@insert username yea but if you gave a 12 year old that problem in the 1800's they wouldn't know how to solve it even if they knew systems of equations as it would've been an original problem not taught in schools
@@warpromo6636 That doesn't make any sense. Once they knew that they can add and subtract equations they can solve it pretty easily. I knew how to add equations and I could do these by myself in seventh grade
Riddle: A father said to his son, Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each? Answer - ruclips.net/video/NcrTWE4w784/видео.html
I have a problem plz tell me can anyone help When Nabil's age was equal to Shubh's present age, Shubhar's age was twice that of Nabil's present age. When Shubhar's age is equal to Nabil's present age, what is the present age of each of them if the sum of the ages of the two of them is 63?
12 and 13 year olds solve these questions easily in India. It's just simple equation what's so hard there. If you can understand the language of question and frame the correct equation it's a piece of cake.
Just imagine the brain power of Newton at that time. When MIT was asking a 7th grade math problem but Newton created formulas which even current generation can't grasp properly.
Everyone is in the comment section flexing, bragging about doing this in grade school, and they don't seem to grasp the simple concept that this is from an admissions test into MIT not a part of the curriculum, meaning that it wouldn't be taught at the university but rather be expected that you already know how to do it. Plus, who's to say it isn't one of the easiest questions on the exam. In other words, congrats on being able to do something you were supposed to have been taught to do in school.
I know right why is it that everyone is flexing on this question but I never see them on all the other ones on this channel?. but anyway, this is very well said even if i was so close to getting it right ( got 46 for some reason)
@@william2344 probably because you took the equation as X-2 = 3*Y in the first one. :) Honest mistake. Same here, first try from the thumbnail. Haven't watched the video.
Riddle: A father said to his son, Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each? Answer - ruclips.net/video/NcrTWE4w784/видео.html
This question is badly posed. It asks “What WERE the ages of each ?” The answer is that they WERE 48 and 16. If the question was instead “What ARE the ages of each” then the answer is that they ARE 50 and 18.
No, the whole narrative is in the past tense, ("A father said to his son..." etc.) and there never is a reference to the present time. Thus, the formulation ""What were the ages..." is the correct one. It refers to the time the father and son happened to have this conversation, which is some time in the past, but not at the present moment.
@@AnssiRai Well I think the question contains three different time zones, and hence needs three different tenses. I understand your reasoning but well-posed questions should not depend upon interpretation, unless of course the interviewer deliberately wants to wrong-foot the interviewee.
@@raspberrieswithcream and @Parth Pandya - the three tenses here would be: "how old were they [at the moment of the conversation]", "how old had they been [two years earlier" and "how old would they be [fourteen years later]"? None of these cases involve the present tense.
Edit: My "shortcut" worked for this problem, but isn't actually a valid method, as noted below. This one was actually easy. I saw the thumbnail and figured it out on the way to the fridge. I used a slightly different approach. The time difference between 2x and 3x age is 16 years. (14+2) So that's x. So when the father was 3x the son's age (2 years ago) son was 16 and father was 48. Add 2 to each to and bingo, you have your current ages. Bingo.
@@truth5030 my direct logic is a 16 year gap (14-(-2)) represents a change in relative ages from 2x to 3x. The difference between them (3x-2x) is x. So x=16. Now I know the age of the son 2 years ago when he was 1/3 the age of the father. Add 2 years to get his current age. The father was 3x 2 years ago, so take 3*16+2 and you get the father's current age of 50.
But just like you, other participants will also be able to solve it very easily, then the competition will become even more tougher... don't u think so? 🤔🤔🤔
50 and 18 is the right answer, but not to the question asked. The question asks ‘what were the ages of each?’, emphasis on the were ie what were the ages of the father and son when the father was three times older than the son. The answer to that is 48 and 16. If the question was ‘what are the ages of each?’ then 50 and 18 would be correct because that is 2 years after the fact.
Man whenever there is "%99 fail" or "harvard admission" kinda thing it ends up to be easy. But when there isnt we get olympiad problems. Is it just me?
I thought of this puzzle as, for what pair of numbers-- in which the larger is 3 times the smaller-- does the larger become only two times as large when you add 16 to both. Naturally I tested on multiples of 16 first, and realized that in the earlier time, the son can be 16 and the father 48, while in the later time, the son is 32 and the father 64. So at the current time, son is 18 and dad is 50. It helps that I myself am 32 years younger than my own father. Indeed, my own father could have posed this riddle to me in 2002.
【arithmetic】 father son gap two years ago ③ ① ② ll in fourteen years ❷ ❶ ❶ therefore father son gap two years ago ③ ① ② ll in fourteen years ④ ② ② ∴①＝2＋14 ＝16
Well, they didn’t study quantum physics or relativity at MIT back then, so I guess this was appropriate for the time? Or maybe this was a sine qua non question to weed out the clueless.
