PROBLEM 1:
Five school chums have 1st names A, B, C, D & E
  with sur names K, L, M, N & P
  who were representative of class F, G, H, I or J .
Using the 5 clues below solve for each ID and class Rep.

1. "I"'s Rep, who isn't "K" or "L", is neither "A" nor "D".
2. "E" isn't "N" and didn't Rep class "G".
3. "E", "H" and "BK" are different class Reps
4. "P" is class "F"'s Rep.
5. "M", who isn;t "A", is class "G"'s Rep.

   From 1:I / K / L & I / A / D[Reps]
   From 2:E (=\N) / G          [Reps]
   From 3:E / H / BK           [Reps]
   From 4:PF                   [Reps]
   From 5:MG / A               [Reps]
   
   NOTE:
   (1)  B=K  from[c]
   (2)  P=F  from[d]
   (3)  M=G  from[e]
   (4)  J=K after H=\K from[c] & I=\K from[a] & FP + GM from (2/3)
   (5)  I=N after FP from[d] & GM from[e] & I =\K =\L from[a]
   (6)  H=L left over!
   (7)  E=P after E=\HL from[c](6) & BK from[c](1) & E=\N + GM from(b)
   (8)  C=N after A=\IN & D=\IN from[a](5) & BJK from(1/4) & EFP from(2/7)
   (9)  A=L=H after A=\INC from[a](5/8) & BKJ from(1/4) & EPF from(2/7) & A=\GM from[e]
   (10) D=M=G left over!
   

See grid below:

	K	L	M	N	P		F	G	H	I	J
A	c	9	e	\6	|||		|||	\3	*	a	|||
B	1	c	c	c	c		\4	\4	c	\4	8
C	c	==	==	8	|||		|||	|||	|||	*	|||
D	c	|||	10	\6	|||		|||	*	|||	a	|||
E	c	\5	\3	b	7		8	b	c	\6	|||
F	d	d	d	d	2
G	e	e	3	e	d
H	\1	6	e	==	d
I	a	a	e	5	d
J	4	==	e	==	d

SOLUTION
1st	SUR	CLASS
A	L	H
B	K	J
C	N	I
D	M	G
E	P	F

A LITTLE MORE DIFFICULT:
PROBLEM 2:
Five school chums have 1^st names A, B, C, D & E
  with sur names K, L, M, N & P
  who were representative of class F, G, H, I & J
  at the age of 19, 20, 21, 22 & 30.
Using the 6 clues below solve for each ID and class Rep.

1. "C", who isn't "P", isn't class "H"'s Rep.
2. "H"'s Rep is 1 year younger than "P",
   but older than "D".
3. "F"'s Rep is younger than "M"
   who is younger than "I"'s Rep,
   who isn't "B" .
4. "E", who isn't "G"'s Rep, is younger than "K".
5. "F"'s Rep isn't "N" and is neither "D"'s nor "G"'s Rep.
6. "J"'s Rep is older than "A" who isn't "K".

   From 1:  C(=\P) / H       [Reps]
   From 2:  D / H | P        [age]
   From 3:  F / M / I(=\B)   [age]
   From 4:  E(=\G) / K       [age]
   From 5:  F(=\N) / D / G   [Reps]
   From 6:  A(=\K) / J       [age]
   
   NOTE I EXPECT ONLY 1 SOLUTION:
   (1)  D(=\G)(=\F) / H / P / C                 [Reps] from[a/b/e]
   (2)  D(=I/J)                                 [Reps] from[a/b/e] 
        i.  but if D=I then D=I=21        
            &  F19 / M20 / ID21 / H22 | P30     [age]  from[b/c]
            a. FC19 / M20 / ID21 / H22 | P30    [age]  from[a]
            b. F19 / MC20 / ID21 / H22 | P30    [age]  from[a]
             1. FEL19/MCG20/IDK21/HNA22|PJB30 OK[age]  from[d/e/f]
             2. FE19 / MC20 / ID21 / HK22 | P30 [age]  from[d]
        ii. but if D=J then A(=\K) / DJ / H | P [age]  from[b/f]
            a. A(=\K) / DJM / H | P             [age]  from[b/f]
            b. A(=\K) / DJ / HM | P             [age]  from[b/f]
   (3)  E(=\G) / K(=\J) / A / J                 [age]  from[d/f]
   
