Hello,
I have two questions.
1.Regarding the above question, Greg mentioned it as a quick solution for mixture problems; however, I couldn’t fully understand it. Could you kindly explain the quick solution in more detail?
The quick solution, i.e, the mixture trick requires more intuition than the algebraic approach.
First, try to estimate the percent of alcohol in the resultant mixture of solution A and C alone.
Weight of solution A : Weight of solution C = 80: 20 = 4 : 1 ( 5 parts )
To estimate the percent of alcohol in the resultant mixture, think about the distance in percentages. Solution A ----> Solution C in terms of alcohol percentage is 10 —> 20. The distance is 10. Divide the distance into 5 parts (since our ratio of weight 4:1 has 5 (4+1) parts).
Then each part will have 2%. So the idea is this distance will also follow the same ratio, but flipped, i.e 1:4. Then in terms of distance, it’s 2%:8%. So you land on 12% which is 2% away from Solution A and 8% percent away from Solution C following the 1:4 ratio. The resultant mixture of Solution A and C will have an alcohol content of 12%.
When solution B is included, we want a final alcohol content of 13%. Solution B has an alcohol content of 14%. Right now we have a 100g solution with 12% alcohol content.
12% → 14% We want 13%, the ratio now is 1:1. Use the same ratio in weight
Weight of Solution B: Weight of mixture of A and C = 1:1
Weight of Solution B : 100g = 1:1
Weight of Solution B = 100g
You can still stick with the algebraic approach because I feel the difference in time taken between the approaches is minimal and there is a higher chance of making a mistake (in my opinion) if you go with this approach.
He simply meant the question asks for the difference in amount expressed in dollars. When we calculated the solution the difference comes down to 19 cents. But we have to express it in dollars so 0.19 is the right answer (not 19 which would expressing the solution in cents)
The nearest cent part means, in case you get an answer like this 0.197 or 0.182. Round it down to 2 decimal places (nearest cent)
Hello,
Regarding the unique approach to solving mixture problems, I would prefer using the standard formula, as the current method feels somewhat confusing. Thank you for your explanation, but it still seems challenging, particularly the final part involving the addition of solution B. I couldn’t fully grasp the rest of the process.
Secondly, if we have a number like 19, we should express it as 0.19. Additionally, if we initially have a number like 0.192, we round it to two decimal places, resulting in 0.19. Similarly, numbers like 0.198 or 0.199 would round up to 0.2, correct?
Yes, assuming that that’s what you need to do in the question itself.
