I just figured that the difference between sitting cat and sleeping cat is (150 - 110) / 2 = 40 / 2 = 20 cm, so if for example both cats were sleeping, it would obviously be 130 cm and the height of 2 sleeping cats cancels itself out, it's basically 130 + a - a.
But how did you get the height difference between sitting abd standing like that? What did you do? You know it's 20 because 150 - - 20 - 20 = 110 but I don't know what you did.
@@maxhagenauer24 So, we know the height measured from the top of a sleeping cat on the floor to the top of another sleeping cat on the table is the same as the height of the table, because that's basically (t - a) + a. The first diagram has the sitting cat (c) on top of the table, so that height (150cm) is t plus the difference in heights between a sitting and a sleeping cat (c - a, or we can call that difference h). So the first diagram tells us t + h = 150. The second diagram has the sitting cat on the floor, so the difference between the two heights is subtracted from the height of the table, or t - h = 110. Subtract the second equation from the first; the t's cancel out and you're left with 2h = 40, which reduces to h = 20. So the difference in heights between a sitting cat and a sleeping cat is 20 cm. What OP did was observe visually that the difference in heights between the two diagrams was twice the distance h, and the difference in heights can be found by subtraction to be 40cm, so h = 20cm. Then it's just a matter of plugging that into either original equation to get 130cm.
@@maxhagenauer24Basically, the cats in the first diagram have their heads as far away as possible given the 2 positions. (sitting and laying) The second diagram shows us that when both cats change position toward each other, the distance decreases by 40. Therefore, since the cats are identical, you can infer that if only one of them changed position then the distance would change by 20.
It is possible that China's education system pushes these challenge problems to elementary level students to find the more advanced students. As you say, there is the out-of-the-box visual solution but also some of the elementary students are more advanced than the others, most likely because they have educated relatives who motivated and tutored the young student in Math. I have seen this in a 3rd grade US math classroom in which I volunteered for a year. The few advanced kids were bored and could have benefited if they skipped a grade or 2 for their math class period. WIth online programs like Kahn academy and YT channels like yours an interested student could easily be several grades ahead and it's a shame our local elementary school could not meet them where they are in their Math learning pathway because they will be better prepared math-wise when they later hit science and engineering classes. If this is China's goal, it's a smart idea. BTW -- I love your YT channel.
I'm glad online learning is an option now. A lot of kids can benefit from independent learning, where just a couple of decades ago you would need to set up appointments and possibly spend thousands of dollars for a tutor.
@@michaelgarrow3239 It is very basic if you have the apparatus to solve it, i.e. solving a set of linear equations by addition/substraction. Solving it without toolbox of such concept advanced (for kids).
While intuition would not be accepted as a valid method at mid level, at the higher level, dressed up with the fancy words of "symmetry" and "linearity", it again becomes a valid method.
@@marcosolo6491 I agree. I thought that average would turn out to be the correct answer but I didn't care to make the effort to prove it. I was happy to see it proved to be correct
Quite common thing in engineering, when you have to measure some part's dimension and its tolerances but don't want to (cannot) consider all tolerances of the setup and measuring devices. You flip the setup/ rotate the shaft/ etc, sum up the total run out and take the half of it. Also helps finding centerlines
Im willing to believe that the fact that this question was asked to elementary school students may just be blatant misdirection or just chinese propaganda. They only specified that they 'asked' the students this question, not if anyone had answered correctly. So we can conclude that, idk. Cool problem tho!!!
For me: visually I could see that the average height of the sleeping/sitting cat would be the same in both diagrams therefore cancelling out and the height of the table is the average of the dimensions given = 130cm
This is what I did intuitively. Upon reflecting I believe you're correct it's just an averaging problem. Total the height deltas, and divide by the number of measurements. Interesting problem.
This reminded me of a few equivalent concepts i forgot about. Basically, if both kittys lay down or stand up together, the difference is equal to the height of the table. Its been too long since I have needed this type of math. The cats get 20 taller from standing so add it to simulate a standing cat in place of a laying one, or subtract to simulate a laying one in place of a standing one.
At first I thought the problem had no definitive solution, but I figured it out after a few minutes. There's no way I would have been able to solve it in 6th grade though, grade 9 or 10 seems much more appropriate for the average student. Still, it's a really cool math problem with an unusual solution!
I just used the midpoint of the difference between the heights of the two cats… So 150 is like an “upper bound” of sorts And 110 is like a “lower bound” of sorts Therefore the difference is 40cm and the midpoint of that is 20cm. 110+20=130cm
🫤or if you wanna just do it in your head notice the difference is 2 cat heads and the difference in length is 40 so 1 cat head equals 20 now take the right image and move the 110 line down by a cat body length since it cancels and then add the cat head. I know the way I wrote it might be confusing but it is the easiest way to do it
Like a few others here I just looked at the difference in size of the cats as a single variable, so basically you have t+diff on one side and t-diff on the other.
This is simpler: From the 1st drawing we know: t+c-a=150 From the 2nd drawing we know: t+a-c=110 Add the two equations together, c and a cancel: 2t=260, hence t=130
standing A//laying down B//Table C no point in overcomplicating with extra T you used: A+150-B = C && B+110-A = C => A+150-B+B+110-A = 2C => 150+110 = 2C => 260 = 2C => C = 130
I just imagined that all of the cats are half-sleeping and half-sitting. Also the problem as presented is symmetrical, so intuitively I couldn't imaging there being any other solution than the average of the two given heights.
Solved from thumbnail, I do have to say that it is quite a tall table. Probably a sidetable of sorts, because most dining tables are 75 cm high. Table = 150 + Sleeping - Sitting Table = 110 + Sitting - Sleeping 2Table = 260 + Sitting - Sitting + Sleeping - Sleeping 2Table = 260 Table = 130
I knew instinctively that it would be the average of the two numbers because the sitting cat and sleeping cat cancel each other out, but seeing the formula was a more exciting way to travel along the answer
More simply, set the table height to t, the squatting cat to x and the standing cat to y. Then, (1st image) t - x + y = 150 = t + (y - x) and (2nd image) t - y + x = 110 = t - (y - x) Now, simply add the two equations to get 2 t = 260 => t = 130
I solved it in two seconds honestly: a) T - sittingCat + standingCat = 150; b) T + sittingCat - sittingCat = 110; If A and B are true, then their sums is also true, so A and B imply: 2T - sittingCat + sittingCat + standingCat - standingCat = 150+110; 2T = 260; T = 130.
