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11 Jan 2022, 09:22
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D
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Difficulty:
55%
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Question Stats:
67% (02:35) correct
33%
(02:40)
wrong
based on 131
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The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is:
A) 100
B) 150
C) 180
D) 200
E) 240
A) 100
B) 150
C) 180
D) 200
E) 240
28 Jan 2022, 07:20
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sujoykrdatta
Percentage of indigo in bottles 1 and 2: 33% and 17% respectively.
A part of the solution (say y cc) from the first bottle is thrown away and replaced by an equal volume (y cc) of the solution from the second bottle
Thus, in the first bottle, there is a mix of 33% indigo (say x cc) with 17% indigo (y cc) => the resultant solution has 21% indigo.
(33x + 17y)/(x + y) = 21
=> 12x = 4y
=> x : y = 1 : 3
Since the total volume is 800cc, we have: y = [3/(1+3)] x 800 = 600 cc
Thus, 600 cc of solution from bottle 2 was moved to bottle 1.
Thus, volume remaining in bottle 1 = 800 - 600 = 200 cc
Watch the video solution here:
Answer D
03 Jul 2024, 09:14
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Given, Percentage of indigo in Bottles 1 = 33% indigo and Bottle 2 = 17% Indigo.
Thus, in the first bottle, there is a mix of 33% indigo (let x cc) with 17% indigo (let y cc) => the resultant solution has 21% indigo.
=> (33x + 17y)/(x + y) = 21
=> 12x = 4y
=> x : y = 1 : 3 ------eq.1
Total volume in each bottle is 800cc,
=> x+y=800 -----eq.2
=> From Eq.1 & Eq.2 we get, y = [3/(1+3)] x 800 = 600 cc
Thus, 600 cc of solution from bottle 2 was moved to bottle 1.
Thus, volume remaining in bottle 1 = 800 - 600 = 200 cc
Answer: D
All The Best for GMAT
Thus, in the first bottle, there is a mix of 33% indigo (let x cc) with 17% indigo (let y cc) => the resultant solution has 21% indigo.
=> (33x + 17y)/(x + y) = 21
=> 12x = 4y
=> x : y = 1 : 3 ------eq.1
Total volume in each bottle is 800cc,
=> x+y=800 -----eq.2
=> From Eq.1 & Eq.2 we get, y = [3/(1+3)] x 800 = 600 cc
Thus, 600 cc of solution from bottle 2 was moved to bottle 1.
Thus, volume remaining in bottle 1 = 800 - 600 = 200 cc
Answer: D
All The Best for GMAT
