Pi is now 5 due to inflation

Having Pi equal five definitely does not help the world go round.

To be Devil's Advocate:
Given that the rest written in Comic Sans, it may be an early elementary school exercise, aimed at teaching kids to do multiplications. In this case, it's tolerable and/or defensible to find a simplification for pi.

That said, making pi equal to 3 would have been more accurate for that...

That's how you know math is OP when you can calculate volumes in parallel worlds where circles don't even looks like circles

They're just rounding up from "3."

Damn inflation is everywhere now, even the transcendentals :'(

This is how the wealthy calculate their tax exemptions.

With π=5 maths break down completely. If π=5, then e^(5i) = -1, meaning -1 = cos(5) + i * sin(5), or -1 ≈ 0.284 - 0.959 i

Idk, if you want to test people on how they understand formulae and order of operations without letting them just punch it into a calculator. The actual math isn't hard, but if you don't get substituting values into an equation then it's not trivial

When you want students to not use calculators but still rate a question for 10 second answer.

"One! Two! FIVE!" "Three, Sire!" "THREE!"

That just means the curvature of spacetime is negative.

Teaching them to to obey dumb instructions from incompetent bosses.
Very useful skill.

Assume the earth is a flat disc . . .

π = 5 for very small values of r.

New New math

My favorite is when doing Order of Magnitude estimations, you are supposed to let pi = 1

Ah, Indiana math.

https://en.wikipedia.org/wiki/Indiana_pi_bill

This physics class?

I would be a smartass and leave Pi as a factor throughout and in the answer. I'm used to doing that in Calculus anyways.

V = πr^2^h

V = π⋅10^2^⋅10

V = π⋅100⋅10

V = π1000

BONUS SOLUTION:

V =∫~0~^10^ A⋅h dh

A = ∫~0~^10^ 2πr dr

V= ∫~0~^10^ ∫~0~^10^ h⋅2πr dr dh

h is a constant for A's integral so we can safely move it into V's integral

V= ∫~0~^10^ h⋅∫~0~^10^ 2πr dr dh

π is a constant so we can safely remove it from A's integral

A = π⋅∫~0~^10^ 2r dr

A = π⋅[r^2^]~0~^10^

A = π⋅( [10^2^] - [0^2^] )

A = π10^2^

A = π100

V = ∫~0~^10^ h⋅π100 dh

π100 is a constant so we can safely remove it from V's integral

V = π100⋅∫~0~^10^ h dh

V = π100⋅[h]~0~^10^

V = π100⋅([10] - [0])

V = π100⋅10

V = π1000

It goes a lot deeper but I'm not bored enough for that, yet.

EDIT: Hang on. I'm wrong with that height integral. Can somebody help remind me?

The answer is substantially different though. 3141 units (rounded correctly) vs 5000. That's a significant margin that even working it out in your head is not forgiving enough to cover.

Unless I fucked up.