Fractional Equations????

dunit0001

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1) An intake pipe can fill a certain tank in 6 hours when the outlet pipe is closed, but with the outlet pipe open it takes 9 hours. How long would it take the outlet pipe to fill an empty tank?

2) Members of the ski club contributed equally to obtain $1800 for a holiday trip. When 6 members found that they could not go, their contributions were refunded and each remaining member then had to pay $10 more to raise the $1800. How many went on the trip?

3) Because of traffic Maria could average only 40 km/h for the first 20% of her trip, but she averaged 75 km/h for the whole trip. What was her average speed for the last 80% of her trip?

4) A number x is the harmonic mean of a and b if 1/x is the average of 1/a and 1/b. Find two positive numbers that differ by 12 and have harmonic mean 5

I don't know how to do these please help?? I am really clueless on how to do these.....

I don't know where to start I know I need a variable but I don't know how to make it all work
 
3) Because of traffic Maria could average only 40 km/h for the first 20% of her trip, but she averaged 75 km/h for the whole trip. What was her average speed for the last 80% of her trip?

as this is my first time making a reply on this forum, someone correct me if i'm wrong please.

We know Maria was averaging 40 km/h for the first 20% of her trip. [40 x 20%]
We know that she averaged x km/h for the remaining 80% of her trip. [x x 80%]
We know those two numbers together give us 75 km/h.

so we add those figures:
(40 x 0.2)+(x x 0.8) = 75 km/h
8 + 0.8x = 75 km/h
-8 + 8 + 0.8x = 75 - 8
0.8x = 67
 
dunit0001 said:
1) An intake pipe can fill a certain tank in 6 hours when the outlet pipe is closed, but with the outlet pipe open it takes 9 hours. How long would it take the outlet pipe to fill an empty tank?

2) Members of the ski club contributed equally to obtain $1800 for a holiday trip. When 6 members found that they could not go, their contributions were refunded and each remaining member then had to pay $10 more to raise the $1800. How many went on the trip?

3) Because of traffic Maria could average only 40 km/h for the first 20% of her trip, but she averaged 75 km/h for the whole trip. What was her average speed for the last 80% of her trip?

4) A number x is the harmonic mean of a and b if 1/x is the average of 1/a and 1/b. Find two positive numbers that differ by 12 and have harmonic mean 5

I don't know how to do these please help?? I am really clueless on how to do these.....

I don't know where to start I know I need a variable but I don't know how to make it all work
3) Because of traffic Maria could average only 40 km/h for the first 20% of her trip, but she averaged 75 km/h for the whole trip. What was her average speed for the last 80% of her trip?
Assume

The distance = d

speed for the last 80% = v

then

time for first 20% = t_1 = 0.2*d/40 hour = d/200 hr

time for last 80% = t_2 = 0.8*d/v hour = 4d/(5v) hr

total time = t_1 + t_2 = d(0.005 + 0.8/v)

total average speed = d/[d(0.005 + 0.8/v)]

Now continue...

From the complexity (and variety) of these problems - it looks like a test or a prep for a test. If you have no clue how to do these problems - you should talk to your teacher and re-evaluate your priorities.
 
dunit0001 said:
1) An intake pipe can fill a certain tank in 6 hours when the outlet pipe is closed, but with the outlet pipe open it takes 9 hours. How long would it take the outlet pipe to fill an empty tank?



I don't know where to start I know I need a variable but I don't know how to make it all work


An "outlet" pipe would take water OUT of the tank, right? So, I don't see how an outlet pipe could POSSIBLY fill the tank. Please check to see if you've typed the question correctly.
 
Hello, dunit0001!

3) Because of traffic Maria could average only 40 km/h for the first 20% of her trip.
But she averaged 75 km/h for the whole trip.
What was her average speed for the last 80% of her trip?

Let \(\displaystyle d\) = distance (length of her trip in kilometers).

She drove \(\displaystyle 0.2d\) km at 40 kph.
. . This took her: \(\displaystyle \:\frac{0.2d}{40\) hours.

She drove \(\displaystyle 0.8d\) km at \(\displaystyle x\) kph.
.This took her: \(\displaystyle \:\frac{0.8d}{x}\) hours.

Hence, her total driving time was: \(\displaystyle \:\frac{0.2d}{40}\,+\,\frac{0.8d}{x}\) hours.


