Can you solve the frog, sheep, horse puzzle?

[@wyattstevens8574]

Even though there was no pause at 0:40, I recognized this was just another "everything's equally multi-counted" problem! 2*answer=10+20+24=54 kg, so answer= 27 kg!

Thanks, I thought the same. The 4 minutes that followed were redundant ...

[@wyattstevens8574]

Everybody's saying "the horse weighed this much, the sheep this much, and the frog this much," but they're actually going "too far!"

[@renedekker9806]

@@wyattstevens8574 Even that is easy to do. Just subtract each of the given 2-animals weights from the total of 27 kg, and you get the weight of the missing animals.

[@Sam_on_YouTube]

Same. And then solving the weights of each one from the weight of all 3 together is trivial.

[@Magst3r1]

Damn, nice

[@stevenz933]

Since the problem does not ask for the weight of each animal individually and recognizing there are two of each animal after adding up all the weights simply dividing the total sum by two will give you the total weight of all three animals. Fun problem!

[@cguy96]

This

[@kevinfinke6927]

Bingo!

[@Vex-MTG]

Even if you needed the weight of each individual animal, this is still the simplest way to do it. Once you know that one of each is 27kg, it's basic subtraction to get the weight of each of the individual animals.

[@RedEyesBlackDragon0]

H = S+10. H+S = 24. H=17, S=7. At that point, you can either add 17 to F+S or add 7 to F+H.

[@aleksandarb]

That got to be a dense frog weighing so much and looking so small.

[@ArabianShark]

It's a porcelain frog... filled with lead... and osmium.

[@CoPoint]

Yes, but not impossible, as it turns out: my first thought was "A little much for a bullfrog...", too - Wikipedia says up to 1.4 kg. But: turns out the heaviest frog isn't the bullfrog, but the Goliath frog, up to 3.3 kg - so those 3 kg aren't too outlandish, after all 😁... But don't get me started about the 7kg sheep and the 17kg horse, now 😄...

[@stuchly1]

A severely underweight horse too

[@CoPoint]

@@stuchly1 Well, there _are_ those mini horses, with a shoulder height of a middle sized dog... Couldn't find any useful data about those, but I'll go out on a limb here: fully grown, even those may break that barrier somewhat easily...

[@Magst3r1]

The sheep and horse are way more impressive

[@phantomlogic6940]

I immediately saw that since (sheep + frog) + (horse + frog) = 30kg while (sheep + horse) = 24kg, that this must mean 2 frogs = 6kg to make up the difference. Thus, 1 frog = 3kg and the combined weight of all three must be 24kg + 3kg = 27kg.

[@tygerion4404]

Huh, neat. I noticed that Horse+Frog was 10 larger than sheep+frog, so the horse was 10 larger than the sheep. I then used this and the fact that H+S=24 to calculate that the horse was 17, and the sheep was 7. At that point, it was easy enough to calculate that the Frog was 3, letting me add 3+7+17 to get 27. Totally different paths to get the same answer... Though, TBF, I may have been influenced by a personal project I've been working on to categorically calculate the EXP/money yields of enemies from a game... Hence my overly complex answer that relied on calculating the individual weights. 😅

[@dalestaveley3810]

Did anyone else solve this in 10 seconds and then watch the video wondering what you might've missed?

[@ChessSacrifice-z7i]

me

[@PeerAdder]

Except I saved time by reading the comments, since it seemed unlikely there was anything to miss.

[@lorenzodiambra5210]

im

[@LeoLiss]

Rare case of MYD videos when I can feel smart.

[@Akshay-Raut]

Lol

[@SotiBen]

Relatable

[@DanMusceac]

You have the sense of humor. Unfortunately presh tallwaker has not.

[@srf2112]

horse ~ 17kg sheep ~ 7kg frog ~ 3kg Total ~ 27kg The horse is 10kg heavier than the sheep. From there it was easy imo.

[@n00blamer]

Even easier if you add up all the weights, and are autistic and see 2 horses, 2 frogs, well 2 of everything.. so the sum is double of what is being asked

[@johnconnor555]

You're right, so mine solved mentally in 10-12 seconds... but to be honest, i love to solve things like this everytime i find them...

