Recruiting test question

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peterdenker

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Apr 7, 2021
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Hi everybody,

I hope I am posting in the right place, but here is a math question I stumbled upon during a recruiting test.

I am aware this is simple for most of the math aces here, but can you help me figure it out and how you got to the result?

Thank you!

20x+11y+40z=6500

16x+11y+38z=6040

20x+9y+36z=5780

16x+13y+36z=6580

then 18x+11y+36z=???
 
peterdenker said:
Hi everybody,

I hope I am posting in the right place, but here is a math question I stumbled upon during a recruiting test.

I am aware this is simple for most of the math aces here, but can you help me figure it out and how you got to the result?

Thank you!

20x+11y+40z=6500

16x+11y+38z=6040

20x+9y+36z=5780

16x+13y+36z=6580

then 18x+11y+36z=???
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
peterdenker said:
I hope I am posting in the right place, but here is a math question I stumbled upon during a recruiting test.

I am aware this is simple for most of the math aces here, but can you help me figure it out and how you got to the result?

Thank you!

20x+11y+40z=6500
16x+11y+38z=6040
20x+9y+36z=5780
16x+13y+36z=6580

then 18x+11y+36z=???
Click to expand...
Since there are four equations in three unknowns, it is possible (perhaps likely) that there is no solution. It is also conceivable that there could be many solutions.

If you trust that there is a unique solution, then I see a quick way to find the value of the given expression by combining just two of the given equations. But this provides no evidence as to whether that answer is really valid! And it could be that this is exactly how they want to trap you!

If you need a genuine solution, with certainty, then go back to what you learned about solving systems of equations. It might be wisest to use just three of the equations, and then check whether your solution works in all four.

By the way, this is definitely not mere arithmetic!
 
Apologies for not explaining why I am stuck.

Also sorry for this question not being pure arithmetic.

Where I am stuck:
So looking at the four given equations, I can see:
4x+2z -> 460 difference
2y+4z -> 720 difference

So that gives me a few hints, but I don't see more patterns to figure out the value of 1x or 1y or 1z.

My whole thinking process and approach seems to be wrong.
 
Have you ever solved a system of three equations in three unknowns? You need to follow a systematic method, not just look for random relationships. (Though my quick method, which doesn't prove there is a solution at all, did rely on that.)

But if you just find a third combination of equations that yields a simpler equation (even if there are still three unknowns), you will have a simpler system making it easier to eliminate variables. Try combining the second and fourth equations, for example.
 
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