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Say you have this problem: “A farm contains horses and chickens. 116 heads and 282 legs. How many chickens are there?” Here’s how to solve any of this type quickly:
116×2 = 232. — Multiply number of heads by 2 (two legs per head for chickens).
282 – 232 = 50. — Subtract legs gotten from part one from total legs.
50 / 2 = 25. — Divide part two by two (difference of legs between chickens and horses)
116 – 25 = 91. — Subtract part three from number of heads for answer.
Answer is 91.
Note that some of these have chickens and THREE-legged horses. In that case, divide by 1 in part three rather than 2.
Even easier: Use algebraic formula:
(Animal Type 1:X)x(Number of Legs on that type) + (Animal Type 2:Y)x(Number of Legs on that type) = Total number of legs i.e.: X\*4 + Y\*2 = # Legs
Also: X+Y = Heads
Riddle Total legs 100, Total Heads 30, Horses(X) & Chickens(Y)
Known true values: X\*4 + Y\*2 = 100, X+Y = 30
Divide he equation 4X + 2Y = 100 by 2 on both sides = 2X+ Y = 50
If X+Y = 30, and 2X+Y = 50, then X must = 20 (subtract X+Y from one side and 30 from the other)
20 horses 10 chickens = 100 legs 30 heads
This formula works regardless of how many ‘legs’ the creature type has. Just plug in the appropriate value in the original equation to match the new criteria.
what i did is this:
lets say you have 3 legged cows and horses..
100 heads and 360 legs..
100 times 4 = 400
400 – 360 = 40
40 × 3 = 120
100 – 40 = 60
60 × 4 = 240
120+240 = 360
simply put, the ammount of heads times the max ammounts of legs possible (if there were only horses in this case), take away the ammount of legs mentioned in the question (400-360 in this case) and you have the answer as 40..
and every time, the question involves 2, 3 or 4 legged animals, and every time, the animal with the least legs is the one asked (atleast most of the time)
and in this case, the correct answer would be, 40 three-legged-cows..
aaaaaand i might not be helping at all due to my rather random way of solving it.. but that’s the way I’ve solved em n it seems to work for me atleast.. and also, I figured it out on my own, rather than asking for help.. so if someone already had the same kind of calculating system as I’ve used, I havent copied you, I figured it out on my own..
and i was too lazy to read through the whole topic so i just wrote an answer
I do it about the same way only a bit shorter.
number of heads x least legged animal = x
(number of legs – x)/difference in number of legs between the two animals=number of most legged animal = y
number of heads – y= number of least legged animals.
example: 63 heads, 190 legs chickens and horses
63 × 2 = 126
(190-126) / (4-2) = 64 / 2 = 32 horses
63 – 32 = 31 chickens
check: 31 × 2 = 62; 32 × 4=128; 62+128=190
if x = chickens, y = horses & 94 heads and 194 legs
x + y = 94
y = 94 – x
2x + 4y = 194
2x + 4(94 – x) = 194
2x + 376 – 4x = 194
-2x = -182
x = 91 chickens
Questions like this is why I fear for the future of mankind.
Using variables is the most basic and easiest way to do these types of problems.
The first post is basically based on variables, I doubt people who can’t even do simple variables can understand it…
It’s middle school math.
And yes there are middle schoolers here, but if an X and a Y make your head spin, you are in for a rough time assuming you want to do ANYTHING technical.
VARIABLES isn’t even a technical word, stop treating it like some medicine. It’s just a representation for your thoughts…
ryuzog you are a moron. Variables in this situation is a waste of time and any mathematician would agree.
By your logic x + y = 7
y = 3
x = 4
therefore 4 + 3 = 7
the question gives you everything you need to know. So why not just write
4 + 3 = 7
No need for that, whatams. Using variables is actually much faster and easier, since you only have to write the formula once, then simply plug in the values. In fact, there are fewer steps involved when using variables than doing it the ‘long’ way. No one is forcing you to use a particular method. Feel free to use whichever one you prefer. However, as a mathematician, I can tell you that using an algebraic formula is much faster and easier, especially for any repetitive tasks.
Get a grip.
Any amount of heads, any amount of legs
Say.. 76 heads, 193 legs .. chickens and horses (2 and 4 feet)
76 × 2 = 152
TOTAL heads \* lower feet
192 – 152 = 40
TOTAL feet – number of possible chicken feet
40 / 2 = 20
difference of possible chicken feet / difference between number of legs
(Only necessary if there’s chicken and horses, with 2 and 3 or 3 and 4 legs it doesn’t matter..)
that’s 20 Horses, just is.
And 76 heads? 56 chicken. Simple?
Lopen this isn’t even about the trivia, and your in high school. For the people who are able to solve problems like this in their head it can be difficult to accept that for other people its easier to use variables although in reality they are actually using the same formulas in thier head with out the same symbolism
Although that could just be a lot of hot air. Its just my reasoning as someone who can do things in my head.
Not that doing algebraic equations isn’t fun or anything…but here’s the links to the tools that solve all the puzzles in this game
Click on the “tools” section at the top if you need help on the color coded one as well.
Easy way to solve. Find difference in number of legs. Find the total if you use max of the most legs. Subtract and divide.
So with “On a farm there are horses and three-legged-cows. There are total of 95 heads and 367 legs. How many horses are on the farm?”
difference between the animals is 1 leg. if all were horses we would have 95 \* 4 = 380 legs. That is 13 more legs than we need. So 13 / 1 = 13 most be cows instead. Meaning the answer is 82 horses are on the farm. For those who do better with variables:
legsA – legsB = legDifference
legsA \* totalHeads = maxLegs
maxLegs – neededLegs = extraLegs
extraLegs / legDifference = how many of the animal with the least legs.