I’m interested in this as maths doesn’t come naturally to me & I have to work bloody hard to ‘get’ equations.
How do you know the answer? Is it just because you have a good feel for numbers or is it trial & error? How can you so easily see the answer?!

I had an eye-opening experience when my mate, whose maths is really bad, asked me for help (although I’m no maths whizz) practising questions for an aptitude test he had to do to become a fireman. The problems involved the manipulation of fractions and I found them straightforward, but what I found harder was explaining to my mate how I got the answers as I had acquired the habit of doing the intermediate steps automatically in my head, so much so that I had to re-explain those steps to myself before I could explain them to him.
With the problem above, someone with an intuition for maths (or who is in the habit) will see the value of the substitution (so – in the notation bobness uses – banging in (A+50) wherever you see a B. It’s the most efficient way to get the answer: it’s the quickest and, having only a couple of lines, it’s the one least likely to produce a mistake
Because, in this case, the numbers here are big fat round ones and the values for A and B are so low, when you look at the two equations: 2A+B =500; B=A+50
if you have a head for it, you can roll these around your noggin to get the answer. It’s an idea to do a quick guesstimate at this stage anyway – I look at 2A+B =500 where the two numbers aren’t of wildly different values and I might speculate without any real effort that A must be somewhere in the 100 to 200 ballpark. (Yes, that is a big ballpark, but if you’re dealing with questions involving decimals or, as my mate was, fractions this can be a big help in avoiding the wrong answer).
But if you haven’t seen one of these for decades, and don’t spot the substitution, you can still do it the long way.
You know B=A+50 and you have another equation that says 2A+B=500
So if you isolate the B in the second equation: B=500-2A, then the two equations are equivalent, as they are both the same as B, so
A+50=500-2A , then
A=(500-2A)-50, then
A=450-2A, then
3A=450
A=150, so B=200
This method involves no technique other than plonking the two equations you’ve been handed alongside one another. The problem with it is, if you haven’t done this stuff for yonks, there’s a lot of lines there in which to make a mistake (hence the value of the guesstimate at the start).
In short, the noggin roller, the seen-these-a-million-times dude and the WTF person all get the right answer, which is the important thing. Additionally, the speed advantages of the different techniques will diminish as the problems get more convoluted.
Dunno if this helps at all..?

All those blimmin’ words and I forgot to say: viz the noggin rolling – if 2A+B=500 and you know B is 50 bigger you can knock the 50 off the 500 to get 450 and then treat 2A+B as 3A to get A=150, which you should be able to do in your head.
Handy this, but beware miscalculation!

Addendum.
My ex went for a job with a telco and as part of the interview they presented her with a mathematical problem. When told the problem I foolishly didn’t say how did you go. Instead I said that’s easy and proceeded to do some simultaneous equations and work the problem out. Ex Mrs Wells wasn’t happy. Needless to say she hadn’t solved the problem not progressed with the job interview.
So yes some people find it easy to see answers . I don’t but do find equation based problems easier than the rest of it.

Seems familiar. I went for a job with one of the then big-5 accounting/consulting firms and had a similar experience in the late 90s. An ‘aptitude test’ which merely asked me to regurgitate high school maths from 30 years prior (early 70s). It so happened that I remembered simultaneous equations among other things and got a question much like this that I could answer, after a minute or two of head scratching at least. It’s a technique that I have used precisely twice since I left high school, once professionally (trying to produce an algorithm defining a test case which would scale up as it happened) and once in this interview….

For the record, my version created without reference was…A=spade, B=club
A=B+50
2B + A = 500
————————-
Swap the A stuff with the equivalent B stuff and start moving stuff around…
2B + B + 50 = 500
3B = 500-50 = 450
B=150

And therefore when you swap B back again, A=B+50=200

That helps a great deal actually- thanks Mr Robot. I’m currently redoing my maths GCSE as I flunked it rather spectacularly back when I was a stroppy teen & we touch on formulae, which I quite like, but when it comes to missing numbers- my head hurts!

Thing is, when you say that I realize there were probably several different little algebra tricks that were involved. Simultaneous Equations was swapping the equivalent bits around (I did it for A, Mr Robot for B), then there’s the rules for moving things from one side of the = to the other, and things you can do with addition, multiplication and subtraction that you can’t do with division and so on…. and here is where I remember that I can only help my teenage son so far (probably up to 2 years ago when he was 13 and last asked).

Stephen G says

Club = 150

Spade = 200

Twang says

DO show your workings G

bobness says

Club = A, spade = B

2A + B = 500

B-50=A which is the same as B=A+50

Substitute in…

2A + (A + 50) = 500

Expand and simplify,

3A = 450

A=150

Thus B = 200.

Twang says

Yes I follow that though I would never have come up with it in a million years. 10/10.

biggles says

Err:

A = B +50

2B +A = 500 (i.e. A = 500 -2B)

So B+50 = 500 – 2B

So 3B = 450

Therefore B = 150 and thus A = 200

??

Junior Wells says

No pen or paper n me but substitute club +50 for spade then you have only one unknown in equation. Should be straight forward after that.

fitterstoke says

All cats have four legs

My dog has four legs

Therefore….

