They got married ...
Three young ladies got married on the same day. Their combined weight was three hundred and ninety-six pounds. Nellie weighed 10lbs. more than Kitty, and Minne weighed 10lbs. more than Nellie. Bridegroom John Brown weighed exactly the weight of his bride. William Jones weighed half again as much as his bride. Charles Robinson weighed twice the weight of his bride. All together, the 6 people weighed half a ton.
What was the full name of the three brides after the wedding ceremony?
What was the full name of the three brides after the wedding ceremony?
posted by Zaux at
2:08 PM
9 Comments:
Mrs Brown
Mrs Jones
Mrs Robinson
regards, Curtis
But seriously,
There is no solution for half a ton as I understand it....so I guess it means a "short" ton
k+10=n
n+10=m
thus 3k+30=396
so k=122, n=132, m=142
j+w+c=604
trial and error
kitty brown
nellie jones
minnie robinson
regards, Curtis
hey Curtis ...
I like the first answer better, but both are correct.
:)
Given:
1) n+k+m = 369
2) n = k+10
3) m = n+10
JB = bride's age
WJ = 3/2 * bride's age
CR = 2 * bride's age
2a) n=k+10 ⇒ k=n-10
Substituting 2a)and 3) into 1)
n+(n-10)+(n+10) = 369
3n = 369
n = 123 lbs.
m = 123 lbs.
k = 113 lbs.
All 6 weigh half ton = 1000 lbs.
3 men weigh 1000 - 369 = 631 lbs.
NOW, I'M STUCK ... (Curtis used brute force) ... can anyone see an algebraic solution!
spotted an error above:
m = 133 lbs.
Zaux, your original problem statement was that the girls totalled "three hundred and ninety-six pounds" : 396, not 369.
k+m+n = 396
n = k+10
m = n+10
k+(k+10)+(k+20) = 306
3k = 366
k = 122
n = 132
m = 142
k+n+m+j+w+c = 1000
j+w+c = 1000-396 = 604
j = his bride
w = his bride * 1.5
c = his bride * 2
at this point I just tried combinations, and the first combination worked:
j&k : j = k = 122
w&n: w = n*1.5 = 198
c&m: c = m*2 = 284
j+w+c = 122+198+284 = 604
you're right Ross ... guess I'm tired ... thanks man
Hi Zaux. No algebraic solution that I can think of, but, by noting that the sum of the men's weigh ends with 4, and looking at a table with all the combinations, you quickly realise that there are only two possible full trial and error additions needed.
Thanks Chris ... I looked at it 30 minutes and could see no math solution.
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