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

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Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 19 Aug 2018, 09:41
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Question Stats:

70% (01:48) correct 30% (01:50) wrong based on 70 sessions

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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 19 Aug 2018, 10:57
Bunuel 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 three times as old as Mary
let the age of Mary be M,
Then L=3M Or, \(M=\frac{L}{3}\)
Given, Mary is twice as old as Natalie
Or, M=2N Or, \(N=\frac{M}{2}=\frac{L}{6}\)

Average age of L,M, and N in terms of L=\(\frac{M+N+L}{3}=\frac{1}{3}(\frac{L}{3}+\frac{L}{6}+L)\)=\(\frac{L}{2}\)

Ans. (B)
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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 20 Aug 2018, 20:42
Solved this by substituting numbers

Lets consider L = 60, then M = 20 , and N = 10

Average age of 3 women = (60+20+10)/3 = 90/3 = 30

Average in terms of L = 60/2 = 30

Ans: B
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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 20 Aug 2018, 20:47
I did basic algebra but would like to know more easy way to do it.

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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 20 Aug 2018, 23:40
\(L=3M\)
\(M=2N\)

now arithmetic mean is:
\(\frac{(L+M+N)}{3}\)
(L+L/3+M/2)/3
(L+L/3+L/(3*2))/3
(L+L/3+L/6)/3
(9L/6)/3
\(\frac{9L}{18}\)
\(\frac{L}{2}\)
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Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 22 Aug 2018, 19:54
Bunuel 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) \(\frac{L}{6}\)

Kritika2008poddar wrote:
I did basic algebra but would like to know more easy way to do it.

Kritika2008poddar , "plugging in" numbers often can be quicker and easier than straight algebra.* In this case, you could:

1) Write the variable definitions

Louise is three times as old as Mary. Mary is twice as old as Natalie.
\(L=3M\)
____\(M=2N\)

2) Assign an age to one person. In years of age, let

\(N=2\). Then
\(M=2N=4\), and
\(L=3M=12\)

3) Find the average age of the three girls
Average age: \(\frac{12+4+2}{3}=6\)

4) Use \(L=12\) and find the option that yields \(6\). You must check ALL options.

(With this method we must check all options because our numbers might work for two answers. In that case, choose a different number and check the two options. Not hard.)

(A) \(\frac{L}{3}\):\(\frac{12}{3}=4\) REJECT

(B) \(\frac{L}{2}\):\(\frac{12}{2}=6\) KEEP

(C) \(\frac{2L}{3}\):\(\frac{24}{3}=8\) REJECT

(D) \(\frac{L}{4}\):\(\frac{12}{4}=3\) REJECT

(E) \(\frac{L}{6}\): \(\frac{12}{6}=2\) REJECT

Answer B

Hope that helps.

[size=85]Most of the time, avoid picking 0 or 1. Pick easy values that fit the question. For material on this topic,
see Bunuel , listing many links to "Number Plugging"; and
ManhattanPrep, outlining how to choose smart values for variables; and
mikemcgarry , discussing whether to use algebra or pick numbers.
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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 20 Oct 2019, 06:59
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Bunuel 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

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Brent
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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.  [#permalink]

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New post 23 Oct 2019, 19:13
Bunuel 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) \(\frac{L}{6}\)


We can let M and N be the ages of Mary and Natalie, respectively. Therefore, we have:

L = 3M

and

M = 2N

Since M = L/3 and N = M/2 = (L/3)/2 = L/6, the average age of the three women, in terms of L, is:

(L + L/3 + L/6)/3

Multiplying the expression by 6/6, we have:

(6L + 2L + L)/18

9L/18

L/2

Answer: B
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Re: Louise is three times as old as Mary. Mary is twice as old as Natalie.   [#permalink] 23 Oct 2019, 19:13
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