A question I got in a test today.

Hey guys,

Today in a test, I got this question to do in five minutes. Here it is from memory:

Johns wife's age is one year younger than John. John has two kids, they are one year apart in age.

John figures out that the product of him, and his wife's age + the sum of both his kids age is equal to 2011.

How old is John; how old is his wife; how old are his kids?

*Note ages are in positive integers.


Try solve it, without a calculator (Pen and paper only) and tell us what you get.

 

I got Johns age as 45, his wife's age as 44. Kid 1's age is 16 and the other 15.

If your interested i solved it using simultaneous equations and substitution methods. Solved in about 3:30.

 

Wow, i lost it.
Did you solve it on you're test?
I aint that smart tbh

 

Perfect answers!

I did, yea. I created the formula p² + p + 2c = 2011, and solved from there.

 

That always works. I also found it out that way but forget to post. :huh:

 

SHIT how old are you, im 17 if your close to my age i can coherently say that your FAAAAR smarter than me

 

How the hell did you solve it with 2 unknown variables? =/

 

Ugh, well the answer was said already so I won't try and work it out my self, but these types of problems always seem to get on my nerves.

 

Here is my favorite math problem.

OT: What grade are you in?

 

J: 45
W: 44
K1: 16
K2: 15

 

Love images like that haha

OT: I love these kinds of questions, shame I'm done with all my math classes after taking Calc 2

 

i must stop playing runescape..

i didnt even get the question..

 

What tier maths was this like i dunno what the call the like level of the subject where your from, because this seems pretty advanced

 

I would consider this a "GCSE Year 11" Level question in the "Harder" region of questions.

EDIT: Substitution + Sim eq's are pretty basic. This is not something you would seen in A level exams.

 

well i got the question and im only in yr 10, im 15. If you know how to do simultaneous equations its easy but if you don't then you've got no hope haha.

 

I really don't know how you can do simultaneous equation with this question. It's not how I solved it. Can anyone show me how'd you do it with the substitution method?

 

If the kids' ages are one year apart, shouldn't it be something like p² - p + 2c - 1 = 2011?

I did that and ended up making a mistake somewhere and got 0. Didn't bother fixing. I marked John's age as X, wife's age as X-1, kid 1 as Y and kid 2 as Y-1:

X*(X-1) + (Y+[Y-1]) = 2011 =>
X^2 - X + 2Y - 1 = 2011 =>
X^2 - X + 2Y = 2012 =>
2Y = 2012 - X^2 + X =>
Y = (2012 - X^2 + X)/2

Then I slap the Y value to X^2 - X + 2Y = 2012 as such:

X^2 - X + 2*([2012-X^2+X]/2) = 2012 =>
X^2 - X + 2012 - X^2 + X = 2012 =>
0=0 fuck everything

Has been like half a year since I did stuff like this, someone fix my math. My interest is piqued but my patience is spent; I assume I made a mistake with assigning the ages but can't be bothered to find my notes on how to do this properly~

 

Started reading this, was like "His 2 kids are 1 year apart" and my head exploded

 

Very good. Wish I was better at making equations. I'm pretty smart, just not good at that.

 

I don't see how this can be solved using any of the simultaneous equations approaches that I know of.

You are given:

J = John, W = wife, A = Kid #1, B = Kid #2

1) W = J - 1

2) A = B - 1

3) J*W + A + B = 2011


Substituting gives:

4) J^2 - J + 2B = 2012


In order to use a simultaneous equations approach, you would need an additional equation that would allow you to relate A or B in terms of J or W.

If you were supposed to use simultaneous equations to solve this, I think you may have forgotten a statement in the question prompt, such as something like:

"John's youngest child is 1/3 his age."

A = J/3

With this information, you could then use the substitution method to solve the equations.

As you've stated the question here, that type of information wasn't given, so the best you can do is guess-and-check, which wouldn't take too long with this problem.

First, I'd narrow down the range of possible years for the parents' ages by doing some rough calculations. I know that 50*49 is going to be too big (50*50 = 2500), and 40*39 is going to be too small (40*40 = 1600). So the answer must be between 40 and 50. Using some more quick calculations, I'd go through the combinations between 40 and 50 and determine if the product was close enough to 2011 to add in the sum of the kids' ages (kids must be younger than the parents, so the sum will always be less than 100).

For example, 41 * 40 = 1640. This is almost 400 away from 2011, which would make each of the kids around 200 years old. Not possible.

42*41 = 1722. No, almost 300 off.

43*42 = 1806. No, still over 200 off.

44*43 = 1892. No, still over 100 off.

45*44 = 1980. Close, add to possible solutions to check.

46*45 = 2070. No. Over 2011.


Before even calculating the kids' ages, you can see that there is only one possible solution for the parents' ages. Just check to see if the kids' ages will work out to be integers, as the problem states they must be (if they don't then there isn't a solution).

1980 + 2B = 2012

-> B = 16

A = B - 1

-> A = 15

Solved. J = 45, W = 44, A = 15, B = 16.

Not very elegant, but I don't see a quicker way to solve the problem. If someone else could share a method that is cleaner than this I would be very interested to see how it is done.