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

Question

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}

PKN · Answer

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)

akadiyan · Answer

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

Kritika2008poddar · Answer

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

Posted from my mobile device

hasnain3047 · Answer

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}

generis · Answer

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.

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.

BrentGMATPrepNow · Answer

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
= 6 L /6 + 2 L /6 +  L /6    
= 9L/6
= 3L/2

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

Answer: B

Cheers,
Brent

ScottTargetTestPrep · Answer

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

bumpbot · Answer

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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.

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