Can you solve an elementary school question from Singapore?
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- เผยแพร่เมื่อ 26 ม.ค. 2025
- This is a challenge question for students. Can you figure it out?
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The trickiest part was finding my glasses, so I could see the grading lines on the scale.
Glad you found your glasses, Velma!
I had a hard time reading it too.
I saw the lines, but didn't realise every line was 0.5, and came up with a negative as my first answer.
My first attempt gave the round ball a mass of -1 because I misread the gradations!!
The grading lines were messing with my head.
The tricky part was not to mistakenly see the markers as pointing at 13 & 7 instead of 14 & 6.
I did and got c=-1, which is plausible if you consider a negative weight to be a buoyant force.
Fell for that too but noticed it wasn't right. I guess that's why you double check.
I kept getting tripped up myself, going "-1, how?"
I "reasoned" the circles were helium baloons tied to the scale weighing -1 KG.
@@matthewmiller6979 I also misread the tick marks on the scale, and thought that this was a lateral thinking exercise, where analog scales (such as these) are really reporting the weight modulo some number (20 in this case). I was able to find weights that worked using that paradigm:
circle = 19 kg
square = 8 kg
My conclusion: the person who designed those scales should never be allowed to design anything again, because they clearly do not understand how to present legible information!
Agreed!
what do you mean? there are obviously more than 4 ticks between 0 and 5, and the one in the middle is highlighted, so, all you have to do is count the lines. counting is a math. it’s offered as a cross cultural math puzzle for anyone who’s not from singapore… why would you expect it to be in a familiar format? experiencing new things is very common when you step outside your own home town. or you could just go back to your local mcdonalds version of whatever you do. nothing will ever change and you’ll always be comfortable.
@@patrickmaline4258 Making the half number ticks identical to the whole number ticks is objectively bad design with poor legibility. It also adds nothing to the problem as a math exercise.
And thanks for the passive aggressive toxicity BTW. That was definitely necessary and makes your arguments much more persuasive and rational.
They even fucked up the half number tick between 5 and 10; they put it on 7 instead of 7.5 lmao
Welcome to metric markings. some use shorter marks for the haves, but the ones I used to use for commercial baking were marked with the same size for whole and half Kgs. On one those were the only markings, on another the 1/10th Kg marks were shorter. in most metric stuff I've used, only full power of 10 units get different sized markings, which makes sense enough, but also makes for difficult ro read scales on things where there is not enough room for clear 1/10th marks, in those cases half marks are all that's available, because you can't do quarters or eighths in metric.
I just doubled the right side to get a total weight of 12, compared to the 14. Both sides would then have at least 2 squares and 2 circles. Then I'd eliminate those and get that one circle is 2 kg.
Yes but I don't think "just" is a good word to use as there are other methods that are just as fast.
@@dalebaker9533 Realistically weight shouldn't be negative
@@dalebaker9533 If one blue square weighed 8kg, then two of them would be 16kg, without any green circles. The left scale disproves that.
@@maxhagenauer24 I mean I thought that this method would be the easiest to illustrate and explain using diagrams
@@purplealarmclocker Its one of them yeah, you could also use a method of moving squares and/or circles to other scales and see it very easily as well.
Since I'm eight and don't know substitution or other methods yet!
Circle and square is 6 so take away a left circle and square leaves 8. Take away another circle and square leaves 2 and leave a circle which will be 2.
I'm from Singapore and this is indeed the expected method for our grade 2 to solve
Well done! That's very intelligent at only eight years old. You have a good, logical brain
@@welcomb Well I'm almost 70 years old, but I have grandchildren and tried to approach it from what they would know. I like this method also because it gets kids ready for the concept of division!
@@jrkormanhow old are you
@@JaneAustenAteMyCat This is how Singapore Primary Math is taught. I taught it to my kids. They don't need algebra to solve most word problems, though they know how to "show your work" for their American school, after arriving at the answer in their heads.
The hardest part of the problem for me was to notice that it wasn’t 7kg and 13kg but 6 and 14...
I solved it for 7 and 13 too lol
Squares are 8 kg and circles are -1 kg... I guess they're balloons filled with helium.