Before and really hated math but when I found your channel a really inspired me to understand math and now I'm a lot better at it I wouldn't say I'm great in it but I wouldn't say I'm terrible either
I can solve this problem in a simpler way by mere using ratios. 2 years ago the ratio between father's age and son 's age will be 3:1 and 14 years later it will be 2:1 But we all know that difference between age of 2 people never change. (if you are 3 years older than your brother then you will be 3 years older to him after 50 years also.) Here we have 2 stituations 1 1)Before 2 years the diff. Wwas 3-1=2 2)After 14 years the diff. will be 2-1=1 We need to equalize the diff so multiply 2 to 2 years ago ratio, i.e. 3:1)×2= 6:2 And multiply 4 to 14 years later ratio i.e. 2:1)×4=8:4 Now the diff is same in both the situations. Now the gap between both the years is 14+2=16 But in terms of ratio it is 2 units. This implies 2units=16 so 1unit=16/2=8. Thus, 2 years ago father's age was 6×8=48 years and son's age was 2×8=16. And 14 years later father's age will be 8×8=64 years and son's age will be 4×8=32. And Their present ages are 48+2=50 and 16+2=18.
Human knowledge has increased so much in just 2 centuries. It never happened in last 1000s of years. The question that was considered extremely hard for mathematicians, is now asked in 8th class
Just because the question was on an MIT entrance exam in the 19th century doesn't mean it was considered "extremely hard for mathematicians" at the time. An entrance exam has to test some basic material too, such as the ability to create the proper equations to solve word problems. This could have been just one of 50 such questions. Moreover, people applying to MIT are high school students, not mathematicians.
@@zzzzzz5411 no bro it's my personal email account actually I join the school class via using this id and teachers told use to write the class name that's why I do so. If you are a student you will understand.🤣🤣🤣🤣
How I got the answer: I made it simpler by using “now” instead of “2 years ago” and “In 16 years” instead of “In 14 years” so I only had to keep track of the 16. Then I did a sample problem. Say they were 10 and 30, it would take 10 years for the father’s age to be only double of his son’s. I realize the starting sons age doubles until he is only double his fathers age, so I figured that the son had to be 16 and father had to be 48 two years ago, so I reached 18 and 50 by adding those years back
super funky that it’s MIT, guess it teaches you how much we’ve improved culturally at certain things over time now simultaneous equations are a super normal thing to do in middle/high school
in order to simplifie one of the equation, i choose my varables being the age 2 years ago so the first one is juste x = 3y and the other one is x + 16 = 2 (y + 16) and we get y = 16 and x = 48 so when we add the 2 last years we also get 18 and 50
I find it easier to make it one variable. 2 years ago: father = 3x and son = x Now : father = 3x + 2 and son = x + 2 14 years later : father = 3x + 16 and son = x + 16 since the father is only twice the son's age in 14 years, we can conclude that : 3x + 16 = 2 ( x + 16 ) 3x + 16 = 2x + 32 x = 16 substitute back into the Now equations and get : father = 50 and son = 18
solution with only one equation: just equate expressions for father's age from both the relations. Ages 2 years back : x, 3x Ages 14 years from now: x+16, 2x +32 Thus : 3x + 16 = 2x + 32 => x = 16 => present ages : 18, 50...
WOOHOO! This was actually one I was able to solve on my own! I guess I'm better at these than at geometry :) Keep up the good work and happy holidays each and all
Without watching the video, just reading the problem, I reasoned as follows: a-2=3(b-2) & a+14=2(b+14). We end up with a=3b-4 & a=2b+14, so 3b=2b+18, hence b=18 and a=50 substituting back into the first two equations: 50-2=3(18-2): 48=48 and 50+14=2(18+14): 64=64 it all checks out. Pretty straightforward.
i can solve this in seconds, here are the steps: 1,Consider: the father is three times than the son today, and will be twice of that in 16 years later 2,Instantly conclude: the son is 16 years old today, and the father is 48 years old( this is because 2(A+A)=3A+A ) 3,but this is what happened two years ago, so the son is 18 and father is 50
I got caught up in interpreting the tense of the question. The Question states: what "were" the ages of each? It didn't say: What "are" the ages of each. So I took this to be the ages of the boy and his father 2 years ago. f=3s, f+2+14=2(s+2+14), f=48, s=16
Son: Dad how old are you ? Dad : Two years ago I was three times as old as you ; but in fourteen years I shall be only twice as old as you. Son : Whatever I dont need that !
