# SOLUTION: Sorry there is no ISBN of this textook. "Will you solve this Word problem please" 1. A, B, and C can finish a job in 6 days. If B & C work together, the job will take 9 days;

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Evaluation -> SOLUTION: Sorry there is no ISBN of this textook. "Will you solve this Word problem please" 1. A, B, and C can finish a job in 6 days. If B & C work together, the job will take 9 days;       Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Word Problems: Evaluation, Substitution Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Evaluation Word Problems Question 85129: Sorry there is no ISBN of this textook. "Will you solve this Word problem please" 1. A, B, and C can finish a job in 6 days. If B & C work together, the job will take 9 days; if A and C work together, the job will take 8 days. In how many days can each man working alone do the job? 2. A bicyclist and a hiker leave at the same place at the same time and travel in the same direction. The bicyclist travel three times as fast as the hiker. at the end of 3 hr they are 24 miles apart. how fast does the hiker travel. 3. A person travelling 4 hr by plane and 25 hr by ship covers 1580 miles. If the speed of the plane had been one-half of the actual speed and the speed of the ship had been one-fourth greater, the person would have travelled only 1315 miles in the same length of time. Find the speed of the plane and the ship. Thank you so much. Answer by ankor@dixie-net.com(15746)   (Show Source): You can put this solution on YOUR website!1. A, B, and C can finish a job in 6 days. If B & C work together, the job will take 9 days; if A and C work together, the job will take 8 days. In how many days can each man working alone do the job? : I was confident that the all solutions would be integers, but it was not the case with c; Here's what I got, using 3 equations: Let the completed job = 1 : Eq1: = 1 : Eq2: = 1 : Eq3: = 1 : Find b in terms of c using eq2: 9c + 9b = bc; multiplied thru by bc 9b - bc = -9c b(9-c) = -9c b = : Find a in terms of c using eq3: 8c + 8a = ac; multiplied thru by ac 8a - ac = -8c a(8-c) = -8c a = : Substitute for a and b in eq1 (invert and multiply the dividing fraction): 6 + 6 + = 1 : + + = 1 : Multiply thru by 72c, get rid of the denominators and you have: -9(48-6c) + -8(54-6c) + 72(6) = 72c : -432 + 54c - 432 + 48c + 432 = 72c : 54c + 48c - 72c - 432 - 432 + 432 = 0 : 30c - 432 = : 30c = +432 : c = 432/30 : c = 14.4 days required for c working alone : Find a using: a = a = a = -115.2/-6.4 a = +18 days for a to do the job alone : Find b using:b = b = b = -129.6/-5.4 b = +24 days for b : We should check the solutions in each original equation: eq1: = 1 = 1; I used a calc : Try checking solutions in eq2 and eq3 : : 2. A bicyclist and a hiker leave at the same place at the same time and travel in the same direction. The bicyclist travel three times as fast as the hiker. at the end of 3 hr they are 24 miles apart. how fast does the hiker travel. : Let s = hikers speed in mph Then 3s = bicyclist speed : Write a distance equation: dist = time * speed bicyclist dist - hiker dist = 24 mi 3(3s) - 3s = 24 9s - 3s = 24 6s = 24 s = 4 mph is the hikers speed, obviously the bicyclist speed the 12 mph : check 3(12) - 3(4) = 24 : : 3. A person traveling 4 hr by plane and 25 hr by ship covers 1580 miles. If the speed of the plane had been one-half of the actual speed and the speed of the ship had been one-fourth greater, the person would have traveled only 1315 miles in the same length of time. Find the speed of the plane and the ship. : let x = actual speed of the plane let y = actual speed of the ship : Dist = time * speed: actual: 4x + 25y = 1580 : If equation: .5(4x) + 1.25(25y) = 1315 2x + 31.25y = 1315 : Mult above equation by 2 and subtract the actual equation: 4x +62.5y = 2630 4x + 25y = 1580 ------------------subtracting eliminates x 0x + 37.5y = 1050 y = 1050/37.5 y = 28 mph, actual ship's speed : Find x using 2x + 31.25y = 1315 2x + 31.25(28) = 1315 2x + 875 = 1315 2x = 1315 - 875 x = 440/2 x = 220 mph, actual speed of the plane : Check solutions in the actual equation: 4(220) + 25(28) = 880 + 700 = 1580