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Math Expert V
Joined: 02 Sep 2009
Posts: 61243
In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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3 00:00

Difficulty:   65% (hard)

Question Stats: 54% (02:15) correct 46% (02:01) wrong based on 69 sessions

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Competition Mode Question

In a certain aquarium, the number of red fish is three times the number of the green fish, and the number of blue fish is half the number of the red fish. If the ratio of green fish to red fish were to be doubled, and the ratio of red fish to blue fish were to be doubled, then the ratio of blue fish to green fish in that aquarium would be

A. 3:2
B. 1:1
C. 1:2
D. 3:4
E. 3:8

_________________
Current Student G
Joined: 11 Feb 2013
Posts: 274
Location: United States (TX)
Schools: Neeley '21 (A$) GMAT 1: 490 Q44 V15 GMAT 2: 690 Q47 V38 GPA: 3.05 WE: Analyst (Commercial Banking) Re: In a certain aquarium, the number of red fish is three times the numbe [#permalink] ### Show Tags 1 E is my answen Given, GREEN:RED=1:3 RED:BLUE=2:1. By doubling the GREEN:RED ratio, it becomes 2:3 By doubling the RED:BLUE ratio, it becomes 4:1 Since RED is the same thing in both ratio, equalizing RED in both ratio (Multiplying GREEN:RED by 4 and RED:BLUE) makes three colors as follows GREEN:RED:BLUE=8:12:3. So, BLUE: GREEN=3:8 Current Student G Joined: 11 Feb 2013 Posts: 274 Location: United States (TX) Schools: Neeley '21 (A$)
GMAT 1: 490 Q44 V15 GMAT 2: 690 Q47 V38 GPA: 3.05
WE: Analyst (Commercial Banking)
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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HOW TO DOUBLE:
JUST MULTIPLY the RATIO by 2.

Given, GREEN: RED= (1/3)
Multiplying by 2, it becomes (1/3)*2=2/3.
So, the NEW RATIO is 2:3
Current Student G
Joined: 11 Feb 2013
Posts: 274
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Schools: Neeley '21 (A\$)
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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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NOT in this question:

HOW TO MAKE HALF:
JUST MULTIPLY the RATIO by (1/2).

Given, GREEN: RED= (1/3)
Multiplying by 1/2, it becomes (1/3)*(1/2)=1/6.
So, the NEW RATIO would be 1/6
Manager  S
Joined: 30 Nov 2017
Posts: 62
GMAT 1: 690 Q49 V35
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
Ans: E
Given:
R=3G
B=R/2

G(n)/R(n)=2/3
R(n)/B(n)=4/1

Therefore, B(n)/G(n)=3/8
Intern  B
Joined: 21 Sep 2016
Posts: 21
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
Assume following notations,
r = red fishes,
g = green fishes,
b = blue fishes,
G = modified nos. of green fishes,
R = modified nos. of red fishes,
B = modified nos. of blue fishes.

Now, as per the question,
r = 3g,
b = r/2
=> r = 3g = 2b
=> g/r = 1/3 = 2b/3r
=> r/b = 2/1

Now g/r ratio is doubled, i.e., now it is 2/3,
And r/b ratio is doubled, i.e., now it is 4/1.

=> G/R = 2/3
and,
=> R/B = 4/1.

=> B/R = 3/12 and R/G = 12/8
=> B : G = 3:8
Option (E).
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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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In a certain aquarium, the number of red fish is three times the number of the green fish, and the number of blue fish is half the number of the red fish. If the ratio of green fish to red fish were to be doubled, and the ratio of red fish to blue fish were to be doubled, then the ratio of blue fish to green fish in that aquarium would be

A. 3:2
B. 1:1
C. 1:2
D. 3:4
E. 3:8

G:R:B
1:3:1.5
2:6:3 [after doubling G:R]
4:12:6 [after doubling R:B]

Now B:G = 6:4 = 3:2

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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
Let the number of red fish, R = 6k

the number of red fish is three times the number of the green fish
—> Number of green fish, G = 2k

the number of blue fish is half the number of the red fish
—> Number of blue fish, B = 3k

Given, G : R = 1 : 3
If the ratio of green fish to red fish were to be doubled
—> G : R = 2 : 3
—> G = 2R/3

Given, R : B = 2 : 1
the ratio of red fish to blue fish were to be doubled
—> R : B = 4 : 1
—> B = R/4

Required ratio = B : G = R/4 : 2R/3
= 12*1/4 : 12*2/3
= 3 : 8

IMO Option E

Posted from my mobile device
Senior Manager  G
Joined: 23 Nov 2018
Posts: 253
GMAT 1: 650 Q49 V28 GPA: 4
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
In a certain aquarium, the number of red fish is three times the number of the green fish, and the number of blue fish is half the number of the red fish. If the ratio of green fish to red fish were to be doubled, and the ratio of red fish to blue fish were to be doubled, then the ratio of blue fish to green fish in that aquarium would be