Looking at the picture you can see that one pic adds the height difference while the other subtracts it. 150-110 = 2x(A-C) =40 Standing cat minus sleeping cat equal 20 Now just add it to 110 + 20 = 130
I figured how to do it a bit more intuitively, if you make a horizontal line towards the oppostie table in each picture, you'd see that both segments which are being intersected are both the head of the standing cat, thus if we do 150-110cm we get that the standing cat is 40cm tall if we assume that the laying cat is half the size of the normal cat and that the head of the standing cat is about the same height as the laying cat based on the picture and the lines interesecting them you could either do: 150 - 40 (standing cat) + 20 (laying cat) = 130 cm or 110 - 20 (laying cat) + 40 (standing cat) = 130 cm
If you really want to think out of the box , you put one table on the other and then you use the third dimension to wrap the two tables like a cylinder, which gives you a circle with circumference 2 tables equals 260cm, done.
Ht = 150 - HC + Hc; Ht = 110 - Hc + Hc. You add them together and get 2Ht = 260 so Ht = 130. Where HC height of standing cat and Hc is height of sleeping cat.
I’m not sure which solution this falls into but I saw that the change in height was 40 cm, divided by 2 for the number of cats changing positions, then add or subtracting that difference to make both cats sitting or lying down, canceling out their influence, equaling the height of the table. Not sure how any of this would work if the cats were of different sizes.
This is even simpler when you ask yourself what would happen if the cats were actually of equal height. The images would be the same and the number would directly show the height of the table. Then imagine that you make one cat X cm taller. On one image X would have to be added to the number but on the other image the same X would have to be subtracted from the initial number (the table height). You're just adding and subtracting the same number resulting in two numbers (150 and 110 in this case) that must be equally distant from the initial number (the table height). So the number we're looking for must be in the middle - arithmetic mean.
It can also be inferred from the data given that the *"SITTING" cat is 20 cm "taller"* than the "lying / sleeping" cat. That's NOT asked in the question, but it's still a fun data inference. :P
X = sleeping cat’s height Y = sitting cat’s height T = table’s height X + 150 = T + y Y + 110 = T + x Rearrange those to align the variables: T = x - y + 150 T = -x + y + 110 Add these 2T = 0x + 0y + 260 T = 130
Diagram 1: 150cm, taller than table by difference of standing to sleeping cat, Diagram 2: 110cm, shorter than table by difference of standing and sleeping cat. Find averages, tah dah. (Didnt take 9 minutes)
For me it was: I have the two extreme values given for a linear problem so the answer about the constant that doesn't change must be the average between the 110 and 150. To double-check: the height of the table would be as measured if both cats were sleeping or standing. if the difference between the standing and sitting cat is d, then the difference between the two problems is 140 or 2d, so d is 20. The height of the table is 110 + 20 or 150 - 20.
A much simpler solution, the difference between the 2 measurements is 40 cm, and so the the height of the table is 110 + (40/2), or 150 - (40/2). 130cm.
I love to use the figurative method because it stays in touch with story in the problem better than the algebraic approach. And I have used it also here, to figure out the heights of the cats and the table. I have represented the two situations as given in the problem and added a rotated representation of the second situation in order to align the two sitting cats. This way it became evident that the height of the sitting cat is 40 cm. Finally, the comparison of the three representations revealed that: -the difference of height between the two situations is also the height of the sleeping cat and is equal to 20 cm; -the height of the table is equal to 150-40+20=110-20+40=90cm. Now I will continue to watch the video to see how Mr. Presh Talwalkar have solved this nice problem.
I noticed that one diagram included a sitting cat and the other diagram included a lying down cat. It's the difference between the two diagrams was the difference between sitting and laying down then you could just take the difference between the two measurements and split it equally. 150 - 110 equals 40 this was 20 added to the sleeping cat and 20 removed from the standing cat You can then add 20 to the lying down cats and take 20 away from the sitting cats measurement and you get the tables measurement of 130
I: y+x=150cm II: y-x=110cm I+II: 2y=260cm => y=130cm y is the table height, x the cat contribution to the total height. the problem is not that hard in my opinion and kids are good at thinking
I just assumed that the cats where all the same size. In my head I had the cat on top of the right table stand up. Splitting the difference between the measurements (40cm) to get 20cm. So if both cats are standing, you can measure the size of the table by measuring between to top of their heads. 110 + 20 = 130.
I solved this by realizing that the difference in the two measurements presented are double the difference in height between a sleeping cat and a sitting cat. Compared to the cats on the left, the cat on the top right has gone from sitting to lying down, and the cat on the bottom right has gone from lying down to sitting up, so the difference in height between sleeping and sitting is subtracted twice from the 150 cm measurement on the left to get 110 cm. 150 - 2x = 110. x = (150 - 110)/2. x = 20 cm. Subtract 20 cm from 150 cm to get the difference between a cat lying on the table and a cat lying on the floor, i.e. the height of the table. 150 - 20 = 130 cm.
You said the cats switched positions, but it could also be that table cat laid down and floor cat stood up. Then you wouldn’t be able to solve unless the cats were said to be the same height. To demonstrate switching positions, you could color standing cat black and leave sleeping cat as tabby
There is a simpler way to solve the problem visually. The "error" in measuring the table height is caused by the difference in the height of the 2 cats. In one case, it makes the measurement too high, in the other, too low. The magnitude of the error is the same, so the correct height is the average of the 2 measurements. 6th grade school students may not know how to calculate average, but they can "pick the number in the middle".
Visually... If you put the 2nd table on the 1st table... ... you know the height from the lower to the upper sleeping cats... (150+110)... ...which is the same as the height of 2 tables... ... So one table is (150+110)/2 = 130
Easier method that is about 2 lines: 1:30 When the cat goes from sleeping to standing, it is x units taller. The measured distance on the left is 150 units. The measured distance on the right is 150 minus the distance x lost from the bottom cat's action, minus the distance, again x, lost from the top cat's action. Thus, 150 - x - x = 110, or 2x = 40, or x = 20. Thus, if in the left picture, the top cat was sleeping, the distance would be 130 units. Take the cats away and shift the measurement's start point down to the ground, and 130 units is the distance from the floor to the table's surface.
At first sight, the problem is a system of equations with two equations and three variables, but since we don't really care how tall the cats are we can rework their difference as a single variable (c - a), which exists in both equations with different signs [(t + (c - a) = 150] & [x - (c - a) = 110].
