All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Current Age Conclusion (Posted on 2013-09-10) Difficulty: 3 of 5
The birthdays of the five friends Al, Ben, Cal, Dan, and Ethan falls on the same day, though they are different ages.

On their mutual birthday, the five friends were discussing their respective current ages. The discussion was as follows:
Dan said to Ben: "I'm nine years older than Ethan."
Ethan said to Ben: "I'm seven years older than Al."
Al said to Ben: "Your age is exactly 1.7 times that of mine."
Ben said to Cal: "Ethan is younger than you."
Cal said to Dan: "The difference between our ages is six years."
Cal said to Al: "I'm ten years older than you."
Cal said to Al: "Ben is younger than Dan."
Ben said to Cal: "The difference between your age and Dan's is the 
same as the difference between Dan's and Ethan's."
It is known that when one of them spoke to someone older, everything they said was true. However, when speaking to someone younger, everything they said was false.

What are the current ages of each person?

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Hybrid solution: part computer, then p'n'p | Comment 1 of 5

DECLARE SUB permute (a$)
CLS
a$ = "abcde": h$ = a$
DO
  a = INSTR(a$, "a")
  b = INSTR(a$, "b")
  c = INSTR(a$, "c")
  d = INSTR(a$, "d")
  e = INSTR(a$, "e")
  good = 1
  IF d < b THEN IF d < e THEN good = 0
  IF b < c THEN IF e > c THEN good = 0
  IF a > c THEN good = 0
  IF b < d THEN good = 0
  IF b < c THEN
     IF SGN(e - d) <> SGN(d - c) THEN good = 0
  END IF
  IF good THEN PRINT a$
  permute a$
LOOP UNTIL a$ = h$

The program disregards specific references to differences in age, and so can produce only a list of eligible orders, from youngest to oldest:

acedb
aecdb
aedbc
aedcb
eacdb
eadbc
eadcb
edabc
edacb
edbac

Dan is younger than Ben in all these scenarios, so Dan is indeed 9 years older than Ethan by statement 1.

Ethan is younger than Ben and is thus 7 years older than Al. This reduces the order possibilities to

acedb
aecdb
aedbc
aedcb

So Al is younger than Ben and Ben is indeed 1.7 times Al's age.

Cal of course is older than Al, but not by the exact ten years he claimed.

                            C                  order            
A    B    E    D    if B<C    if C<D     if B<C    if C<D      if order is aedcb
10   17   17   26    35
20   34   27   36    45         30       aebdc      aecbd
30   51   37   46    55         40       aedbc *    aecdb *       46 < C < 51 *
40   68   47   56    65 x       50

The first row is impossible, with two ages of 17.
The x against 65 for C is a contradiction of the column-title assumption, and thus a contradiction.

The orders marked * are within the list of possibilities.

Let's check the possibilities:

A=30
E=37
D=46
B=51
C=55

A=30
E=37
C=40  This set contradicts Ben lying (as older) to Cal
D=46         saying Ethan is younger than Cal.
B=51

A=30
E=37  This set contradicts Ben lying (as older) to Cal
D=46         saying Ethan is younger than Cal.
C={47 .. 50}
B=51

Al is 30.
Ethan is 37.
Dan is 46.
Ben is 51.
Cal is 55.

 


  Posted by Charlie on 2013-09-10 17:33:12
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information