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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?
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9
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Clue
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
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
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
I AM 19 YEARS OLD AND MY SISTER IS 9. LET X = MY SISTER'S AGE AND Y = MY AGE. Y = X + 10 Y + 1 = 2(X + 1) Y = 2X + 1 SUBSTITUTING THIS RESULT IN OUR FIRST EQUATION, WE HAVE: 2X + 1 = X + 10 X = 9 SO Y = 19 WHEN MY SISTER WAS 5, I WAS 3 TIMES OLDER THAN SHE WAS
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