I didn't think to add them together. I thought to make both equal to 't' t = 150 + a - c t = 110 + c - a Make em equal to each other and see what to do from there 150 + a - c = 110 + c - a Get constants to one side and variables to other and 40 = 2c - 2a 40 = 2 ( c - a ) 20 = c - a Go back an see if i can do something with this and saw that i can with the second equation t = 110 + ( c - a ) t = 110 + (20) t = 130
An easier way to solve this puzzle is by using Quantum Mechanics, and replacing both cats by Schrödinger's cats. Schrödinger's cat are in a superposition of the sleeping (a) and sitting state (b): |phi> = c|a> + c|b>, with c being (1/2)^2 Since both cat states |a> and |b> are equal likely, we can easily calculate the height of the table from the expected height-difference between both Schrödinger's cats, which is the arithmetic mean of the two measurements: t = (150 cm + 110 cm)/2 = 130 cm. q.e.d Pretty easy that way, isn't it?
If you sum both you get the measurement for a table without a sitting cat plus a table with a sitting cat. The two sitting cat nullifies themself, getting you two table is equal to 260.
Here is a harder scenario: A student is asked to measure the height of a table, but not allowed to measure it directly. Instead, the student is given a bag with 20 objects in it, each with a unique height. The student must pick randomly from the bag 2 items. One of the 2 items the student chooses to place on the floor, the other on the table. The student is then allowed to measure the distance between the top of the two objects. From these measurements the student must calculate the true height. How many random pulls from the bag will guarantee that the student can determine the true height of the table? In other words, what is the minimum of the maximum number of pulls required?
I really think you should explain the rationale behind adding the two equations together. Otherwise, one might just think this is a "trick" (as you described it). It's not something I have ever been taught. I would have gone with two simultaneous equations and rearranged to eliminate the cat variables, leaving just the table, which is effectively what the second solution did, albeit based on a visual representation.
Grades 9 and 10 are effectively high school level in my country. I was enrolled in a special science high school and we were already doing pre-calculus problems at the equivalent grade levels. Admittedly, regular HS in our country would be only doing algebra and this should be a problem they could be presented with.
If you look at the picture at 1:57 you will notice that the difference of 150cm and 110cm is just the height difference between standing and sitting cat times 2, and it's evenly distributed on the top and bottom. So 150cm - 110cm = 40cm, divide that by 2 is 20cm. That is the difference in height between standing cat and sitting cat. If you measure from top of sitting cat on floor to the top of sitting cat on table, that is equivalent to the height of the table. Swap the standing cat in left picture to sitting cat, minus 20cm from 150cm. Table height is 130cm.
I went straight to solution with intuition. We have only two numbers to play with. Basic operations to do with them is adding, subtracting, which doesn't look right. Must be something slightly more funcy, like arythmetic mean. Yeah, must be the mean. The two numbers are uniform physical quantities and result is also uniform. That is we are given two lengths and asking for a length. We're not asked e.g. an area. If so we could argue that multiply would be the sensible operation with just two given numbers. But since we are just manipulating lengths to get a length, must restrict sensible operations to just those stated. Adding, subtracting, bysecting, lengths into lengths... mean just came naturally.
Here's how I did it in my head, with no algebra. Suppose the cat on the left went to sleep. That would reduce the 150 measurement by the difference in heights of the sleeping and sitting cat. And that measurement would now be exactly the height of the table. Now suppose the sleeping cat on the left sits up. That would reduce the measurement by the same amount again, and we're at 110cm. So the table height is exactly half way between 150 and 110, where both cats are sleeping. That is, 130 cm.
So, I did t=150-c+a and t=110-a+c (Height of table equals top cat height - given height + bottom cat height) Then, 150-c+a=110-a+c, 2a-2c=40, a-c=20 Then, in the one with the sleeping cat on top, if I substitute the sleeping cat for the standing cat, I add 20 ((the difference between the sleeping and standing cat heights) to the given height, 110, giving 130. Since this is the distance between two cats of the same height, one on the floor and one on the table, they cancel out and the table's height is 130cm.
The difference between the numbers (40 cm) is twice the difference between the heights of the cats (once above and once below). Replace the cat sleeping on the table with the standing cat, and the new distance (110+20=130 cm) is the height of the table. A smart 3rd-grader should be able to solve this. For a 9th-grader (algebra 1), this is a second week exercise in setting up equations, maybe.
My approach, using similar algebra, was to note that, in the left diagram, the 150cm can be considered as the sum of the height of the sitting cat and the distance from the top of the sleeping cat to the table. The former is of course C and the latter is T-A. The gives equation 1 as T - A + C = 150. The corresponding observation in the other diagram gives equation 2 as T - C + A = 110. These equations can the simply be added to make C and A disappear, giving 2T=260. Now this T=130 can be plugged back into the two equations. But can we solve for C and A? No! As we might expect from having two equations with three unknowns, we can't solve for them all, even though we were able to solve for T. So what happens when we try to plug in 130 for T? Then equation 1 becomes 130-A+C=150 and equation 2 becomes 130-C+A=110. The first becomes C-A=20 and the second becomes A-C=-20. They're equivalent. We can't figure out how tall each cat is, only that the sitting one is 20cm taller than the sleeping one.
150 - 130 = 40 which implies that the sitting cat is 20 taller than the sleeping cat. On the left side, if you replace the sitting cat with a sleeping cat, the difference in height would be 130 which implies that the table is 130. You could also replace sleeping cat with sitting cat or do these replacements on the right side.
Another proof that the individual cat height is non-revelant is to see the same problem, but this time with a sitting cat and a cat standing on its rear paws. Now the sitting cat is the shortest, the standing cat is 20 cm taller and the measures remain the same at 110 and 150 cm.
Before trying to calculate anything, my gut told me that the answer would simply be the average of the two heights, and that intuition turned out to be correct. Think about it this way: let's say you have a slider that controls the heights of the cats so that as one cat gets shorter, the other one gets taller by the same amount. If you move the slider all the way to the left so that the heights are like in the first picture, then the difference between their heights is 150 cm. If you move the slider all the way to the right where it's like in the second picture, then their difference is 110 cm. But what happens if you move the slider exactly halfway? The distance between the two cats' heads will be 130 cm, because the slider changes that difference linearly. But also, the two cats will be exactly the same height, which means the distance between their heads is exactly the height of the table! Thus, the table is 130 cm tall, because it's the average of the two distances.
so i feel i cheesed it and was like "what if I make sleeping cat 0 it should work, because it just have to be less the the standing cat". That at least simplifes it to just the cant standing and setting being the differents but still not trivial for elementary students to solve it.