I started the same way as Presh Talwaker, but I suddenly found an easier solution. My solution: "forget" 2 years ago (for a while): 3x = y now remember that 2 years + in 14 years, he'll be twice as old as his son, so: 2x + 16 = y 3x = 2x + 16 Therefore 2 years ago, the son was 16 and father was 48 -> now 18 and 50. Correct me if I'm wrong.
Me, who solved using substitution instead of elimination: Darn it, that took longer than it should’ve. Well at least I could solve ONE of these puzzles
Why do they use the preterit "were" instead of the present : "what are the ages of each?". I might be overthinking but for me it can be misleading. Like do you answer with their ages today or two years ago?
The problem starts with "A father *SAID* to his son...", how do you know how long ago the conversation was? You can only deduce their ages at the time of the conversation (or at some date relative to when the conversation occurred). From the information given you can't deduce anything about their ages today nor two years ago. Admittedly the question using "were" does not make clear for when you should calculate there ages, but your answer can only be relative to the date of conversation. The correct answer is 18+c and 50+c.
I remember posting a riddle of sort of similar nature on QuizUp but nobody got in right, though it might have been confusing wording. "12 years ago, I was twice as old as people born 12 years ago were when I was 12. How old am I now?"
I don't expect they wouldbhave thought students would fail to solve this puzzle. Rather, they were likely listening to them talk it through to get a sense of their speed and thought process.
So two years ago, the father was 48 and the son was 16. In 14 years, the father will be 64 and the son will be 32. Thus, the father was 32 years old when his son was born.
I got a riddle for you. If the father's age in 2020 is 25, his son's birthdate in the 25th of October, 2020; What will be the age of his son in the New Year of 2021?
To those who think this is very simple, yes it is very easy because it was a question from 1876 admission test of MIT . And that time simultaneous linear equations were like calculus at present
age of father: X age of son: Y equation (1): X-2 = 3(Y-2) equation (2): X+14 = 2(Y+14) simplify (1) to get (3): X = 3Y-4 use (3) to substitute out X in (2) and get (4): 3Y-4+14 = 2(Y+14) simplify (4) to get (5): Y = 18 use (5) to substitute out Y in (3) and get (6): X = 3*18-4 simplify (6) to get (7): X = 50 by (5) and (7) we have the solution: the father is 50 years old and the son is 18.
I thought of it like this: Adding up the 2 years and the 14 years, you get 16 years between the years when the father was 3 times as old and when the father is twice as old. When the father is 3 times as old as the son, the age difference is 2 times the son's age and when the father is twice as old as the son, the age difference is the same as the son's age, so the son has to be twice as old in the second scenario as he is in the first. So this means the son's age 2 years ago was 16 and in 14 years, he will be 32, and the age difference between the father and son is 32 years. Since the son was 16 2 years ago, he must be 18 now, and the father is 32 years older so 50.
I worked out 16 and 44 to be their current age. 2 years ago they'd be 14 and 42. 42 is 3x larger than 14. And in 14 years they'll be 28 and 56. 56 is 2x bigger than 28. Am I wrong or is this answer also correct?
If their current ages are 16 and 44, Then in 14 years they'll be 30 and 58. And 58 is NOT 2× bigger than 30. So I guess you made a silly mistake there.....
Hi mind your decision Can you solve it for me. I have a irregular hexagon whose three sides are equal and also the remaining three sides are equal. The first three sides are equal to 3.88meter and the remaining three sides are equal to 0.5meter. Calculate the area of hexagon. Hoping for your response.
I did it mentally 😅 I think in India we hv these kind of que. in class 6,7,8 😅 but the que. is from 1876 😂😂😂 So thats ok 👍 I donot need to fly in air 😂
F-2 = 3(S-2) ==> two years ago F = 3S-6+2 F = 3S-4 ---- (1) F+14 = 2(S+14) ==> 14 years from now F = 2S+28-14 F = 2S+14 ---- (2) Equate (1) and (2): 3S-4 = 2S+14 S = 18 And solve for F with (1): F = 3(18)-4 = 54-4 F = 50 Two years ago they were 48 and 16 (✓) and in 14 years they'll be 64 and 32 (✓).
"what were the ages" .. ( WHEN ? ) AGE at time equations are stated or when it was "2 years ago" ( if I was not allowed to ask, I'd have provided both .
Real nos : Real Complex nos : Real + imaginary Real ---> ♾️ Imaginary ---> ♾️ Hence, no of complex nos = ♾️ + ♾️ = ♾️ Hence, no of complex nos = no of real nos = nos of complex nos Conclusion : All Infinite sets of nos are equal. @I am somebody, no of even = no of odd = no of complex = no of rational = irrational = imaginary = quantum = dark = ..........😡 My Q : Is is true that "N + 1 = W"
No one picked up on the question which is "What were their ages?" Not "What are their ages?" Since the only reference to the past is two years ago, the answer to the question must be that the son was 16 and the father was 48.