[@markphc99]

Should read: This problem has simple people stumped , because it was the easiest one you've ever posted

[@Admiralgrusbil]

I just did it in my head trying to find numbers that fit and it took me ~1 minute. As soon as a more clever method was mentioned, I saw it and its elegant!

[@geoninja8971]

I think I solved this directly inspection in seconds, probably only due to the masterful tuition I've received on this channel over the past several years! :) I'm going to get my sons to solve this, it's a cool problem. This sort of stuff comes up on the maths comps/challenges they do at school....

[@wyattstevens8574]

@@geoninja8971 I did the same thing! I don't know why everyone's figuring out all 3 unique weights.

[@EdwardCurrent]

Big agree. It's so interesting that with experience, you no longer have to solve some of these problems rigorously -- you can just go straight to the answer. In this case I noticed that the difference between the sheep and horse is 10 kg, so the sheep + (10 + the sheep) = 24, so the sheep must be 7, so the horse must be 17, which makes the frog 3.

[@wyattstevens8574]

@@geoninja8971 Me too! I just looked at the platforms, realized it's an "everything's double-counted" question, and then went (10+20+24) kg= 54 kg= 2*answer, so answer=27 kg!

[@shohamsen8986]

(10+20+24)/2=27

[@julesboth1103]

Simplest answer without having to worry about anything else. Nice and effective.

[@MichaelJLambie]

Boss Move!

[@JDrapic]

Pre-solution attempt: problem can be written as f+s=10, f+h=20, s+h=24. Combine to f+s+f+h+s+h=10+20+24 -> Grouped 2f+2s+2h=54 -> f+s+h=27. And to take it a step further, f=3, s=7, h=17 (worked out by subtracting each of the starting equations from the final equation).

[@f5673-t1h]

There are two ways to solve problems: The backwards direction, and the forwards direction. The first approach to this puzzle was the backwards; you solve for all the variables, then get the quantity you want. The second is the forwards; you immediately get the quantity you want from what you're given. The backwards way, while tried-and-true and works for most typical problems, presents its own issues. You're solving for variables by trying to undo operations/functions (hence the "backwards"), and that sometimes requires undoing functions that aren't injective, meaning they map multiple values to the same value, which you can see is an issue for example when taking roots and getting extraneous solutions. It's just more overhead work. The forwards way goes directly to the answer without doing the extra work of undoing anything nor solving for variables, but it's not applicable to typical problems. It's usally a good idea to try it first since it's easier.

[@JA_Balbuena]

As a zoologist I can not only solve the problem but tell that the frog is most probably a goliath frog Conraura goliath 😊

[@positvelymaybe]

Surely the simplest solution, and imo the most obvious, is to firstly add the combined weight of the sheep and frog (10 kgs) and the horse + frog ( 20 kg) ; total 30 kgs. Then deduct the stated combined weight of horse + sheep (24 kgs) . The remainder is 6 kgs which is the weight of 2 x frogs ; therefore a single frog is 3kgs. From this you can calculate the combined weight of frog ( 3 kgs) + horse (17 kgs) + sheep ( 7 kgs) = 27 kgs . Ta dah !

[@mantasr]

27 kg. I love intuitive math. Difference between sheep and horse is 10 kg. So you need two numbers that add up to 24, but are different by 10. You can start at 12+12 and go until you reach it. 11+13, 10+14 etc. 7+17 kg. Means 7+3 is 10 (works for sheep), 17+3 is 20 (works for horse).

[@elliotlesser4373]

The second way you did it at the end of the video is the way I did it immediately. I was like wait a second, I don't even need to find the individual weight of each animal if I can do it this way.

