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0 / 60 seg.
There are 12 kids in a classroom. 6 kids are wearing socks and 4 are wearing shoes. 3 kids are wearing both. How many are bare feet?
5
K
_
D
S
_
R
_
B
_
R
_
F
_
_
T
.
A
S
W
_
K
N
_
W
,
3
K
_
D
S
_
R
_
W
_
_
R
_
N
G
B
_
T
H
.
S
_
,
_
N
L
Y
3
K
_
D
S
_
R
_
W
_
_
R
_
N
G
_
N
L
Y
S
_
C
K
S
(
6
-
3
=
3
)
,
_
N
D
1
_
S
W
_
_
R
_
N
G
_
N
L
Y
S
H
_
_
S
(
4
-
3
=
1
)
.
S
_
,
_
N
T
_
T
_
L
,
3
+
3
+
1
=
7
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N
_
W
,
1
2
K
_
D
S
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R
_
T
H
_
R
_
,
S
_
,
1
2
-
7
=
5
Clue
204 SQUARES. THE CHESSBOARD IS MADE UP OF 8×8 SQUARES. WE HAVE TO DETERMINE HOW MANY SQUARES OF ANY DIMENSION FROM 1×1 TO 8×8 THERE ARE ON A CHESS BOARD. THE KEY IS TO DETERMINE HOW MANY POSSIBLE POSITIONS THERE ARE FOR EACH TYPE OF SQUARE. A 2×2 SQUARE, FOR EXAMPLE, CAN BE LOCATED IN 7 LOCATIONS HORIZONTALLY AND 7 LOCATIONS VERTICALLY. EG. IN 49 DIFFERENT POSITIONS. SIZEHORIZONTAL POSITIONSVERTICAL POSITIONSSQUARES 1×18864 2×27749 3×36636 4×45525 5×54416 6×6339 7×7224 8×8111 TOTAL204 IN TOTAL THERE ARE 204 POSITIONS. THIS IS THE SUM OF THE NUMBER OF POSSIBLE POSITIONS FOR ALL THE DIFFERENT SIZED SQUARES
THE PERSON WAS BORN IN 2005 B.C. (BEFORE CHRIST). THEREFORE, HE WAS 5 YEARS OLD IN 2000 B.C, 10 IN 1995 B.C, AND 15 IN 1990 B.C
THERE ARE 9 SHEEP AND 18 CHICKENS. LET S BE THE NUMBER OF SHEEP AND C BE THE NUMBER OF CHICKENS. SO: 2S = C 5S + 3C = 99 WE CAN REPHRASE THE FIRST EQUATION, SO: 6S - 3C = 0 AND THEN WE CAN ADD THIS TO THE SECOND EQUATION, WHICH YIELDS: 11S = 99 BY SOLVING FOR S, WE FIND THAT S EQUALS 9. BY SUBSTITUTING BACK IN ONE OF THE ORIGINAL EQUATIONS, WE FIND THAT C EQUALS 18. SO THERE ARE NINE SHEEP AND EIGHTEEN CHICKENS
5 KIDS ARE BARE FEET. AS WE KNOW, 3 KIDS ARE WEARING BOTH. SO, ONLY 3 KIDS ARE WEARING ONLY SOCKS (6 - 3 = 3), AND 1 IS WEARING ONLY SHOES (4 - 3 = 1). SO, IN TOTAL, 3 + 3 + 1 = 7. NOW, 12 KIDS ARE THERE, SO, 12 - 7 = 5
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