2X + 110 = 150 2X = 40 X = 20 150 - X = 130 I'm not great at math, but this is how I did it. Where X is the difference between a sleeping and sitting cat.
Went at it in a completely different way i isolated 'L'aying to equal 110cm+'S'tanding-'T'able, and substituted in to the other equation getting T+S=150cm+110cm+S-T. Moving unknowns over to the first side i get T+T+S-S=260cm, and after cancelling we get 2T=260cm => T=130cm
Both of these solutions seem way too complicated. All you have to realize is that the difference in measurement between the identical cats in two different positions totals 40 cm, meaning each side is contributing half that (20cm) when measured, and the rest falls into place.
It’s the average between the two height scenarios. Then the cat height is the same, and only the difference matters. It took 2 seconds. I don’t see why this is considered hard by any stretch.
Heght difference between sleeping cat and sitting cat is: c-a This height difference (c-a) is equal to the difference in total height of each of the two situations depicted, or: (c-a)=(150+a)-(110+c) Solves as: 2c-2a=150-110 2(c-a)=40 (c-a)=20 Substitute that back into either equation posed in the video, you get your height for "t". Isn't the most interesting part of the problem the fact that there is an infinite number of possible heights for the cat???
I figured that in the second picture, the cat on the floor had grown by twenty and the cat on the table had shrunk by twenty, given that the difference is forty and the standing cat grows by just as much as the sitting cat shrinks. So then if the standing cat in the second picture lays down, it will increase the measurement to 130, which would be the height of the table because you are measuring from the top of one sleeping cat on the floor to the top of another sleeping cat on the table. Cancel out the cats, and you get the table.
This is also elementary school level in Eastern Europe. Around 5th grade, I'd presume. What's with the American educational system that they need to be 9th or 10th grade to solve it, and at a competition no less! Competition problems are usually much harder than normal homework problems.
I solved a slightly different, visual way. In both cases, lower the measuring line by the height of the sleeping cat. On the Left, it rests on the floor and the difference is the standing cat body. On the right, it is missing the standing cat body in order to reach the floor. The difference between the 2 lines is 2 standing cat bodies. In math, the difference between the 2 lines is 40cm (150-110), so each standing cat body is 20cm. 110cm + 1 standing cat body is 130cm. 150 - 1 standing cat body is 130cm. The table is 130cm.
All I did was solve for the sleeping cat. Either works though. They went from peak distance in the left image down to shortest distance. The reduction was a distance of 40. Since both cats moved, we can divide 40 by 2. That means the distance between 2 sleeping cats is 150-20 so the table is 130. I like the first answer better than my method though.
I did it a slightly different way. Since we are treating the height of the sitting cat as a fixed quantity, and the height of the sleeping cat as a fixed quantity, it's clear that (for instance) measuring from the top of a sleeping cat on the floor to the top of a sleeping cat on the table would equal the height of the table. In one diagram, we have that height PLUS the difference in heights of a sitting vs sleeping cat (c - a) and in the other diagram we have that height MINUS the difference in heights (c - a). Let's call this difference (c - a) by its own variable, h. So in one case we have t + h = 150, and in the other we have t - h = 110. By subtracting the second equation from the first, we get 2h = 40, or the height difference between a sitting and sleeping cat is 20cm. Going back to either of the original equations t + h or t - h and plugging in our solution for h gives us t = 130cm, which was easier to do in my head than it was to write out.
Notice, also, that the height of the table is also the average height between the two. Taking the average essentially cancels out the difference in height between the two cats.
Variables c a t; nice!
I thought that was funny too
@@seentome9783 i totally missed that 😹
I just figured that the difference between sitting cat and sleeping cat is (150 - 110) / 2 = 40 / 2 = 20 cm, so if for example both cats were sleeping, it would obviously be 130 cm and the height of 2 sleeping cats cancels itself out, it's basically 130 + a - a.
Yeah, that's what I did too.
But how did you get the height difference between sitting abd standing like that? What did you do? You know it's 20 because 150 - - 20 - 20 = 110 but I don't know what you did.
@@maxhagenauer24 So, we know the height measured from the top of a sleeping cat on the floor to the top of another sleeping cat on the table is the same as the height of the table, because that's basically (t - a) + a. The first diagram has the sitting cat (c) on top of the table, so that height (150cm) is t plus the difference in heights between a sitting and a sleeping cat (c - a, or we can call that difference h). So the first diagram tells us t + h = 150. The second diagram has the sitting cat on the floor, so the difference between the two heights is subtracted from the height of the table, or t - h = 110. Subtract the second equation from the first; the t's cancel out and you're left with 2h = 40, which reduces to h = 20. So the difference in heights between a sitting cat and a sleeping cat is 20 cm. What OP did was observe visually that the difference in heights between the two diagrams was twice the distance h, and the difference in heights can be found by subtraction to be 40cm, so h = 20cm. Then it's just a matter of plugging that into either original equation to get 130cm.
Yea this is more or less how I approached it too. This is probably the kind of reasoning they expected when giving it to younger students in China.
@@maxhagenauer24Basically, the cats in the first diagram have their heads as far away as possible given the 2 positions. (sitting and laying)
The second diagram shows us that when both cats change position toward each other, the distance decreases by 40. Therefore, since the cats are identical, you can infer that if only one of them changed position then the distance would change by 20.
Easy breezy, Z is sleeping and S is standing
T+Z-S=110
T-Z+S=150, add both lines together
2T=260
T=130
and shorter: (110+150)/2 = 130
I did it this way (and with same variable names too). It feels more intuitive.
That's *not* the answer. The answer is *130 centimeters.*
@@forcelifeforce you must be fun at parties
I did the same as well.
@@sadilhas9118 I mean it could be 130 dead orphans according to my teacher so the unit is important
The cats cause measuring errors, equal but opposed, so they average out to being correct.
Meowsuring errors?
Maybe I guess, but I'm no expert, for all I know they might be meshpurring errors.
Great idea! 👏
Who else remembers this problem back when the sleeping cat was a turtle?
It is possible that China's education system pushes these challenge problems to elementary level students to find the more advanced students. As you say, there is the out-of-the-box visual solution but also some of the elementary students are more advanced than the others, most likely because they have educated relatives who motivated and tutored the young student in Math. I have seen this in a 3rd grade US math classroom in which I volunteered for a year. The few advanced kids were bored and could have benefited if they skipped a grade or 2 for their math class period. WIth online programs like Kahn academy and YT channels like yours an interested student could easily be several grades ahead and it's a shame our local elementary school could not meet them where they are in their Math learning pathway because they will be better prepared math-wise when they later hit science and engineering classes. If this is China's goal, it's a smart idea. BTW -- I love your YT channel.