[@jettsteel7305]

Put the top left and right ones together, so two frogs, a sheep and a horse is 10+20, or 30kg... Remove the 24kg of the bottom left and that gives us the weight of two frogs: 6kg, and that means one frog is 6÷2 or 3kg... Then, just add the 3kg to the 24kg (Horse+Sheep), and you get the last one: 27kg You don't even *need* the individual weights of the Horse and Sheep for this, but of course, you can still calculate them if you want (Sheep first, from the top left: 10-3 = 7kg, and then the horse next from bottom left: 24-7 = 17kg, or horse first from top right: 20-3 = 17kg) *TLDR; Frog = 3kg, Sheep = 7kg, Horse = 17kg, therefore, total = 27kg*

[@mr._jgp]

x + y = 10 x + z = 20 y + z = 24 Just add the three of them: 2x + 2y + 2z = 54 That means that x + y + z = 27 And that's the answer to the problem

[@HaydenTheEeeeeeeeevilEukaryote]

I did f + s = 10kg f + h = 20kg so all we did was swap the sheep for a horse and it went up 10kg, so we know that h = s + 10kg. We also know that h + s = 24kg, so h also equals h = 24kg - s. With two equations that h equals, we can substitute. s + 10kg = 24kg - s add S on both sides then subtract 10 on both sides to isolate it to 2s = 14kg, then divide by 2 to get s = 7. from here i just filled in the rest one by one to get 27 in the end i guess this was just subtracting the equations like you showed, which is really interesting how you can do that, i had no idea.

[@prasathgj]

0:40 its gonna be 3kg frog; 7kg sheep and 17kg horse👍🏼

[@tubular618]

Technically this solves for the mass of the animals, not the weight. The weight would be in Newtons. 😎

[@tyagagerson]

omg ain't no way this s##tty question stumped the world 😅

[@oooodaxteroooo]

I struggle with your explanation. After trying around a bit, i got it straight the simple way. Three pieces of information, three variables. So the idea is taking the constraints of the first two and the using them in the third to solve for one variable. So clearly, we want to describe the sheeps (B) and horses (C) weight by the frog (A). I A + B = 10 II A+ C = 20 III B + C = 24 B from I: B = 10 - A C from II: C = 20 - A Both in III: 10 - A + 20 - A = 24 30 - 2A = 24 | -24 6 - 2A = 0 | +2A 6 = 2A 3 = A The rest follows. A simple how and a clear why. 👍

[@trevorgreenough6141]

I did it differently, comes to show how different perspectives can have different processes: f+s=10 f+h=20 s+h=24 f+s+f+h=30=2f+s+h 2f+s+h-s+h=30-24 2f=6 f=2/6=3 f+s+h=3+24=27

[@robertnett9793]

I did it the other way around - if you look at f+s, f+h and s+h - you notice that you can add all together leading to 2f+2s+2h = 54 - deviding by two f+s+h = 27. From there you can subtract the weight of each pair of animals to get the value of the third one.

[@marcusscience23]

What’s more, after you find the total weight (f + s + h) = 27 kg using the second method, you can then find the weights of each individual animal just by subtracting the equation where it’s absent: To find f, subtract: s + h = 24 kg f = (f + s + h) - (s + h) = 27 kg - 24 kg = 3 kg To find s, subtract: f + h = 20 kg s = (f + s + h) - (f + h) = 27 kg - 20 kg = 7 kg To find h, subtract: s + f = 10 kg h = (f + s + h) - (f + s) = 27 kg - 10 kg = 17 kg

[@MichaelPaoli]

An easy do it in your head: units of kg, f, s, and h for frog, sheep, and horse respectively, solve for f+s+h, given: f+s=10, f+h=20, s+h=24 (s+h=24)-(f+h=20)=(s-f=4),s=f+4 (f+s=10)=(f+f+4=10),2f=6,f=3 f+s=10,3+s=10,s=7 s+h=24,7+h=24,h=17 f+s+h=3+7+17=27

[@mooncowtube]

Even if we go down the algebraic approach, at 01:30 there's no need to solve for each of the variables, because all we need is the total. Add the three equations: 2(f+s+h) = 54, so f+s+h=27. And *that's* the answer!

[@red-sparrow.819]

I'd expected to see how to solve a complex problem with a simple solution. Turn out I got how to solve a simple problem with a complex solution.

[@EdwardCurrent]

There are usually easier ways to get to the solution. Usually Presh goes step by rigorous step, which is good way to present if for a general audience. In this case he also presented a simpler solution, which is helpful.

