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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
.
N
_
W
,
1
2
K
_
D
S
_
R
_
T
H
_
R
_
,
S
_
,
1
2
-
7
=
5
Clue
TERRY IS 5 YEARS OLD. LET'S TRANSLATE WORDS TO MATH. "ALICE WAS FIVE YEARS OLDER THAN TERRY IS NOW" TRANSLATES TO: A = 5 + T, WHERE A IS THE AGE OF ALICE AND T IS THE AGE OF TERRY. NOW TRANSLATE AGAIN. "TERRY IS HALF AS OLD AS ALICE WAS" BECOMES : T = (1⁄2)A. SOLVING THE TWO EQUATIONS WILL GIVE T = 5
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
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
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
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