Mixture problem

The answer needs to be between 25% and 40%, and because we're using more of the 40% solution, the answer must be closer to 40%, so only one answer makes sense, 37.5%. That's assuming both vessels are full, something the question needs to mention (but doesn't).

If you want to use alligation, you can draw a number line with the percentages for each component, and the overall percentage A in the middle:

---25-------d---------A---e--40----

Because the parts are in a 6 to 2, or 3 to 1 ratio, the distances I've labeled d and e are in a 3 to 1 ratio. So d is 3/4 of the total distance between 25 and 40, so is (3/4)(15) = 45/4 = 11.25, and A = 36.25.

I suppose that's "approximately" 37.5, but that's a very bizarre "correct" answer to this question when a number like 36 is a much better approximation. What is the source?

Mo2men wrote:A vessel of capacity 2 liter has 25% of alcohol and another vessel of capacity 6 liter had 40% alcohol. Both liquids were poured out in one vessel.. What is approximately the new concentration of Mixture?

A) 20%
B) 30%
C) 37.5%
D) 50%
E) 66.67%

OA: C

(2 x 0.25 + 6 x 0.4)/(2 + 6) = (0.5 + 2.4)/8 = 2.9/8 ≈ 3/8 = 0.375 = 37.5%

Alternate Solution:

We start with 2 liters of 25% alcohol and add to it 6 liters of 40% solution, and the result is 8 liters of x% solution. Let’s express this information as an equation:

2(0.25) + 6(0.40) = 8(x)

0.5 +2.4 = 8x

5 + 24 = 80x

29/80 = x

We can approximate this answer with 30/80 = ⅜ which is about 37.5%

Answer: C