# Factory Work Flow

by Andrew
(New York)

The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. In two hours, 60 workers can produce 240 widgets and 300 whoosits. In three hours, 50 workers can produce 150 widgets and m whoosits. Find m.

### Comments for Factory Work Flow

 Jan 08, 2012 Factory Work Flow by: Staff --------------------------------------------------------- Part II To begin solving for m, solve equations 1 and 2 to determine the values of x and y. 300x + 200y = 100 240x + 300y = 120 Multiply both sides of equation 2 by 2/3 (2/3) * (240x + 300y) = (2/3) * (120) (2/3) * (240x) + (2/3) * (300y) = (2/3) * (120) 160x + 200y = 80 Subtract the modified version of equation 2 from equation 1 300x + 200y = 100 -(160x + 200y = 80) ------------------------- 300x - 160x + 200y -200y = 100 - 80 (300x - 160x) + (200y -200y) = 100 - 80 (140x) + (0) = 20 140x + 0 = 20 140x = 20 Divide each side of the equation by 140 140x/140) = 20/140 x*(140/140) = 20/140 x*(1) = 20/140 x = 20/140 x = 1/7 (it takes 1/7 of a man-hour to produce 1 widget) substitute 1/7 for x in either equation 1 or equation 2, and then solve for y 240x + 300y = 120 240*(1/7) + 300y = 120 Multiply each side of the equation by 7 7*[240*(1/7) + 300y] = 7*120 7*[240*(1/7)] + 7*[300y] = 7*120 240 + 2100y = 840 Subtract 240 from each side of the equation 240 - 240 + 2100y = 840 - 240 0 + 2100y = 600 2100y = 600 Divide each side of the equation by 2100 2100y/2100 = 600/2100 y*(2100/2100) = 600/2100 y*(1) = 600/2100 y = 600/2100 y = 2/7 (it takes 2/7 of a man-hour to produce 1 whoosit) Solve for m using equation 3. Substitute 1/7 for x and 2/7 for y. 150x + my = 150 150*(1/7) + m*(2/7) = 150 Multiply each side of the equation by 7 7*[150*(1/7) + m*(2/7)] = 7*150 7*[150*(1/7)] + 7*[m*(2/7)] = 7*150 150*(7/7) + *[m*2*(7/7)] = 7*150 150*(1) + [m*2*(1)] = 7*150 150 + 2m = 1050 Subtract 150 from each side of the equation 150 - 150 + 2m = 1050 - 150 0 + 2m = 900 2m = 900 Divide each side of the equation by 2 2m/2 = 900/2 m*(2/2) = 900/2 m*(1) = 450 m = 450 >>>>>>>>>>>>>>> The final answer is: m = 450 whoosits Thanks for writing. Staff www.solving-math-problems.com

 Jan 08, 2012 Factory Work Flow by: Staff Part I Question: by Andrew (New York) The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. In two hours, 60 workers can produce 240 widgets and 300 whoosits. In three hours, 50 workers can produce 150 widgets and m whoosits. Find m. Answer: The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. The total number of man-hours used by 100 workers working for 1 hour = 100 man-hours These 100 man-hours are divided between the production of widgets and whoosits. (Man-hours used for widget Production) + (Man-hours used for whoosits Production) = 100 x = the number of hours it takes to produce 1 widget (hours per widget). y = the number of hours it takes to produce 1 woosit (hours per woosit). Therefore, the division of the hours between widgets and woosits can be expressed as the following equation: 300x + 200y = 100 In two hours, 60 workers can produce 240 widgets and 300 whoosits. The total number of man-hours used by 60 workers working for 2 hours = 2*60 = 120 man-hours The division of the hours between widgets and woosits can be expressed as the following equation: 240x + 300y = 120 In three hours, 50 workers can produce 150 widgets and m whoosits. Find m. The total number of man-hours used by 50 workers working for 2 hours = 3*50 = 150 man-hours The division of the hours between widgets and woosits can be expressed as the following equation: 150x + my = 150 The system of three equations which must be solved is: Equation 1: 300x + 200y = 100 Equation 2: 240x + 300y = 120 Equation 3: 150x + my = 150 ---------------------------------------------------------