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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
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
FIVE. THERE ARE ONLY FOUR COLORS, SO FIVE SOCKS WILL GUARANTEE THAT TWO WILL BE THE SAME COLOR
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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