need help on this question (1 Viewer)

[naq69]

Cambridge mathematics 3unit year 11
Exercise 1G

11.g)
A certain tank can be filled by two pipes in 80minutes. The larger pipe by itself can fill the tank in 2 hours less then smaller pipe by its self. How long does each pipe take to fill the tank on its own.

Any help would be appreciated
Thanks

 

[life92]

Okay firstly, lets look at a similar example, or a more common type of question.

If pipe A can fill a tank in 3 hours and pipe B can fill the same tank in 4 hours, how long does it take for the two pipes working together to fill the tank.

Rate of pipe A = 1/3 of the total per hour
Rate of pipe B = 1/4 of the total per hour

Therefore, rate of pipe A and B = 1/4 + 1/3 of the total per hour
= 7/12
now 7/12 * x = 1
therefore, x = 12/7, i.e. the time is 12/7 hours.

Using the same type of logic...

Now if we let x = time it takes for pipe A to fill the tank in hours
and therefore x-2 = time it takes for pipe B to fill the tank in hours

Rate of pipe A = 1/x
Rate of pipe B = 1/(x-2)

Now, [1/x + 1/(x-2)] = combined rate
combined rate = (2x-2)/ [x(x-2)] of the total per hour
now the time it takes for them combined is equal to 4/3 hours (80 minutes)

therefore, (2x-2)/ [x(x-2)] . 4/3 = 1
8x-8 = 3x^2-6x
3x^2-14x+8=0
(3x-2)(x-4)=0
Therefore x = 4 or 2/3
However, if the time it takes for pipe B = x-2, then x must = 4
Therefore, the time it takes for pipe A = 4 hours, and the time for pipe B = 2hours.

 

[fullonoob]

+1 to life. hit the degree button to change it into minutes

 

[naq69]

thanks
was stuck in dat question
only 1 person in the class acutually got the answer by themselves
but still thanks
really appreciated

 

[11MX11]

was it me?

 

[moonsuyoung]

Okay firstly, lets look at a similar example, or a more common type of question.

If pipe A can fill a tank in 3 hours and pipe B can fill the same tank in 4 hours, how long does it take for the two pipes working together to fill the tank.

Rate of pipe A = 1/3 of the total per hour
Rate of pipe B = 1/4 of the total per hour

Therefore, rate of pipe A and B = 1/4 + 1/3 of the total per hour
= 7/12
now 7/12 * x = 1
therefore, x = 12/7, i.e. the time is 12/7 hours.

Using the same type of logic...

Now if we let x = time it takes for pipe A to fill the tank in hours
and therefore x-2 = time it takes for pipe B to fill the tank in hours

Rate of pipe A = 1/x
Rate of pipe B = 1/(x-2)

Now, [1/x + 1/(x-2)] = combined rate
combined rate = (2x-2)/ [x(x-2)] of the total per hour
now the time it takes for them combined is equal to 4/3 hours (80 minutes)

therefore, (2x-2)/ [x(x-2)] . 4/3 = 1
8x-8 = 3x^2-6x
3x^2-14x+8=0
(3x-2)(x-4)=0
Therefore x = 4 or 2/3
However, if the time it takes for pipe B = x-2, then x must = 4
Therefore, the time it takes for pipe A = 4 hours, and the time for pipe B = 2hours.

I understand everything except for the

"therefore, (2x-2)/ [x(x-2)] . 4/3 = 1"

Why would it equal "1?" What is "1?' Could someone please explain to me

 

[Bronx_Tips]

was it me?

yes it was

 

[ZakaryJayNicholls]

I understand everything except for the

"therefore, (2x-2)/ [x(x-2)] . 4/3 = 1"

Why would it equal "1?" What is "1?' Could someone please explain to me

In the first example 7/12 was the combined-pipe fraction of a tank that could be filled in one hour.

7/12 times x equals 1 allowed you to find the time taken in hours (x) in order to fill the tank.

In the second part, we now know time taken is 80 minutes (which is 4/3 hours), and the combined-pipe fraction of a tank which could be filled in 1 hour is the expression containing the unknown (eg (2x-2)/ [x(x-2)]), this, when multiplied by the time taken in hours (eg 4/3), must be equal to 1.

This is because the rate (as a fraction of the full tank, in terms of full tanks per hour) multiplied by the time (in hours) is always equal to one so long as you are trying to fill one tank completely.

 

[moonsuyoung]

In the first example 7/12 was the combined-pipe fraction of a tank that could be filled in one hour.

7/12 times x equals 1 allowed you to find the time taken in hours (x) in order to fill the tank.

In the second part, we now know time taken is 80 minutes (which is 4/3 hours), and the combined-pipe fraction of a tank which could be filled in 1 hour is the expression containing the unknown (eg (2x-2)/ [x(x-2)]), this, when multiplied by the time taken in hours (eg 4/3), must be equal to 1.

This is because the rate (as a fraction of the full tank, in terms of full tanks per hour) multiplied by the time (in hours) is always equal to one so long as you are trying to fill one tank completely.

Ohhh that makes sense, thank you so much!