It took me 10 seconds to solve it. This isn’t advanced.
we think that the cats are the same size? The quiz does not say that the cat that was sleeping below should jump up to sleep on the table.
I'm glad online learning is an option now. A lot of kids can benefit from independent learning, where just a couple of decades ago you would need to set up appointments and possibly spend thousands of dollars for a tutor.
@ school is unnecessary.
@@michaelgarrow3239 It is very basic if you have the apparatus to solve it, i.e. solving a set of linear equations by addition/substraction. Solving it without toolbox of such concept advanced (for kids).
For me it was just intuitive that it would be the average between the 2 measurements
Me, too.
While intuition would not be accepted as a valid method at mid level, at the higher level, dressed up with the fancy words of "symmetry" and "linearity", it again becomes a valid method.
@@marcosolo6491 I agree. I thought that average would turn out to be the correct answer but I didn't care to make the effort to prove it. I was happy to see it proved to be correct
i was expecting that, but, wasn't sure if my logic was good...thanks for saying this.
Given…
S = stand up
L = lie down
T = table
…we have…
S + T - L = 150
-S+T + L = 110
…adding yields…
0 + 2T + 0 = 260
2T = 260
T = (260/2)
*T = 130*
lol i literally just did the problem out of curiosity and did it the exact same way
@@ageometrydashlevel9830 This solution seems readily apparent. I'm not sure why the problem justified a 9-minute video, featuring fancy graphics.
Quite common thing in engineering, when you have to measure some part's dimension and its tolerances but don't want to (cannot) consider all tolerances of the setup and measuring devices. You flip the setup/ rotate the shaft/ etc, sum up the total run out and take the half of it. Also helps finding centerlines
Im willing to believe that the fact that this question was asked to elementary school students may just be blatant misdirection or just chinese propaganda. They only specified that they 'asked' the students this question, not if anyone had answered correctly. So we can conclude that, idk. Cool problem tho!!!
For me: visually I could see that the average height of the sleeping/sitting cat would be the same in both diagrams therefore cancelling out and the height of the table is the average of the dimensions given = 130cm
This is what I did intuitively. Upon reflecting I believe you're correct it's just an averaging problem. Total the height deltas, and divide by the number of measurements. Interesting problem.
Same
Same here.
I just saw each cat's delta to be 20 cm, which you can add or subtract to get 130.
x+h-y=150
y+h-x=110
2h=260
h=130
This reminded me of a few equivalent concepts i forgot about. Basically, if both kittys lay down or stand up together, the difference is equal to the height of the table. Its been too long since I have needed this type of math. The cats get 20 taller from standing so add it to simulate a standing cat in place of a laying one, or subtract to simulate a laying one in place of a standing one.
That's how I did it!
At first I thought the problem had no definitive solution, but I figured it out after a few minutes. There's no way I would have been able to solve it in 6th grade though, grade 9 or 10 seems much more appropriate for the average student. Still, it's a really cool math problem with an unusual solution!
I feel like this was uploaded yesterday.
I feel like I share your feeling. 😄
Strange though, same video, different title, why? 🤔
It is also reuploaded from 6 years ago.
this is also i felt, i was also planning to see this video yesterday
Deja vu is a sign there has been an alteration made in the Matrix…
from the description area under the video: "I re-uploaded this video to fix a typo at 8:15. "
I just used the midpoint of the difference between the heights of the two cats…
So 150 is like an “upper bound” of sorts
And 110 is like a “lower bound” of sorts
Therefore the difference is 40cm and the midpoint of that is 20cm.
110+20=130cm
🫤or if you wanna just do it in your head notice the difference is 2 cat heads and the difference in length is 40 so 1 cat head equals 20 now take the right image and move the 110 line down by a cat body length since it cancels and then add the cat head.
I know the way I wrote it might be confusing but it is the easiest way to do it
Imo that was what 4th grade students were supposed to do. All other methods using equations are for higher grade students.
Like a few others here I just looked at the difference in size of the cats as a single variable, so basically you have t+diff on one side and t-diff on the other.
This is simpler:
From the 1st drawing we know: t+c-a=150
From the 2nd drawing we know: t+a-c=110
Add the two equations together, c and a cancel: 2t=260, hence t=130
The very same problem was already posted on this channel 6 years ago: ua-cam.com/video/BPRueCu3fXU/v-deo.html
standing A//laying down B//Table C
no point in overcomplicating with extra T you used: A+150-B = C && B+110-A = C => A+150-B+B+110-A = 2C => 150+110 = 2C => 260 = 2C => C = 130
that's 3 variables, only you used the 'b' instead of the t. OP made it spell out cat as a joke..
@@Jacob-W-5570 I didnt imply that I used less variables just that I go in alphabet order, to not over complicate.
I just imagined that all of the cats are half-sleeping and half-sitting. Also the problem as presented is symmetrical, so intuitively I couldn't imaging there being any other solution than the average of the two given heights.
This is normal elementary school question. Idk why you put that in the title as if it was some anormal thing
I redrew the picture in the notes app and used a simple system to figure out and was pleasantly surprised to see I was right.
A grade 10 competition problem given to elementary kids to solve is actually crazy.
Just stack two pictures together , up and down. Its height is two table with +a -a, so the table tall (150+110)/2= 130.
Solved from thumbnail, I do have to say that it is quite a tall table. Probably a sidetable of sorts, because most dining tables are 75 cm high.
Table = 150 + Sleeping - Sitting
Table = 110 + Sitting - Sleeping
2Table = 260 + Sitting - Sitting + Sleeping - Sleeping
2Table = 260
Table = 130
I knew instinctively that it would be the average of the two numbers because the sitting cat and sleeping cat cancel each other out, but seeing the formula was a more exciting way to travel along the answer
Yes. I thought giving both the mathematical explanation and a visual explanation will help develop intuition when it comes to similar problems.
The positive difference between sittning and laying cat, delta c
Table is 150- delta c and 110+delta c
Symmetri gives table, midpoint, 130
More simply, set the table height to t,
the squatting cat to x
and the standing cat to y.
