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| Abramowitz And Bernhardt Families (Posted on 2006-06-21) |
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Two families lived next door to each other the Abramowitzs and the Bernhardts. The total ages of the four members of the Abramowitz family amounted to one hundred years, and the total ages of the four members of the Bernhardt family also amounted to the same.
It was found in the case of each family that the sum obtained by adding the squares of each of the children's ages to the square of the mother's age equaled the square of the father's age. In the case of the Abramowitz family, however, Emmylou was one year older than her brother Eric, whereas Francine Bernhardt was two years older than her brother Frank.
What was the age of each of the eight individuals?
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Submitted by K Sengupta
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Rating: 3.5000 (2 votes)
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Solution:
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Let (a,b,c,d) =(Eric, Emmylou, Mrs. Abramowitz, Mr.Abramowitz) , in order of Age
and (e,f,g,h) =(Frank, Francine, Mrs. Bernhardt, Mr. Bernhardt), in order of Age
In terms of provisions of the problem under reference:
(I) a + b + c + d = e + f + g + h = 100;
(II) a^2 + b^2 + c^2 = d^2
(III) e^2 + f^2 + g^2 = h^2
(IV) b = a + 1 and f = e + 2
We know that,
4*(ap)^2 + 4*((a+1)p)^2 + (2*a^2 + 2*a +1 p^2)^2 = (2*a^2 + 2*a +1 + p^2)^2
or, a^2 + (a+1)^2 + c^2 = d^2;
where, c = (2*a^2 + 2*a +1 p^2)/(2p) and d = (2*a^2 + 2*a +1 + p^2)/(2p)
Now, a+b+c+d = 100
Or, a+(a+1)+ (2*a^2 + 2*a +1 p^2)/(2p) + (2*a^2 + 2*a +1 + p^2)/(2p) = 100
Or, (2*a^2 + 2*a +1)/p = 99 2*a
Or, 2*a^2 + 2a(1+p)-(99p-1) = 0
Or, a = (-2(1+p) +/- sqrt(D))/4; where D = 4(p^2 + 200*p -1)
For p=5, we observe that D=4096, which is a perfect square, so that :
a = (-2*6 +/- 64)/4 = 13
b = a+1 = 14
c = (2*a^2 + 2*a +1 p^2)/(2p) = (365-25)/(2*5) = 34
d = (2*a^2 + 2*a +1 + p^2)/(2p) = (365+25)/(2*5) = 39.
We know that,
4*((eq)^2) + 4*((e+1)*q))^2 +(2*e^2 +4*e + 4 - q^2)^2= (2*e^2 +4*e +4 + q^2)^2 ;
Or, e^2 + (e +2)^2 + g^2 = h^2;
where g = (2*e^2 +4*e + 4 - q^2)/(2q) and h = (2*e^2 +4*e +4 + q^2)/(2q);
By the problem;
e + (e+2) + g + h =100
or, (2*e^2 +4*e +4)/q = 98 2*e
or, 2*e^2 + 2e(2+q) (98q 4) = 0
or, e = (-2(2+q) +/- sqrt(D))/4; where D = 4*(2+q)^2 +8(98q -4) = 4(q^2 +200*q -4)
For q=2, we observe that, D = 1600; giving :
e = (-8 +/- 40)/4 = 8, so that :
f = e+2 = 10;
g = (2*e^2 +4*e + 4 - q^2)/(2q) = 160/4 = 40;
h =(2*e^2 + 4*e +4 + q^2)/(2q)= 168/4 = 42.
Hence, (Eric, Emmylou, Mrs. Abramowitz, Mr.Abramowitz) = (13, 14, 34, 39) and,
(Frank, Francine, Mrs. Bernhardt, Mr. Bernhardt) = (8, 10, 40, 42)
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ALTERNATE SOLUTION:
The computer assisted solution by Charlie is given here.
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