Who Is My Husband?
In splitting up $1000, a group of three married couples agrees upon the following plan:
The wives receive a total of $396, of which Mary gets $10 more than Diane, and Ellen gets $10 more than Mary.
Bill Brown gets twice as much as his wife, Henry Hobson gets the same as his wife, and John Jones gets 50 percent more than his wife.
What are the full names of the three wives?
Note: Assume that the wives have the same last names as their husbands.
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7 solutions
(x + 10 ) + x + (x+20) = 396 3x = 396 x= 122 = D x + 10 = 132 =M x + 20 =142 = E 1000-396 =604 H HOB =x= 122 John Jones = 2*142 = 284 Bill Brown = 132 + 132/2 = 198 122+142+284=604
But why divisible by 4 ?
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Because since all numbers are even, the husband's number must be even.
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I'm really Sorry. But I really don't seem like getting a clue of the even and the husband and the number 4 :(
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@Dhvinay Pv – I showed the numbers for the women: 122, 132 and 142. All even. Given that the husbands are equal to the women or 0.5 or 2 of their number. The twice and equal are even. So, 5 of the six are even. The sixth must be even.
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@Hatim Zaghloul – Wow.. That is some serious application of a simple concept! Brilliant thinking.. And Thanks for the patience :)
Because 1.5 is 6/4?
Here is another good problem that you may try. https://brilliant.org/problems/who-gets-the-buck-2/ It has been qualified for level 2 algebra but the no. of solvers required is yet not met.
The wives are Mary ( m ) , Diane ( d ) , and Ellen ( e ).
m = d + 1 0
e = m + 1 0
Then..
m + d + e = 3 9 6
( d + 1 0 ) + d + ( m + 1 0 ) = 3 9 6
2 d + 2 0 + m = 3 9 6
2 d + 2 0 + ( d + 1 0 ) = 3 9 6
3 d = 3 6 6
d = 1 2 2
m = 1 3 2
e = 1 4 2
Then I try one by one and finally got that..
6 0 4 = 2 e + 1 . 5 m + d
So, there should be Ellen Brown .
So in this solution everyone is getting answers by hit and trial Me too Anyone here who had equation and then landed to solution of whose husband whom?
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yeah, some time passed, but i just came to this problem today :) maybe i'll add some idea about not guessing the surnames by trial and error: 142b+132h+122j=604 b+h+j=4,5 and also we know that 4,5=1+1,5+2 so we find that h=1,5 otherwise we will deal with odd number. after leaving with only 2 unknowns, we find b=2 and j=1
actually, at first i did it like this: wives' average=132 husbands' total=604 e+m+d=4,5 so I checked 132×4,5=594 and searched for relation between 10 and -10 as differences from average and their weights which being different resulted in 10=604-594: 10e-10d=10 leading to e-d=1 and thus e=2 and d=1 (and m=1,5). and - yes, from multiple choice answers it was sufficient to find Ellen Brown.
not sure whether it was understandable but hope it'll help someone :)
I have another idea.After gotting what wives have ,we just need to multiply 132,142,122 by 1.5 respectively,and then we check their units .And add up 2+4(units digits of the other two numbers ×1 or 2).Fortunately,132×1.5 is the answer just what we want.And we know who is Mary's husband.
So the women received x-10, x, and x+10, so x-10 + x + x+10 = 396, thus, x = 132. Specifically, Mary received 132.
The husbands then got a total of 604. It was stated that one of them had twice his wife's, one had the same amount as his wife, and Mr. Jones had 50%more than his wife.
Looking at the ones digit of the amounts that each husband received, one would've gotten 4 as the ones digit (twice the amount of all ladies would've ended with 4), the second one 2 as the ones digit (all the amounts given to the ladies ended with 2), and Mr. Jones either a 3 (if his wife was either the one who got 122 or 142), or 8 (if his wife was Mary).
However, the only way the sum of the money the husbands will get will be 604 is for Mary to be Mrs. Jones (since the sum of the amounts should end with 4, and that wouldn't be possible if Mr. Jones married the other two ladies). Thus, the only plausible combination among the choices was the one with Mary Jones.
I was about to post a similar solution.. But even i couldn't have explained so well.. Good work..
all the wives is having((x + 10 ) + x + (x+20) = 396),hence x=122,so wives is having 122,132,142 now we have to adjust 604 more in the husbands & at twice of any number last digit will be 4, if husband earns same last digit will be 2,so wed need last digit 8 after doing 1.5 of number,which comes only at 132
Solve for the wives' shares, keeping in mind that they must add up to 396. Diane = x = 122 Mary = x+10 = 132 Ellen = x+20 = 142 3x + 30 = 396 Therefore, x = 122
Since John gets 1.5 times his wife's share, her number must be one divisible by 4- henceforth, that must be Mary. And, since only one option contains a Mary Jones, that must mean that that is the answer.
Shame John's last name isn't Watson. :)
x + 10 + x + x - 10 = 396 x = 132
Husbands get 1000 - 396 = 604.
If you check the different combinations given in the options, you'll find the answer.
x+x+10+x+20=396 x=122, hence others are 132, 142
other husbands share x , x+10, x=20 in the ratio 1, 1.5, 2 times their wifes... 4.5 times x can be eliminated from the balance 604 to simplify the problem to distributing 0,10,20 to multiples of 1, 1,5, 2 so that the sum is 55
let ellen =M+10, diane=M-10 and mary =M
M+M-10+M+10= 396'
3M=396
M=132. by substituting into the previous equations we get ellen=142, diane=122.
husbands=1000-396=604.
we know that one of the husbands are 1.5 more thn his wife, 1 with same and one with twice as much.So we know that there is at least a (2*2)+2 in the ones digit. so we just need to combine each of the numbers( 122, 132 ,142) and multiply with the husbands ( 1.5, 1, 2) to know which wife is married to who.
It is really simple given the choices. We solve for the wives' shares (122, 132, and 142). Given that all numbers were even and one husband gets 1.5 times his wife's share, his wife must have a share that is divisible by 4. Only Mary had this share (132). So Mary must be Mary Jones.
We can verify that the rest works out.