@@nicholasscott3287 i made this mistake, too :D
@@nicholasscott3287 exotic matter spheres!!
Same
Am I the only one that got confused by the thick line between 5 and 10 not being in the middle and taking way longer due to that?
Yeah, I'm willing to bet most adults baffled by this problem misunderstood the lines and didn't take the time to count out what the tick marks stood for.
Yeeeah, I got caught on that part too. I keep trying to figure out how two sets plus a circle weighed less than a single set. Surely the circles had to be positive weight!
Green circle weighs -1 kilogram and the blue square weighs blue square weighs 8 kilograms. green circle =1 Kilogram. If +=7. That would mean if you were to double it you would have 14. But you have and that adds up to 13.
Yes, the scales are horribly designed and don't convey the intended meaning clearly.
Exactly, it should be on 7.5 not on 7
I imagined removing a square and a circle from the left scale, which makes it go down by 6 kg. Then remove another square and circle, which makes it go down by 6 more kg. There will then be a single green circle on the scale, and you can read off its weight.
This is probably the 'intended' solution that an 8 year old should come up with - I went with the system of equations route
we had such problems around this age. i assume as an introduction to express numbers with letters. and yes in the most cases it was easy like "remove everything you see on one side from another side". or do it twice.
and eventually problems became harder so you solve for circles and squares instead of whole scales
I did the same, and when he started writing out equations, I was thinking about how difficult that would be for most second grade students.
I'm usually pretty lazy, so I try to figure things out in my head before running to a notepad. This method is way simpler, and most people can even do it in their head.
Only in a math textbook you can have markings on scales so confusing - who in the real world would ever think of making 10 marks between 0 and 5 and making the middle one bolder? So, what? 2.5 is somehow more important than 1, 2, 3 and 4? That guy would be fired from his job as a designer immediately.
I don't find it confusing. Don't know how to read a clock?
@@yackawaytube clocks go by ones. each tick = one second. on this each tick = 2.5. completely different since before even starting the problem you have to figure out what each tick represents.
There are 10 marks if you disregard the other 40. Each mark is 100 g, with every 5th more prominent. 2.5 is the halfway point between 0 and 5. This is a flaw of the decimal system. In duodecimal the marks would have been at 3, halfway from 0 and 6.
@@yackawaytube what does have reading clock has to do with this? Look at your clock, where do you have marking in the middle between 12 and 3? There is none, because it doesn't make sense - you have 1 and 2 marked there. You don't make any sense whatsoever, sir.
@@yackawaytube My clock doesnt have a major mark at 2.5 minutes.
The scale being in 1/2 kg increments threw me off initially.
I expect that is the challenge. The students need to check the scale carefully and not make assumptions. Just like you I made the single unit assumption.
Hardest part was counting the markings. Otherwise trivial.
That AI thing was torturous.
Took me 10sec to figure out the 6-->12 if doubled, 2 left over from 14, so boom, green=2.
For some reasons correctly reading the scale was harder than solving it. (I read 7 and 13 instead of 6)
Subtracting the right scale two times from the left solves it, if you manage to read the scale correctly.
Blame old age and eyesight deficiency. That’s my excuse!!
Alternate 7 & 13 solution for misreaders: some weighing scales can handle multiple revolutions; there's no indication on these that 20 is max or modulo.
Here, if the right scale shows 27 kg and the left scale shows 73 kg then green would be 19 kg.
Suppose one square is equal to x and one circle is equal to y. You can get a system of equations: 2x + 3y = 13 and x + y = 7. Solving the system of equations for x and y, and you get x=8 and y=-1, which totally works.
Yeah, I did the 7 and 13 one in my head. But then I came back to complain that 3.5+ revolutions is not realistic for a weighing scale in the real world. The most I've seen is about 1.5, and the 2nd rotation is also clearly marked.
This was easy. Since a square and circle combined weigh six kilograms and the total weight of three circles and two squares is 14 kilograms, a circle weighs two kilograms.
Lol can you explain your process?
@@purplealarmclockeryou can use a linear equation system.
