SOLUTION: 1. pipe A can fill a tank in 5 h. pipe B can fill it in 2 h less time than it takes pipe C, a drainpipe, to empty the tank. with all three pipes open, it takes 3 h to fill the tank

Question 172726This question is from textbook Algebra and Trigonomety
: 1. pipe A can fill a tank in 5 h. pipe B can fill it in 2 h less time than it takes pipe C, a drainpipe, to empty the tank. with all three pipes open, it takes 3 h to fill the tank. how long would it take pipe C to empty it?
2. elvin drove halfway from Ashton to Dover at 40 mi/h and the rest of the way at 60 mi/h. what was his average speed for the whole trip? (Hint: let the distance for the whole trip be, say, 100 mi.)
3. the reciprocal of half a number increased by half the reciprocal of the number is 1/2. find the number.
4. find two positive numbers that differ by 8 and whose reciprocals differ by 1/6. This question is from textbook Algebra and Trigonomety

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

1)
Let x=time, in hours, that it takes pipe C to empty the tank
Then pipe C empties the tank at the rate of 1/x tank per hour
Pipe A fills at the rate of 1/5 tank per hour
We are told that it takes pipe B (x-2) hours to fill the tank
So pipe B fills at the rate of 1/(x-2) tank per hour
With all three pipes open, they fill at the rate of 1/3 tank per hour
So our equation to solve is:
1/5 +1/(x-2)-1/x=1/3 multiply each term by 15x(x-2)
3x(x-2)+15x-15(x-2)=5x(x-2) get rid of parens
3x^2-6x+15x-15x+30=5x^2-10x subtract 5x^2 from and add 10x to both sides
3x^2-5x^2-6x+15x-15x+30+10x=5x^2-5x^2-10x+10x collect like terms
-2x^2+4x+30=0 divide each term by -2
x^2-2x-15=0----------------------------quadratic in standard form and it can be factored
(x-5)(x+3)=0
x-5=0
x=5 hours------------------------------answer
and
x+3=0
x=-3-----------------------------disregard negative value; hours in this case are positive
CK
1/5+1/(5-2)-1/5=1/3
1/5+1/3-1/5=1/3
1/3=1/3
2)
Distance(d)=Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Average speed=(Total distance)/(Total time)
Let d=distance from Ashton to Dover (you can use any distance because the distance cancels out in the end. In fact, it might be a little easier to use, say, 100 mi)
Time required to drive the first half of the trip=(d/2)/40=d/80
Time required to drive the second half of the trip=(d/2)/60=d/120
Total time =d/80 + d/120=(3d+2d)/240=5d/240=d/48 hr
So, average speed=d/(d/48)=d*(48/d)=48 mi/hr-----------------ans
3)
Let x=the number
The reciprocal of the number is 1/x;
half the reciprocal of the number is 1/2x
Half the number =x/2
The reciprocal of half the number =2/x
So, our equation to solve is:
2/x (the reciprocal of half the number)+ 1/2x(half the reciprocal of the number) equals 1/2, or
2/x +1/2x = 1/2 multiply each term by 2x
4+1=x or
x=5
CK
reciprocal of half the number =2/5
half the reciprocal of the number=2/10
So, 2/5 + 2/10=1/2
4/10 + 1/10=1/2
1/2=1/2
4)
Let x= one number
1/x=reciprocal of the number
Then x-8=the other number
1/(x-8)= reciprocal
So, our equation to solve is:
1/(x-8) - 1/x=1/6---eq1
(note: although x>(x-8) the reciprocal of x is smaller than the reciprocal of (x-8))
Multiply each term in eq1 above by 6x(x-8)
6x-6(x-8)=x(x-8) get rid of parens
6x-6x+48=x^2-8x subtract x^2 from and add 8x to each side
6x-6x+48-x^2+8x=x^2-x^2-8x+8x collect like terms
-x^2+8x+48=0 multiply each term by -1
x^2-8x-48=0-------------------quadratic in standard form and it can be factored:
(x-12)(x+4)=0
x=12-----------------------one number
x=-4--------------------------disregard negative value
x-8=12-8=4------------------------the other number
CK
1/4-1/12=1/6
3/12-1/12=1/6
2/12=1/6
1/6=1/6
Does this help???---------------------ptaylor

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