[@Martinrd72]

That’s one heavy frog!

[@EaglePicking]

It's got lead poisoning.

[@BrettMKW]

I did this by finding all the weights with respect to the weight of the frog which I called F. Since F + H = 20 and S + H = 24, we can pretty quickly see that S = F + 4. Since F + S = 10 we can now substitute and say F + F + 4 = 10, which gives us F = 3, and then it's easy to see S = 7 and H = 17.

[@richardashby6981]

Solved it using the first method because I didn't consider the faster one. Lesson learned.

[@abcde_5949]

Solved in couple of seconds. Sheep is 4kg heavier than frog and both weigh 10gk, therefore it's 3 and 7. Add 3 to 24 and done.

[@Angi_Mathochist]

The easy approach was immediately obvious to me. Just add the three equations together, you have 2 of each, so divide by 2.

[@finnaddict]

That second method was genuinely clever! 😊

[@jdtubaman]

I just saw the difference of the horse with the frog and horse with the sheep as 4. I then deduced that that was the weight of the sheep, and 10-4 was 3, the weight of the frog. While it was simple enough to work for this. However, I do love learning the real way of proving this without going on observational experience with such questions!

[@Gideon_Judges6]

I did it in my head via: bottom left minus top left, then add that result to the top right, this gives you the mass of 2 horses = 34 kg, 34/2=17, then add the horse's mass to the top left that's missing a horse for the combined mass. s+h-(f+s)=24-10 -> h-f=14 h-f+h+f=2h=14+20=34 -> 2h=34 -> h=17 f+s+h=10+17=27kg

[@ryanschnarr9998]

Fun way, take top to add them and subtract the bottom one, you are left with 2 frogs and 6 pounds, divide by 2, 3, add the the third, 27.

[@GraveUypo]

yup, from the thumbnail. sheep 7, horse 17, frog 3

[@TerjeMathisen]

I didn't see the "everything twice" trick, I simply took the second, subtracted the first and added the third: This left (frog+horse)-(frog+sheep)+(horse+sheep) = 2xhorse = 34, from which I got the 3/7/17 individual weights.

[@hikari1690]

After someone reminded me of how to answer the standing cat, table and crouching cat problem (sorry I can't remember who you are) I now approach these puzzles by the 2nd method rather than the 1st. Which is a lot faster. So thank you kind stranger on the internet who I forgot the name of for making me a little smarter(or atleast efficient?) than I already was 😊

[@fmixthings]

The difference between F+H and S+H is 4kg, therefore the difference between F and S is 4kg. 10-4=6, divided by 2 is 3kg. S + H + 3=27kg

[@bennc8559]

Naming the squares FS, FH, SH ( fs+hs((FS+HS)-SH))-/2 looks complicated but that was my in the head solution as I saw that adding the first two squares together gave me almost what I was looking for just had an extra frog. I could then subtract 24Kg from that because it would leave me just the two frogs. Then divide the frogs by two and add that frogs weight to the second box to get the final. I wouldn't say the second method in the video is more clever. Each method is more suited to a different thinking path and all our brains see the problem and solutions differently.

[@kkulkulkan5472]

I just did it visually using figure 1 + 2 - 3 = weight of 2 frogs. 10+20-24 =6, so each frog is 6 / 2=3. Just add one frog to figure 3 and you get all three at 24+3=27 total.

[@InPursuitOfCuriosity]

Another rare occasion where I get the correct answer on this channel. Thanks for throwing in questions here and there that you don't need to be a math olympiad enthusiast to solve.

[@Ynook]

Or we could use a similar trick you used in the 'cat and table' video. The weight of the sheep is (24-10)/2, then we add that to the 20 kg.

[@s3b714]

I had a nice intuition here where 10+24=34 would be the total weight plus one extra sheep, while 20 is the total without sheep, so the mean of 20 and 34 is the total weight

[@UODZU-P]

Initial thoughts - sum all 3 cases together. Each animal is counted twice.Divide by 2 to find weight of all 3 animals together. Subtract any other case to find weight of individual animal.