It doesn't have to be complicated. Suppose two years ago the son's age is x, then the father's age is 3x+2. 3x+2+14=2*(x+2+14) Solve for x. x=16 16+2=18 The son's age is 18. 3x+2=50. The father's age is 50.
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
Was a challenging problem at that time but now, grade 7 and above students will be able to solve it.
That's how, math has evolved
nah, hasn't really evolved. if the student knew systems of equations, but not the problem itself, they wouldn't have been able to solve it.
@insert username yea but if you gave a 12 year old that problem in the 1800's they wouldn't know how to solve it even if they knew systems of equations as it would've been an original problem not taught in schools
Watch this RUclips video for the correct answer of this riddle.
ruclips.net/video/NcrTWE4w784/видео.html
@@warpromo6636 That doesn't make any sense. Once they knew that they can add and subtract equations they can solve it pretty easily. I knew how to add equations and I could do these by myself in seventh grade
Actually class 6 students will solve it too
When I heard it came from MIT I'm like wut this is so simple, but when I heard the year I was like ahh no wonder
Riddle: A father said to his son, Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each?
Answer - ruclips.net/video/NcrTWE4w784/видео.html
1876 ;)
I mean this is simply about naming the variables and adding the equations. This is taught in grade 7.
@@achyuththouta6957 see back in 1876
This might have been considered one of the hardest question
I have a problem plz tell me can anyone help When Nabil's age was equal to Shubh's present age, Shubhar's age was twice that of Nabil's present age. When Shubhar's age is equal to Nabil's present age, what is the present age of each of them if the sum of the ages of the two of them is 63?
Indians outta nowhere: "We learnt solving this kind of problems before we learnt how to walk and talk."
Yeah bro, I have been solving this type of problem since My class 6 days
12 and 13 year olds solve these questions easily in India. It's just simple equation what's so hard there. If you can understand the language of question and frame the correct equation it's a piece of cake.
True that for 7th- 8th grade students
Yeah, this problem can be solved easily by most of the children in Asia when they are 12 or 13 years old.
In 10th or 9th grade we have done this
A puzzle without Gogou's theorem is not a puzzle
Where them triangles at?
Watch this RUclips video for the correct answer of this riddle.
ruclips.net/video/NcrTWE4w784/видео.html
What is Gogous theorem
🤣🤣🤣🤣
@@ashurawat1721 It's what Mr. Tallwalker has taken to calling the Pythagorean Theorem, for ... reasons.
Just imagine the brain power of Newton at that time. When MIT was asking a 7th grade math problem but Newton created formulas which even current generation can't grasp properly.
@@RealNeelamPal Still is.
Everyone is in the comment section flexing, bragging about doing this in grade school, and they don't seem to grasp the simple concept that this is from an admissions test into MIT not a part of the curriculum, meaning that it wouldn't be taught at the university but rather be expected that you already know how to do it. Plus, who's to say it isn't one of the easiest questions on the exam. In other words, congrats on being able to do something you were supposed to have been taught to do in school.
Well said!
I know right why is it that everyone is flexing on this question but I never see them on all the other ones on this channel?. but anyway, this is very well said even if i was so close to getting it right ( got 46 for some reason)
Watch this RUclips video for the correct answer of this riddle.
ruclips.net/video/NcrTWE4w784/видео.html
@@william2344 probably because you took the equation as X-2 = 3*Y in the first one.
:)
Honest mistake. Same here, first try from the thumbnail. Haven't watched the video.
@@kishoretejaswi1948 thank you!
I just realized I'm 18 and my father is 50. What a coincidence !!
@Farhan: Then I think two years ago your father was three times as old as you and in 14 years just two times.
so give your father this problem and you'll find out if he graduated MIT or is some high school dropout ...
@@danmimis4576 destroyed in seconds
Next year I’ll be 18 and my father will be 50 xDD
@@danmimis4576 Roasted!!!
Now these kind of problems are taught in schools
Yes definitely
Not anymore
In 6th class
Riddle: A father said to his son, Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each?
Answer - ruclips.net/video/NcrTWE4w784/видео.html
@@raman2192 I watched a video like that before I watched this one.
ruclips.net/video/_RWsx9gU_XI/видео.html
This question is badly posed. It asks “What WERE the ages of each ?” The answer is that they WERE 48 and 16. If the question was instead “What ARE the ages of each” then the answer is that they ARE 50 and 18.
Hmmm kinda makes sense tho
No, the whole narrative is in the past tense, ("A father said to his son..." etc.) and there never is a reference to the present time. Thus, the formulation ""What were the ages..." is the correct one. It refers to the time the father and son happened to have this conversation, which is some time in the past, but not at the present moment.