1. 3x+2y=14
2. x+y=6
Then there are different options to solve this. One would be:
(-2)* 2.
1. 2x+3y=14
2. (-2)x-2y=-12
Then you add them and get
y=2
You could also use the second equation to say x=6-y and set it into the first equation
@@purplealarmclocker
Simple: two squares and two rounds are obviously twice as heavy as one square and one round. Thus that must be 12 kg. As two squares and THREE rounds weight 14, one rounds must be 2 kg.
@@purplealarmclocker If a circle and a square weigh 6 kilograms, two circles and two squares will weigh 12 kilograms. But we know that two squares and three circles weigh 14 kilograms, so by adding one circle you added 2 kilograms
Pretty much 2 ways to solve it easily.
1) Double the right side, which would be equal to 12. The left side has 1 more circle and equals 14, thus circle = 2.
2) Triple the right side, which would be equal to 18. The left side has 1 less square and equals 14, thus square = 4 and therefore circle equals 2.
Or the third option: one square and one round is 6, so simply check all options of two weights that can weight 6 in total. Try whole numbers first, you'll get the right answer within 30 seconds.
@@annekekramer3835 True this works, but if the weights were fractions this wouldn't work. The other 2 ways are guaranteed to yield the correct result.
The second method is how I got there within the first 15 seconds of the video and now I feel like a genius.
I love how the 8 year old solution makes so much sense but my brain immediately went to solve it using algebra and I didn't even see the easy solution
The alternative solution is actually genius
my first instinct is to take away a square and a circle twice so that a circle is left
so the required mass is (14-6-6) kg = 2 kg
I'm from singapore but i'm not sure if I learned that when i was 8 years old. I can solve the question now, but i'm pretty sure i learned these types of questions when I was 9 years old rather than 8 years old. This is most likely because every once in a while, Singapore changes their syllabus, and the new syllabus is always harder than the old syllabus, so maybe my syllabus when i was 8 years old was different than the syllabus for 8 year olds now.
I think we get so wrapped up with equations that we overlook a simple solution children would use, which is far easier. My go to was the equation.
The problem is that we have not been taught to do math from the standpoint of someone who posesses a Profound Understanding of Fundamental Mathematics (PUFM). In Singapore Primary Math, students are given this powerful kind of understanding.
Ocham's razor.
Algebra is probably not taught to 8 year olds. So the way to solve this is to observe that on the right we have a square and a circle and if on the left we take a circle and a square twice we get one circle, so if we do 14 - 6 - 6 = 2 (we get the weight of a circle).
Yes we use sets in P2
Sets = We actually draw a circle with the items in each
Sets = Hidden Algebra
P6 Algebra = Basic Notation
So in summary: Technically yes but no
1. C’mon, you aren’t expecting non-integers for small kids…
And since one quarter represents 5, don’t tell me the right arrow points to 7 (almost halfway between 5 and 10) - it’s also a test of number sense on a number line
2. Yes, standard way of learning how to solve a pair of simul. linear equations (addition only) for kids doing Olympiad math kinda training…
green=14-6*2=2
blue=6-2=4
and this can be done in a split second in the head for everyone…
Well, what makes it tricky is the circles turning out to be helium balloons. ;)
I tried to do it without algebra because you said it was for kids. It was pretty easy to guess and check. You can see if your guess is getting closer or further, and also which direction you need to adjust it. I think this method is underappreciated. It's not as fast or elegant, but is simple enough for a kid to understand and very general purpose. Many real world engineering problems don't have algebraic solutions and are basically solved this way.
Used solving for a system of equations to find the answer, then thought - "How would an elementary student solve this?" I then tried a table, and was able to get the answer after the second line. Would love to see an elementary student thinking through this, I think they would have fun if the were no time or grade constraints.😁
Was wondering because it seemed so easy - only to find out that I didn't properly read the scale ... too bad because the rest was so easy.
The left scale is twice the right scale plus one circle. So double the right scale and what's left is the circle.
I can probably do it after spending 10 hours, but im too lazy to spend 10 hours on this
I think this is the 2kg "kitchen scale" design. Typically used to measure ingredients like flour. Which is why it is divided per 50 grams per small line. So the answer is 200 grams.
