# Algebra Work Problem - it takes John 6 hours to complete a project and Peet 5 hours

if it takes john 6 hours to complete a project and peet 5 hours to complete a project how long will it take them if they work together

need formula

### Comments for Algebra Work Problem - it takes John 6 hours to complete a project and Peet 5 hours

 Jun 04, 2012 Algebra Work Problem by: Staff Part I Question: if it takes john 6 hours to complete a project and peet 5 hours to complete a project how long will it take them if they work together need formula Answer: The solution 1. Start with an equation showing the “Total Work Completed”: (work completed by John) + (work completed by Peet) = Total Work Completed For the sake of brevity: John_work + Peet_work = Total_work 2. THIS IS THE KEY. Convert the equation shown above into FRACTIONS. John_work + Peet_work = Total_work Divide each side of the equation by Total_work (John_work + Peet_work) / (Total_work) = (Total_work) / (Total_work) (John_work) / (Total_work) + (Peet_work) / (Total_work) = 1 As you can see, the fraction of the entire project completed by John + the fraction of the entire project completed by Peet = 1 Again, for the sake of brevity: John_fraction + Peet_fraction = 1 a. What is John’s fraction of the project? The portion of the project completed by John depends upon two things: 1) How fast John works, and 2) how long John works. How fast John works: Since John can complete the project in 6 hours; he completes 1/6 of the project per hour. John’s speed is 1/6 per hour. How long John works: time in hours = unknown = t₁ John’s fraction of the project = (John’s Speed) * (John’s time spent working) John_fraction = (1/6)*t₁ (explanation: if John works 1 hour, his fraction of the project completed is 1/6; if John works 2 hours, his fraction of the project completed is 2/6, and so on.) the equation now becomes: John_fraction + Peet_fraction = 1 (1/6)*t₁+ Peet_fraction = 1 b. What fraction of the project does Peet complete? The portion of the project completed by Peet depends upon two things: 1) How fast Peet works, and 2) how long Peet works. How fast Peet works: Since Peet can complete the project in 5 hours, he completes 1/5 of the project per hour. Peet’s speed is 1/5 per hour. How long Peet works: time in hours = unknown = t₂ Peet’s fraction of the project = Speed * time Peet_fraction = (1/5)*t₂ (explanation: if Peet works 1 hour, his fraction of the project completed is 1/5; if Peet works 2 hours, his fraction of the project completed is 2/5, and so on.) 3. the equation now becomes: John_fraction + Peet_fraction = 1 (1/6)*t₁ + (1/5)*t₂ = 1 Evaluation of t₁ and t₂ Both John and Peet work exactly the same amount of time. They begin the project together, and both continue working until the entire project is complete. t₁ = t₂ = t = total time to complete the project ------------------------------------------------------------

 Jun 04, 2012 Algebra Work Problem by: Staff ------------------------------------------------------------ Part II The equation now becomes: (1/6)*t₁ + (1/5)*t₂ = 1 (1/6)*t + (1/5)*t = 1 Solve for t (the total time to complete the project) (1/6)*t + (1/5)*t = 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 5 ------------------------------------------------ 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 5 = 5¹ - List every prime number listed as a factor of either 6 or 5. Prime Number Factors Listed: 2, 3, 5 - List these same prime factors again, but include the highest exponent listed for each Prime Number Factors with highest exponent: 2¹, 3¹,5¹ Multiply the factors with the highest exponent to compute the LCM: LCM = 2¹ * 3¹ * 5¹ = 30 ------------------------------------------------ Convert the denominator for both fractions on the left side of the equation to 30 (1/6)*t + (1/5)*t = 1 (5/5)*(1/6)*t + (6/6)*(1/5)*t = 1 (5*1*t)/(5*6) + (6*1*t)/(6*5)= 1 (5t) / (30) + (6t)/(30)= 1 Multiply both sides of the equation by 30 to remove the 30 as the denominator for the fractions on the left side of the equation 30 * [(5t) / (30) + (6t)/(30)] = 30 * 1 (5t) * (30) / (30) + (6t) * (30) / (30) = 30 * 1 (5t) * 1 + (6t) * 1 = 30 * 1 5t + 6t = 30 Combine like terms (5t + 6t) = 30 (11t) = 30 11t = 30 Divide each side of the equation by 11 11t = 30 11t / 11 = 30 / 11 t * (11 / 11) = 30 / 11 t * (1) = 30 / 11 t = 30 / 11 t = 2 8/11 hours t = 2 + (8/11)*60 t = 2 hours 43.6364 minutes >>>the final answer to the question is: t = 2 hours 44 minutes (rounded to the nearest minute) 4. check your answer by substituting 2 8/11 hours for t in the original equation: (1/6)*t + (1/5)*t = 1 (1/6)*(2 8/11 hours) + (1/5)*(2 8/11 hours) = 1 (1/6)*(30/11 hours) + (1/5)*(30/11 hours) = 1 (1/6)*(30/11 hours)*11 + (1/5)*(30/11 hours)*11 = 1*11 (1/6)*(30)*(11/11) hours) + (1/5)*(30)*(11/11 hours) = 11 (1/6)*(30)*(1) + (1/5)*(30)*(1) = 11 (30/6) + (30/5) = 11 (5) + (6) = 11 11 = 11, OK → t = 2 8/11 hours is a valid solution Thanks for writing. Staff www.solving-math-problems.com