   
   Entering above data a-f into grid below:
   

	K	L	M	N	P		19	20	21	22	30		F	G	H	I	J
A	f	.	.	.	.		.	.	.	.	f		.	.	.	.	f
B	.	.	.	.	.		.	.	.	.	.		.	.	.	c	.
C	.	.	.	.	a		.	.	.	.	.		.	.	a	.	.
D	.	.	.	.	b		X1	.	.	b	b		e	e	b	.	.
E	d	.	.	.	.		.	.	.	.	d		.	d	.	.	.
F	.	.	c	e	X4		.	.	.	c	c
G	.	.	.	.	.		.	.	.	.	.
H	.	.	.	.	b		b	X2	a	(1)	b
I	.	.	.	.	.		c	c	.	.	.
J	.	.	.	.	.		f	.	.	.	.
19	d	.	c	.	b
20	.	.	.	.	b
21	.	.	.	.	X3
22	.	.	.	.	.
30	.	.	c	.	.

NOTE:
D19: 
Since (=) e/e/b///  &  (=) ///b/c/f
Thus D19 has no solution 
on super imposing ==> X1. 
And since D / H / P [age] from [c]
  then H20 is eliminated ==> X2
  and P21 is eliminated ==> X3.
Now, like D19, 
  FP has no solution ==> X4;
  - later \1a from [c]
Now, since H=\19 [b]; H=\20 [X2]; 
  H=\21 [a]; & H=\30 [b];  
thus H=22.
Now we can start solving the grid,
  below!

	K	L	M	N	P		19	20	21	22	30		F	G	H	I	J
A	f	\7	\7	(7b)	|||		==	==	|||	(7)	f		\7	\7	(7c)	\7	f
B	\2	\2	\2	\2	(2a)		==	==	|||	==	(2)		.	.	\2	c	.
C	\6	.	.	\5	a		.	.	|||	\1	\1a		.	.	a	.	.
D	(6b)	\4	==	\5	b		X1	\1	(1b)	b	b		e	e	b	.	.
E	d	.	.	|||	|||		.	.	|||	d\6	d		.	d	|||	.	.
F	==	(4a)	c	e	X4		(3b)	\3,c	\3,c	c	c
G	\6	.	.	\5	.		|||	(8)	|||	|||	|||
H	==	==	==	(5a)	b		b	X2	a	(1)	b
I	(6a)	==	==	\5	==		c	c	(3)	|||	\2
J	\6	.	.	\5	.		f	\7,f	|||	|||	(7a)
19	d	(4)	c	\3b	b
20	==	==	(3A)	==	b
21	(6)	|||	\3	|||	X3
22	X5	|||	\3	(5)	\1.b
30	==	==	c	==	(1a)

NOTE:

(1a) is not "C"  from [a]; 
leading to (2) and (2a).
K=\22=(1) has no solution ==> X5
(2) is not "I"  from [c]; 
leading to (3)
and c leads to (3a) & 
(3b) from [age].
(3b) is not "N"  from [e]; 
leading to (4) and (4a).
(5) & (5a) obvious leading to (5a).
(6) obvious leading to (6a) & (6b).
(1) is not "C" from [a] 
and (6) is not E22 from[d]
leading to ==> (7)=22=A.

I leave the rest of 

  the solution to you!

