T H _
T _ T _ L
W _ _ G H T
_ F
T H _
D _ G ,
C _ T
_ N D
R _ B B _ T
_ S
2 7 K G .
T H _
D _ G ,
C _ T
_ N D
R _ B B _ T
W _ _ G H
1 7 K G ,
7 K G
_ N D
3 K G
R _ S P _ C T _ V _ L Y .
L _ T
D = D _ G ,
C = C _ T ,
R = R _ B B _ T .
W _
C _ N
C _ M _
W _ T H
T H _
F _ L L _ W _ N G
_ Q _ _ T _ _ N S :
C
+
R
=
1 0
D
+
R
=
2 0
D
+
C
=
2 4
F R _ M
1 S T
_ Q _ _ T _ _ N ,
W _
G _ T :
R
=
1 0
- C
S _ B S T _ T _ T _
T H _ T
_ N T _
2 N D
_ Q _ _ T _ _ N
_ N D
W _
G _ T :
D
+
( 1 0
-
C )
=
2 0
S _ L V _ N G
F _ R
D
G _ V _ S :
D
=
C
+
1 0
S _ B S T _ T _ T _
T H _ T
_ N T _
3 R D
_ Q _ _ T _ _ N
_ N D
W _
G _ T :
( C
+
1 0 )
+
C
=
2 4
S _ L V _ N G
F _ R
C
G _ V _ S
7 .
W _
C _ N
T H _ N
S _ B S T _ T _ T _
C = 7
_ N
T H _
_ T H _ R
_ Q _ _ T _ _ N S
T _
G _ T
T H _
W _ _ G H T
_ F
D
_ N D
R Clue
TAKE 6 BALLS AND SPLIT THEM INTO 2 GROUPS OF 3 BALLS EACH. USE THE SCALE TO WEIGH THE 2 GROUPS. IF ONE OF THE GROUPS IS HEAVIER THAN THE OTHER GROUP, PICK 2 BALLS FROM THE HEAVIER GROUP AND USE THE SCALE TO WEIGH THEM. IF THEY ARE EQUAL IN WEIGHT, THE OTHER BALL IN THAT GROUP IS THE HEAVY ONE. IF BOTH THE GROUPS ARE EQUAL IN WEIGHT, THEN PICK THE REMAINING 2 BALLS AND USE THE SCALE TO DETERMINE THE HEAVIER BALL SABRINA HAD $50 AND SAMANTHA HAD $30. THIS CAN BE SOLVED BY SETTING UP TWO SIMULTANEOUS EQUATIONS, BUT IT'S EASIER JUST TO WORK BACKWARDS. AT EACH STEP, THE PERSON WHO IS RECEIVING THE MONEY GETS THE AMOUNT THAT THEY WERE ALREADY HOLDING. IN OTHER WORDS, THEY DOUBLE THEIR MONEY. THEREFORE, JUST BEFORE THE LAST EXCHANGE, SAMANTHA MUST HAVE HAD $40 AND WAS GIVEN ANOTHER $40 TO GET TO HER TOTAL OF $80. JUST BEFORE THE SECOND EXCHANGE, SABRINA MUST HAVE HAD HALF OF HER $40. THEREFORE, SABRINA HAD $20 AND SAMANTHA HAD $60. AFTER THE FIRST EXCHANGE SAMANTHA DOUBLED HER MONEY, SO SHE MUST HAVE HAD $30 BEFORE THE EXCHANGE, LEAVING SABRINA WITH $50 AT THE START THE MINIMUM NUMBER OF WEIGHTS REQUIRED IS FIVE AND THESE SHOULD WEIGHT 1, 3, 9, 27 AND 81 POUNDS. THE MERCHANT HAS TO USE A BALANCE WEIGHING SCALE TO DO THE JOB. TO WEIGH 2 POUNDS, HE'LL HAVE TO PUT THE 3 POUND WEIGHT ON ONE PAN AND 1 POUND WEIGHT ON THE OTHER PAN. TO WEIGH 5 POUNDS, HE'LL HAVE TO PUT THE 9 POUND WEIGHT ON ONE PAN AND 1 AND 3 POUND WEIGHTS ON THE OTHER PAN THE TOTAL WEIGHT OF THE DOG, CAT AND RABBIT IS 27KG. THE DOG, CAT AND RABBIT WEIGH 17KG, 7KG AND 3KG RESPECTIVELY. LET D=DOG, C=CAT, R=RABBIT. WE CAN COME WITH THE FOLLOWING EQUATIONS: C + R = 10 D + R = 20 D + C = 24 FROM 1ST EQUATION, WE GET: R = 10 -C SUBSTITUTE THAT INTO 2ND EQUATION AND WE GET: D + (10 - C) = 20 SOLVING FOR D GIVES: D = C + 10 SUBSTITUTE THAT INTO 3RD EQUATION AND WE GET: (C + 10) + C = 24 SOLVING FOR C GIVES 7. WE CAN THEN SUBSTITUTE C=7 IN THE OTHER EQUATIONS TO GET THE WEIGHT OF D AND R