A lot of people are arguing over the correct answer to this emoji math algebra problem involving horses, horseshoes, and boots. A Facebook post with this problem has over 500,000 comments. Can you figure it out? In this video I explain the correct answer.
Can you figure it out?
Some viewer comments:
twwc960 said:
It’s amazing how wildly different the difficulty levels are in your various problems. I’m thinking of last week’s geometry problem where you had to construct some very non-obvious isosceles triangles to find that an angle x was 30°. Then there’s this simple problem, arguably much simpler than the somewhat similar flower problem posted a few weeks ago where one flower was missing a petal or something and we had to guess that means it had 4/5 the value of the five petal version of the flower. This problem is trivial in comparison to either of those.
TheFinalRevelation said:
The problem I have with these types of problems is that you have to assume that the symbol “picture of two horseshoes” has double the value of the symbol “picture of one horseshoe”. To me there is nothing indicating that is the case. The equation should read 2 times horseshoe symbol + 2 times boot symbol = 2. That is the only way in which there is sufficient information to solve the system of equations. For example Van Gogh’s vase with fifteen sunflowers sold for about 25m pounds. Do assume that each sunflower is worth $1.67m pounds? An economist might argue that the marginal value of a second horseshoe is less than that of the first horseshoe, unless you have a horse that walks on two legs, in which case the second horseshoe has greater value than the first as I can now shoe the horse!
Sandokiri said:
This is a similar problem to the “Chinese Flower Math” viral problem… and has the same fundamental issue: single-boot and single-horseshoe are undefined variables. Since we cannot define the relationship between S and s (horseshoes), and B and b (boots,) we get this. 3h = 30: h=10. So we know that a horse is 10. 2S + 10 = 18; S=4. So we know that the “pair of horseshoes” is 4. 4-B = 2; B=2. So we know that the “pair of boots” is 2. 10s + b = n, solve for n. We have introduced two new variables. We cannot simply say that S=2s, because S could stand in any number of relationships to s, or none at all. Even assigning a relationship (say, s=S!) does not allow us to say that b=B!, because the double boot has a different colour and orientation compared to the double horseshoe. I know this is a bit of overanalysis, but yeah.
