I bring receipts
I ran the numbers. I used U.S. federal minimum wages (1968 and today) and current spot gold to compute how many troy ounces a full-time minimum-wage worker could buy per week in 1968 vs today, and then calculated the annual income today that would be required to match 1968’s gold-buying power.
Key source numbers I used
- Federal minimum wage (effective Feb 1, 1968): $1.60 / hr.
- Federal minimum wage today (federal floor, still in effect in 2025): $7.25 / hr.
- Spot gold price (used for “today”): $3,648.25 per troy ounce (spot; source sampled Sept 2025).
- Historical 1968 gold market prices: after the London gold-pool stresses 1968 saw market prices above the $35 official parity; daily fixes cluster around ~$38–$40/oz for much of 1968 (I used $38.50/oz as a representative 1968 market average).
Calculations (step-by-step)
Assumptions: full-time = 40 hours/week, 52 weeks/year. Gold measured in troy ounces.
1968 (using $1.60/hr and $38.50/oz):
- Weekly earnings = $1.60 × 40 = $64.00.
- Ounces of gold/week = $64.00 ÷ $38.50 ≈ 1.6623 oz/week.
Today (federal minimum $7.25/hr and spot gold $3,648.25/oz):
- Weekly earnings = $7.25 × 40 = $290.00.
- Ounces of gold/week = $290.00 ÷ $3,648.25 ≈ 0.07949 oz/week.
Relative buying power:
- 1968 oz/week ÷ 2025 oz/week ≈ 1.6623 ÷ 0.07949 ≈ 20.9.
So a minimum-wage worker in 1968 could buy ~21× more gold per week than a federal minimum-wage worker can buy today (using the numbers above).
How large would yearly income
today
need to be to match 1968’s gold-buying power?
Take the 1968 annual gold quantity (1.6623 oz/week × 52 weeks = 86.44 oz/year). At today’s spot price:
- Required annual income today = 86.44 oz × $3,648.25/oz ≈ $315,370 per year.
Equivalently, you’d need about $6,065 per week (which ×52 ≈ $315,370) to buy the same gold annually a 1968 minimum-wage earner could.