/t[S] + d/t[K]) * 2 .





Distributing the denominators,





= 2 * (d*t[K] + d*t[S]) / t[S]t[K]





= 2 * d * (t[K] + t[S])/t[S]t[K]





but therefore





(t[K] + t[S])/t[K]t[S] = 1/2





substituting t[S] = t[K] + 3





(2t[K] + 3)/t[K](t[K]+3) = 1/2





2t[K] + 3 = 1/2 * t[K](t[K]+3)





2t[K] + 3 = 1/2*t[K]^2 + 3/2t[K]





multiply everywhere by 2





4t[K] + 6 = t[K]^2 + 3t[K]





t[K]^2 - t[K] - 6 = 0





(t[K]+2)(t[K]-3) = 0





Therefore the solutions of t[K] are -2 and 3. Reject -2 since can't have negative time.





t[K] = 3 hrs, t[S] = 3+3 = 6 hrs.
well, you know the following:





x= Sara's time


y=Kate's time





x -3 = y


x+y=2








so replace the y in the second equation by the y in the first.





x+(y) = 2


x+(x-3) = 2


2x = 2+ 3


2x= 5


x= 2.5





Sara = 2.5 hours





Kate = x=3 = 2.5 + 2.5 = 5.5