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Louise is three times as old as Mary. Mary is twice as old a
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19 Aug 2018, 03:41
19 Aug 2018, 03:41
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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) \(\frac{L}{6}\)
(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 a
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21 Aug 2018, 11:36
21 Aug 2018, 11:36
1
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) \(\frac{L}{6}\)
(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
Re: Louise is three times as old as Mary. Mary is twice as old a
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13 Oct 2019, 15:12
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) \(\frac{L}{6}\)
(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
