I came up with this riddle which is struggling me.

Could you provide any hint or help?

A number of women proposed once each, of whom one-eighth were widows. In consequence, a number of men were to be married of whom one-eleventh were widowers. Of the proposals made to widowers, one-fifth were declined. All the widows were accepted. Thirty-five forty-fourths of the widows married bachelors. One thousand two hundred and twenty-one spinsters were declined by bachelors. The number of spinsters accepted by bachelors was seven times the number of widows accepted by bachelors. Those are all the particulars that I was able to obtain. Now, how many women proposed?

What I tried was to give $x$ as women and $y$ as men.

Now, I took the numerical hint of the $1221$ spinsters as

$$1221 = \text{women} - \text{accepted widows} - \text{rejected widows} - \text{accepted spinster}$$

Where we can read that women = $x$ and widows = $1/8 x$ and accepted spinsters = $7$ times the accepted widows.

Now the accepted widows can be seen as the sum between those who married bachelors and those who married widowers.

Those who married bachelors were told to be $35/44$ of the total widows that is $35/44 \cdot 1/8 x$.

But now I'm stuck with the number of widows who married widowers..

Any help?

Thank you!

  • 2
    $\begingroup$ For non-native english speakers, what's a spinster? $\endgroup$ – 5xum Jul 21 '17 at 9:22
  • 2
    $\begingroup$ @5xum It's a woman who never got married! $\endgroup$ – Turing Jul 21 '17 at 9:24
  • $\begingroup$ Ah, OK. So a women is either a widow or a spinster, but not both? $\endgroup$ – 5xum Jul 21 '17 at 9:24
  • $\begingroup$ And what about men? Is a widower a bachelor or not? $\endgroup$ – 5xum Jul 21 '17 at 9:25
  • $\begingroup$ @5xum Exactly. Basically you have women = $x$. Of those, one eight of them are widows and the others are spinsters. $\endgroup$ – Turing Jul 21 '17 at 9:25

Let $a,b,c,d$ be the numbers of widows, respectively accepted by widowers, rejected by widowers, accepted by bachelors and rejected by bachelors.

Let $e,f,g,h$ be the numbers of spinsters, respectively accepted by widowers, rejected by widowers, accepted by bachelors and rejected by bachelors.

The eight equations are:

$b=f=0 ; h=1221; a+b+c+d+e+f+g+h=x$


$d=(a+c+d)/5; 35/44(a+e)=e; 7e=g$

Algebric manipulations yield:










Ouf !

  • 1
    $\begingroup$ Wow that was PERFECT!! I tried to create such a system of equations, but I failed in writing down two of them, now I understood my error! Thank you again! $\endgroup$ – Turing Jul 21 '17 at 13:07

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