SOLUTION: Hi Water is leaking fron a tank at a constant rate . At 230pm the tank is 3/5 full .At 325pm the tank was 2/7 full. At what time will the tank be empty. Thanks

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Question 1206959: Hi
Water is leaking fron a tank at a constant rate . At 230pm the tank is 3/5 full .At 325pm the tank was 2/7 full. At what time will the tank be empty.
Thanks
Found 5 solutions by mananth, math_tutor2020, ikleyn, josgarithmetic, Edwin McCravy:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
ater is leaking fron a tank at a constant rate . At 230pm the tank is 3/5 full .At 325pm the tank was 2/7 full. At what time will the tank be empty.
from 2,30 pm to 3.25 pm 55 minutes
In 55 minutes tank drained = 3/5 -2/7 Take LCM
= 21/35 - 10/35 = 11/35
11/35 of the tank drained in 55 minutes
1 tank draining will take (35/11)*55=175 minutes
175 minutes = 2 hours 55 minutes
you can find the actual time by adding this time
Thanks tutor @tutor2020 to point out my error. noted.

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 4:15 PM

Explanation

x = number of minutes after 2:30 PM
y = fraction representing how full the tank is.
y = 0 means the tank is completely empty.
y = 1 means the tank is completely full.

The water is leaking at a constant rate. Which will mean we can use a linear equation.

We have the two points on this line which are (0,3/5) and (55, 2/7)
The 55 refers to 55 minutes after 2:30 PM to arrive at 3:25 PM.

Note: The jump from 2:30 to 3:30 is 1 hour, aka 60 minutes, so we subtract off 5 minutes to go from 2:30 to arrive at 3:25.

I'll skip a few steps but the equation of the line through those points mentioned is y = (-1/175)x + 3/5

We want to know when the tank will be empty, so we want to determine x when y = 0.

y = (-1/175)x + 3/5
0 = (-1/175)x + 3/5
(1/175)x = 3/5
x = 175*(3/5)
x = 105
It takes 105 minutes to go from 3/5 full to completely empty.

105 min = 60 min + 45 min
105 min = 1 hr + 45 min

Now add this duration to 2:30 to see where we end up.
From 2:30 to 3:30 is 1 hour.
From 3:30 to 4:00 is 30 minutes.
From 4:00 to 4:15 is another 15 minutes (30+15 = 45 minutes total)

Or...
The jump from 2:30 PM to 4:30 PM is 2 hours
Rewind the clock 15 minutes to go from 4:30 PM to 4:15 PM
Rewinding those 15 minutes means the 2 hour duration is adjusted to 1 hour+45 minutes.


Therefore, the time gap from 2:30 PM to 4:15 PM is 1 hour+45 minutes aka 105 minutes.

--------------------------------------------------------------------------

Tutor @mananth made a mistake.
The tank doesn't start off 100% full, so we don't need to drain 100% of the contents.
Instead we need to drain 3/5 = 0.6 = 60% of the contents.

If you solved (11/35)/55 = 1/x then you'll get x = 175
While this is a correct solution to this equation, the equation itself is set up wrong.
The "1" on the right hand side should be 3/5 to represent the fractional amount we need to drain.
The "1" on the right hand side represents 100% full.

If you solved (11/35)/55 = (3/5)/x, then you'll get x = 105 which then leads to 4:15 PM as mentioned above.

