Algebra Problem: Sally can paint a house in four hours....

[MizzHoneyDue]

Sally can paint a house in 4 hours. John can paint it in 6. How long will it take them together to paint the house?

 

[goingnutz]

I'm not positive on this, but I think it works out like this

1/4 + 1/6 = 1/x

24x(1/4)+24x(1/6)=24x(1/x)

6x+4x= 24
10x= 24

x= 24/10 or 2 4/10 hours

I hope this is right, and it helps you. If not, sorry, I tried.

 

[MizzHoneyDue]

I'm not positive on this, but I think it works out like this

1/4 + 1/6 = 1/x

24x(1/4)+24x(1/6)=24x(1/x)

6x+4x= 24
10x= 24

x= 24/10 or 2 4/10 hours

I hope this is right, and it helps you. If not, sorry, I tried.

Thank you.

 

[jonboy]

Yes you are correct goingnutz. Good job by muliplying by 24 to make it much easier

\(\displaystyle \L \frac{1}{4} +\frac{1}{6}=\frac{1}{x}\)

\(\displaystyle \L \frac{5}{12}=\frac{1}{x}\)

\(\displaystyle \L 12=5x\)

\(\displaystyle \L 2.4=x\)

\(\displaystyle \L .4\bullet60=24 mins\)

So the time will be 2 hours and 24 minutes (do not leave the 4/10 hours in your answer convert it to minutes by multiplying by 60 to get minutes unless the problem states otherwise).

 

[TchrWill]

Re: Algebra Problem: Sally can paint a house in four hours..

Sally can paint a house in 4 hours. John can paint it in 6. How long will it take them together to paint the house?

Problems of this type are solvable by either of the following methods.


<< If it takes me 2 hours to paint a room and you 3 hours, ow long will it take to paint it together? >>

Method 1:

1--A can paint the house in 5 hours.
2--B can paint the house in 3 hours.
3--A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.
4--B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.
5--Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.
6--Therefore, the time required for both of them to paint the 1 house is 1 house/(A+B)/AB houses/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.

Note - T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

Method 2:

Consider the following diagram -

.........._______________ _________________
..........I B / /\
..........I * / I
..........I * / I
..........Iy * / I
..........I * / I
..........I*****x****** I
..........I / * (c)
..........I(c-y) / * I
..........I / * I
..........I / * I
..........I / * I
..........I / * I
..........I/___________________* ________\/__
A

1--Let c represent the area of the house to be painted.
2--Let A = the number of hours it takes A to paint the house.
3--Let B = the number of hours it takes B to paint the house.
4--A and B start painting at the same point but proceed in opposite directions around the house.
5--Eventually they meet in x hours, each having painted an area proportional to their individual painting rates.
6--A will have painted y square feet and B will have painted (c-y) square feet.
7--From the figure, A/c = x/y or Ay = cx.
8--Similarly, B/c = x/(c-y) or by = bc - cx.
9--From 7 & 8, y = cx/a = (bc - cx)/b from which x = AB/(A+B), one half of the harmonic mean of A and B.

I think this should give you enough of a clue as to how to solve your particular problem.