Passing the Architectural Equipment Engineer exam by self-study (Drawing a psychrometric chart, past questions on the psychrometric chart)

Now, today I think I'll try solving a problem. First, before solving the problem, I'll always draw a diagram. For the mixed air, look at the mixed air, find the dry-bulb temperature, absolute humidity, and enthalpy at point 3. So, this is point 3. The outdoor air volume is like this: 4, the return air volume is 6. So, outdoor air is $5 \times 0.4$, which is $2 +$ indoor air, which is $6 \text{ m}^3/\text{min}$, so $21 \times 0.6$ gives $14.6^\circ$. Is that right? And the absolute humidity is the same, just put it in the same place. Ah, I need to draw this. I need to draw it and calculate based on the numbers. So, first, what I can do is find the indoor sensible heat ratio. The indoor sensible heat ratio is found by calculating the sensible heat load divided by the latent heat load. This is $50 + 15$ divided by $50$. If I quickly punch this into the calculator, it's $0.77$, rounding up to $0.77$. Now, for point 3, point 3 is to find the supply air temperature. So, the supply air volume, I don't know the exact formula because I'm not a major in this field, I know nothing else, so I only know the formula $Q = cm\Delta T$. Using $Q = cm\Delta T$, $M$ is the ventilation rate, $6,000$ plus $4,000$, so the total input is $10,000$. The outdoor air volume is $4,000$, and $6,000$ goes in here and $4,000$ goes in here, so the total entering the room will be $10,000 \text{ m}^3/\text{h}$. The supply air temperature, $50 \text{ kW}$ is from the formula $Q = cm\Delta T$, so $50 \text{ kW}$ is $50 \text{ kW}$ equals $10,000 \text{ m}^3$. I don't know why the units aren't shown in the book, but you have to study by always writing the units like this so you can guess correctly even if you don't know. An hour is one hour, so $\text{kW}$ is $\text{Joules}/\text{second}$. So, this must be $3,600$ seconds. I always solve it like this. Just $3$.