Her average speed is: \(\displaystyle \:\frac{\text{Total distance}}{\text{Total time}} \;=\;\L\frac{d}{\frac{0.2d}{40}\.+\,\frac{0.8d}{x}} \;=\;\frac{40x}{0.2x\,+\,32}\)\(\displaystyle \text{ km/h}\)

There is our equation! . . . \(\displaystyle \L\;\;\frac{40x}{0.2x\,+\,32} \:=\:75\)

Then we have: \(\displaystyle \:40x \:=\:75(0.2x\,+\,32)\)

. . \(\displaystyle 40x \:=\:15x\,+\,2400\)

. . \(\displaystyle 25x\:=\:2400\)

. . \(\displaystyle x \:=\:\fbox{96\text{ km/h}}\)


 
Hello, dunit0001!

Here's #2 . . .


2) Members of the ski club contributed equally to obtain $1800 for a holiday trip.
When 6 members found that they could not go, their contributions were refunded
and each remaining member then had to pay $10 more to raise the $1800.
How many went on the trip?

\(\displaystyle N\) people paid \(\displaystyle x\) dollars each: \(\displaystyle \:Nx \:=\:1800\;\;\Rightarrow\;\;x\:=\:\frac{1800}{N}\;\)[1]

When 6 members dropped out, only \(\displaystyle N-6\) went on the trip.
. . and they each paid \(\displaystyle x\,+\,10\) dollars.
. . . . \(\displaystyle (N\,-\,6)(x\,+\,10) \:=\:1800\;\)[2]

Substitute [1] into [2]: \(\displaystyle \:(N \,-\,6)\left(\frac{1800}{N}\,+\,10\right) \:=\:1800\)

. . and we have: \(\displaystyle \:1800\,+\,10N\,-\,\frac{10800}{N}\,-\,60\:=\:1800\;\;\Rightarrow\;\;10N\,-\,60\,-\,\frac{10800}{N} \:=\:0\)

Multiply by \(\displaystyle \frac{N}{10}:\;\;N^2\,-\,6N\,-\,1080\:=\:0\)

This factors: \(\displaystyle \:(N\,-\,36)(N\,+\,30)\:=\:0\)

. . and has roots: \(\displaystyle \:N \:=\:36,\:-30\)


Therefore, 36 members originally signed up for the trip.

. . but only \(\displaystyle N\,-\,6\:=\:\fbox{30}\) members went on the trip.

 
Hello again, dunit0001!

4) A number \(\displaystyle x\) is the harmonic mean of \(\displaystyle a\) and \(\displaystyle b\) if \(\displaystyle \frac{1}{x}\) is the average of \(\displaystyle \frac{1}{a}\) and \(\displaystyle \frac{1}{b}.\)

Find two positive numbers that differ by 12 and have harmonic mean 5.

Let's translate that ugly definition . . .

\(\displaystyle \L\frac{1}{x} \;=\;\frac{\frac{1}{a}\,+\,\frac{1}{b}}{2} \;=\;\frac{a\,+\,b}{2ab} \;\;\Rightarrow\;\;x \;=\;\frac{2ab}{a\,+\,b}\)


We have two numbers that differ by 12.
. . Let \(\displaystyle a\) = the smaller number,
. . \(\displaystyle a\,+\,12\) = the larger number.

Their harmonic mean is: \(\displaystyle \L\:\frac{2a(a\,+\,12)}{a\,+\,(a\,+\,12)} \;=\;\frac{2a(a\,+\,12)}{2a\,+\,12} \;=\;\frac{2a(a\,+\,12)}{2(a\,+\,6)} \;=\;\frac{a(a\,+\,12)}{a\,+\,6}\)

We're told that the harmonic mean is 5: \(\displaystyle \L\:\frac{a(a\,+\,12)}{a\,+\,6} \:=\:5\)

We have: \(\displaystyle \:a(a\,+\,12)\;=\;5(a\,+\,6)\;\;\Rightarrow\;\;a^2\,+\,12a\;=\;5a\,+\,30\)

This is a quadratic: \(\displaystyle \:a^2\,+\,7a\,-\,30\;=\;0\)

. . which factors: \(\displaystyle \:(a\,-\,3)(a\,+\,10)\;=\;0\)

. . and has roots: \(\displaystyle \:a\;=\;3,\,-10\)


Therefore, the two positive numbers are: \(\displaystyle \:\fbox{3 \text{ and }15}\)
 
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