Mathematician Presh Talwalkar frequently digs up samples of classic college entrance exams so his followers
can see how they'd fare in the days before standardized testing.
Recently, Talwalkar unearthed a question from an 1876 algebra exam given to applicants at the Massachusetts Institute of Technology.
Can you solve it?
A father said to his son, "Two years ago I was three times as old as you; but in fourteen years I shall be only twice as old as you. What were the ages of each?"
Solution to the Problem:
The father is 50 years old and the son is 18 years old.
Let F = age of the father now.
Let S = age of the son now.
Then we can write two equations:
F - 2 = 3 (S - 2)
F + 14 = 2 (S + 14)
Distributing, we get:
F - 2 = 3S - 6
F + 14 = 2S + 28
From the first equation, F = 3S - 4 which we can substitute into the second equation: (3S - 4) + 14 = 2S + 28
S = 18.
Therefore, F = 3 (18) - 4 = 50.
Let F = age of the father now.
Let S = age of the son now.
Then we can write two equations:
F - 2 = 3 (S - 2)
F + 14 = 2 (S + 14)
Distributing, we get:
F - 2 = 3S - 6
F + 14 = 2S + 28
From the first equation, F = 3S - 4 which we can substitute into the second equation: (3S - 4) + 14 = 2S + 28
S = 18.
Therefore, F = 3 (18) - 4 = 50.
Correctly solved by:
| 1. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 2. Veena Mg | Bangalore, Karnataka, India |
| 3. Ivy Joseph | Pune, Maharashtra, India |
| 4. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 5. Ryan Huff |
Central High School, Grand Junction, Colorado |
| 6. Mohamed Sheriff (MEDDORA) | Freetown, Western Area Urban District, Sierra Leone, West Africa |
| 7. Brijesh Dave | Mumbai City, Maharashtra, India |
| 8. Kelly Stubblefield | Mobile, Alabama |
| 9. Alan Rench | Armstrong Creek, Wisconsin |
| 10. Ritwik Chaudhuri | Shantiniketan, West Bengal, India |