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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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82% (01:29) correct 17% (01:38) wrong based on 34 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] New post 13 Oct 2019, 15:12
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) \(\frac{L}{6}\)


L = 3M ; M = 2N ; L = L all of are in terms of their given value.

Now L = 3M = 3* 2N(where M=2N) = 6N So, L=6N;

Therefore, L=6N ; M=2N ; L=N(lets assume L=N); and there sum of 6N+2N+N=9N
Now we need to divide by 3 as they want AVERAGE So 9N/3=3N
BUT we know that L=6N/3N = L/2
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Re: Louise is three times as old as Mary. Mary is twice as old a   [#permalink] 13 Oct 2019, 15:12
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