[@shmuelpollak3272]

Not only easy but also very well known. Seems like Presh has finished "solving the world's problems one by one"...

[@kev_whatev]

We know that f+s=10, and f+h=20, so h=s+10. If s+h=24, the only values that satisfy h=s+10 are h=17 and s=7, which tells us f=3

[@machonsote918]

Three equations, three unkowns.

[@abdopower5913]

I didn't watch much video in this channel, but This the first time I solve a problem here 😂: X(Frog) Y(Sheep) Z(Horse) X + Y = 10 X + W = 20 W + Y = 24 X = 10 - Y X = 20 - W Y = 24 - W 20 - W + 24 - W = 10 44 - 10 = 2W 34 = 2W W= 17 X = 3 Y=7

[@easy_s3351]

f+s=10 so f=10-s. f+h=20 so 10-s+h=20 which means h=10+s. h+s=24 so 10+s+s=24 which means s=7. Which means f=3 and h=17, so f+s+h=27. Big fat frog, small sheep and very tiny horse.

[@Dr_piFrog]

Easy, write the three equations and add the three. One sees then that the total weight is simple 57/4 , therefore frog + sheep + horse = 27 kg Horse = 17 kg, Sheep = 7 kg, and the lowly Frog = 3 kg.

[@jacobgoldman5780]

Ok my solution at first glance: If we add up the three weights we see there are two of each animal (frog on the top two scales, horse on top right and bottom left scale, and sheep on left scales) so if we take half the sum of the weight we get half of two of each animal or one of each animal total so the answer is 1/2(10+20+24)=1/2(54)=27 kg.

[@briant7265]

First two, the horse weighs 10 more than the sheep. Third, sheep 7, horse 17 (24/2 +/- 5). First, frog+sheep=10. Add the horse (17) for 27.

[@st-gelaisrene3287]

I just saw with the 2 first images in order that the horse weighted 10kg more than the sheep. So in the third image it was 24kg-10kg=14kg equally divided in the 2 so 7kg for the sheep, 7+10kg for the horse (17kg ) Then the frog weighted 3kg and the addition was simple.

[@rhiivideos]

I didn’t do all of this math. I just mentally ran down the numbers from 10 to 1 until they fit. Quickly realized that only 3,7,17 could fit all 3 scenarios.

[@olivvapor4873]

It took me 7 seconds to get the answer … but I was certainly lucky on this one ! 😅 I added the 3 first plates just to see … and they were 2 of each for 54kg ; So the last plate would be 27kg 😊

[@pellepels50]

A horse of 17 kg , a lamb of 7 and a frog of 3. Odd creatures

[@qqma4791]

Since you weigh each animal twice, add up the amount of weight for each time and divide by 2. 10 + 20 + 24. This equals 54. 54 divided by 2 = 27. 27 kgs for every animal weighed at the same time. You don’t actually need to calculate their individual weights.

[@ve2zzz]

The perfect question for this problem would be: What is the total weight of the three animals WITHOUT knowing the individual animal weight ? Althrough this "outside of the box" method may render some data errors unnoticed. (greatly exaggerated) example: F + S = 10 H + F = 20 H + S = 100 that does not make sense unless the frog weighs -35 kg !!!

[@nedmerrill5705]

Note that in the sum of the known weighings there are two of each animal. Simply sum the given weights and divide by 2. No algebra needed.

[@samlee5549]

The wright of all three is just the weight of the first three plates (which works out to 2 horses 2 sheep and 2 frogs =54kg) and halve it to get 27kg.

[@AzureKyle]

I solved it by adding all three problems together, getting 2f+2s+2h=54, then divided by 2 to get f+s+h=27

[@randomxgen6167]

I took the sorta-long way and subtracted a frog and horse from the sheep and frog to get a sheep and anti-horse. Then added a sheep and horse to the sheep and anti-horse to get two sheep and an annihilated horse, so effectively just two sheep. From there, it was regular non-destructive math to get the total. But adding everything together to get two of every animal and then just halving it is the kind of intuition I still need to develop. Thanks for continuing to fill my feed with neat little brain exercises.

