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Difficulty:
55%
(hard)
Question Stats:
71% (02:41) correct
29%
(03:08)
wrong
based on 356
sessions
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Three completely filled identical vessels, contain fruit juice, soda and water in the ratio 1 : 3 : 5, 6 : 2 : 3 and 7 : 4 : 1. If the contents of all the three vessels are poured into an empty container, find the percentage of water in that container.
(A) 50%
(B) 38%
(C) 28.125%
(D) 25%
(E) 20%
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(A) 50%
(B) 38%
(C) 28.125%
(D) 25%
(E) 20%
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Posts: 13
I think the choices are correct. You should add the water percentage of each other : 5+3+1 = 9 and the total mix is 32. 9/32 x 100 = 28.125 The answer is C
General Discussion
Posts: 4,849
Hi Bunuel, can you please check if the answer options are correct.
Let the volume of each container = V
When the contents of the 3 containers are put together, total Volume = 3V
Total Water = Water from 1 + Water from 2 + Water from 3
= \(\frac{5}{9}*V + \frac{3}{11}*V + \frac{1}{12}*V = \frac{361}{396}*V\)
Then \(\frac{Percentage}{100} * Total = Portion\)
%\( = \frac{\frac{361}{396}*V}{3V} * 100 = 30.39\)
Arun Kumar
Let the volume of each container = V
When the contents of the 3 containers are put together, total Volume = 3V
Total Water = Water from 1 + Water from 2 + Water from 3
= \(\frac{5}{9}*V + \frac{3}{11}*V + \frac{1}{12}*V = \frac{361}{396}*V\)
Then \(\frac{Percentage}{100} * Total = Portion\)
%\( = \frac{\frac{361}{396}*V}{3V} * 100 = 30.39\)
Arun Kumar
ScottTargetTestPrep
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Location: United States (CA)
Expert reply
Posts: 22,738
Bunuel
Solution:
We could let the capacity of each identical vessel be 396 liters. The number 396 is the least common multiple of 9, 11 and 12 where each of these numbers is the sum of the numbers in each of the three ratios, respectively.
Now, let the first vessel contain x liters of fruit juice, 3x liters of soda and 5x liters of water. We have:
x + 3x + 5x = 396
9x = 396
x = 44
So there are 5(44) = 220 liters of water in the first vessel.
Let the second vessel contain 6y liters of fruit juice, 2y liters of soda and 3y liters of water. We have:
6y + 2y + 3y = 396
11y = 396
y = 36
So there are 3(36) = 108 liters of water in the second vessel.
Let the third vessel contain 7z liters of fruit juice, 4z liters of soda and z liters of water. We have:
7z + 4z + 1z = 396
12z = 396
z = 33
So there are 33 liters of water in the third vessel.
When the contents of all the three vessels are poured into an empty container, the container now has a total of 396 x 3 = 1188 liters of content, of which 220 + 108 + 33 = 361 liters are water. Therefore, water constitutes 361/1188 ≈ 30.4% of the contents in the container.
Answer: 30.4%
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J1:S1:W1 = 1:3:5 = 9k
J2:S2:W2 = 6:2:3 = 11k
J3:S3:W3 = 7:4:1 = 12k
Total quantity = 9k+11k+12k= 32k
Total water quantity= 5+3+1 =9k
Percent of water in total = (9k/32k)*100 = 28.125%
J2:S2:W2 = 6:2:3 = 11k
J3:S3:W3 = 7:4:1 = 12k
Total quantity = 9k+11k+12k= 32k
Total water quantity= 5+3+1 =9k
Percent of water in total = (9k/32k)*100 = 28.125%
danbrown9445
Joined: 01 Feb 2021
Last visit: 11 Nov 2023
Posts: 46
GMAT 1: 600 Q36 V38
GMAT 2: 710 Q49 V36
Products:
GMAT 2: 710 Q49 V36
Posts: 46
Adambhau
The question says that the vessels are identical,so the capacity of each should be the same. Scotts answer seems correct.
Posts: 107
Three completely filled identical vessels, contain fruit juice, soda and water in the ratio 1 : 3 : 5, 6 : 2 : 3 and 7 : 4 : 1. If the contents of all the three vessels are poured into an empty container, find the percentage of water in that container.
(A) 50%
(B) 38%
(C) 28.125%
(D) 25%
(E) 20%
My approach to such question first glance at the word *identical* and then the different ratios, sum the parts in each ratio and take LCM
1 + 3 + 5 = 9 Water % in V1 = \(5/9\) * 396 = 220
6 + 2 + 3 = 11 Water % in V2 =\( 3/11\) * 396 = 108
7 + 4 + 1 = 12 Water % in V3 = \(1/12 \)* 396 = 33
LCM Of 9,11,12 = 396 Sum= 361
\(361/361*3\) * 100 = Approx 30%
C
(A) 50%
(B) 38%
(C) 28.125%
(D) 25%
(E) 20%
My approach to such question first glance at the word *identical* and then the different ratios, sum the parts in each ratio and take LCM
1 + 3 + 5 = 9 Water % in V1 = \(5/9\) * 396 = 220
6 + 2 + 3 = 11 Water % in V2 =\( 3/11\) * 396 = 108
7 + 4 + 1 = 12 Water % in V3 = \(1/12 \)* 396 = 33
LCM Of 9,11,12 = 396 Sum= 361
\(361/361*3\) * 100 = Approx 30%
C
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