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The ages of three people are such that the age of one person is twice

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The ages of three people are such that the age of one person is twice  [#permalink]

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New post 15 May 2017, 07:22
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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

(A) 3
(B) 6
(C) 9
(D) 11
(E) 18

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Re: The ages of three people are such that the age of one person is twice  [#permalink]

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New post 15 May 2017, 07:48
A= 3x
b=6x
c = 2x
a+b+c = 11x = 33 => x = 3
so least age = 6
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Re: The ages of three people are such that the age of one person is twice  [#permalink]

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New post 15 May 2017, 07:57
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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.

Answer:

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New post 15 May 2017, 13:02
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Let A, B & C be the ages of the three people.

Given,
A=2B
A=3C
A+B+C=33

C=?

A+B+C=33
A+A/2+A/3=33
11A/6=33
A=18

C=A/3=18/3=6

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Re: The ages of three people are such that the age of one person is twice  [#permalink]

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New post 20 May 2017, 02:19
Same method as others have posted - but just to avoid fractions, we could take the age of eldest as a multiple of 2 & 3 both.

Let age of first person = 6x,
then age of second person = 3x
and age of third person = 2x

Sum = 6x+3x+2x = 33
Solve to get x=3. Thus age of youngest = 2x = 2*3 = 6.
hence answer is B
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Re: The ages of three people are such that the age of one person is twice  [#permalink]

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New post 30 Jun 2018, 08:23
You can use a model to make it easier to visualize.

Since, age of person 1 is 3 times of person 3, we have 4 units and age of person 1's age is twice of person 2 which is the equivalent of 1.5 units.

5.5 units = 33 years
1 unit = 6 years (Person 3's age)

Answer : B
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Re: The ages of three people are such that the age of one person is twice &nbs [#permalink] 30 Jun 2018, 08:23
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