Hard Puzzle - Most Difficult Age Puzzle

Hardest Puzzle based on Age: Find the ages
#hard #puzzle
- Adam is as old as Bob will be when Adam will be twice as old as Bob was when Adam's age was half the sum of their present ages.
- Bob is as old as Adam was when Bob was half the age he will be 10 years from now.
The puzzle is: How old are Adam and Bob?
(The actual puzzle mentioned the names Julia and John)

It's one the most difficult puzzles based on age... but it's definitely worth thinking over the solution for as much time as you need.

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[EnglishWithEnes]

Hard part was being a non-native English speaker. So it took a lot to understand, but simplifying and then generalizing always works. It took under 25 minutes for me.

[cool-aquarian]

nice.
I solved it using multiple variables for Adam and Bob’s ages at different times: A, B, A1, B1, A2, B2 and A3, B3.
A-B=A1-B1=A2-B2=A3-B3

[fardeenansari]

Yeeee! I solved it on my own. It took 20 minutes 🥵🥵🥵 and I use the same process you do 😊🙃. Thank you for such tricky puzzles❤.

[trnfncb]

If a and b are the current ages of Adam and Bob, one can rephrase:
1. When age of A was b, age of B was (b+10)/2.
2. When age of A was (a+b)/2, age of B was x.
3. When age of A will be 2x, age of B will be a.
Age difference d is the constant. From 1. we get d=(b-10)/2. From 2. we get x=(a+b)/2-d=(a+10)/2. From 3. we get a+10=a+d, i.e. d=10. Therefore (b-10)/2=10, or b=30 and a=40.

[decoder]

Awesome content.
Gained a good amount of knowledge on age problems.

[youtubegoogle]

My process was a bit too complicated,
So it required around 25-30 minutes,
But I got correct answer on my own. 👍

By the way,
Thanks for simpler solution.

[jimmykitty]

Intersecting Puzzle ♥️🌿🌿
JazakALLAAHU Khairan Brother 🥰
I've solved a similar problem form our General Mathematics Text Book in Bangladesh 🇧🇩 ♥️♥️

[girejaggu]

I've a major doubt....so in the first statement it says that Adam will be twice as old as Bob was when Adam's age was half the sum of their present ages
So if their present ages are 40 and 30
Adam should be Twice as old as Bob when Adam is 35....but that's not happening is 25 when Adam is 35 and I'm stuck here

[juxtamedullarynephron]

solved it on my own, but with a slight different approach. I used that the difference of their ages will be constant no matter how old they were and thus i could understand fastly.

[RisetotheEquation]

Nailed it, great puzzle. It's just a spider web of verbal hurdles; the math is the easy part!

[olerask]

Nice puzzle. It was easy. Solved it similarly to the video. 🙂

[gopanneyyar]

Ammar, it appears hard because you are trying to solve some part of the statements while translating it into algebraic equation. Instead, layout as many number of equations separately with those temporary variables that represents 'will be'(f for future) and 'was'(p for past).

Adam is as old as Bob will be:-
A = B + f (E1)

when Adam will be twice as old as Bob was:-
A + f = 2(B - p) (E2)

when Adam's age was half the sum of their present ages:-
A - p = (A + B) / 2
ie, A + B = 2(A - p) (E3)

Now, subtracting E3 - E2, we can get rid of 'p', but it will still have 'f'.
2(A - B) = B - f (E4)

E1 + E4 will give 3A - 2B = 2B
ie 3A = 4B

First part is over.

[reuvenlewis]

I solved it with three equations.
A = B + X (where A is present age of Adam, B is present age of Bob and X is the delta)
Then the second sentence gives: B = (B+10)/2 + X which simplifies to B = 10 + 2X
The first sentence gives A = 2((2B + X)/2 - X) - X which simplifies to A = 2B - 2X
When you put the first equation into the third, you remove A and find that B = 3X
Put that back into the second and find that X = 10
Now solve the second to find that B = 30
Now solve the first to find that A = 40

[Susp]

Using 3 variables makes it much easier, a, b & d for difference of ages which is always same

So like this we get d=10
And from any statement, we can find ages are 30 and 40

[Niko-fjmo]

I solved it using 5 variables: A, B, x, y, z.

Then we have 5 equations:
i) A - B - y = 0
ii) A - B - z = 0
iii) A - B - 2x = 0
iv) A -2B + 2x + y = 0
v) B - 2z = 10

From the first three equations, it's obvious that y = z = 2x.

If we replace "x" in the last two equations:

iv) A - 2B + 4x = 0
v) B - 4x = 10

Thus we reduce the system to 3 equations (iii, iv and v) and 3 variables (A, B, x). If you solve it you get: A = 40, B = 30, x = 5.

[moozeeck]

It's very tricky to understand phrasing because will be/was is used in same sentences and one can get in a mind loop.
It can be solved using two equations (like I solved it), but setting up equations can also be tricky.
Nice one!

[andreafollo]

This is a tricky algebra word problem and good for learning. I'm not sure I would call this a puzzle though.

[anil]

He has perfectly explained the solution

[madIndianHindu]

It took me 15 minutes, and at the end of equation i got negative age 😂, after watching your explanation i got to know about solving from back approach, Excellent Explanation! Keep it up bro 👍

[deep]

I took about 30-35 mins. to solve this. Really hard puzzle but one of the best brain exercises