[@arunplay100]

say a=frog, b=sheep, c=horse a+b=10 b+c=24 a+c =20 Add 1st and 2nd equation. 2b+a+c=34 --> 2b+(a+c) = 34 2b+20=34 ---> b=7 Since a+c=20, a+b+c=27

[@johnconnor555]

Your frog is heavier than my cat...😜🤪😂😂

[@avatarofaiyel]

when you look at the three examples, across the three, each critter is represented twice. So, add together the three example weights and divide by 2, ending up with 27kg. From that, you can determine the weight of each creature if you so choose (and I did) by simply subtracting the weight of the picture without that animal from the total weight. so you get: Frog 3kg Sheep 7kg Horse 17kg

[@nitailevi8001]

I don't often solve these but this one was very easy to do in your head. From the weights, the horse is 10kg more than the sheep and 14kg more than the frog, so the sheep is 4kg more than the frog, must be 7kg and 3kg, then the horse is 17kg, total is 27kg.

[@briancarreno2779]

Also, there is another method, since there is a frog in both the 10kg and 20kg, that means the horse weighs 10kg more than the sheep, so h=s+10, so this means f+(s+10)=20 f+s=10 and s+(s+10)=24, so 2s+10=24 2s=14 s=7, then h=7+10=17, finally f+7=10 f=3, so 7+17+3=27kg

[@TheSheKitty]

Now when you explain it like that, it's helpful 😊

[@christopherg2347]

Nice, my brain managed to figure out both solutions. Which were actually the same solution - one visual, one algebraic.

[@peterz53]

Sheep plus frog plus horse plus frog equal 30. Ie. Sheep plus horse plus 2 frogs equal 30 Horse plus sheep equal 24, so 2 frogs equal 6. Subtract 3 for one frog to get 27.

[@PoppySuzumi1223]

What I was first thinking is using a system of 3 equations ( 3 unknowns) to solve this, even not to mention the matrix way.

[@nataliakhomenko5325]

We all enjoy overkill!

[@feedbackzaloop]

Bit disappointed not to see Excel regression solver.

[@taflo1981]

That frog has to be a goliath frog.

[@Tolbens]

a+b=10 a+c=20 > c=20-a b+c=24 > c=24-b ... 20-a=24-b> b-a=4 a+b=10 + --------------- 2b=14 > b=7 and a=3 So the little pony is 17. 17+3+7=27

[@barneybarney3982]

the classic, simply looking but great example that shows why math is so great. because you can count with weights of each animal twice, even if you have only one of each. result is (10+20+24)/2

[@stevekerp1]

Pretty straightforward - when the frog (upper right picture) is replace by the sheep (lower left) the weight increases by 4 kg. This means the sheep weighs 4 kg more than the frog. Take this info to first graphic (upper left) and you discover the frog weighs 3 kg.

[@spinnwebe_]

I like a set of parallel equations as much as the next guy, but this seems like an easier way to go: we can see from diagrams 2 and 3 that the sheep’s weight is frog + 4. Then diagram 1 tells us they’re 3 and 7, and diagram 2 tells us the horse is 17., so the total is 27.

[@SergioBracali]

It's Just 3 equations and 3 variables, nothing difficult. But it's even easier: you sum the weights of the 3 couples and divede by 2

[@ArabianShark]

Solved it in my head before Presh started declaring variables and what they represent. Presh has trained me far too well for these problems... 😋

[@PlasteredDragon]

On the pause. Looks like there are any multiple way of solving this, so here is how I did it. 1: F+S = 10 2: F+H = 20 3: S+H = 24 Since the combined weight of all possible pairs is given we can easily express any variable in terms of any other variable. From #2 and #3 we can see that adding sheep to horse yields 4 more kg than adding frog to horse, therefore S=F+4 and F=S-4. From #1 and #2 we can see that adding sheep to frog yields 10 less kg than adding horse to frog, therefore H=S+10 and S=H-10. From #1 and #3 we can see that adding frog to sheep yields 14 kg less than adding horse to sheep, therefore H=F+14 and F=H-14. 4: S=F+4 5: F=S-4 6: H=S+10 7: S=H-10 8: H=F+14 9: F=H-14 But we don't need all of these identities to solve the puzzle. Since #4 tells us that S=F+4, then we can substitute F+4 for S in #1. F+F+4=10 --> 2F + 4 = 10 --> 2F = 6 --> F = 3. Now that we know F is 3, we can plug it into #4 to get S. S=3+4 --> S=7. Similarly we can plug it into #8 to get H. H=3+14 --> H=17. So the final solution for the weight of all three is 3 + 7 + 17 or 27 kg. Now to unpause and see how wrong I am.