You either duble the right side and conpare the two weights,leaving 1 green as the dofference, or remove from left side what you know the right side weight.
I think most of you got it wrong. You have to think like an 8 year old who doesn't know fractions. Maybe they know half a kilo. They only know to add what numbers make 6. 1+5 and 2+4. (1+1+1) + (5+5+5) is 18kg which is too much. So the answer is 2 kg and 4kg.
The problem being an elementary problem made the problem very simple tbh, because it made algebra out of the equation.
Another (possibly even simpler) solution for younger children would just involve _successive subtraction:_
One square and one circle is 6 kg. So remove one square and one circle from the left scale, and it becomes 14 - 6 = 8. Then remove one square and one circle _again_ and it becomes 8 - 6 = 2, and you have one circle remaining, so one circle must be 2 kg.
I took a different method from both.
I looked at the difference between the two measurements. It was 8. It was also 1 square and two circles removed. Comparing the one square and 2 circles to one square and one circle, there's a difference of 2, and a difference of one circle.
It solves itself at that point.
I actually algebraically solved the way how you proposed the second solution.
As we have two equations:
2b + 3g = 14
b + g = 6
Multiply the second equation by 2 (choose the multiplier to make b have matching coefficient):
2b + 3g = 14
2b + 2g = 12
Now subtract second equation from first:
g = 2
At least this simple method was tough in my school.
Let:
□ = weight of circle
○ = weight of square
1st: □ + □ + ○ + ○ + ○ = 14
2nd: □ + ○ = 6
Notice if you re-arrange 1st, you'll have 2 pairs of □ + ○:
□ + ○ + □ + ○ + ○ = 14
And □ + ○ = 6, so:
6 + 6 + ○ = 14
○ = 2
"
□ = weight of *circle*
○ = weight of *square*
"
That's some wild mislabeling there.
1:05 " ... or people might make fun of you because you don't know how to start this question" ?! please tell me that our mostly helpful s.t.e.m. community has not sunk down to the level of the twitter/trolls around the rest of the internet 😱
When I grew up most shops hadn't switched to digital weighing scales, so everyone knew how to read analogue weighing scales. Apparently it's a skill people take for granted, because primary school kids 20 years ago would definitely have been able to solve this.
I admire your skills. Very nice solution.
half of 14 is 7, which is 1 square and 1 1/2 circles, the other side is 6, take that away from the half and you're left with half a circle =1, so a circle equals 2.
This one was easier than I expected. The way it is presenting looks daunting, but in the end it's just 14-12.
Left : 2a + 3b = 14
Right : a + b = 6
Left - 2 x Right = 2a+3b - 2(a+b) = 14 - 2x6
b = 2
This seems perfectly reasonable for 8-year olds. In most countries they will have had three full years of education by that point. The left has double the right plus an extra ball. Double the right would be 12kg, so the extra ball weighs 2kg. Algebra is clearly over the top and much slower here.
Bonus question.. solve for possible mass values of the green ball, for whole numbers m, n > 0 and m >= n, such that the readings are 20m + 14 and 20n + 6
After I determined what weights the scales were displaying, I made a guess of 2 and plugged it in to "trial and error" it. Turned out that I "trail and succeeded" it.
Damn those scales are hard to read. I think they read 14 and 6 kg respectively but I keep doubting it because you'd think the thicker white lines between the numbers would mean the halfay mark in which case it wouldn't be 14 and 6 but 14 and 6.25 kg.
The final conclusion is okay, but I actually think it's more likely for the kids to subtract a circle and square from each side (and the associated weight) until they have just a single circle remaining.
so today I learn that when I don't solve math rigorously, I think like an eight year old... lmao!
Once you have ascertained that each mark around the dials represents 500g. (!), you can see one blue and one green together weigh 6kg. and two blues and three greens weigh 14kg. If you took one blue and one green off the left scales, there would now be one blue and two greens weighing 8kg. And if you removed the remaining blue and one of the greens, leaving just one green, the reading would drop by another 6kg. to 2kg. Which happens to be exactly what we want.