Then, (1st image) t - x + y = 150 = t + (y - x)
and (2nd image) t - y + x = 110 = t - (y - x)
Now, simply add the two equations to get 2 t = 260 => t = 130
I solved it in two seconds honestly:
a) T - sittingCat + standingCat = 150;
b) T + sittingCat - sittingCat = 110;
If A and B are true, then their sums is also true, so A and B imply:
2T - sittingCat + sittingCat + standingCat - standingCat = 150+110;
2T = 260;
T = 130.
I solved it a bit differently. I started with 150-c+a=t and 110-a+c=t which simplifies directly to 260 = 2t and t=130.
I like this one a lot.
H + T - L = 150 with H = high cat L = low cat and T = table
L + T - H = 110
------------------------- +
2T = 260 so T = 130 cm
I saw that the difference between standing and sitting was 20, then just put them both in the same pose by adding/ subtracting 20.
Looking at the picture you can see that one pic adds the height difference while the other subtracts it. 150-110 = 2x(A-C) =40
Standing cat minus sleeping cat equal 20
Now just add it to 110 + 20 = 130
I figured how to do it a bit more intuitively, if you make a horizontal line towards the oppostie table in each picture, you'd see that both segments which are being intersected are both the head of the standing cat, thus if we do 150-110cm we get that the standing cat is 40cm tall if we assume that the laying cat is half the size of the normal cat and that the head of the standing cat is about the same height as the laying cat based on the picture and the lines interesecting them you could either do:
150 - 40 (standing cat) + 20 (laying cat) = 130 cm
or
110 - 20 (laying cat) + 40 (standing cat) = 130 cm
If you really want to think out of the box , you put one table on the other and then you use the third dimension to wrap the two tables like a cylinder, which gives you a circle with circumference 2 tables equals 260cm, done.
"We have to solve for the value of height of table, which is unknown, so lets assume the unknown variable to be as 't' "
Sad 'x' noices :😢
Ht = 150 - HC + Hc; Ht = 110 - Hc + Hc. You add them together and get 2Ht = 260 so Ht = 130. Where HC height of standing cat and Hc is height of sleeping cat.
I’m not sure which solution this falls into but I saw that the change in height was 40 cm, divided by 2 for the number of cats changing positions, then add or subtracting that difference to make both cats sitting or lying down, canceling out their influence, equaling the height of the table. Not sure how any of this would work if the cats were of different sizes.
This is even simpler when you ask yourself what would happen if the cats were actually of equal height. The images would be the same and the number would directly show the height of the table. Then imagine that you make one cat X cm taller. On one image X would have to be added to the number but on the other image the same X would have to be subtracted from the initial number (the table height). You're just adding and subtracting the same number resulting in two numbers (150 and 110 in this case) that must be equally distant from the initial number (the table height). So the number we're looking for must be in the middle - arithmetic mean.
It can also be inferred from the data given that the *"SITTING" cat is 20 cm "taller"* than the "lying / sleeping" cat.
That's NOT asked in the question, but it's still a fun data inference.
:P
It was immediately obvious to me that its the average, but I think this would be rough for elementary school students
a = sleeping
b = sitting
t = table
b +t =150 +a
a +t =110 +b
b =150 +a -t
b =-110 +a +t
150 +a -t
=-110 +a +t
150 +110
=260
=2t
t =130
X = sleeping cat’s height
Y = sitting cat’s height
T = table’s height
X + 150 = T + y
Y + 110 = T + x
Rearrange those to align the variables:
T = x - y + 150
T = -x + y + 110
Add these
2T = 0x + 0y + 260
T = 130
Diagram 1: 150cm, taller than table by difference of standing to sleeping cat, Diagram 2: 110cm, shorter than table by difference of standing and sleeping cat. Find averages, tah dah. (Didnt take 9 minutes)
It would be impossible to know the height of the table if Schrödinger's had not yet opened his box.
Take the distance from the sitting cat on the floor to the sitting cat on the table and divide by two.
There is a sleeping cat in between.
For me it was: I have the two extreme values given for a linear problem so the answer about the constant that doesn't change must be the average between the 110 and 150.
To double-check: the height of the table would be as measured if both cats were sleeping or standing. if the difference between the standing and sitting cat is d, then the difference between the two problems is 140 or 2d, so d is 20. The height of the table is 110 + 20 or 150 - 20.
A much simpler solution, the difference between the 2 measurements is 40 cm, and so the the height of the table is 110 + (40/2), or 150 - (40/2). 130cm.
I love to use the figurative method because it stays in touch with story in the problem better than the algebraic approach. And I have used it also here, to figure out the heights of the cats and the table.
I have represented the two situations as given in the problem and added a rotated representation of the second situation in order to align the two sitting cats. This way it became evident that the height of the sitting cat is 40 cm.
Finally, the comparison of the three representations revealed that:
-the difference of height between the two situations is also the height of the sleeping cat and is equal to 20 cm;
-the height of the table is equal to 150-40+20=110-20+40=90cm.
Now I will continue to watch the video to see how Mr. Presh Talwalkar have solved this nice problem.
I noticed that one diagram included a sitting cat and the other diagram included a lying down cat. It's the difference between the two diagrams was the difference between sitting and laying down then you could just take the difference between the two measurements and split it equally. 150 - 110 equals 40 this was 20 added to the sleeping cat and 20 removed from the standing cat
You can then add 20 to the lying down cats and take 20 away from the sitting cats measurement and you get the tables measurement of 130
I: y+x=150cm
II: y-x=110cm
I+II: 2y=260cm => y=130cm
y is the table height, x the cat contribution to the total height. the problem is not that hard in my opinion and kids are good at thinking
I just assumed that the cats where all the same size. In my head I had the cat on top of the right table stand up. Splitting the difference between the measurements (40cm) to get 20cm. So if both cats are standing, you can measure the size of the table by measuring between to top of their heads. 110 + 20 = 130.
DVDFab
I solved this by realizing that the difference in the two measurements presented are double the difference in height between a sleeping cat and a sitting cat. Compared to the cats on the left, the cat on the top right has gone from sitting to lying down, and the cat on the bottom right has gone from lying down to sitting up, so the difference in height between sleeping and sitting is subtracted twice from the 150 cm measurement on the left to get 110 cm. 150 - 2x = 110. x = (150 - 110)/2. x = 20 cm. Subtract 20 cm from 150 cm to get the difference between a cat lying on the table and a cat lying on the floor, i.e. the height of the table. 150 - 20 = 130 cm.