Let number of red fish= R
number of green fish=G
number of blue fish= B

R=3G
R=2B

double the ratio of green fish: red fish = >

G:R=1/3*2

G:R=2/3

R=3/2G-----1

double the ratio of Red fish: blue fish--->
R/B= 2*2

R:B=4

R=4B------2

equating 1 and 2

4B=3/2G

B/G=3/8

therefore E is the correct option
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Manager  G
Joined: 31 Oct 2015
Posts: 95
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
As per Question $$\frac{R}{G}=\frac{3}{1}$$ & $$\frac{B}{R}=\frac{1}{2}$$

In other words $$\frac{G}{R}=\frac{1}{3}$$ & $$\frac{R}{B}=\frac{2}{1}$$

After doubling (i.e multiplying ratio by 2),

$$\frac{G}{R}=\frac{2}{3}$$ & $$\frac{R}{B}=\frac{4}{1}$$

Red is common in both ratios and the LCM of 4 & 3 is 12.

Therefore the above ratio can be written as $$\frac{G}{R}=\frac{8}{12}$$ & $$\frac{R}{B}=\frac{12}{3}$$

Combining the ratios , G:R:B = 8:12:3

Therefore B:G = 3:8
Director  P
Joined: 25 Jul 2018
Posts: 543
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
In a certain aquarium:
—> red= 3*green —> $$\frac{green}{red}= \frac{1}{3}$$

—> blue=$$(\frac{1}{2})*red$$ —> $$\frac{red}{blue}= \frac{2}{1}$$

If the ratio of green fish to red fish were to be doubled
—> $$\frac{green}{red}= \frac{2}{3}$$

If the ratio of red fish to blue fish were to be doubled
—> $$\frac{red}{blue}= \frac{4}{1}$$

What would be the ratio of blue fish to green fish in that aquarium???

blue =$$( \frac{1}{4})* red$$
green = $$(\frac{2}{3})*red$$

$$(\frac{blue}{green})= (\frac{1}{4})*red/ (\frac{2}{3})*red= \frac{3}{8}$$

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In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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given
g/r=2/3
r/b=2/1
g:r:b
2:3:
:4:1
8:12:3
B/G= 3/8

IMO E

In a certain aquarium, the number of red fish is three times the number of the green fish, and the number of blue fish is half the number of the red fish. If the ratio of green fish to red fish were to be doubled, and the ratio of red fish to blue fish were to be doubled, then the ratio of blue fish to green fish in that aquarium would be

A. 3:2
B. 1:1
C. 1:2
D. 3:4
E. 3:8

Originally posted by Archit3110 on 30 Oct 2019, 01:34.
Last edited by Archit3110 on 30 Oct 2019, 22:20, edited 1 time in total.
Manager  G
Joined: 17 Mar 2019
Posts: 133
Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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1
Let Red Fish be RF, Green Fish GF and Blue Fish BF

RF=3GF

GF/RF=1:3 and the ratio is doubled so it will be 2/3

BF=1/2RF and the ratio will be RF/BF=4:1

GF RF BF
2 3
4 1

GF*4 and BF*3

There fore BF/GF is 3: 8

IMO E
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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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Another plug the numbers way of doing this

Let Redfish =180

Thus Green = 60, Blue =90

R:G:B = 180:60:90

Iteration 1
G:R is doubled (which essentially means G is doubled)
R:G:B = 180:120:90

Iteration 2
R:B is doubled (which essentially means R is doubled)...but since In Iteration 1 even g was doubled wrt R..we need to double it too
R:G:B =360:240:90

Required ratio B:G =90 240 = 3:8

Hence E
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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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Bunuel wrote:

Competition Mode Question

In a certain aquarium, the number of red fish is three times the number of the green fish, and the number of blue fish is half the number of the red fish. If the ratio of green fish to red fish were to be doubled, and the ratio of red fish to blue fish were to be doubled, then the ratio of blue fish to green fish in that aquarium would be

A. 3:2
B. 1:1
C. 1:2
D. 3:4
E. 3:8

R = 3G

B = (1/2)R
R = 2B

R = 3G = 2B
R/6 = G/2 = B/3

Let R = 6
G = 2
B = 3
Total = 10

Now to double red to blue fish. New combination will be
R = 6*2 = 12
G = ?(as this ratio has also be changed, so we will calculate it in next phase)
B = 3

Now to double green to red fish. New combination will be
R = 6*2 = 12
G = 2*2(as red is doubled)*2(as ration of green to red is doubled) = 8
B = 3

So, Ratio B/G = 3/8 . Answer E
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Re: In a certain aquarium, the number of red fish is three times the numbe  [#permalink]

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R = 3G
B = (1/2)R -> R = 2B

R = 3G = 2B
R/6 = G/2 = B/3
R:G:B = 6:2:3

R:B is doubled maintaining same ratio of R:G.
R:G:B = 12:4:3

G:R is doubled maintaining the doubled R:B ratio
R:G:B = 12:8:3 Re: In a certain aquarium, the number of red fish is three times the numbe   [#permalink] 30 Oct 2019, 23:17
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