@@AnssiRai you see the problem here is english doesn't have "past" and "more past" that's why that can be interpreted in both ways
@@AnssiRai Well I think the question contains three different time zones, and hence needs three different tenses. I understand your reasoning but well-posed questions should not depend upon interpretation, unless of course the interviewer deliberately wants to wrong-foot the interviewee.
@@raspberrieswithcream and @Parth Pandya - the three tenses here would be: "how old were they [at the moment of the conversation]", "how old had they been [two years earlier" and "how old would they be [fourteen years later]"? None of these cases involve the present tense.
Edit: My "shortcut" worked for this problem, but isn't actually a valid method, as noted below.
This one was actually easy. I saw the thumbnail and figured it out on the way to the fridge. I used a slightly different approach. The time difference between 2x and 3x age is 16 years. (14+2) So that's x. So when the father was 3x the son's age (2 years ago) son was 16 and father was 48. Add 2 to each to and bingo, you have your current ages. Bingo.
same me also.. I solved before watch the video
But it should be 2(x+14)-3(x-2)=16 , what's your direct logic i can't understand brother.
@@truth5030 my direct logic is a 16 year gap (14-(-2)) represents a change in relative ages from 2x to 3x. The difference between them (3x-2x) is x. So x=16.
Now I know the age of the son 2 years ago when he was 1/3 the age of the father. Add 2 years to get his current age. The father was 3x 2 years ago, so take 3*16+2 and you get the father's current age of 50.
Same but after I overcomplicated with other formula, was very easy
@@ajhieb Be careful, I think you've skipped a step. Your reasoning wouldn't work properly if the factors were 3x and 4x for example.
I was taught these types of problems in grade 8, but I still love solving them! Thanks Presh!
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
Bro, I wish the questions now are as easy as the questions from before.
Seconded
But just like you, other participants will also be able to solve it very easily, then the competition will become even more tougher... don't u think so? 🤔🤔🤔
These questions were taught to us in 7 class in INDIA😂😂😂
Not in 1970s bro
@@gauravtawri8803 right
exactly
50 and 18 is the right answer, but not to the question asked. The question asks ‘what were the ages of each?’, emphasis on the were ie what were the ages of the father and son when the father was three times older than the son. The answer to that is 48 and 16. If the question was ‘what are the ages of each?’ then 50 and 18 would be correct because that is 2 years after the fact.
Man whenever there is "%99 fail" or "harvard admission" kinda thing it ends up to be easy. But when there isnt we get olympiad problems. Is it just me?
I thought of this puzzle as, for what pair of numbers-- in which the larger is 3 times the smaller-- does the larger become only two times as large when you add 16 to both. Naturally I tested on multiples of 16 first, and realized that in the earlier time, the son can be 16 and the father 48, while in the later time, the son is 32 and the father 64. So at the current time, son is 18 and dad is 50. It helps that I myself am 32 years younger than my own father. Indeed, my own father could have posed this riddle to me in 2002.
We get sums like this in class 10th lulw
Yeah.
Which are very easy....
@@divyamgoyal774 yep
Yeassss
how are you guys preparing for boards?
【arithmetic】
father son gap
two years ago ③ ① ②
ll
in fourteen years ❷ ❶ ❶
therefore
father son gap
two years ago ③ ① ②
ll
in fourteen years ④ ② ②
∴①＝2＋14
＝16
Well, they didn’t study quantum physics or relativity at MIT back then, so I guess this was appropriate for the time? Or maybe this was a sine qua non question to weed out the clueless.
They did study calculus however.
Before and really hated math but when I found your channel a really inspired me to understand math and now I'm a lot better at it I wouldn't say I'm great in it but I wouldn't say I'm terrible either
I can solve this problem in a simpler way by mere using ratios.
2 years ago the ratio between father's age and son 's age will be 3:1 and 14 years later it will be 2:1
But we all know that difference between age of 2 people never change.
(if you are 3 years older than your brother then you will be 3 years older to him after 50 years also.)
Here we have 2 stituations
1
1)Before 2 years the diff. Wwas 3-1=2
2)After 14 years the diff. will be 2-1=1
We need to equalize the diff so multiply 2 to 2 years ago ratio, i.e.
3:1)×2= 6:2
And multiply 4 to 14 years later ratio i.e. 2:1)×4=8:4
Now the diff is same in both the situations.
Now the gap between both the years is 14+2=16
But in terms of ratio it is 2 units.
This implies 2units=16 so 1unit=16/2=8.
Thus, 2 years ago father's age was 6×8=48 years and son's age was 2×8=16.
