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 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 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 ANDREW HAVE BEEN TEACHING FOR 21 YEARS. BEYONCE HAVE BEEN TEACHING FOR 15 YEARS. CURT HAVE BEEN TEACHING FOR 7 YEARS. LET ANDREW = A, BEYONCE = B AND CURT = C, THEN: A + B = 36 B + C = 22 A + C = 28 FROM EQUATION (2), WE GET: B = 22 - C FROM EQUATION (3), WE GET: A = 28 - C SUBSTITUTE A AND B INTO EQUATION (1): (28 - C) + (22 - C) = 36 SOLVING FOR C GIVES 7. SUBSTITUTE (C = 7) INTO EQUATION(2) AND (3)