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Three children, Doug, Darla, and Dave, all collect marbles. The number

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Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 28 Aug 2017, 23:37
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A
B
C
D
E

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Question Stats:

73% (02:13) correct 27% (02:20) wrong based on 123 sessions

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Three children, Doug, Darla, and Dave, all collect marbles. The number of Doug’s marbles is 2/5 the number of Darla’s marbles. However, Doug has 4 times as many marbles as Dave does. The number of Darla’s marbles is how many times the average of Doug and Dave’s marbles?

A. 1/4
B. 1/2
C. 3/2
D. 2
E. 4

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Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 29 Aug 2017, 01:21
If the number of Doug's marbles is \(\frac{2}{5}\)(40%) of the number of Darla's marbles,
Assume Darla has 10 marbles, making the number of marbles Doug has 4.
It has also been given that the Doug has 4 times as many marbles as Dave.
Therefore, Dave will have 1 marble.

The average of Doug and Dave's marbles is \(\frac{(1+4)}{2} = \frac{5}{2}\)

Since we have been asked how many times is the number of Darla's marbles to the average,

it must be \(\frac{10}{{\frac{5}{2}}\) = \(\frac{10*2}{5}\) = 4(Option E)
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Re: Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 29 Aug 2017, 01:35
Bunuel wrote:
Three children, Doug, Darla, and Dave, all collect marbles. The number of Doug’s marbles is 2/5 the number of Darla’s marbles. However, Doug has 4 times as many marbles as Dave does. The number of Darla’s marbles is how many times the average of Doug and Dave’s marbles?

A. 1/4
B. 1/2
C. 3/2
D. 2
E. 4


let's assume Darla has 20 marbles. So, Dough has =(2/5)*20= 8 marbles and Dave has 2 Mabels
the average of Dough and Dave marbles is 5. So, Darla has 4 times the average of Doug and Dave’s marbles.
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Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 29 Aug 2017, 19:19
Bunuel wrote:
Three children, Doug, Darla, and Dave, all collect marbles. The number of Doug’s marbles is 2/5 the number of Darla’s marbles. However, Doug has 4 times as many marbles as Dave does. The number of Darla’s marbles is how many times the average of Doug and Dave’s marbles?

A. 1/4
B. 1/2
C. 3/2
D. 2
E. 4

I picked numbers, my math went awry, and I switched to algebra.

Doug is the common element.

Darla = A
Doug = B
Dave = C

B's marbles = \(\frac{2}{5}\)A, so A = \(\frac{5}{2}\)B

B has 4 times as many marbles as C.

B = 4C, and

C = \(\frac{B}{4}\)

Average of B and C:

\((\frac{4}{4}\)B + \(\frac{B}{4})\) *\(\frac{1}{2}\) = \(\frac{5}{8}B\)

The number of marbles that A has is how many times the average of B and C’s marbles?

\(\frac{\frac{5}{2}B}{\frac{5}{8}B}\) =

\(\frac{5}{2}\) * \(\frac{8}{5}\) = 4

Answer E
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Re: Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 01 Sep 2017, 12:11
Bunuel wrote:
Three children, Doug, Darla, and Dave, all collect marbles. The number of Doug’s marbles is 2/5 the number of Darla’s marbles. However, Doug has 4 times as many marbles as Dave does. The number of Darla’s marbles is how many times the average of Doug and Dave’s marbles?

A. 1/4
B. 1/2
C. 3/2
D. 2
E. 4


We can let Doug’s marbles = A, Darla’s marbles = B, and Dave’s marbles = C. Thus:

A = (2/5)B

(5/2)A = B

B = 5A/2

and

A = 4C

A/4 = C

Let’s find the average number of Doug’s and Dave’s marbles: (A + A/4)/2 = (5A/4)/2 = 5A/8. Thus, the number of Darla’s marbles is (5A/2)/(5A/8) = 5A/2 x 8/5A = 4 times the average number of Doug’s and Dave’s marbles.

Answer: E
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Re: Three children, Doug, Darla, and Dave, all collect marbles. The number  [#permalink]

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New post 20 Feb 2019, 08:21
Since the names starts with D, i am using their last letter Doug= G, Darla= A Dave= E

Using Propotions
5G=2A -> G:A =2:5 x2 =4:10
G=4E -> G:E = 4:1
Make G = 4 to combine the two equations

G:A:E = 4:10:1

Average of G and E = (G+E)/2 = 5/2

how many times is G=10 will be equal to 5/2=2.5
-> 10/2.5= 4 = E
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Re: Three children, Doug, Darla, and Dave, all collect marbles. The number   [#permalink] 20 Feb 2019, 08:21
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