Discouraged by word problems: work probs, length, etc

monica13

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Oct 6, 2006
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:cry: It has been along time since i have done these types of problems but i am studying for a nusing entrance exam and got stumbled..

1. Steven can mix 20 drinks in 5 min, Sue can mix 20 drinks in 10 min., and Jack can mix 20 drinks in 15min. How much time will it take if they all 3 work together to mix 20 drinks?

I think that the problem is set like this: 5x + 10x + 15x =20 but i am clueless

2. If Sam can do a job in 4 days that Lisa does in 6 days and Tom can do in 2 days how long would it take then if they worked together?

I have no idea how to start this.

3. Jim has 5 strings and his belt broke. His waist is 36 inches so the string needs to be at least 4 inches longer than his waist but no more than 6 inches more. What is the best answer?
3 4/5, 3 2/3, 3 3/8, 3 1/4, 2 1/2

I think it is 3 3/8 which = 40.5 buit i dont remember how i got the answer.

I reeally appreciate the help, i just need some refreshing and this is in the study guide online for my entrance exam but no way to see how they arrived at the answer! crazy huh?
 
Re: Discouraged, help!!

monica13 said:
:cry: It has been along time since i have done these types of problems but i am studying for a nusing entrance exam and got stumbled..
1. Steven can mix 20 drinks in 5 min,Sue can mix 20 drinks in 10 min., Jack can mix 20 drinks in 15min. How much time will it take if they all 3 work together to mix 20 drinks?

I think that the problem is set like this: 5x + 10x + 15x =20 but i am clueless

2. If Sam can do a job in 4 days that Lisa does in 6 days and Tom can do in 2 days how long would it take then if they worked together.

I have no idea how to start this.
This should enable you to solve your combined work problems.

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*****x****** ....................................{
..........I............./....*................................(c)
..........I(c-y)..../.........*...............................{
..........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.


Three people version

It takes Alan and Carl 40 hours to paint a house, Bill and Carl 80 hours to paint the house, and Alan and Bill 60 hours to paint the house. How long, to the nearest minute, will it take each working alone to paint the house and how long will it take all three of them working together to paint the house?

1--The combined time of two efforts is derived from one half the harmonic mean of the two individual times or Tc = AB/(A + B), A and B being the individual times of each participant.
2--Therefore, we can write
AC/(A + C) = 40 or AC = 40A + 40C (a)
BC/(B + C) = 80 or BC = 80B + 80C (b)
AB/(A + B) = 60 or AB = 60A + 60B (c)
3--From (a) and (c), 40C/(C - 40) = 60B/(B - 60)
4--Cross multiplying, 40BC - 2400C = 60BC - 2400B or BC = 120(B - C)
5--Equating to (b) yields 120(B - C) = 80(B + C)
6--Expanding and simplifying, 40B = 200C or B = 5C
7--Substituting into (b), 5C^2 = 400C + 80C = 480C making 5C = 480 or C = 96.
8--Therefore, B = 480 and A = 68.571
9--The combined working time of three individual efforts is derived from Tc = ABC/(AB + AC + BC)
10--Therefore, the combined time for all three to paint the house is
Tc = 68.571(480)96/[(68.571x480) + (68.571x96) + 480x96)) = 36.923 hours = 36 hr - 55.377 min
 
monica13 said:
1. Steven can mix 20 drinks in 5 min, Sue can mix 20 drinks in 10 min., and Jack can mix 20 drinks in 15min. How much time will it take if they all 3 work together to mix 20 drinks?
In 1 minute, Steven does 1/5 of job, Sue does 1/10 and Jack does 1/15;
so in 1 minute, 1/5 + 1/10 + 1/15 = 11/30 of job gets done:
so full job gets done in 30/11 = 2 8/11 minutes.

Other one works the same way.

I don't understand your belt problem: did you type it ALL?
Seems to be information missing...
 
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