Blood Flow

[adrakdavra]

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If blood pressure doubled and the resistance to blood flow increased by 50%, the amount of blood by the heart would have:

Ans : Increased by 1/3

Formula used Co= P/VR

I did not understand the math .... I am sure it is an easy math, but I am not seeing it.

 

[wiloghby]

I just did this problem last week... I bet it's from AAMC 8, if I had to guess.

So, starting with C_o = P/V_r as our initial conditions, if we let pressure = 2P and resistance = 1.5V_r, you get.

C_o = 2P/1.5V_r

If you convert 2/1.5 to a lowest terms fraction you get: C_o = (4/3)*P/V_r.

From this, you can see that the cardiac output is now 4/3 times what it was originally, because it was originally just P/V_r. So if it is now 4/3 times higher than it was originally, that means it has gone UP by 1/3 compared to where it started. (3/3 + 1/3 = 4/3).

If you prefer to think in percentages, the blood flow after the change is 4/3 or 133% of what it started at. That's an increase of 33%, or 1/3.

 

[adrakdavra]

From this, you can see that the cardiac output is now 4/3 times what it was originally, because it was originally just P/V_r.


This the part that i don't understand , why not to say that 4/3 is the answer, why do we have to assume the original is 1 then subtract and all that , most of the problems that I did before I did not have to do the extra steps, when do we do the old vs new and when do we just solve the problem with the old new concept it is confusing

 

[NeverBackDown]

From this, you can see that the cardiac output is now 4/3 times what it was originally, because it was originally just P/V_r.


This the part that i don't understand , why not to say that 4/3 is the answer, why do we have to assume the original is 1 then subtract and all that , most of the problems that I did before I did not have to do the extra steps, when do we do the old vs new and when do we just solve the problem with the old new concept it is confusing

I think the trickyness is in the wording of the answers. if it said:

4/3 the original output then you'd be right

but it says "increases" so you need to find the amount it increases by, which is 1/3. Nothing on the MCAT is straight forward...lol, expect the unexpected.

 

[adrakdavra]

I think the trickyness is in the wording of the answers. if it said:

4/3 the original output then you'd be right

but it says "increases" so you need to find the amount it increases by, which is 1/3. Nothing on the MCAT is straight forward...lol, expect the unexpected.

You are right it is the wording, however some of the MCAT Qs are straight forward, It is the tricky Qs that makes it fun lol .

 

[pacer]

Was this CO= P/VR equation given (in the passage)?

I get that flow = (delta P)/R and CO = SV * HR but how do you get to the equation being used to solve this problem.

 

[adrakdavra]

Was this CO= P/VR equation given (in the passage)?

I get that flow = (delta P)/R and CO = SV * HR but how do you get to the equation being used to solve this problem.

It was given in the passage.