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Louise is three times as old as Mary. Mary is twice as old a

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Louise is three times as old as Mary. Mary is twice as old a [#permalink] New post 19 Aug 2018, 03:41
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Question Stats:

75% (01:24) correct 25% (01:25) wrong based on 8 sessions
Louise is three times as old as Mary. Mary is twice as old as Natalie. If Louise is L years old, what is the average (arithmetic mean) age of the three women, in terms of L?

(A) \(\frac{L}{3}\)
(B) \(\frac{L}{2}\)
(C) \(\frac{2L}{3}\)
(D) \(\frac{L}{4}\)
(E) \(\frac{L}{6}\)
[Reveal] Spoiler: OA

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Re: Louise is three times as old as Mary. Mary is twice as old a [#permalink] New post 21 Aug 2018, 11:36
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sandy wrote:
Louise is three times as old as Mary. Mary is twice as old as Natalie. If Louise is L years old, what is the average (arithmetic mean) age of the three women, in terms of L?

(A) \(\frac{L}{3}\)
(B) \(\frac{L}{2}\)
(C) \(\frac{2L}{3}\)
(D) \(\frac{L}{4}\)
(E) \(\frac{L}{6}\)


GIVEN: Louise is L years old

Louise is three times as old as Mary
This also means that Mary is 1/3 as old as Louise
Louise's age = L
So, Mary's age = (1/3)L = L/3

Mary is twice as old as Natalie.
This also means that Natalie is 1/2 as old as Mary
Mary's age = L/3
So, Natalie's age = (1/2)(L/3) = L/6

What is the average (arithmetic mean) age of the three women, in terms of L?
Average age = (sum of all 3 ages)/3

Sum of all 3 ages = L + L/3 + L/6
= 6L/6 + 2L/6 + L/6 [I rewrote each fraction with a common denominator of 6]
= 9L/6
= 3L/2

So, average age = (3L/2)/3
= L/2

Answer: B

Cheers,
Brent
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Re: Louise is three times as old as Mary. Mary is twice as old a   [#permalink] 21 Aug 2018, 11:36
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Louise is three times as old as Mary. Mary is twice as old a

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