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0 / 60 seg.
There are two brothers whose combined age is eleven years. One is ten years older than the other. What are their ages?
T
H
_
Y
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R
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1
0
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5
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N
D
0
.
5
Y
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_
R
S
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L
D
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A
C
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M
M
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N
W
R
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N
G
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N
S
W
_
R
_
S
1
0
_
N
D
1
.
T
H
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S
_
S
W
R
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N
G
B
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C
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S
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T
H
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D
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F
F
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R
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N
C
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N
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G
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W
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L
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B
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9
,
N
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T
1
0
Clue
THE PROBABILITY IS 1/2. THERE ARE INITIALLY 4 POSSIBILITIES: GIRL-GIRL, GIRL-BOY, BOY-GIRL AND BOY-BOY. SINCE THE OLDER CHILD IS A BOY, WE CAN RULE OUT THE GIRL-GIRL AND GIRL-BOY COMBINATIONS
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
JACK IS 28, JOHN IS 21. LET A AND B BE JACK'S AND JOHN'S AGE RESPECTIVELY. SO, A + B = 49 THE DIFFERENCE IN THEIR AGES (A - B), ALWAYS REMAIN THE SAME. "WHEN JACK WAS AS OLD AS JOHN IS NOW" MEANS JACK'S AGE WAS B AND JOHN'S AGE HAD TO BE: B - (A - B) = 2B -A JACK IS NOW TWICE THE AGE, SO: A = 2(2B - A) A = 4B - 2A 3A = 4B SUBSTITUTING (A + B = 49), WE CAN GET B = 21 AND A = 28
THEY ARE 10.5 AND 0.5 YEARS OLD. A COMMON WRONG ANSWER IS 10 AND 1. THIS IS WRONG BECAUSE THE DIFFERENCE IN AGE WOULD BE 9, NOT 10
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