Saturday, April 28, 2007

MATH PROBL;EM OF THE DAY

The ages of three people are such that the age of one person is twice the age of the second person and
three times the age of the third person. If the sum of the ages of the three people is 33, then the age of
the youngest person is
(A) 3
(B) 6
(C) 9
(D) 11
(E) 18
at

7 comments:

  1. AnonymousApril 28, 2007 at 10:46 AM

    b

    ReplyDelete
  2. AnonymousApril 28, 2007 at 10:57 AM

    X+Y+Z=33
    X=2Y 3Z=2Y
    X=3Z
    2Y+Y+2Y/3=33
    =>6Y+3Y+2Y/3 =33
    =>11Y=99
    =>Y=9(COFFEE)
    THEREFORE X=18(THUMBS UP)
    AND Z=6(SMALL BOY...DRINK MAAZA)
    GUNDASWAMI FROM VELLORE

    ReplyDelete
  3. UnknownApril 28, 2007 at 10:56 PM

    B

    ReplyDelete
  4. NagsMay 2, 2007 at 6:16 AM

    B

    ReplyDelete
  5. AdminMay 3, 2007 at 12:24 PM

    Let a represent the age of the oldest person, b the age of the age of second person, and c the age of
    youngest person. The age of first person is twice the age of the second person and three times the age of the
    third person. This can be expressed as a = 2b and a = 3c. Solving these equations for b and c yields b = a/2
    and c = a/3. The sum of the ages of the three people is a + b + c = 33. Substituting for b and c in this
    equation, we get
    a + a/2 + a/3 = 33
    6a + 3a + 2a = 198 by multiplying both sides by 6
    11a = 198
    a = 198/11 = 18 by dividing both sides by 11
    Since c = a/3, we get
    c = a/3 = 18/3 = 6
    The answer is (B).

    ReplyDelete
  6. AnonymousSeptember 12, 2008 at 5:32 AM

    B

    ReplyDelete
  7. AnonymousSeptember 18, 2008 at 2:47 AM

    B IS THE CORRECT ANSWER

    ReplyDelete