What’s the correct answer to this arithmetic problem?

Question

Alexis is playing a game in which blue chips are worth 5 points and red chips are worth 3 points. If Alexis has 9 chips for a total of 21 points and the game has no other color chips, then

A. $B + R = 21$

$\quad5B + 3R = 9$

B. $B + R = 21$

$\quad3B + 5R = 9$

C. $B + R = 9$

$\quad5B + 3R = 21$

D. $B + R = 9$

$\quad3B + 5R = 21$

I think the answer is (C.) This is a Arithmetic: Transalation math problem.

Answer 1

I don't have enough rep to comment

I've Googled "Arithmetic Transalation" and "Arithmetic Translation" and haven't come across any special branch of maths that makes this problem non-trivial.

Each blue = 5pts

Each red = 3pts

Let B = number of blue chips

Let R = number of red chips

5B + 3R = 21 and B+R = 9 are straightforward to get from the wording

But these equations are only satisfied when B=-3 and R=12.

Is there something else that Nij and I are missing?

Answer 2

The intended answer is clearly (C).

In addition, it is clear that the person who posed this problem did not think it through.

The minimum chip value is $3.$ If Alexis has $9$ chips then (under any reasonable real-world interpretation) the minimum point value held by Alexis is $27$, achieved when Alexis holds $9$ red chips. Trading red chips for blue chips would only increase the point value and could not decrease it to $21.$

Solving the equations without considering what it means to "have a chip," we find that Alexis "has" $12$ red chips and $-3$ blue chips.

That's a very strange way to have $9$ chips.

To quote from an old story about lawyers, that question should be taken out and shot.