an irrigation pump can do a job 3 times faster than another. If the y work together it takes 9 hours to complete the job. How long would it take the larger pump by itself
takes time t for pump 1 the big one
takes time 3 t for pump 2
so
pump 1 does 1/t jobs/hr
pump 2 does 1/3t jobs/hr
(1/t + 1/3t) = 4/3t jobs/hr together
(4/3t jobs/hr)9hr = 1 job
12 hours = t
1/t + 1/(3t) = 1/9
To find the time it would take for the larger pump to complete the job by itself, we can first determine the rate at which each pump works if they were working alone.
Let's assume the smaller pump takes X hours to complete the job on its own. Since the larger pump can do the job 3 times faster than the smaller one, it would take the larger pump X/3 hours to complete the job by itself.
Next, let's calculate the combined work rate when both pumps are working together. The combined rate can be calculated using the formula: (a + b) = 1/t, where "a" and "b" are the individual work rates of the smaller and larger pumps, and "t" is the time taken when they work together.
Given that it takes 9 hours for both pumps to complete the job together, the combined rate is 1/9. Substituting the individual rates, we have: (1/X + 1/(X/3)) = 1/9.
To solve this equation, we can get rid of the denominator by multiplying both sides of the equation by X(X/3)(9). This yields: 9(X/3) + 9(X) = X(X/3)(9).
Simplifying further, we get: 3X + 9X = X^2. Combining like terms, we have: 12X = X^2.
Dividing both sides by X, we have: 12 = X.
Therefore, the smaller pump takes 12 hours to complete the job by itself.
Now, let's find the time it would take for the larger pump to complete the job alone. Since we determined earlier that the larger pump can do the job 3 times faster than the smaller pump, the larger pump would take 12/3 = 4 hours to complete the job by itself.