# Math Help - A French Week

1. ## A French Week

For about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?

(a) $1 \mbox{ French Week} = 1 \mbox{ French Week} \left(\frac{10 \mbox{ days}}{1 \mbox{ French Week}}\right)\left(\frac{10 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{100 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{100 \mbox{ s}}{1 \mbox{ min}}\right)$

$=10^6 \mbox{ seconds}$

$1 \mbox{ Normal Week} = 1 \mbox{ Normal Week} \left(\frac{7 \mbox{ days}}{1 \mbox{ Normal Week}}\right)\left(\frac{24 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{60 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{60 \mbox{ s}}{1 \mbox{ min}}\right)$

$=604800 \mbox{ seconds}$

So the ratio is 1.65, but the answers say that it is 1.43 ?

Also, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.

2. Originally Posted by DivideBy0
For about 10 years after the French Revolution, the French government attempted to base measures of time on multiple of ten: One week consisted of 10 days, one day consisted of 10 hours, one hour consisted of 100 minutes, and one minute consisted of 100 seconds. What are the ratios of (a) the French decimal week to the standard week and (b) the French decimal second to the standard second?

(a) $1 \mbox{ French Week} = 1 \mbox{ French Week} \left(\frac{10 \mbox{ days}}{1 \mbox{ French Week}}\right)\left(\frac{10 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{100 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{100 \mbox{ s}}{1 \mbox{ min}}\right)$

$=10^6 \mbox{ seconds}$

$1 \mbox{ Normal Week} = 1 \mbox{ Normal Week} \left(\frac{7 \mbox{ days}}{1 \mbox{ Normal Week}}\right)\left(\frac{24 \mbox{ hrs}}{1 \mbox{ day}}\right)$ $\left(\frac{60 \mbox{ min}}{1 \mbox{ hr}}\right)\left(\frac{60 \mbox{ s}}{1 \mbox{ min}}\right)$

$=604800 \mbox{ seconds}$

So the ratio is 1.65, but the answers say that it is 1.43 ?

Also, in the first question I assumed that the 'French' second was the same as the 'normal' second, so wouldn't my answer to the second question make the answer to my first question invalid? Thanks.
Hello,

the ratio between the weeks is:

$\frac{\text{french week}}{\text{normal week}}=\frac{10}7 \approx 1.42857...$

to b):
$1 \text{ french day}= 10 \ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:

$\frac{10 \ h \cdot 100\ min \cdot 100\ sec}{24\ h \cdot 60\ min \cdot 60\ sec} \approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.

3. Originally Posted by earboth
Hello,

the ratio between the weeks is:

$\frac{\text{french week}}{\text{normal week}}=\frac{10}7 \approx 1.42857...$

to b):
$1 \text{ french day}= 10 \ h$ I assume that a new french day lasts as long as a normal day. Then you'll get the ratio:

$\frac{10 \ h \cdot 100\ min \cdot 100\ sec}{24\ h \cdot 60\ min \cdot 60\ sec} \approx 11.574...$ That means one of the new french seconds is a very short time interval. So better check my calculations.
Why wouldn't DivideByZero's calculations be correct? When comparing two terms of measurement, you must relate them somehow

4. Hello, DivideBy0!

A fascinating problem . . . took me a few moments . . .

For about 10 years after the French Revolution, the French government
attempted to base measures of time on multiple of ten:
One week consisted of 10 days, one day consisted of 10 hours,
one hour consisted of 100 minutes, and one minute consisted of 100 seconds.

What are the ratios of (a) the French decimal week to the standard week
and (b) the French decimal second to the standard second?

$\text{1 French Week }\:=\:\frac{\text{1 French Week}}{1} \times \frac{\text{10 days}}{\text{1 French Week}} \times \frac{\text{10 hrs}}{\text{1 day}}$ $\times \frac{\text{100 min}}{\text{1 hr}} \times \frac{\text{100 s}}{\text{1 min}}$
. . $=\;10^6 \text{ seconds}$

$\text{1 Normal Week} \;= \;\frac{\text{1 Normal Week}}{1} \times \frac{\text{7 days}}{\text{1 Normal Week}} \times \frac{\text{24 hrs}}{\text{1 day}} \times \frac{\text{60 min}}{\text{1 hr}} \times \frac{\text{60 s}}{\text{1 min}}$
. . $=\;604800 \text{ seconds}$

All of this is absolutely correct!
Note that (nearly) all the units are unequal . . . weeks, hours, minutes, seconds.

The only unit that both systems agree upon is the length of one day.
. . (Perhaps measured from sunrise to sunrise.)

$a)\;\;\frac{\text{1 Frech week}}{\text{1 Normal week}} \;=\;\frac{\text{10 days}}{\text{7 days}} \;\approx\; 1.43$

$b)\;\;\frac{\text{1 French day}}{\text{1 Normal day}} \; = \; \frac{10^5\text{ French sec}}{86,400\text{ Normal sec}} \;=\;1.1574
$

. . The French second is about 1½ jiffys longer than a Normal second.

5. Thanks, I don't know why I thought days could be of different lengths.