You said the cats switched positions, but it could also be that table cat laid down and floor cat stood up. Then you wouldn’t be able to solve unless the cats were said to be the same height. To demonstrate switching positions, you could color standing cat black and leave sleeping cat as tabby
There is a simpler way to solve the problem visually. The "error" in measuring the table height is caused by the difference in the height of the 2 cats. In one case, it makes the measurement too high, in the other, too low. The magnitude of the error is the same, so the correct height is the average of the 2 measurements. 6th grade school students may not know how to calculate average, but they can "pick the number in the middle".
I remeber that I did a similar problem when I was in the middle school (10-13), about 30 years ago, in Italy
I just figure it must be the average, since 150=t+unkown_quantity and 110=t-that_same_quantity (which is the hight difference of the 2 cats)
If you stack the two tables you can just add the two heights 😊
Visually...
If you put the 2nd table on the 1st table...
... you know the height from the lower to the upper sleeping cats... (150+110)...
...which is the same as the height of 2 tables...
... So one table is (150+110)/2 = 130
Easier method that is about 2 lines:
1:30 When the cat goes from sleeping to standing, it is x units taller. The measured distance on the left is 150 units. The measured distance on the right is 150 minus the distance x lost from the bottom cat's action, minus the distance, again x, lost from the top cat's action. Thus, 150 - x - x = 110, or 2x = 40, or x = 20. Thus, if in the left picture, the top cat was sleeping, the distance would be 130 units. Take the cats away and shift the measurement's start point down to the ground, and 130 units is the distance from the floor to the table's surface.
That's exactly how I did it, find cat sleeping vs sitting height distance and subtract or add.
At first sight, the problem is a system of equations with two equations and three variables, but since we don't really care how tall the cats are we can rework their difference as a single variable (c - a), which exists in both equations with different signs [(t + (c - a) = 150] & [x - (c - a) = 110].
I didn't think to add them together.
I thought to make both equal to 't'
t = 150 + a - c
t = 110 + c - a
Make em equal to each other and see what to do from there
150 + a - c = 110 + c - a
Get constants to one side and variables to other and
40 = 2c - 2a
40 = 2 ( c - a )
20 = c - a
Go back an see if i can do something with this and saw that i can with the second equation
t = 110 + ( c - a )
t = 110 + (20)
t = 130
An easier way to solve this puzzle is by using Quantum Mechanics, and replacing both cats by Schrödinger's cats.
Schrödinger's cat are in a superposition of the sleeping (a) and sitting state (b): |phi> = c|a> + c|b>, with c being (1/2)^2
Since both cat states |a> and |b> are equal likely, we can easily calculate the height of the table from the expected height-difference between both Schrödinger's cats, which is the arithmetic mean of the two measurements: t = (150 cm + 110 cm)/2 = 130 cm. q.e.d
Pretty easy that way, isn't it?
BTW: I forgot to mention, that the quantum states of both Schrödinger's cats is entangled. Otherwise the math wouldn't work out.
If you sum both you get the measurement for a table without a sitting cat plus a table with a sitting cat. The two sitting cat nullifies themself, getting you two table is equal to 260.
Here is a harder scenario:
A student is asked to measure the height of a table, but not allowed to measure it directly. Instead, the student is given a bag with 20 objects in it, each with a unique height. The student must pick randomly from the bag 2 items. One of the 2 items the student chooses to place on the floor, the other on the table. The student is then allowed to measure the distance between the top of the two objects. From these measurements the student must calculate the true height. How many random pulls from the bag will guarantee that the student can determine the true height of the table? In other words, what is the minimum of the maximum number of pulls required?
I really think you should explain the rationale behind adding the two equations together. Otherwise, one might just think this is a "trick" (as you described it). It's not something I have ever been taught. I would have gone with two simultaneous equations and rearranged to eliminate the cat variables, leaving just the table, which is effectively what the second solution did, albeit based on a visual representation.
Grades 9 and 10 are effectively high school level in my country. I was enrolled in a special science high school and we were already doing pre-calculus problems at the equivalent grade levels. Admittedly, regular HS in our country would be only doing algebra and this should be a problem they could be presented with.
If you look at the picture at 1:57 you will notice that the difference of 150cm and 110cm is just the height difference between standing and sitting cat times 2, and it's evenly distributed on the top and bottom.
So 150cm - 110cm = 40cm, divide that by 2 is 20cm. That is the difference in height between standing cat and sitting cat.
If you measure from top of sitting cat on floor to the top of sitting cat on table, that is equivalent to the height of the table. Swap the standing cat in left picture to sitting cat, minus 20cm from 150cm.
Table height is 130cm.
I went straight to solution with intuition.
We have only two numbers to play with. Basic operations to do with them is adding, subtracting, which doesn't look right. Must be something slightly more funcy, like arythmetic mean. Yeah, must be the mean.
The two numbers are uniform physical quantities and result is also uniform. That is we are given two lengths and asking for a length. We're not asked e.g. an area. If so we could argue that multiply would be the sensible operation with just two given numbers. But since we are just manipulating lengths to get a length, must restrict sensible operations to just those stated. Adding, subtracting, bysecting, lengths into lengths... mean just came naturally.
Here's how I did it in my head, with no algebra. Suppose the cat on the left went to sleep. That would reduce the 150 measurement by the difference in heights of the sleeping and sitting cat. And that measurement would now be exactly the height of the table. Now suppose the sleeping cat on the left sits up. That would reduce the measurement by the same amount again, and we're at 110cm. So the table height is exactly half way between 150 and 110, where both cats are sleeping. That is, 130 cm.
So, I did t=150-c+a and t=110-a+c (Height of table equals top cat height - given height + bottom cat height)
Then, 150-c+a=110-a+c, 2a-2c=40, a-c=20
Then, in the one with the sleeping cat on top, if I substitute the sleeping cat for the standing cat, I add 20 ((the difference between the sleeping and standing cat heights) to the given height, 110, giving 130. Since this is the distance between two cats of the same height, one on the floor and one on the table, they cancel out and the table's height is 130cm.
The difference between the numbers (40 cm) is twice the difference between the heights of the cats (once above and once below). Replace the cat sleeping on the table with the standing cat, and the new distance (110+20=130 cm) is the height of the table.
A smart 3rd-grader should be able to solve this. For a 9th-grader (algebra 1), this is a second week exercise in setting up equations, maybe.
My approach, using similar algebra, was to note that, in the left diagram, the 150cm can be considered as the sum of the height of the sitting cat and the distance from the top of the sleeping cat to the table. The former is of course C and the latter is T-A. The gives equation 1 as
T - A + C = 150. The corresponding observation in the other diagram gives equation 2 as
T - C + A = 110. These equations can the simply be added to make C and A disappear, giving 2T=260.
