## Problem Descriptions and Solutions

- Charlie's Birdhouses Growth
- Karen's Christmas Party Guests
- Uranium-234 Decay
- Ornithine Decarboxylase Decay

**Problem Description:** Charlie's Birdhouses experiences an annual growth rate of 10%. In 2010, Charlie made 120 birdhouses. Determine the number of birdhouses he will make in 2015.

**Solution:** The exponential model for the number of birdhouses (Xj) each year is given by

X_(j+1)=1.1×X_j

And X_2010=120. Therefore, X_2015=120×〖1.1〗^5=193.26

Rounding to two decimal places, the expected number of birdhouses in 2015 is approximately 193.

Year |
Number of Birdhouses |

2010 | 120.00 |

2011 | 132.00 |

2012 | 145.20 |

2013 | 159.72 |

2014 | 175.69 |

2015 | 193.26 |

**Table 1: Number of Birdhouses vs. the year
**

**Problem Description:** Karen's Christmas party started in 1990 with 12 guests, increasing by 5% every other year. Find the expected number of guests for her 2012 Christmas party.

**Solution:** The number of guests (Nj) follows a geometric progression

N_j=1.05×N_(j-1)

The number of guests for 2012 (2012N2012) is found to be 35.1.

Year | No. of Guests |
---|---|

1990 | 12 |

1991 | 12.60 |

1992 | 13.23 |

1993 | 13.89 |

1994 | 14.59 |

1995 | 15.32 |

1996 | 16.08 |

1997 | 16.89 |

1998 | 17.73 |

1999 | 18.62 |

2000 | 19.55 |

2001 | 20.52 |

2002 | 21.55 |

2003 | 22.63 |

2004 | 23.76 |

2005 | 24.95 |

2006 | 26.19 |

2007 | 27.50 |

2008 | 28.88 |

2009 | 30.32 |

2010 | 31.84 |

2011 | 33.43 |

2012 | 35.10 |

**Table 2:** Number of guests vs. the year for party guests

**Problem Description: **Uranium-238, when mined for nuclear reactors, produces Uranium-234 with a half-life of 24 days. Calculate the remaining amount on December 31, 2011, if there were initially 24.5g on January 1, 2011.

**Solution:** Using the radioactive decay formula, the remaining amount (N) is found to be 0.00064g on December 31, 2011.

N = N_0× (1/2)^(t/T)

N = N0 * (1/2)^(t/T)= 24.5 * (1/2)^15.21= 0.00064 g

**Problem Description:** The enzyme ornithine decarboxylase in the human body has a half-life of 11 minutes. If 131mg is present, determine the amount remaining after 1 hour.

**Solution: **Applying the radioactive decay formula, approximately 2.98mg of ornithine decarboxylase will remain after 1 hour.

N = N_0× (1/2)^(t/T)

N = N0 * (1/2)^(t/T)= 131 * (1/2)^(60/11)≈ 2.98 mg

Therefore, after 1 hour, approximately 2.98 mg of ornithine decarboxylase will be left from the initial 131 mg.