logo for solving-math-problems.com
leftimage for solving-math-problems.com

Painting A Room - Work Problem

by Andrew
(New York)











































It looked like an easy problem when I initially saw it. However, I simply have no clue how to even start it.

Working alone, Roger can paint a room in 6 hours. Working alone, Chris can paint the
same room in 8 hours. If they begin the job together, and Chris takes an hour off for
lunch, determine the amount of time, in hours and minutes, that it will take for the room
to be painted. Round your answer to the nearest minute.

Comments for Painting A Room - Work Problem

Click here to add your own comments

Sep 25, 2011
Painting a Room - Work Problem
by: Staff


------------------------------------------------------------

Part II

substitute t for t₁ and (t – 1) for t₂

(1/6)*t₁ + (1/8)*t₂ = 1

(1/6)*t + (1/8)*(t - 1) = 1



Solve for t (the total time to complete the job)

(1/6)*t + (1/8)*(t - 1) = 1


Use the distributive law to expand (1/8)*(t - 1)

(1/6)*t + (1/8)*t - (1/8) = 1


Convert each fraction on the left side of the equation to the same common denominator.

The new common denominator can be the least common multiple (LCM) of the two original denominators: 6 and 8

------------------------------------------------
To find the LCM, you can use the prime factorization method

- Factor the numbers 6 and 8 into prime factors. Use exponents to show how many times each prime factor is listed.

6 = 2¹*3¹
8 = 2³

- List every prime number listed as a factor of either 6 or 8.

Prime Number Factors Listed: 2, 3

- List these same prime factors again, but include the highest exponent listed for each
Prime Number Factors with highest exponent: 2³, 3¹

Multiply the factors with the highest exponent to compute the LCM:

LCM = 2³ * 3¹ = 24
------------------------------------------------


Convert the denominator for all three fractions on the left side of the equation to 24


(1/6)*t + (1/8)*t - (1/8) = 1


(4/4)*(1/6)*t + (3/3)*(1/8)*t - (3/3)*(1/8) = 1

(4*1*t)/(4*6) + (3*1*t)/(3*8) - (3*1)/(3*8) = 1

(4t) / (24) + (3t)/(24) - (3) / (24) = 1


Multiply both sides of the equation by 24 to remove the 24 as the denominator for the fractions on the left side of the equation


24 * [(4t) / (24) + (3t)/(24) - (3) / (24)] = 24 * 1

(4t) * (24) / (24) + (3t) * (24) / (24) - (3) * (24) / (24) = 24 * 1

(4t) * 1 + (3t) * 1 - (3) * 1 = 24 * 1

4t + 3t - 3 = 24


Combine like terms

(4t + 3t) - 3 = 24

(7t) - 3 = 24

7t - 3 = 24


Add 3 to each side of the equation

7t - 3 + 3 = 24 + 3

7t + 0 = 24 + 3

7t = 24 + 3

7t = 27


Divide each side of the equation by 7

7t = 27

7t / 7 = 27 / 7

t * (7 / 7) = 27 / 7

t * (1) = 27 / 7

t = 27 / 7

t = 3 6/7 hours

t = 3 + (6/7)*60

t = 3 hours 51.4286 minutes

the final answer to the question is: t = 3 hours 51 minutes (rounded to the nearest minute)



4. check your answer by substituting 3 6/7 hours for t in the original equation:


(1/6)*t + (1/8)*(t - 1) = 1

(1/6)*(3 6/7 hours) + (1/8)*( 3 6/7 hours - 1 hour) = 1

(1/6)*(3 6/7 hours) + (1/8)*( 2 6/7 hours ) = 1

(1/6)*(27/7 hours) + (1/8)*( 20/7 hours ) = 1

(1/6)*(27/7 hours)*7 + (1/8)*( 20/7 hours )*7 = 1*7

(1/6)*(27)*(7/7) + (1/8)*( 20)*(7/7) = 1*7

(1/6)*(27)*(1) + (1/8)*( 20)*(1) = 1*7

(27/6) + (20/8) = 7

(27/6) * 24 + (20/8)* 24 = 7 * 24

(27) * (24/6) + (20) * (24/8) = 7 * 24

(27) * (4) + (20) * (3) = 7 * 24

108 + 60 = 168

168 = 168, OK → t = 3 6/7 hours is a valid solution





Thanks for writing.

Staff
www.solving-math-problems.com


Sep 25, 2011
Painting a Room - Work Problem
by: Staff


Part I

The question:

by Andrew
(New York)


It looked like an easy problem when I initially saw it. However, I simply have no clue how to even start it.

Working alone, Roger can paint a room in 6 hours. Working alone, Chris can paint the
same room in 8 hours. If they begin the job together, and Chris takes an hour off for
lunch, determine the amount of time, in hours and minutes, that it will take for the room
to be painted. Round your answer to the nearest minute.


The answer:


The solution


1. Start with an equation showing the “Total Work Completed”:

(work completed by Roger) + (work completed by Chris) = Total Work Completed


For the sake of brevity:

Roger_work + Chris_work = Total_work


2. THIS IS THE KEY. Convert the equation shown above into FRACTIONS.

Roger_work + Chris_work = Total_work


Divide each side of the equation by Total_work

(Roger_work + Chris_work) / (Total_work) = (Total_work) / (Total_work)

(Roger_work) / (Total_work) + (Chris_work) / (Total_work) = 1


As you can see, the fraction of the entire job completed by Roger + the fraction of the entire job completed by Chris = 1

Again, for the sake of brevity:

Roger_fraction + Chris_fraction = 1


a. What is Roger’s fraction of the job?

The portion of the job completed by Roger depends upon two things: 1) How fast Roger works, and 2) how long Roger works.

How fast Roger works: Since Roger can complete the job in 6 hours; he completes 1/6 of the job per hour. Roger’s speed is 1/6 per hour.

How long Roger works: time in hours = unknown = t₁

Roger’s fraction of the job = (Roger’s Speed) * (Roger’s time spent working)

Roger_fraction = (1/6)*t₁

(explanation: if Roger works 1 hour, his fraction of the job completed is 1/6; if Roger works 2 hours, his fraction of the job completed is 2/6, and so on.)

b. What fraction of the job does Chris complete?

The portion of the job completed by Chris depends upon two things: 1) How fast Chris works, and 2) how long Chris works.

How fast Chris works: Since Chris can complete the job in 8 hours, he completes 1/8 of the job per hour. Chris’s speed is 1/8 per hour.

How long Chris works: time in hours = unknown = t₂

Chris’s fraction of the job = Speed * time

Chris_fraction = (1/8)*t₂

(explanation: if Chris works 1 hour, his fraction of the job completed is 1/8; if Chris works 2 hours, his fraction of the job completed is 2/8, and so on.)


3. the equation now becomes:

Roger_fraction + Chris_fraction = 1

(1/6)*t₁ + (1/8)*t₂ = 1


Evaluation of t₁ and t₂

They begin the job together. Roger works the entire time, but Chris takes an hour off for
lunch.

t₁ = total time to complete job (Roger does not take a break) = t

t₂ = total time to complete job - 1 hour (Chris takes a 1 hour break for lunch) = t – 1

------------------------------------------------------------

Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.



Copyright © 2008-2015. All rights reserved. Solving-Math-Problems.com