And
14 years later father's age will be 8×8=64 years and son's age will be 4×8=32.
And
Their present ages are 48+2=50 and 16+2=18.
Human knowledge has increased so much in just 2 centuries. It never happened in last 1000s of years. The question that was considered extremely hard for mathematicians, is now asked in 8th class
Just because the question was on an MIT entrance exam in the 19th century doesn't mean it was considered "extremely hard for mathematicians" at the time. An entrance exam has to test some basic material too, such as the ability to create the proper equations to solve word problems. This could have been just one of 50 such questions. Moreover, people applying to MIT are high school students, not mathematicians.
This is the easiest question I found in this channel. This type of question I have been used it in my school exam.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ..
This was my maths book question of class 10th ncert 🤣🤣🤣🤣
Are you using ur schools email😂
@@zzzzzz5411 no bro it's my personal email account actually I join the school class via using this id and teachers told use to write the class name that's why I do so.
If you are a student you will understand.🤣🤣🤣🤣
@@nelser1160 Bhai tereko kya hoa
Watch this RUclips video for the correct answer of this riddle.
ruclips.net/video/NcrTWE4w784/видео.html
Congrats in Advanced MYD FOR 2M SUBS🤗❤🙏
How I got the answer: I made it simpler by using “now” instead of “2 years ago” and “In 16 years” instead of “In 14 years” so I only had to keep track of the 16. Then I did a sample problem. Say they were 10 and 30, it would take 10 years for the father’s age to be only double of his son’s. I realize the starting sons age doubles until he is only double his fathers age, so I figured that the son had to be 16 and father had to be 48 two years ago, so I reached 18 and 50 by adding those years back
super funky that it’s MIT, guess it teaches you how much we’ve improved culturally at certain things over time
now simultaneous equations are a super normal thing to do in middle/high school
Nice fun simple problem! The way Presh aggressively said 50 at 2:14 made me laugh tho lol. 😂
in order to simplifie one of the equation, i choose my varables being the age 2 years ago
so the first one is juste x = 3y
and the other one is x + 16 = 2 (y + 16)
and we get y = 16 and x = 48
so when we add the 2 last years we also get 18 and 50
@@UTube4MNi also do it in my mind
I find it easier to make it one variable.
2 years ago: father = 3x and son = x
Now : father = 3x + 2 and son = x + 2
14 years later : father = 3x + 16 and son = x + 16
since the father is only twice the son's age in 14 years, we can conclude that : 3x + 16 = 2 ( x + 16 )
3x + 16 = 2x + 32
x = 16
substitute back into the Now equations and get : father = 50 and son = 18
Yep I did the same
Why was the question is here ? What did I miss ?
Itna aasan ( so simple in hindi )
" I solved mit question "
Now I will tell my friends ☝️
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
solution with only one equation: just equate expressions for father's age from both the relations.
Ages 2 years back : x, 3x
Ages 14 years from now: x+16, 2x +32
Thus : 3x + 16 = 2x + 32
=> x = 16
=> present ages : 18, 50...
WOOHOO! This was actually one I was able to solve on my own! I guess I'm better at these than at geometry :) Keep up the good work and happy holidays each and all
😉 Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
Hi, even a 7th grader can solve it
@@Neel-sama 😝
Loved this puzzle! :D
Without watching the video, just reading the problem, I reasoned as follows:
a-2=3(b-2) & a+14=2(b+14). We end up with a=3b-4 & a=2b+14, so 3b=2b+18, hence b=18 and a=50 substituting back into the first two equations:
50-2=3(18-2): 48=48 and 50+14=2(18+14): 64=64 it all checks out. Pretty straightforward.
Now little kids are allowed to subscribe this channel
For the first time I solved a 'Mind Your Decision' problem before seeing the video.
i can solve this in seconds, here are the steps:
1,Consider: the father is three times than the son today, and will be twice of that in 16 years later
2,Instantly conclude: the son is 16 years old today, and the father is 48 years old( this is because 2(A+A)=3A+A )
3,but this is what happened two years ago, so the son is 18 and father is 50
This is a much more interesting answer.
Now, how would someone teach it for other age riddles?
@@צילהירחי-ח9מ maybe already?
First time I solved a problem from Mind Your Decisions 🙂🙂🙂
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ....
I got caught up in interpreting the tense of the question. The Question states: what "were" the ages of each? It didn't say: What "are" the ages of each. So I took this to be the ages of the boy and his father 2 years ago. f=3s, f+2+14=2(s+2+14), f=48, s=16
Son: Dad how old are you ?
Dad : Two years ago I was three times as old as you ; but in fourteen years I shall be only twice as old as you.
Son : Whatever I dont need that !
