7 machines at a certain factory operate at the same constant rate . if 6 of these machines , operating simultaneously , take 42 hours to fill a certain production order , how many fewer hours does it take all 7 machines , operating simultaneously , to fill the same production order ?
the total work is 6 * 42 = 252 machine hours the time required for seven machines is 252 / 7 = 36 hours , thus 6 fewer hours . the answer is d .
subtract(42, divide(multiply(6, 42), 7))