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There are 3 Box A, B, C with full of marbles contain White and Red

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Intern
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Joined: 05 Dec 2017
Posts: 21
GPA: 3.35
There are 3 Box A, B, C with full of marbles contain White and Red  [#permalink]

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New post 05 Apr 2018, 13:03
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

53% (01:56) correct 47% (03:17) wrong based on 26 sessions

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There are 3 Box A, B, C with full of marbles contain White and Red marbles only. Box A, which has Red to White marbles in ratio of 1:2. Box B, with twice as much as capacity of A has Red to white marbles in 2:1 ratio. Box C, with thrice as much as capacity of box B has Red to white marbles in ratio of 1:2. If all the marbles from these 3 Box's are kept in another box D, What will be the ratio of White to Red marbles in box D?
A. \(\frac{3}{5}\)

B. \(\frac{2}{5}\)

C. \(\frac{5}{2}\)

D. \(\frac{11}{16}\)

E. \(\frac{16}{11}\)
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Re: There are 3 Box A, B, C with full of marbles contain White and Red  [#permalink]

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New post 05 Apr 2018, 16:40
Let capacity of A = A
so, capacity of B = 2A
capacity of C = 6A

# of red in A: \(\frac{1}{3}A\), # of white in A: \(\frac{2}{3}A\)

# of red in B: \(\frac{2}{3}2A\), # of white in B: \(\frac{1}{3}2A\)

# of red in C: \(\frac{1}{3}6A\), # of white in C: \(\frac{2}{3}6A\)

# of red in D: \(\frac{1}{3}A + \frac{2}{3}2A+\frac{1}{3}6A=\frac{11}{3}A\), # of white in D: \(\frac{2}{3}A+\frac{1}{3}2A+\frac{2}{3}6A=\frac{16}{3}A\)

\(\frac{white in D}{red in D} = \frac{16}{3}A/\frac{11}{3}A=\frac{16}{11}\)

Answer: E
GMAT Club Bot
Re: There are 3 Box A, B, C with full of marbles contain White and Red &nbs [#permalink] 05 Apr 2018, 16:40
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There are 3 Box A, B, C with full of marbles contain White and Red

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