Junior Wells says

If it was

All four legged animals are cats

My dog has four legs

Therefore my dog is a cat

But it wasn’t

So it isn’t necessarily.

Sewer Robot says

..therefore your dog is equipped to chase cats?

Sewer Robot says

btw: the average cat will have less than four legs because some cats are deficient in the leg department to the tune of one (or God forbid, more…)

Jackthebiscuit says

Club = 150 / Spade = 200.

Worked out mentally so I have no working out to show (honestly)

andielou says

I’m interested in this as maths doesn’t come naturally to me & I have to work bloody hard to ‘get’ equations.

How do you know the answer? Is it just because you have a good feel for numbers or is it trial & error? How can you so easily see the answer?!

Sewer Robot says

I had an eye-opening experience when my mate, whose maths is really bad, asked me for help (although I’m no maths whizz) practising questions for an aptitude test he had to do to become a fireman. The problems involved the manipulation of fractions and I found them straightforward, but what I found harder was explaining to my mate how I got the answers as I had acquired the habit of doing the intermediate steps automatically in my head, so much so that I had to re-explain those steps to myself before I could explain them to him.

With the problem above, someone with an intuition for maths (or who is in the habit) will see the value of the substitution (so – in the notation bobness uses – banging in (A+50) wherever you see a B. It’s the most efficient way to get the answer: it’s the quickest and, having only a couple of lines, it’s the one least likely to produce a mistake

Because, in this case, the numbers here are big fat round ones and the values for A and B are so low, when you look at the two equations: 2A+B =500; B=A+50

if you have a head for it, you can roll these around your noggin to get the answer. It’s an idea to do a quick guesstimate at this stage anyway – I look at 2A+B =500 where the two numbers aren’t of wildly different values and I might speculate without any real effort that A must be somewhere in the 100 to 200 ballpark. (Yes, that is a big ballpark, but if you’re dealing with questions involving decimals or, as my mate was, fractions this can be a big help in avoiding the wrong answer).

But if you haven’t seen one of these for decades, and don’t spot the substitution, you can still do it the long way.

You know B=A+50 and you have another equation that says 2A+B=500

So if you isolate the B in the second equation: B=500-2A, then the two equations are equivalent, as they are both the same as B, so

A+50=500-2A , then

A=(500-2A)-50, then

A=450-2A, then

3A=450

A=150, so B=200

This method involves no technique other than plonking the two equations you’ve been handed alongside one another. The problem with it is, if you haven’t done this stuff for yonks, there’s a lot of lines there in which to make a mistake (hence the value of the guesstimate at the start).

In short, the noggin roller, the seen-these-a-million-times dude and the WTF person all get the right answer, which is the important thing. Additionally, the speed advantages of the different techniques will diminish as the problems get more convoluted.

Dunno if this helps at all..?

Sewer Robot says

All those blimmin’ words and I forgot to say: viz the noggin rolling – if 2A+B=500 and you know B is 50 bigger you can knock the 50 off the 500 to get 450 and then treat 2A+B as 3A to get A=150, which you should be able to do in your head.

Handy this, but beware miscalculation!

Junior Wells says

Equations ,simultaneous equations finding one unknown ,finding 2 unknowns is part of the maths curriculum in middle school in Australia.

Junior Wells says

Addendum.

My ex went for a job with a telco and as part of the interview they presented her with a mathematical problem. When told the problem I foolishly didn’t say how did you go. Instead I said that’s easy and proceeded to do some simultaneous equations and work the problem out. Ex Mrs Wells wasn’t happy. Needless to say she hadn’t solved the problem not progressed with the job interview.

So yes some people find it easy to see answers . I don’t but do find equation based problems easier than the rest of it.

Harold Holt says

Seems familiar. I went for a job with one of the then big-5 accounting/consulting firms and had a similar experience in the late 90s. An ‘aptitude test’ which merely asked me to regurgitate high school maths from 30 years prior (early 70s). It so happened that I remembered simultaneous equations among other things and got a question much like this that I could answer, after a minute or two of head scratching at least. It’s a technique that I have used precisely twice since I left high school, once professionally (trying to produce an algorithm defining a test case which would scale up as it happened) and once in this interview….

For the record, my version created without reference was…A=spade, B=club

A=B+50

2B + A = 500

————————-

Swap the A stuff with the equivalent B stuff and start moving stuff around…

2B + B + 50 = 500

3B = 500-50 = 450

B=150

And therefore when you swap B back again, A=B+50=200

andielou says

That helps a great deal actually- thanks Mr Robot. I’m currently redoing my maths GCSE as I flunked it rather spectacularly back when I was a stroppy teen & we touch on formulae, which I quite like, but when it comes to missing numbers- my head hurts!

Harold Holt says

Thing is, when you say that I realize there were probably several different little algebra tricks that were involved. Simultaneous Equations was swapping the equivalent bits around (I did it for A, Mr Robot for B), then there’s the rules for moving things from one side of the = to the other, and things you can do with addition, multiplication and subtraction that you can’t do with division and so on…. and here is where I remember that I can only help my teenage son so far (probably up to 2 years ago when he was 13 and last asked).

Harold Holt says

…..And Bobness and Biggles et al…..