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In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 21:11
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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
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Re: In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 21:33
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



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Re: In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 22:12
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



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Re: In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 22:42
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
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29 Oct 2019, 23:36
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
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Re: In a certain aquarium, the number of red fish is three times the numbe
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30 Oct 2019, 00:33
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=3/2G1 double the ratio of Red fish: blue fish> R/B= 2*2 R:B=4 R=4B2 equating 1 and 2 4B=3/2G B/G=3/8 therefore E is the correct option
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Re: In a certain aquarium, the number of red fish is three times the numbe
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30 Oct 2019, 01:10
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



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Re: In a certain aquarium, the number of red fish is three times the numbe
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30 Oct 2019, 01:23
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 = \((\frac{2}{3})*red\)
\((\frac{blue}{green})= (\frac{1}{4})*red/ (\frac{2}{3})*red= \frac{3}{8}\)
The answer is E



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Re: In a certain aquarium, the number of red fish is three times the numbe
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30 Oct 2019, 11:37
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
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29 Oct 2019, 21:38
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



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Re: In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 21:39
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



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Re: In a certain aquarium, the number of red fish is three times the numbe
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29 Oct 2019, 23:12
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 Answer A.
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In a certain aquarium, the number of red fish is three times the numbe
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Updated on: 30 Oct 2019, 23:20
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, 02:34.
Last edited by Archit3110 on 30 Oct 2019, 23:20, edited 1 time in total.



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Re: In a certain aquarium, the number of red fish is three times the numbe
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30 Oct 2019, 21:45
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
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31 Oct 2019, 00:12
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
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31 Oct 2019, 00:17
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
B:G = 3:8 Answer E




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