I How can John reduce his debt by paying into a joint account with Kelly?

  • I
  • Thread starter Thread starter Buuuka86
  • Start date Start date
  • Tags Tags
    Calculations Joint
Buuuka86
Messages
2
Reaction score
0
TL;DR Summary
Joint account calculations
John owes Kelly £500.

John and Kelly have a joint account. Each of them pays £200 into the account every month, to spend on joint expenses.

If Kelly pays only half of her monthly contribution into the joint account (so £100), and John pays the usual (£200) how much will John's debt (£500) be reduced by?

Just trying to work out my finances ;)
 
Mathematics news on Phys.org
Buuuka86 said:
Summary:: Joint account calculations

John owes Kelly £500.

John and Kelly have a joint account. Each of them pays £200 into the account every month, to spend on joint expenses.

If Kelly pays only half of her monthly contribution into the joint account (so £100), and John pays the usual (£200) how much will John's debt (£500) be reduced by?

Just trying to work out my finances ;)
Is this a schoolwork question?

What have you tried to do to come up with an answer?
 
It's a real life question :) I'm Kelly and my husband is John. I think the debt would be reduced by £50 every month but I'd love your help!
 
I agree.
You get the same result if you both pay $150 into the account and John gives you $50.
 
  • Like
Likes Buuuka86
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top