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0 / 60 seg.
Three kinds of apples are mixed randomly in a box. How many apples must you take out to be sure of having at least two apples of one kind?
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Clue
SHE NEEDS TO GIVE YOU 5 APPLES. YOU BOTH NEED AT LEAST 5 APPLES TO BEGIN WITH, BUT APART FROM THAT IT DOESN'T MATTER EXACTLY HOW MANY YOU EACH HAVE. WHEN SHE GIVES YOU 5 YOU WILL HAVE 10 MORE THAN HER BECAUSE SHE WILL LOSE 5 AND YOU WILL GAIN 5, RESULTING IN A NET DIFFERENCE OF 10. FOR EXAMPLE, IF BOTH OF YOU HAVE 20 APPLES AND SHE GIVES YOU 5 OF HERS, SHE WILL BE LEFT WITH 15 AND YOU WILL NOW HAVE 25, PRECISELY 10 MORE THAN SHE HAS
FOUR. IF THE APPLES ARE A, B, AND C, IT'S POSSIBLE THAT YOU COULD TAKE OUT ONE A, ONE B AND ONE C WITH THREE APPLES, BUT THE FOURTH APPLE MUST BE ONE OF THE THREE KINDS
JIM AND WANDA HAVE 7 AND 5 APPLES RESPECTIVELY. TO SOLVE USING MATHS, LET J AND W BE THE NUMBER OF APPLES OF JIM AND WANDA RESPECTIVELY. IF JIM GIVES WANDA AN APPLE, THEY WILL BOTH HAVE THE SAME NUMBER OF APPLES: W + 1 = J - 1 IF WANDA GIVES JIM AN APPLE, JIM WILL HAVE TWICE AS MANY AS WANDA: 2(W - 1) = J + 1 SOLVING THE EQUATIONS WILL GIVE J=7 AND W=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
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