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The ages of three people are such that the age of one person is twice
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15 May 2017, 06:22

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78% (02:07) correct 22% (01:56) wrong based on 132 sessions

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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

Re: The ages of three people are such that the age of one person is twice
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15 May 2017, 06:57

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rohan2345 wrote:

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

There are a few ways to solve this question. Let x = age of first person mentioned in the question

The age of one person is twice the age of the second person and three times the age of the third person We're saying that x = the age of the first person. Since this person is TWICE as old as the second person, we can say that the age of the second person is HALF the age of the first person. So, x/2 = age of second person. Likewise, x/3 = age of third person

The sum of the ages of the three people is 33 So, x + x/2 + x/3 = 33 To eliminate the fractions, multiply both sides by 6 to get: 6x + 3x + 2x = (33)(6) [in a second, you'll see why I didn't bother to evaluate 33 times 6] Simplify: 11x = (33)(6) Divide both sides by 11 to get: x = (33)(6)/11 = (3)(6) = 18

Be careful. The question asks for the age of the YOUNGEST person. The ages were assigned as x, x/2 and x/3

So, if x = 18, then the x/2 = 18/2 = 9, and x/3 = 18/3 = 6 So, the three ages are 18, 9 and 6 So, the youngest person is 6 years old.