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

Question

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.

Answer 1

It depends what color they are using.... and how much the paint cost...

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

Sally paints a million/4 abode in an hour John paints a million/6 abode in an hour S + J paint a million/4 + a million/6 = 5/12 abode in an hour So S + J can paint a house in 12/5 hours = 2 2/5 hours = 2 hr 24 min

Answer 6

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

Answer 7

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

Answer 8

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 9

sara - 6hrs. alone
kate - 3hrs. alone