Pressure Calculation for a Container with Water and Olive Oil

[yoshiko]

I have solved this problem but am not sure where I'm going wrong here...


Suppose you pour water into a container until it reaches a depth of 20cm . Next, you carefully pour in a 9.7cm thickness of olive oil so that it floats on top of the water.

What's the pressure at the bottom of the container? Express anser using two significant figures.

P-P_atm= ? Pa

P_1= P_atm + (rho)(g)(h)
= 1.013 x 10^5 Pa + (920kg/m^3)(9.81m/s^2)(0.097m)
= 102175.44 Pa

P=P_1 + (P_water)(g)(h)
= 1.013 x 10^5 Pa + (1000kg/m^3)(9.81m/s^2)(0.20cm)
= 104137 Pa

I keep getting this answer wrong.. I do not know what I am doing wrong. Please help!
I've tried already 1.04x10^5
1.0x10^5
and the above exact number.

 

[Doc Al]

The pressure at the bottom is the sum of:
- atomospheric pressure
- pressure due to .097 m of olive oil
- pressure due to .20 m of water

 

[yoshiko]

Thanks! I have discovered what I did wrong...I was not supposed to include atmospheric pressure in the first place. Thank you, Doc Al.