Pool

If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?

Result

t1 =  19.54 h
t2 =  13.54 h

Solution:

1t1+1t2=18 t2=t1−6 1/t1+1/(t1−6)=1/8   8⋅ (x−6)+8⋅ x=(x−6)⋅ x −x2+22x−48=0 x2−22x+48=0  a=1;b=−22;c=48 D=b2−4ac=222−4⋅1⋅48=292 D>0  x1,2=−b±D2a=22±2922=22±2732 x1,2=11±8.54400374532 x1=19.5440037453 x2=2.45599625468   Factored form of the equation:  (x−19.5440037453)(x−2.45599625468)=0 t1>0 t1=x1=19.544≐19.544=19.54  h \dfrac{ 1 }{ t_{ 1 } } +\dfrac{ 1 }{ t_{ 2 } } = \dfrac{ 1 }{ 8 } \ \\ t_{ 2 } = t_{ 1 } - 6 \ \\ 1/t_{ 1 } + 1/(t_{ 1 }-6) = 1/8 \ \\ \ \\ \ \\ 8 \cdot \ (x-6) + 8 \cdot \ x = (x-6) \cdot \ x \ \\ -x^2 +22x -48 = 0 \ \\ x^2 -22x +48 = 0 \ \\ \ \\ a = 1; b = -22; c = 48 \ \\ D = b^2 - 4ac = 22^2 - 4\cdot 1 \cdot 48 = 292 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 22 \pm \sqrt{ 292 } }{ 2 } = \dfrac{ 22 \pm 2 \sqrt{ 73 } }{ 2 } \ \\ x_{1,2} = 11 \pm 8.54400374532 \ \\ x_{1} = 19.5440037453 \ \\ x_{2} = 2.45599625468 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -19.5440037453) (x -2.45599625468) = 0 \ \\ t_{ 1 }>0 \ \\ t_{ 1 } = x_{ 1 } = 19.544 \doteq 19.544 = 19.54 \ \text { h }

Checkout calculation with our calculator of quadratic equations.

t2>0 t2=t1−6=19.544−6=67750=13.54=13.54  h t_{ 2 }>0 \ \\ t_{ 2 } = t_{ 1 } - 6 = 19.544 - 6 = \dfrac{ 677 }{ 50 } = 13.54 = 13.54 \ \text { h }

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Showing 3 comments:
#
Math student
1/t1+1/(t1-10)=1/18
multiply each term by18(t1)(t1-10)
that results in
18(t1-10)+18t1=t1(t1)(t1)-10t1
using the quadratic formula results in t1=-49.6 and 3.63
ubless i made a mistake, your calculations need reexamination!!!        Correct me, please.

10 months ago  2 Likes
#
Dr Math
right side of equation is wrong - should be t1*(t1-10) = t12 - 10*t1 now t13-10t1

#
Math student
the problems seems to have changed - - - t2 is now equal t1-6

therefore 1/t1+1/(t1-6)=1/18
multiplying each term by18(t1)(t1-6) ==== 18(t1-6)+18t1=t1(t1-6), simplifying further 18t1-108+18t1=t12-6t1
or 0=t12-6t1-18t1+108
graphing y=18(t1-6)+18t1-t1(t1-6) results in t1=39.25 hours and t2=39.25-6=33.25 hours (same as your NEW answer!!!!

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