Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
I'll start you off...
Define your variables: I'd call the bigger pump $\displaystyle b$ and the smaller pump $\displaystyle s$
$\displaystyle b=$the number of tanks per hour the bigger pump can empty
$\displaystyle s=$the number of tanks per hour the smaller pump can empty
Both of those numbers will be fractions.
Now, since you only wanted a hint, I'm only going to give you the first equation and you'll have to find the rest...
$\displaystyle b+s=\frac{1\text{tank}}{5\text{hours}}$Two pumps of different sizes working together can empty a fuel tank is 5 hours.
Do you need any more help?
The next equation is somewhat weird.
If $\displaystyle b=\frac{\text{tanks}}{\text{hours}}$
than $\displaystyle \frac{1}{b}=\frac{\text{hours}}{\text{tanks}}$
So in fact:
$\displaystyle \frac{1}{b}=$the number of hours to empty a tank
So we know that: $\displaystyle \frac{1}{b}-\frac{1}{s}=4$
Hello, blueridge!
Here's another approach . . .
Together, they can do the job in 5 hours.Two pumps of different sizes working together can empty a fuel tank is 5 hours.
The larger pump can empty this tank in 4 hours less than the smaller one.
If the larger one is out of order, how long will it take the smaller one to do the job alone?
. . In one hour, they can do $\displaystyle \frac{1}{5}$ of the job. .[1]
The smaller pump can do the job in $\displaystyle x$ hours. .[Note that: .$\displaystyle x > 4$.]
. . In one hour, it can do $\displaystyle \frac{1}{x}$ of the job.
The larger pump takes 4 hours less; it takes $\displaystyle x - 4$ hours.
. . In one hour, it can do $\displaystyle \frac{1}{x-4}$ of the job.
Together, in one hour, they can do: .$\displaystyle \frac{1}{x} + \frac{1}{x-4}$ of the job. .[2]
But [1] and [2] describe the same thing:
. . the fraction of the job done in one hour.
There is our equaton! . . . .$\displaystyle \boxed{\frac{1}{x} + \frac{1}{x-4} \:=\:\frac{1}{5}}$
Multiply by the common denominator: $\displaystyle 5x(x - 4)$
. . $\displaystyle 5(x - 4) + 5x \:=\:x(x-4)$
. . which simplifies to the quadratic: .$\displaystyle x^2 - 14x + 20 \:=\:0$
The Quadratic Formula gives us: .$\displaystyle x \;=\;\frac{14\pm\sqrt{116}}{2} \;=\;7 \pm\sqrt{29} \;\approx\;\{1.6,\:12.4\}$
Since $\displaystyle x > 4$. the solution is: .$\displaystyle x = 12.4$
Therefore, the smaller pump will take about 12.4 hours working alone.