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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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19 Aug 2018, 09:41
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25% (medium)

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83% (01:23) correct 17% (00:58) wrong based on 29 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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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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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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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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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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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

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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Louise is three times as old as Mary. Mary is twice as old as Natalie. &nbs [#permalink] 22 Aug 2018, 19:54
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