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Three children, Doug, Darla, and Dave, all collect marbles. The number
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28 Aug 2017, 23:37
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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
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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
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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
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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
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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
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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




Re: Three children, Doug, Darla, and Dave, all collect marbles. The number
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20 Feb 2019, 08:21