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
Water is leaking fron a tank at a constant rate. At 230pm the tank is 3/5 full.
At 325pm the tank was 2/7 full. At what time will the tank be empty.
~~~~~~~~~~~~~~~~~~~~~~ 

The volume of water leaked from the tank from 2:30 pm to 3:25 pm is the difference

      3%2F5 - 2%2F7 = simplify using common denominator = %283%2A7%29%2F35+-+%282%2A5%29%2F35 = 21%2F35+-+10%2F35 = 11%2F35


of the total tank volume.  The time elapsed from 2:30 pm to 3:25 pm is 55 minutes.


So, every 5 minutes, the leaking volume is  1%2F35  of the tank volume.


At 3:25 pm the tank is/was  2%2F7 = 10%2F35  full.


Hence, the tank will be empty in 5*10 = 50 minutes after 3:25 pm, which is 4:15 pm.    ANSWER

Solved.

The solution requires a bit of manipulations with fractions,  a bit of common sense,
basic skills of understanding and manipulating with reading clock face,  and a bit of accuracy.

Nothing more.   Even using equations is not necessary.



Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
x, minutes of time past 2:30 pm
y, amount of tank full

Points are
2:30 pm, (0, 3/5)
3:25 pm, (55, 2/7)

Equation can start as y=%28%282%2F7-3%2F5%29%2F55%29x%2B3%2F5
.
steps lead to
.
y=-x%2F%285%2A35%29%2B3%2F5.

For how much time, x, past 2:30 pm is tank empty, y=0?
x%2F%283%2A35%29=3%2F5

x=105 minutes
which is 1 hour 45 minutes...

Add that amount of time to 2:30 pm on the clock.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
To make the fractions easy to use, suppose the tank holds 35 gallons, or
35 of any unit of volume you choose.

At 2:30, the tank contains 21 gallons. (3/5 of 35)
55 minutes later (at 3:25) the tank contains 10 gallons. (2/7 of 35).

So it leaked out 11 gallons in 55 minutes. (21-10)
How long will it take to leak out those remaining 10 gallons?

Set up the proportion: (gallons over minutes equals gallons over minutes):

matrix%281%2C2%2C11%2Cgallons%29%2Fmatrix%281%2C2%2C55%2Cminutes%29%22%22=%22%22matrix%281%2C2%2C10%2Cgallons%29%2Fmatrix%281%2C2%2Cx%2Cminutes%29

11x%22%22=%22%22550

x%22%22=%22%2250

So 50 minutes after 3:25 is 4:15. 

----------------------------------------

Here's an alternate way to get the answer without changing any of the
original information (not the easiest way! LOL)

Let x = the number of hours after 12 noon or midnight
Let y = the fraction of fullness the tank is at time x.

We want to know what x is when y = 0.

When its 2%2630%2F60 hours past 12 noon or midnight, it's 3%2F5ths full.
When its 3%2625%2F60 hours past 12 noon or midnight, it's 2%2F7ths full.

So we want the equation of the line through the points:

%28matrix%281%2C3%2C2%2630%2F60%2C%22%2C%22%2C3%2F5%29%29 and %28matrix%281%2C3%2C3%2625%2F60%2C%22%2C%22%2C2%2F7%29%29
which is
%28matrix%281%2C3%2C2%261%2F2%2C%22%2C%22%2C3%2F5%29%29 and %28matrix%281%2C3%2C3%265%2F12%2C%22%2C%22%2C2%2F7%29%29
which is
%28matrix%281%2C3%2C5%2F2%2C%22%2C%22%2C3%2F5%29%29 and %28matrix%281%2C3%2C41%2F12%2C%22%2C%22%2C2%2F7%29%29

Slope formula:
m%22%22=%22%22%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29
where (x1,y1) = %28matrix%281%2C3%2C5%2F2%2C%22%2C%22%2C3%2F5%29%29
and where (x2,y2) = %28matrix%281%2C3%2C41%2F12%2C%22%2C%22%2C2%2F7%29%29

m%22%22=%22%22%282%2F7-3%2F5%29%2F%2841%2F12-5%2F2%29
Multiply top and bottom by 12%2A7%2A5=420

m%22%22=%22%22%28120-252%29%2F%281435-1050%29%22%22=%22%22%28-132%29%2F385%29%22%22=%22%22-12%2F35

Point-slope formula:
y-y%5B1%5D%22%22=%22%22m%28x-x%5B1%5D%29
where (x1,y1) = %22%22=%22%22%28matrix%281%2C3%2C5%2F2%2C%22%2C%22%2C3%2F5%29%29
y-3%2F5%22%22=%22%22expr%28-12%2F35%29%28x-5%2F2%29

Substituting 0 for y, to find when tank is empty:

-3%2F5%22%22=%22%22expr%28-12%2F35%29%28x-5%2F2%29

Multiply through by 35
-21%22%22=%22%22-12%28x-5%2F2%29

-21%22%22=%22%22-12x%2B30%29

-51%22%22=%22%22-12x%29

%28-51%29%2F%28-12%29%22%22=%22%22x

17%2F4%22%22=%22%22x

4%261%2F4%22%22=%22%22x

So the tank became empty at 4 1/4 hours after 12 noon or midnight.

1/4 of an hour is 15 minutes

So the instant the tank became empty was at 4:15 PM or AM

Edwin