1. Subect "I" is held 3 days before "A"'s session
     which is held in a room #
     that is twice "N"'s Rep room #.
2. "C"'s Rep room # is higher than WED's Rep room #.
3. Subject "G" is held before Rep "L"'s,
     but after Rep "B"'s.
4. Room 250, which isn't Rep'd by "C",
     holds its session the day before Rep "D"'s.
     who holds his session the day before Rep "H"'s.
     who holds his session the day before Rep "P"'s.
5. The session in Room 150 is held before Rep "M"'s.
     who holds his session atleast 2 days before Rep "E"'s.
6. Rep "J" holds his session before THUR's
     which has a room # twice that of "J"'s Rep room #

   From 1:I | | | A (=2 x N)  [days]
   From 2:'WED' / C           [#]
   From 3:B / G / L           [days]
   From 4:250(=\C) | D | H | P[days]
   From 5:150 / M | / E       [days]
   From 6:J / 'THUR' (=2 x J) [days]
   
   NOTE:
   
   THUR's room # can't be "A" (=2xN)
   
    else it has 2 numbers 
   
    200 & 250 from [f, a]
   SHOWING ALL PERMUTATIONS BELOW:
   

   PERMUTATION 1 MON TUE WED THU 2x FRI 250 D H P I   xA A 150 M   E xE no solution for MON

   .

   PERMUTATION 2 MON TUE WED THU 2x FRI 250 D H P I   xA A 150 M=/C   E xE =================== C B D E A =================== B G L xL the solution

   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

   SOLUTION
   1ST	SUR	DAY	SUBJECT	ROOM #
   C	K	MON	J	150
   B	M	TUE	I	250
   D	N(xJ)	WED	G	100
   E	L	THU	H	300=2xJ
   A(xJ)	P	FRI	F	200

   
   SOMETIMES IT'S EASIER TO USE CIRCLES
   WHEN THE QUESTION IS VAGUE!
   PROBLEM 4:
   Mr. A, B, C, D & E are exotic fish owners
     while their wives own a cat;
     each of which killed a fish on a weekend.
   Using the 2 clues below solve for each killer and prey.

   1. Mrs "A"'s cat killed the fish of the husband
        of the owner of the cat which killed "E"'s fish,
        while Mr "A"'s fish was killed by Mrs "B"'s cat.
   2. Mr "D"'s fish was killed by the cat
        owned by the wife of the man
        whose fish was killed by Mrs "C"'s cat.

   
   
   NOTE:
     Y=\B since A=\D
     Y=\A since B=\C
     THUS Y=E and X=C etc.
     ARROWS SHOW WICH CAT KILLED WHICH FISH PROBLEM 5:
   5 racing sloop owners A, B, C, D & E
     all have two daughters with the same
     but different initials.
   Using the 4 clues below solve for each owner,
     sloop, daughters and the winners
     aboard sloop "C".

   1. Each race sloop owner named his sloop
        after the daughter whose sister
        has the same initials as his own daughter
        - the 2 owners and their 4 daughters
          have 5 different initials.
   2. "E"'s daughters are "B" & "C".
   3. "C"'s sloop the "A" has a daughter "D".
   4. "B"owns sloop "E".

   
   NOTE:
     From 1:
     A typical arrangement for owner O of sloop Z,
     with daughters x & y, and
     a second owner S, with daughters y & z,
     is shown below, FIG. 1
   
   From 2:
     E, his car Z, and his 2 daughters c & b
     give 2 possibilities, as shown below.
   
   From 3:
     C who owns sloop A and has a daughter d,
     give 2 possibilities, as shown below.
   
   From adding FIG. 2 to FIG. 4:
     one gets the impossible solution
     with owner and daughter
     having the same initial, below.
   
   From adding FIG. 3 to FIG. 4 and
     adding B owns sloop E:
     one gets the solution, below.
   
   

   SOLUTION
   OWNER	SLOOP	SIBS
   A	B	E, C
   B	E	D, A
   C	A	B, D
   D	C	A, E
   E	D	C, B

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