(This is exactly equivalent to manipulating simultaneous equations -- doubling "b + g = 6" and subtracting it from "2b + 3g = 14" to get "2b - 2b + 3g - 2g = 14 - 12", which then simplifies to "g = 2". Having some physical props you can actually manipulate just makes the connection between the real world and the mathematical equations that describe what is going on more obvious.)
Maybe the green circles are balloons filled with helium🤷♂️
I asked GPT o1.. It read the weights as 15 and 5, but after I corrected to 14 and 6, it answered correctly:
Answer (in English):
Let’s redefine the problem with the corrected measurements you provided:
• On the first scale: 2 blue squares (B) and 3 green circles (G) weigh 14 kilograms. This gives us the equation:
2B + 3G = 14
• On the second scale: 1 blue square (B) and 1 green circle (G) weigh 6 kilograms. This gives us the equation:
B + G = 6
From the second equation, we can express B in terms of G:
B = 6 - G
Substitute this into the first equation:
2(6 - G) + 3G = 14
12 - 2G + 3G = 14
12 + G = 14
G = 2
With G = 2, substitute back to find B:
B + 2 = 6
B = 4
Result: Each blue square weighs 4 kilograms, and each green circle weighs 2 kilograms.
i wish my elementary books had these levels of questions.
The idea that Gemini helped by "identifying" the image does not make much sense to me.
I thought the problem was about overclocking the scales. Like solving it in mod 20.
Then for k,n NAT>=0
Square = 20(n - 2k) + 2
Circle= 20(3k - n) + 4
With 3k >= n >= 2k
I dont know if this is the only accettable answer because you dont know how many rounds the arrows made on each scale. The equations should be 2s + 3c = nKg + 14 and s + c = mKg + 6 with n > m, n and m are integers and both are divisible by 20
The real trick here is why have we four strong mute tick marks at 2.5, 7.5, 12.5 and 17.0 ? 🤔
that's 7.0 not 7.5. Yeah this diagram is awful, you might as well say the scales are just broken and the problem can't be solved.
Easy. Double the right hand image is one less green circle than in the left hand image. But it would then read 12 kg, compared to the left image reading 14 kg, so one green circle = 2kg
Quick&Easy. There is twice the amount of items on the left scale, plus the one item with the unknown weight. Just double the weight of the right scale and substract the result from the left scale. So one ball weights 2 kg. There isn't even a need to write anything down.
Simple solution could be: if the blue square and the green cicrle are 6 kg weight then 2 blue squares plus 2 green circles are 12 kg. 14 - 12 = 2 (in the left pan)
If I’m reading the scales correctly the 2 squares and 3 circles weighs 14 kg and 1 of each is 6 kg. If 1 of each weighs 6 kg than double that (2 of each) weighs 12 kg. So we have 2 squares and 2 circles weighs 12 kg and to get 2 squares 3 circles we need just one more circle or 14-12kg means a circle weighs 2
kg.
I have to congratulate myself, this is the first problem on this challenge i've solved totally before i've watched video. Including head-on algebraic solution and tricky simple solution, that can be used by 8-year old students and in exactly that order :) The only thing I missed is determination value of division. It just didn't occur to me that this could be problematic.
I calculated it assuming it's 13 and 7 on scales, in this case S = 8 and O = -1 which make no sense, but then I assumed those scales can handle overweight, so basically the arrow just spins the multiple times. So when it is pointed 7 on the second scale, it could be actually 27, one whole spin (20) and one partial (7) spin.
In this case
S+O=27
2S+3O=73
which brings us to S=8 and O=19
I hoped this was the answer, but yea, the author of the question tricks the other way :0
Presh, I found it much faster to simply subtract (S+C=6) from (2S+3C=14) => (S+2C=8). If S+C=6 and S+2C=8, then C MUST = 2 [THE ANSWER], and therefore S MUST = 4
Once you figure out the markings indicate 6 and 14, it's no longer math, exactly, but simple elimination. If one square and one circle is 6, then removing two squares and two circles equals 6x2 or 12, leaving one circle which must equal 2 since you just removed 12 and the weight is 14. No lengthy math problem is needed to solve this.