Now this T=130 can be plugged back into the two equations. But can we solve for C and A? No! As we might expect from having two equations with three unknowns, we can't solve for them all, even though we were able to solve for T. So what happens when we try to plug in 130 for T?
Then equation 1 becomes 130-A+C=150 and equation 2 becomes 130-C+A=110. The first becomes C-A=20 and the second becomes A-C=-20.
They're equivalent. We can't figure out how tall each cat is, only that the sitting one is 20cm taller than the sleeping one.
150 - 130 = 40 which implies that the sitting cat is 20 taller than the sleeping cat. On the left side, if you replace the sitting cat with a sleeping cat, the difference in height would be 130 which implies that the table is 130. You could also replace sleeping cat with sitting cat or do these replacements on the right side.
Another proof that the individual cat height is non-revelant is to see the same problem, but this time with a sitting cat and a cat standing on its rear paws.
Now the sitting cat is the shortest, the standing cat is 20 cm taller and the measures remain the same at 110 and 150 cm.
Before trying to calculate anything, my gut told me that the answer would simply be the average of the two heights, and that intuition turned out to be correct.
Think about it this way: let's say you have a slider that controls the heights of the cats so that as one cat gets shorter, the other one gets taller by the same amount.
If you move the slider all the way to the left so that the heights are like in the first picture, then the difference between their heights is 150 cm. If you move the slider all the way to the right where it's like in the second picture, then their difference is 110 cm. But what happens if you move the slider exactly halfway?
The distance between the two cats' heads will be 130 cm, because the slider changes that difference linearly. But also, the two cats will be exactly the same height, which means the distance between their heads is exactly the height of the table!
Thus, the table is 130 cm tall, because it's the average of the two distances.
I think this problem was presented already long time ago in older video
Follow-up question: How many items did the cats knock off the table in the time it took you to find the answer?
That's assuming both cats are the same height in both positions, which I can tell you from experience, that number can differ tremendously.
so i feel i cheesed it and was like "what if I make sleeping cat 0 it should work, because it just have to be less the the standing cat". That at least simplifes it to just the cant standing and setting being the differents but still not trivial for elementary students to solve it.
I had this figured out before the video could spin up. It's way to easy to do in your hear.
2X + 110 = 150
2X = 40
X = 20
150 - X = 130
I'm not great at math, but this is how I did it. Where X is the difference between a sleeping and sitting cat.
Went at it in a completely different way
i isolated 'L'aying to equal 110cm+'S'tanding-'T'able, and substituted in to the other equation getting T+S=150cm+110cm+S-T. Moving unknowns over to the first side i get T+T+S-S=260cm, and after cancelling we get 2T=260cm => T=130cm
The answer is ~ 4.3 cats or 50 cans of tuna
Both of these solutions seem way too complicated. All you have to realize is that the difference in measurement between the identical cats in two different positions totals 40 cm, meaning each side is contributing half that (20cm) when measured, and the rest falls into place.
Before watching the solution: awake cat=A, sleeping cat=S, table=T. T+A-S=150, T-A+S=110 -> T+A-S+T-A+S=150+110 -> 2T=260 -> T=130
7:16 the goal of this problem is to not use any variables at all, just logic. Assigning “c” and “t” variables overcomplicates the problem.
Variables are ... C, A,T !
Flashback to a video of 6y ago:
Viral logic test from China BPRueCu3fXU
"At first glance this problem seems quite impossible" uh I already did it.
It’s the average between the two height scenarios. Then the cat height is the same, and only the difference matters. It took 2 seconds.
I don’t see why this is considered hard by any stretch.
Heght difference between sleeping cat and sitting cat is: c-a
This height difference (c-a) is equal to the difference in total height of each of the two situations depicted, or: (c-a)=(150+a)-(110+c)
Solves as:
2c-2a=150-110
2(c-a)=40
(c-a)=20
Substitute that back into either equation posed in the video, you get your height for "t".
Isn't the most interesting part of the problem the fact that there is an infinite number of possible heights for the cat???
I figured that in the second picture, the cat on the floor had grown by twenty and the cat on the table had shrunk by twenty, given that the difference is forty and the standing cat grows by just as much as the sitting cat shrinks. So then if the standing cat in the second picture lays down, it will increase the measurement to 130, which would be the height of the table because you are measuring from the top of one sleeping cat on the floor to the top of another sleeping cat on the table. Cancel out the cats, and you get the table.
I saw this and instinctively ignored the cats, added the two numbers, and divided by 2. Simple!
This is also elementary school level in Eastern Europe. Around 5th grade, I'd presume. What's with the American educational system that they need to be 9th or 10th grade to solve it, and at a competition no less! Competition problems are usually much harder than normal homework problems.
I solved a slightly different, visual way. In both cases, lower the measuring line by the height of the sleeping cat. On the Left, it rests on the floor and the difference is the standing cat body. On the right, it is missing the standing cat body in order to reach the floor. The difference between the 2 lines is 2 standing cat bodies. In math, the difference between the 2 lines is 40cm (150-110), so each standing cat body is 20cm. 110cm + 1 standing cat body is 130cm. 150 - 1 standing cat body is 130cm. The table is 130cm.
I think you’d be hard pressed to find a kangaroo that can answer the question!
All I did was solve for the sleeping cat. Either works though.
They went from peak distance in the left image down to shortest distance. The reduction was a distance of 40. Since both cats moved, we can divide 40 by 2. That means the distance between 2 sleeping cats is 150-20 so the table is 130.
I like the first answer better than my method though.
I did it a slightly different way. Since we are treating the height of the sitting cat as a fixed quantity, and the height of the sleeping cat as a fixed quantity, it's clear that (for instance) measuring from the top of a sleeping cat on the floor to the top of a sleeping cat on the table would equal the height of the table. In one diagram, we have that height PLUS the difference in heights of a sitting vs sleeping cat (c - a) and in the other diagram we have that height MINUS the difference in heights (c - a). Let's call this difference (c - a) by its own variable, h. So in one case we have t + h = 150, and in the other we have t - h = 110. By subtracting the second equation from the first, we get 2h = 40, or the height difference between a sitting and sleeping cat is 20cm. Going back to either of the original equations t + h or t - h and plugging in our solution for h gives us t = 130cm, which was easier to do in my head than it was to write out.
Notice, also, that the height of the table is also the average height between the two. Taking the average essentially cancels out the difference in height between the two cats.