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Difficulty:
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67% (03:23) correct
33%
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Three containers are filled with a mixture of A and B each. The volume of the mixture in the three containers is in the ratio 1 : 2 : 3. The ratio of A and B in the three containers are 3 : 2, 2 : 5 and 5 : 3 respectively. The content of the containers are mixed in a single container of sufficient capacity. What is the ratio of A and B in the final mixture?
(A) 853 : 827
(B) 53 : 48
(C) 53 : 27
(D) 53 : 25
(E) 53 : 23
_________________
(A) 853 : 827
(B) 53 : 48
(C) 53 : 27
(D) 53 : 25
(E) 53 : 23
_________________
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Bunuel
Let us concentrate the quantity of A in the mixture.
1) 3:2
The ratio of this is \(\frac{1}{1+2+3}\) or \(\frac{1}{6}\).
So the quantity of A = \(\frac{3}{5}*\frac{1}{6}=\frac{1}{10}\)
2) 2:5
Similarly the quantity of A= \(\frac{2}{7}*\frac{2}{6}=\frac{2}{21}\)
3) 5:3
Similarly the quantity of A = \(\frac{5}{8}*\frac{3}{6}=\frac{5}{16}\)
Quantity of A as a fraction of total = \(\frac{1}{10}+\frac{2}{21}+\frac{5}{16}=\frac{21*16+2*10*16+5*21*10}{10*16*21}=\frac{1706}{3360}=\frac{853}{1680}\)
A:(A+B)=853:1680
So A:B=853-(1680-853)=853:827
A
_________________
General Discussion
nityakaul02
Joined: 26 Aug 2019
Last visit: 26 Aug 2026
Posts: 30
Location: India
Concentration: Strategy, Leadership
Schools: Rotman '24
GMAT 1: 550 Q37 V34
GMAT 2: 360 Q21 V19
GMAT 3: 400 Q27 V19
GPA: 4
WE:Brand Management (Consumer Packaged Goods)
please elaborate in a simpler way
Bunuel
Let us use some values for the volume of mix.
The different ratios we are dealing with are 1:2:3, 3:2, 2:5 and 5:3.
Since we have to divide total mixture as per the ratios, let us take product of (1+2+3,3+2,2+5,5+3)=6*5*7*8=1680
So the total is 1680 l.
1:2:3=280:560:840
1) First container
Total 280 in ratio of A:B as 3:2=\(3*\frac{280}{3+2}\): \(2*\frac{280}{3+2}\)=168:112
2 ) Second container
Total 560 in ratio of A:B as 2:5=\(2*\frac{560}{5+2}\): \(5*\frac{560}{5+2}\)=160:400
3 ) Third container
Total 420l in ratio of A:B as 5:3=\(5*\frac{840}{3+5}\): \(3*\frac{840}{3+5}\)=525:315
So A = 168+160+525=853
B=1680-853=827
A
_________________
Quote:
So... given that the three containers had volumes in the ratio 1: 2: 3, their total volume should come up to (1+2+3) = 6x.
Total amount of A in the three containers = 3/5(x) + 2/7(2x) + 5/8(3x) = \(\frac{853}{280}\)x
Total amount of B in the three containers = 2/5(x) + 5/7(2x) + 3/8(3x) = \(\frac{827}{280}\)x
Ratio of A : B = 853 : 827
Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 04 Sep 2026
Posts: 6,129
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
Given: Three containers are filled with a mixture of A and B each. The volume of the mixture in the three containers is in the ratio 1 : 2 : 3. The ratio of A and B in the three containers are 3 : 2, 2 : 5 and 5 : 3 respectively. The content of the containers are mixed in a single container of sufficient capacity.
Asked: What is the ratio of A and B in the final mixture?
Let the volume of mixture in the three containers be x, 2x & 3x respectively.
Container 1:
Total = x
A = 3x/5
B = 2x/5
Container 2:
Total = 2x
A = 4x/7
B = 10x/7
Container 3:
Total = 3x
A = 15x/8
B = 9x/8
Final Single Container:
Total = x + 2x + 3x = 6x
A = 3x/5 + 4x/7 + 15x/8 = 853x/280
A/Total = 853x/280*6x = 853/1680
B/Total = 1 - A/Total = 827/1680
A/B = 853/827
A:B = 853:827
IMO A
Asked: What is the ratio of A and B in the final mixture?
Let the volume of mixture in the three containers be x, 2x & 3x respectively.
Container 1:
Total = x
A = 3x/5
B = 2x/5
Container 2:
Total = 2x
A = 4x/7
B = 10x/7
Container 3:
Total = 3x
A = 15x/8
B = 9x/8
Final Single Container:
Total = x + 2x + 3x = 6x
A = 3x/5 + 4x/7 + 15x/8 = 853x/280
A/Total = 853x/280*6x = 853/1680
B/Total = 1 - A/Total = 827/1680
A/B = 853/827
A:B = 853:827
IMO A
BongBideshini
Joined: 25 May 2021
Last visit: 04 Sep 2026
Posts: 390
Location: India
Concentration: Entrepreneurship, Marketing
Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27
GPA: 3.8
WE:Engineering (Government)
Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27
Posts: 390
heavy calculation. It took me around 4 mins.
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