The second method also has an algebraic equivalent. Go back to the original two equations and subtract twice the second equation from the first. The answer just pops out.
I agree that any algebraic approach is far beyond what a second-grader could handle (even in Singapore). A visual solution was almost certainly what the problem author had in mind.
So simple. Solved in less than a minute
the point of this exercise is to get the student to pay attention to the gradations (½ kg each) & think that half-kilo amounts are admissible in the answer.
having done that, the notion of halving/doubling is introduced, & while it doesn't figure in the actual answer, it leads the student to think of doubling the amounts in the right-side pan: two of each shape totalling 12kg.
the answer just falls out, then.
Seamed easy to me so I'll post my answer and see the rest of the video after. if 🟦🟢= 6, then 🟦🟢🟢= 14-6 = 8. Then take away the 🟦🟢= 6 again to get the answer 🟢= 2kg. 2kg is my answer
Probably the way an elementary school student would do. Love it. My brain would immediately do the basic algebra.
The solution to the problem is so simple, but the author has used all his powers to make it as highly complicated as possible.
Just by visual it is seen that the items on the left scale is twice the items on the right scale plus an additional c. Hence if you remove 2 times the items on right scale from the left scale you will be left with only 1c.
(2s + 3c) - (1s + 1c) - (1s + 1c) = 1c
If (2s + 3c) = 14 and
(1s + 1c) = 6 then
14 - 6 - 6 = 2
Therefore 1c = 2.
When I saw the problem, the two equations from the first method suddenly appeared in my mind just like that, and I thought: what a lame problem. But then Presh told about the chilldren's point of view to this problem, and my mind went WHOW! How simple life can be if you look at things through the eyes of children! 😃
The gradation marks were designed to trip up the students. The solution is easy to find by trial and error, which is probably what the kids were expected to do.
This gave me a boost of confidence. I had to repeat classes twice because of math, but it took me less than a minute., This little ape is stroooong 💪
Circle=2. You can double the weight of square+circle and subtract it from 2 square+3 circle. I did not figure out the exact values, but I did figure out each tick is .5 and the lower right bold line is incorrect.
I just did it by subtracting the equations until I solved for circle, then worked it back up to get the square. I think if you applied the same logic think of them as weights it is basically the same, which would be fine for my 9 year old grand-daughter.
SSCCC = 14
SC = 6
SSCCC - SC = SCC = 14 - 6 = 8
SCC - SC = 8 - 6 = C = 2
solved in my head using a (simplified/better?) version of the final approach. Take 1 green circle off the left balance. Observe the left balance should now be twice the weight registered on the right balance (which would be 12kg). So the removed circle had to be 2kg.
Yes
4:55 S = 4, O = 2, second scale: 4+2=6, first scale: 4+4=8, 8=2S, 2+2+2=6, 3O, 8+6=14, 2S+3O=14. Solved at 4:55
The thing with the scale markings is giving me flashbacks to what pissed me off about elementary school math. They would give you just enough of these questions with tricks in them (like the whole and half unit tick marks being the same length, as in the video) that you would start questioning every problem you saw. On its own, this isn't a bad thing, as it forces critical thinking and examining your assumptions. However, I also remember problems that had mistakes in them that I thought were part of the "trick" only to get the problem wrong because I didn't interpret the problem incorrectly, as intended. I remember catching things like there only being 9 divisions between units on a scale, so my answer was 4/9, but the "correct" answer was 2/5. 😤
I noticed similar issues with my son's homework the other day. He was computing the surface areas of triangular prisms. The worksheet had all three edges of the triangular faces labeled with lengths, but also had them marked as right triangles. Just from a glance, I could tell that the side lengths didn't work out for a right triangle. One of them was even a 1-2-4 triangle. 😂 I get that the teacher was just trying to get them to follow the algorithm of computing the area of each face then summing the areas without needing to know the Pythagorean theorem, but at least give them something that can plausibly be a triangular prism.
Well, you certainly showed A way to do it. But there's an incredibly easier way:
double the equation s+c=6 and you get 2s+2c=12.