Son 18, father 50,
"কেশব চন্দ্র নাগ এর পাটিগাণিত"
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
I started the same way as Presh Talwaker, but I suddenly found an easier solution.
My solution:
"forget" 2 years ago (for a while):
3x = y
now remember that 2 years + in 14 years, he'll be twice as old as his son, so:
2x + 16 = y
3x = 2x + 16
Therefore 2 years ago, the son was 16 and father was 48 -> now 18 and 50.
Correct me if I'm wrong.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ...
Me, who solved using substitution instead of elimination:
Darn it, that took longer than it should’ve. Well at least I could solve ONE of these puzzles
This is a 4th standard question. Mean a 4rth standard student can do this type of questions
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
This was a simple one! Great video. :)
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
yippie! now i can go to MIT.
Why do they use the preterit "were" instead of the present : "what are the ages of each?".
I might be overthinking but for me it can be misleading.
Like do you answer with their ages today or two years ago?
The problem starts with "A father *SAID* to his son...", how do you know how long ago the conversation was? You can only deduce their ages at the time of the conversation (or at some date relative to when the conversation occurred). From the information given you can't deduce anything about their ages today nor two years ago. Admittedly the question using "were" does not make clear for when you should calculate there ages, but your answer can only be relative to the date of conversation. The correct answer is 18+c and 50+c.
I remember posting a riddle of sort of similar nature on QuizUp but nobody got in right, though it might have been confusing wording.
"12 years ago, I was twice as old as people born 12 years ago were when I was 12. How old am I now?"
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ....
Thank god! One problem that i have solved correctly from this channel😌😌(for Indians it was a easy one😁)
Lovely bit of simultaneous equations
I don't expect they wouldbhave thought students would fail to solve this puzzle. Rather, they were likely listening to them talk it through to get a sense of their speed and thought process.
So two years ago, the father was 48 and the son was 16. In 14 years, the father will be 64 and the son will be 32. Thus, the father was 32 years old when his son was born.
I hate these type of puzzled but it is good to learn.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ..
Could you please tell me the name of the software that you use to make your videos on RUclips?
That's easily one of the easiest question on ages. But we'll explained ❤️❤️❤️👍👍👍
I got a riddle for you.
If the father's age in 2020 is 25, his son's birthdate in the 25th of October, 2020; What will be the age of his son in the New Year of 2021?
@@zuregayeta7183 2 months 6 days
To those who think this is very simple, yes it is very easy because it was a question from 1876 admission test of MIT . And that time simultaneous linear equations were like calculus at present
what are the ages of each I think?? not were
That was easy... We learn to solve these sort of maths in primary schools in Bangladesh.
1. Father - 2 = 3*(Son - 2)
2. Father + 14 = 2*(Son + 14)
Subtract 2 from 1
Father 50, son 18
That's MIT?
For what ages is this "puzzle“ intended?
Thanks for the video!!
son is 18. Dad is 50
x is Dad and y is the son. x-2=3(y-2) and x+14=2(y+14) in the end x= 50, y =18
remind me of 28 years ago when I was in 8th Grade. Thanks
It would be easier to consider the ‘2 years ago’ as the 0 time, and the ‘14 years later’ as the time 16 years later
you can be a father only after you reach puberty, so a father cannot be 2 year old
Though now we might consider it just an easy question for preteens to solve, it took almost 125 years to be that way. Ain't that too long?
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers .
age of father: X
age of son: Y
equation (1): X-2 = 3(Y-2)
equation (2): X+14 = 2(Y+14)
simplify (1) to get (3): X = 3Y-4
use (3) to substitute out X in (2) and get (4): 3Y-4+14 = 2(Y+14)
simplify (4) to get (5): Y = 18
use (5) to substitute out Y in (3) and get (6): X = 3*18-4
simplify (6) to get (7): X = 50
by (5) and (7) we have the solution: the father is 50 years old and the son is 18.
It's the same aftab question from linear equation in two variables
16×3 = 48
+16 = 64
16+16 = 32
*2 = 64
16+2 = 18
48+2 = 50
Ah! I remember solving like these kind of father son ages question in class 6.
But I liked the basicity of MIT
Father -50
Son-18
wow shows how much more powerful math tools became. Were engineering concepts like matrices still struggling to make mainstream maths at the time?
Hey sir presh
Please can you tell me which software you use to make your videos.???
I thought of it like this: Adding up the 2 years and the 14 years, you get 16 years between the years when the father was 3 times as old and when the father is twice as old. When the father is 3 times as old as the son, the age difference is 2 times the son's age and when the father is twice as old as the son, the age difference is the same as the son's age, so the son has to be twice as old in the second scenario as he is in the first. So this means the son's age 2 years ago was 16 and in 14 years, he will be 32, and the age difference between the father and son is 32 years. Since the son was 16 2 years ago, he must be 18 now, and the father is 32 years older so 50.
