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How do I solve this problem?
- Thread starter ms1981
- Start date
If Sally can paint a house in 4 hours, and John can paint the same house in hour 6 hour, how long will it take for both of them to paint the house together?
With different numbers:
If it takes one person 5 hours to paint a room and another person 3 hours, how long will it take to paint the room working together?
Method 1:
1--A can paint a room in 5 hours.
2--B can paint a room in 3 hours.
3--A's rate of painting is 1 room per A hours (5 hours) or 1/A (1/5) room/hour.
4--B's rate of painting is 1 room per B hours (3 hours) or 1/B (1/3) room/hour.
5--Their combined rate of painting is therefore 1/A + 1/B = (A+B)/AB = (1/5 + 1/3) = (8/15) rooms /hour.
6--Therefore, the time required for both of them to paint the 1 room working together is 1 room/(A+B)/AB rooms/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.
Note - Generally speaking (if the derivation is not specifically required), if it takes one person A units of time and another person B units of time to complete a specific task working alone, the time it takes them both to complete the task working together is T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.
You might like to derive the equivalant expression involving 3 people working alone and together which results in T = ABC/(AB + AC + BC).
Method 2:
Consider the following diagram -
..........I<----------B---------->I
..........I_______________I_________________
..........I........................../........................../\
..........I..*...................../............................I
..........I.....*................/..............................I
..........Iy.......*........../................................I
..........I................../..................................I
..........I*****x******/ ...................................I
..........I............./....*................................(c)
..........I(c-y)..../.........*...............................I
..........I......../...............*...........................I.
..........I....../....................*........................I
..........I..../.........................*.....................I
..........I../..............................*..................l
.........I./...................................*...............\/__
.........I<-----------------A-------------->I
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.
With different numbers:
If it takes one person 5 hours to paint a room and another person 3 hours, how long will it take to paint the room working together?
Method 1:
1--A can paint a room in 5 hours.
2--B can paint a room in 3 hours.
3--A's rate of painting is 1 room per A hours (5 hours) or 1/A (1/5) room/hour.
4--B's rate of painting is 1 room per B hours (3 hours) or 1/B (1/3) room/hour.
5--Their combined rate of painting is therefore 1/A + 1/B = (A+B)/AB = (1/5 + 1/3) = (8/15) rooms /hour.
6--Therefore, the time required for both of them to paint the 1 room working together is 1 room/(A+B)/AB rooms/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.
Note - Generally speaking (if the derivation is not specifically required), if it takes one person A units of time and another person B units of time to complete a specific task working alone, the time it takes them both to complete the task working together is T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.
You might like to derive the equivalant expression involving 3 people working alone and together which results in T = ABC/(AB + AC + BC).
Method 2:
Consider the following diagram -
..........I<----------B---------->I
..........I_______________I_________________
..........I........................../........................../\
..........I..*...................../............................I
..........I.....*................/..............................I
..........Iy.......*........../................................I
..........I................../..................................I
..........I*****x******/ ...................................I
..........I............./....*................................(c)
..........I(c-y)..../.........*...............................I
..........I......../...............*...........................I.
..........I....../....................*........................I
..........I..../.........................*.....................I
..........I../..............................*..................l
.........I./...................................*...............\/__
.........I<-----------------A-------------->I
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.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Sally \ and \ John: \ \frac{x}{4}+\frac{x}{6} \ = \ 1, \ x \ = \ 2\frac{2}{5} \ hrs.\)
\(\displaystyle Method \ Three: \ \frac{x}{5}+\frac{x}{3} \ = \ 1, \ x \ = \ 1\frac{7}{8} \ hrs.\)
\(\displaystyle Method \ Three: \ \frac{x}{5}+\frac{x}{3} \ = \ 1, \ x \ = \ 1\frac{7}{8} \ hrs.\)
takariendire
New member
- Joined
- Jun 9, 2010
- Messages
- 7
I recorded an answer to your question with BlueTeach :
http://www.blueteach.com/v/P4NRY8Q9F5w3LfrwW2
Hope this helps,
cheers
http://www.blueteach.com/v/P4NRY8Q9F5w3LfrwW2
Hope this helps,
cheers
tTwo cars leave town at the same time traveling in opposite directions.One car travels at a rate of 60 miles per hour.The other car travels at a rate of 71 miles per hour.How long will it take them to arive at a point where they are 650 miles apart
D
Deleted member 4993
Guest
takoria14 said:tTwo cars leave town at the same time traveling in opposite directions.One car travels at a rate of 60 miles per hour.The other car travels at a rate of 71 miles per hour.How long will it take them to arive at a point where they are 650 miles apart
Are you sure it is trevelling at 71 mph and not 70 mph?
The path of a baseball is given by f(x)=-.0032x^2+x+3; where f(x) is the height, in feet, of the baseball and x is the horizontal distance, in feet, from home plate.
a. How far from home plate is the ball when it hits the ground?
b. What is the height of the ball as it passes over the pitcher's mound, which is about 64 ft. from home plate?
c. What is the maximum height of the baseball?
please help and i need step by step....thanks
a. How far from home plate is the ball when it hits the ground?
b. What is the height of the ball as it passes over the pitcher's mound, which is about 64 ft. from home plate?
c. What is the maximum height of the baseball?
please help and i need step by step....thanks