[@michaeldakin1474]

I haven’t (yet) watched the video, but… From the first two images with the frog, I concluded that the horse is 10kg heavier than the sheep. Then, from the image with just the sheep and the horse, I calculated s + (s+10) = 24, so s=7. Back to the image with the frog and the sheep, I subtracted the sheep’s weight from 10 to get the frog’s weight (3kg). Finally, I simply added the frog’s weight to the combined weight of the sheep and the horse to get the combined weight of all three: 27kg.

[@gef56]

It's a waste of time to solve for the individual weights of the horse or sheep since their combined weight is already given at 24 kg; just add the frog's weight to that. For the frog's weight, just subtract s+h=24 from s+2f+h=30 to get 2f=6, then divide by 2 to get f=3. Another easy approach to solve for the frog's weight is to notice that the sheep weighs 4 kg more than the frog (top right to bottom left), so s=f+4. Since s is now defined in terms of f, we have s+f=10 => f+4+f=10 => 2f+4=10 => f=3. This is definitely one of the easier problems on this channel.

[@nordicexile7378]

The reason the frog is so heavy: It's made of "green matter". (Like "dark matter", but green.) Then there's the sheep and the horse being so light... banzai farm animals!

[@Foggeer-von-Dreitveld]

These horses and sheep are air freight approved!

[@Misteribel]

That's some tiny horse and giant frog 🐸 😂 🐎

[@robertodm6986]

Very educational approach :-) - Thank you

[@FCStone1708]

The 2nd way is really clever 😊

[@godsakezz]

Let x be frog, y be sheep, z be horse. x+y=10 (1) x+z=20 (2) y+z=24 (3) Rearrange and subst. (3) into (2), ie. z=24-y into (2), we get x+24-y=20 --> x-y=-4 Negative, this means y is bigger than x. So, from eq (1) we know x can take values of 4,3,2,1, y can take values of 6,7,8,9 and z can take values of 16,17,18,19 From eq (3), y+z=24. Possible cases: 1.) y=6, z=18 2.) y=7, z=17 3.) y=8, z=16 Case 1: When y=6, x+y=10, so x has to be 4. When x is 4, from eq(2), z has to be 16, but z is not 16 in case 1. So not applicable. Case 2: When y=7, x=3. When x is 3, z=20-3=17.This combination is applicable. Case 3: Not applicable. Thus, x+y+z=3+7+17=27. A tedious method to share haha.

[@poldelepel]

Man that is a HUGE frog! Probably a goliath frog...

[@WilliamLious]

How is this difficult? F+S=10, F+H=20, & S+H=24. Sum all (F+S)+(F+H)+(S+H)=10+20+24. F+S+H=54/2=27.

[@petrikVSempty]

f&g (10) + f&h (20) = 2f&s&h (30) - s&h (24) = 6 ÷ 2 (2f from the 1st equation) = 3 kg frog

[@game-guy-1]

This is too easy for this channel

[@ve2zzz]

This problem is the best way to introduce someone, not math-loving, to Presh Talwalkar

[@WhoStoleMyAlias]

Perfect example of how kids run out of time during exams, overcomplicating the question by aiming to answer more than is being asked for and thereby creating a higher possibility of making a mistake. Nice that they turn out to be round numbers, but what if the frog and sheep combined had weighed in at 9 instead of 10?

[@spencereades]

3 kilograms sounds like a lot for a frog, but it actually checks out in the case of the goliath frog of west Africa.

[@markwallen6570]

Surprisingly simple, really!

[@9Point8]

Went right to combining the three equations. With calculator would be a quick 3x3 matrix to solve all three.