Work Word Problems | MathHelp.com
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McCauley can paint a house in 10 hours, while it takes Clayton 15 hours. If they work together, how long will it take them to paint the house? To solve this kind of a problem, which is called a work problem, it’s important to understand the following idea. Since McCauley can paint a house in 10 hours, we know that in 1 hour, McCauley can paint 1/10 of the house. And in 2 hours, McCauley can paint 2/10 of the house. Therefore, in t hours, McCauley can paint t tenths of the house. And since it takes Clayton 15 hours to paint the house, in t hours, Clayton can paint t fifteenths of the house. Pause the audio for a moment if you need time to understand this idea…Now, to solve the problem, we use the following formula: part of job done by McCauley + part of job done by Clayton = 1 job done. And we’re asked how long will it take them to paint the house, so we’re looking for the time, or t. Remember that in t hours, McCauley can paint t/10 of the house, so the part of the job done by McCauley is t over 10. And in t hours, Clayton can paint t/15 of the house, so the part of the job done by Clayton is t over 15. Now, we have the equation t/10 + t/15 = 1. To solve this equation for t, we first get rid of the fractions by multiplying both sides of the equation by the common denominator of 10 and 15, which is 30. Distributing on the left side, 30 times t over 10 is 30t over 10, which simplifies to 3t, and 30 times positive t over 15 is positive 30t over 15, which simplifies to positive 2t. And on the right, 1 times 30 is 30. So we have 3t + 2t = 30, or 5t = 30, and dividing both sides by 5, t = 6. So if Clayton and McCauley work together, they can paint the house in 6 hours. Finally, it’s a good idea to check your answer. If they work together for 6 hours, then McCauley paints 6/10 of the house, and Clayton paints 6/15 of the house, so we have 6/10 + 6/15 = 1. And reducing on the left side, we have 3/5 + 2/5 = 1, which simplifies to 5/5 = 1, which is a true statement, so our answer checks
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