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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # 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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Difficulty:   35% (medium)

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

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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) $$\frac{L}{6}$$

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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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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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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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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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$$L=3M$$
$$M=2N$$

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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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

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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Top Contributor
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

Cheers,
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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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

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