Can You Fill These Yellow Boxes | Two Methods! Challenging Puzzle!

Solve this math puzzle! Fast and easy explanation by PreMath.com

[@terakhan6656]

My route: b-d=2 and c+d=15 added together, to get b+c=17. Subtracting a+c=12 from that, to get b-a=5. Rearranged to b=a+5. We know that a+b=15, and the only number pair that have a difference of 5 and total of 15 is 5 and 10, and b is 5 greater than a, so a=5 and b=10. Which makes c=7 so that a+c=12, and d=8, so that b-d=2 and c+d=15. a=5, b=10, c=7, d=8

[@MrFrmartin]

Super Easy once you set up the 4 equations

[@MarieAnne.]

My method was to use just one variable. Top left = x x + Top right = 15 → Top right = 15-x x + Bottom left = 12 → Bottom left = 12-x (12-x) + Bottom right = 15 → Bottom right = x+3 Now use right column to find x: (15-x) - (x+3) = 2 12 - 2x = 2 -2x = -10 x = 5 15-x = 10 12-x = 7 x+3 = 8

[@alexanderchuyko4889]

Yes. This way does look shorter and easier!

[@RashmisABCD]

Awesome presentation beautifully done and shown this video thanks for sharing such a great video have a fantastic week lk127

[@govindashit6524]

I understand two method very carefully. Thanks.

[@mustafizrahman2822]

Answers are 5 10 7 8 It is easy to me now! Thanks for giving this type of question.

[@PreMath]

Well done Rahman dear! Thanks for the feedback. You are awesome

[@AnshumanTyagi]

This one is quite easy. Any way good explanation

[@mnazar3601]

Top left = 5 ,Top Right = 10 , Bottom left = 7 ,Bottom Right = 8...

[@moth36]

Thank you very much for sharing this nice video

[@wesleysuen4140]

It’s too complicated to use two unknowns to start with. Use only one! Start with (any) one unknown. Use (any) three equations to express the other three values in terms of this unknown. Use the remaining equation to relate the values. And we’re ready to solve for the unknown, and hence the rest. For example, a b=15-a c=12-a d=15-c=3+a Then (15-a)-(3+a)=2. Solving, we have a=5, and so b=10, c=7 and d=8.

[@PreMath]

Thanks Wesley for the feedback. You are the best

[@mustafizrahman2822]

Great solving.

[@shapourmehrkar-asl1845]

As the sums of each pair on the rows are the same, then we can come up with the answer very quickly. From the second column the difference being 2, we can deduce that the deference of the first column should be 2 as well but the sum of the first column being 12 it mean these two should be 5 and 7 but the 5 should be on the top and the second element in the top row should be bigger otherwise the answer to the second column would have been a negative number not positive. As a result the answer on the two on the top would be 5 and 10. On the second row the first one would be 7 and the second one on the second row would 8. Just logic and no equations.

[@supermathlete101]

I did your second method a little differently. I ended up using the second vertical part as b-d=2 -> b=2+d. For the rest I added the horizontal values together and used the first vertical value like you did to get 18, then I used my b value and solved for d which was 8 and everything fell into place! Good video! Keep up the great work!

[@mnazar3601]

Answr top left = 5 , top right =10 , bottom left =7 bottom right = 8

[@highlyeducatedtrucker]

Just for laughs, I turned this problem into a 4x4 matrix and then used reduced row echelon to calculate the answer. You will be pleased to know that the answers are the same! You will also be pleased to know that your way is much easier. 😁

[@PreMath]

Incredible! You are very resourceful. Thanks for sharing my friend. I'm glad you liked it! Your feedback is always appreciated. Kind regards :)

[@gonepallisatya660]

🤯🤯😵‍💫very hard question anna

[@VolkGreg]

a = 15-b = 12-c d = 15-c = b-2 b = c+3 = 17-c 2c = 14 c = 7 b = 10 a = 5 d = 8

[@242math]

got it bro, thanks for the challenge

[@Indiangaming-f5p]

I think second method is too easy

[@iqmathsciencelogicalreason2770]

Appreciable

[@legios07]

My solution before watching the video : A+B=15 / C+D=15 / A+C=12 / B-D=2 --> A+B+C-D=14 --> C-D=-1 --> 2C=14 --> C=7 --> D=8 --> B=10 --> A=5

[@obchardon]

When you have a linear system of equation I prefer to use the Gauss-Jordan elimination, it is more general and help you to keep the things clear. But cool KZbin channel with some fun math problem :)

[@jeminginjemingin780]

I am fans of premath from Indonesia👍

[@PreMath]

Thank you so much dear for taking the time to leave this comment. I'm glad you liked it! Your feedback is always appreciated. Please keep supporting my channel. Love and prayers from the USA!

[@sreenivasbaas7750]

7+8 5+10

[@PreMath]

Super Thanks dear Baas

[@pa28cfi]

I haven't watched the video, but from the thumbnail, the only way to have the same result of 15 with a difference of two in the right boxes, is to have a difference of 2 between the numbers totaling 12. There's only 1 solution. It took me longer to type than to solve.

[@aashsyed1277]

Thank you so so so you

[@PreMath]

You are so welcome Aash dear Thank you so much for taking the time to leave this comment. I'm glad you liked it! You are amazing.

[@spyguyspyguy123]

Here one of the sign is - in these four boxes, what if all boxes are + Example, If a + b = 12, c + d = 15, a + c = 19, b + d = 8, find a, b, c, d? I know the answer but could not reach at it using equations. Can anybody solve step by step.

[@ayeshchanuka52]

5+10/7+8

[@ramdast6360]

5 10 7 8

[@PreMath]

Super Ramdas dear Thanks for the feedback. You are awesome

[@ajoybhattacharya8880]

I venture to solve the problem studying the thumbnail before watching the video. How a mind dull at math solves without losing track in abstractions step 1 a b + c d = 30/4 a 7.5. b 7.5 c 7.5. d 7.5 step 2 a + c = 12 hence the shedding as I find b -- d so; a gives b 2.5 c gives d 0.5 and rests are simple yes! You can Laugh a hearty laugh for this ridiculous tracking

[@APUS_NUNN]

What happend? I saw the boxes and saw the solution.

[@PreMath]

Brilliant Apus!

[@aliebrahimpor4402]

😢😢😢🥺🥺🥺😱😱😱🥳🥳🥳❤

[@ronduz1281]

🤙🤙👍