The son was 16 the father 48. The son is now 18 and the father is now 50. In 14 years the son will be 32 and the father 64.
a-2=1/3(b-2)
a+14=1/2(b+14)
3a-4=b
2a+14=b
a=18
50=b
16×3=48
48/3=16
Well the difference of ages will be same at any period, so a simple understanding of ratio can help to solve the problem in much less time
Oh, If it would be 1800, I would be easily admitted to MIT. But now, I have just sent my application and I'm not sure if I would be in...
I am a 6th grade Indian student and fortunately found it so easy😄😄
Father is 38 years old and son 12 years old.
So much easy this type of questions now ask in 5 th class
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Ans. Father 50yrs and Son 18 yrs
I worked out 16 and 44 to be their current age. 2 years ago they'd be 14 and 42. 42 is 3x larger than 14. And in 14 years they'll be 28 and 56. 56 is 2x bigger than 28. Am I wrong or is this answer also correct?
If their current ages are 16 and 44,
Then in 14 years they'll be 30 and 58.
And 58 is NOT 2× bigger than 30.
So I guess you made a silly mistake there.....
@@pratibhanarvekar7534 yeah I added 14 years after subtracting the 2 years. That was my bad. Thank you.
@@emmanuelvinod4037 Most welcome, Humans are meant to make & correct their mistakes, just don't forget to keep on learning.
Please make video on 2019 IMO Problem 6.
Sir make video on jee advance sums
Hi mind your decision Can you solve it for me. I have a irregular hexagon whose three sides are equal and also the remaining three sides are equal. The first three sides are equal to 3.88meter and the remaining three sides are equal to 0.5meter. Calculate the area of hexagon. Hoping for your response.
You can also email to this new channel e-mail: yourmathsguy1910@gmail.com
Channel tag #onlymathlovers
I did it mentally 😅 I think in India we hv these kind of que. in class 6,7,8 😅 but the que. is from 1876 😂😂😂 So thats ok 👍 I donot need to fly in air 😂
Ths question could be solved by one variable also. You have to Let only sons age.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ....
I wish MIT test today would be as easy as the 1870s.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #onlymathlovers ...
who helps me prove that 1/2
F-2 = 3(S-2) ==> two years ago
F = 3S-6+2
F = 3S-4 ---- (1)
F+14 = 2(S+14) ==> 14 years from now
F = 2S+28-14
F = 2S+14 ---- (2)
Equate (1) and (2):
3S-4 = 2S+14
S = 18
And solve for F with (1):
F = 3(18)-4 = 54-4
F = 50
Two years ago they were 48 and 16 (✓) and in 14 years they'll be 64 and 32 (✓).
Whats the name of the program you use to made thise videos
x-2=3(y-2)
x+14=2(y+14)
x-2=3y-6
x+14=2y+28
x=3y-4
x=2y+14
"what were the ages" .. ( WHEN ? ) AGE at time equations are stated or when it was "2 years ago" ( if I was not allowed to ask, I'd have provided both .
18+c and 50+c for all c, such that: 0 < c < (today() - date_of_converation)
Yes!! I was Right!, (50&18) before watching video! Feels Good^ 👍
If I was alive in 1876, I would have entered MIT at an age of 12.
Same here
You would not. The spread of information then is not like now.
you are under 13
Me too hahaha
Math Question: Is there more real numbers or complex numbers?
Same according to Cantor's one-to-one definition
Isn't it?
Real nos : Real
Complex nos : Real + imaginary
Real ---> ♾️
Imaginary ---> ♾️
Hence, no of complex nos = ♾️ + ♾️ = ♾️
Hence, no of complex nos = no of real nos = nos of complex nos
Conclusion : All Infinite sets of nos are equal. @I am somebody, no of even = no of odd = no of complex = no of rational = irrational = imaginary = quantum = dark = ..........😡
My Q : Is is true that "N + 1 = W"
C is ismorphic to R^2. There exists a bijection from R^2 to R, so |C| = |R^2| = |R|.
@@awindwaker4130 a 6th grader will answer like 6th grader.. haha
@@FootLettuce i^2=-1
I’m so happy I finally solved one 😁
Matthew McConaughey travelling in the fifth dimension:
No one picked up on the question which is "What were their ages?" Not "What are their ages?" Since the only reference to the past is two years ago, the answer to the question must be that the son was 16 and the father was 48.
It doesn't have to be complicated. Suppose two years ago the son's age is x, then the father's age is 3x+2. 3x+2+14=2*(x+2+14) Solve for x. x=16 16+2=18 The son's age is 18. 3x+2=50. The father's age is 50.