So: if 2s+2c=12
and 2s+3c=14
Then you can subtract (which is more intuitive with the pictures, thus making this easier0 and get
c=2
You forgot to add that the guy who built a scale with 1/2 tick marks is bonked in the head.
Since circle + square = 6 (requires a close look, but that’s the hardest part of the problem), removing 2 circles and 2 squares from the left side (14 -12) leaves 1 circle = 2kg. Voila!
Subtraction is a lot more intuitive here. Just subtract 1S,1O twice. You get 14-12=2
solved under a min
There was no need for algebra here, as all you have to do is take away two of each from the left side, and you already know the weight of one of each.
File this one under "solved at a glance."
#me2
I didn't even think about the 2nd solution. Which seems more clever and fast.
Technically, the correct answer would be 2+20n, where n is a nonnegative integer, assuming the scale can go over 20
actually it's simple; we have s+c= 6kg and 2S+3c=14
so if we double the first one we will get 2S+2C=12kg
so 2S+3C - 2S+2C= 14 - 12 = 2kg
I was wondering why a video was made about this on this channel. Now that I see that if 2s + 3c = 14, and s + c = 6, then that implies 2s + 2c = 12, where we can just subtract the equations to get c = 2. I didn't see the quick way.
X + Y = 6
2X + 3Y = 14
2X - X + 3Y - Y = 14 - 6
X + 2Y = 8
X - X + 2Y - Y = 8 - 6
Y = 2
Straight forward from there.
Probably an easier way to do it. Looking forward to seeing how you solve it.
count the marks to five to get ten marks for every 5kg then you remove the tick marks of the right from the left twice it makes four marks left for the circle making it 2kg for circle and 5kg for square.
Since we want to know C, and we have 2 equations 2S + 3C = 14 and S + C = 6, it is dead simple to eliminate S and leave one C by subtracting the latter equation twice from the first, immediately providing the answer C = 2.
Was really curious to hear the non-algebraic way to solve, because every idea I had came back to algebra. Although I’d argue your non-algebra approach (at 7:00) is basically algebra anyway. Most curious that AI was so useless.
DUH! We have been trying to solve the problem as adults.
A child would take one green ball off either scale. The amount of loss is the weight if the ball. QED.
Even easier, remove the blue square from the right hand scale, then read directly the weight of the green ball. 👍
I did it pretty much exactly the same way you did it in the video: solved it the hard way first, then took a second look and said to myself, "Oh cool, there's an easy visual shortcut."
Hm. “Hard” is a personal metric. I immediately saw a pair of simultaneous linear equations, which required a multiplication by 2 and a subtraction to get the answer. Since I didn’t have to get creative, that was as far as I went. Problem is that since I knew how to solve this ab initio, I didn’t think how an 8 year old might have tackled it.
@@johnlister Fair point! Forty years ago, as a would-be mathematics major, "hard way" might have been a bad way to phrase it, but I never completed the degree and am way out of practice, as my life went down other paths.
So, square plus circle equals six. If each one is 3 kg, 3×5 is 15, which is incorrect, the one with more weighs 14 kg. Since I want the next thing to be less, I made the squares 4 and the circles 2. 4+4+2+2+2 does indeed equal 14 so the circle must equal 2. I’m not typically one to use guess and check, but that’s how a little kid would solve it
Why is the fourth tick after 5 on the scales bolded, not the fifth?
When I was a second grader, my reasoning would have been that the answer should be an integer (because I didn't yet know about fractions and decimals, well, not really).
3+3 works on the right but not the left (that would be 15).
So try 4+2.
4+4+2+2+2=14.
So C=2.
I have since been told that this is "using arithmetic to solve algebra," and is not permitted.
This is really an 8th grade math problem. The solution would be that a square weighs 4 kg and a circle weighs 2 kg.
2S+3C=14 : S+C=6 : multiply second formula by 2 it becomes :-
2S+2C=12 : subtract from first formula gives :-
C=2 : Simple simultaneous equation but without my glasses to count the divisions, it would have been
impossible. 🤓
When you know algebra, sometimes you can just complicate the simplest of problems.