Math puzzle.

I'm getting 26.7

 

I couldn't have said it better. Tools have always worked a lot better for me than algebra.

 

Someone knows their math 5.2 gal antifreeze 4.8 gal water is what I got also

 

Think of it this way: it started with 5 gallons each of water and coolant. Lost a gallon of each, then added 2 gallons of water, so 4 gallons of coolant & 6 gallons of water. Draining 2 gallons of that mixture (40% antifreeze) leaves you 3.2 gallons of coolant and 4.8 gallons of water.

So adding 2 gallons of pure coolant will take you up to a 52% mixture. 5.2/4.8 coolant/water.

 

It’s not going to matter…. 10% antifreeze concentration will not freeze solid enough to damage anything down to around 0f…. Did a little experiment several years ago when we had a cold snap. If I remember right at around 10 below the top inch or so in the bucket froze into a stiff slurry idk if it would have damaged anything. as mentioned above you really need to get that wet cylinder dried up or your gonna have rust rings in the liner at each piston ring. Might wipe off but most likely will ruin the liner.

 

2 goes in 10, 5 x. Which makes it 5 gal. Water,5 gal concentrate.you loses 2 gals.that one gal water and one gal.concentrate.you add 2 gals.water.you have now 4 gal. Concentrate 6 gals.water...lol