Problem: running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 1 w widgets ?
Rationale: let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 1 * 6 = 6 days to produce 1 w widgets . answer : b
annotated_formula: divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2)