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Aug 17th, 2002, 05:33 PM
#1
Thread Starter
Hyperactive Member
percent riddle.
Igor,Stanislav, and Vladimir are walking down the street. They each have a certain whole number of rubles with them(currency). It is known that Igor has 240% of what Vladimir has, and that Igor has 25% of the difference of what Vladimir and Stanislav have. It is known that Igor's money squared is less than 20% of Stanislav's money squared, and that Stanislav has more money than Vladimir's money squared * 2. At least how much does each one have? Can this be solved? First one with an answer+explanation wins.
Last edited by snakeeyes1000; Aug 17th, 2002 at 05:46 PM.
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Aug 17th, 2002, 10:16 PM
#2
Fanatic Member
looks like its just a system of equations and inequalities. i'll use the first letter in their name to represent each individual's cash, so we have:
i=2.4v :1
i=(v+s)/4 :2
i^2<(s^2)/5 :3
s>v^2 :4
1 & 2: 2.4v=(v+s)/4
2.15v=s/4
8.6v=s :5
4 & 5: v<8.6 :6
5 & 1: i=12s/43
SBS
i^2=(12/43)^2*s^2 :7
3&7: (12/43)^2*s^2<(s^2)/5
12/43<1/5
but the inequality is false. so it can't happen.
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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Aug 17th, 2002, 10:22 PM
#3
Addicted Member
Let Igor have i rubles, let Stanislav have s rubles, and let Vladmir have v rubles. (i,s,v) are positive integers.
i=(240%)*v
i=(12/5)*v (1)
i=(25%)*(s-v)
i=(1/4)*(s-v) (2)
i^2 < (20%)*(s^2)
i^2 < (1/5)*(s^2) (3)
s > 2*(v^2) (4)
Equate (1) and (2):
(12/5)*v=(1/4)*s-(1/4)*v
s=(53/5)*v (5)
Subst (5) into (4):
(53/5)*v > 2*(v^2)
2*(v^2)-(53/5)*v < 0
10*(v^2)-53*v < 0
v*(10*v-53) < 0
Since v > 0, 10*v-53 < 0
10*v<53
v<53/10
Since v is an integer, v</=5
From (5): s=(53/5)*v,
Thus, v=5, s=53
Subst v into (1):
i=12
Thus, Igor has 12 rubles, Stanislov has 53 rubles, and Vladmir has 5 rubles.
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Aug 17th, 2002, 10:25 PM
#4
Addicted Member
Bugz:
It is actually i=(s-v)/4, as stated in the problem:
Igor has 25% of the difference of what Vladimir and Stanislav have
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Aug 18th, 2002, 06:20 AM
#5
Thread Starter
Hyperactive Member
excellent kalkewl8ter. I am very impressed. Bugz, don't worry about it. You show me up in here most of the time anyway.
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Aug 18th, 2002, 08:40 AM
#6
Fanatic Member
oops thats what i get for posting when i usually sleep
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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