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There are three vessels of equal capacity .Vessel A contains Milk and
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Updated on: 11 May 2015, 01:57
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There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio 4:3 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel C contains Milk and water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the ratio of milk and water ? A) 9:6 B) 3:2 C) 118:126 D) 193:122 E) 201:132
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Originally posted by gmatcracker24 on 10 Oct 2011, 03:09.
Last edited by Bunuel on 11 May 2015, 01:57, edited 1 time in total.
Renamed the topic and edited the question.



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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 04:36
gmatcracker24 wrote: There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio 4:3 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel B contains Milk and water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the ratio of milk and water ?
A)9:6 B)3:2 C)118:126 D)193:122 E)201:132
Try solving this question and share your approach .
I will share a efficient way to solve such kind of questions ,once i see some approaches. 4:3=>4x+3x=7x 2:1=>2y+1y=3y 3:2=>3z+2z=5z 7x=3y=5z Take LCM of 7,3,5=105 Or simply; x=3*5=15 y=7*5=35 z=3*7=21 So, Ratio of Milk:Water= (4x+2y+3z)/(3x+y+2z)=193/122 Ans: "D"
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 05:12
fluke wrote: gmatcracker24 wrote: There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio 4:3 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel B contains Milk and water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the ratio of milk and water ?
A)9:6 B)3:2 C)118:126 D)193:122 E)201:132
Try solving this question and share your approach .
I will share a efficient way to solve such kind of questions ,once i see some approaches. 4:3=>4x+3x=7x 2:1=>2y+1y=3y 3:2=>3z+2z=5z 7x=3y=5z Take LCM of 7,3,5=105 Or simply; x=3*5=15 y=7*5=35 z=3*7=21 So, Ratio of Milk:Water= (4x+2y+3z)/(3x+y+2z)=193/122 Ans: "D" Perfect Fluke...Bang on!! This was the approach i was taking abt +1
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 08:52
gmatcracker24 wrote: There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio 4:3 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel B contains Milk and water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the ratio of milk and water ?
A)9:6 B)3:2 C)118:126 D)193:122 E)201:132
Try solving this question and share your approach .
I will share a efficient way to solve such kind of questions ,once i see some approaches. Here is my approach, very similar to fluke's, just a different thought process of arriving at the same calculation. The three vessels are of equal capacity (which we assume to be full since it is a PS question. I would prefer clearer direction though). The milk:water ratio is given for each as 4:3, 2:1 and 3:2. The reason we cannot just add the milk parts (4+2+3) and water parts (3+1+2) together is that they represent different fractions of the whole volume i.e. in the first vessel, milk is 4 parts out of a total of 7 parts while in the second vessel, milk is 2 parts out of a total of 3 parts etc. If we make the total number of parts in each ratio equal, then we can just add the 'milk parts' together and all the 'water parts' together. Ratios: 4:3 (Total 7 parts) 2:1 (Total 3 parts) 3:2 (Total 5 parts) How can we make the total number of parts equal in the 3 cases? By taking the LCM. 4*15 : 3*15 = 60:45 (Total parts 105) 2*35 : 1*35 = 70:35 (Total parts 105) 3*21 : 2*21 = 63:42 (Total parts 105) When we mix the three solutions, the total number of parts of milk is 60+70+63 = 193 the total number of parts of water = 45+35+42 = 122 The required ratio = 193:122
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 10:42
Can someone tell what is wrong with my method, which is close to the OA: Say A=B=C=100 ml Therefore, Milk in A= 57 ml (57%) B= 67 ml (66.66%) C= 60 ml (60%) Total Milk= 57+67+60= 184 ml out of 300 ml and the rest is Water = 116 ml So, the ratio is 184:116



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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 11:01
BDSunDevil wrote: Can someone tell what is wrong with my method, which is close to the OA: Say A=B=C=100 ml Therefore, Milk in A= 57 ml (57%) B= 67 ml (66.66%) C= 60 ml (60%) Total Milk= 57+67+60= 184 ml out of 300 ml and the rest is Water = 116 ml So, the ratio is 184:116 Nothing wrong. Just that this is an approximation.
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 15:13
To fluke and karishma, why did you multiply 3*5 for x and so for x and z?? i just did not get it thanks for your explanation
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 Oct 2011, 16:54
ABC 4X:3X2Y:1Y3Z:2Z
final mixture of A,B and C will have M and W in the ratio ((4/7)+(2/3)+(3/5)) : (3/7)+(1/3)+(2/3)
=(4*15+2*35+3*21)/(7*3*5) : (3*15+1*35+2*21)/(7*3*5)
= 193: 122



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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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11 Oct 2011, 01:14
manalq8 wrote: To fluke and karishma, why did you multiply 3*5 for x and so for x and z?? i just did not get it thanks for your explanation The basic idea here is to take the LCM.That's the reason 3*5 is multiplied for x and so on.Hope it's clear now.
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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11 Oct 2011, 03:41
I am guessing ones you get the LCM, you divide the LCM by the number of parts the ship carrys then multiply the resultant number to the single part. For Ship 1, 4:3...7 parts 105/7=15 4*15=60 3*15=45 therefore 60:45
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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11 Oct 2011, 09:06
manalq8 wrote: To fluke and karishma, why did you multiply 3*5 for x and so for x and z?? i just did not get it thanks for your explanation Vessel 1 > 4:3 (Total 7 parts) Vessel 2 > 2:1 (Total 3 parts) Vessel 3 > 3:2 (Total 5 parts) Right now, we can't just add the parts of milk together since the parts are not the same. 1 part of vessel 1 is smaller than 1 part of vessel 2 since the same volume in vessel 1 is divided into 7 parts but in only 3 parts in vessel 2. If we can make one part of each vessel equal, we can just add the parts. What is the LCM of 7, 3 and 5? It will be 7*3*5 (=105) since all the numbers are prime numbers. When we multiply the entire ratio by the same number, the ratio remains unchanged. Vessel 1 > We need a total of 105 parts instead of 7 so we multiply by 15. The ratio becomes 4*15 : 3*15 = 60:45 Vessel 2 > We need a total of 105 parts instead of 3 so we multiply by 35. The ratio becomes 2*35 : 1*35 = 70:35 Vessel 3 > We need a total of 105 parts instead of 5 so we multiply by 21. The ratio becomes 3*21 : 2*21 = 63:42 Now we can just add the parts.
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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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10 May 2015, 00:39
I solved this problem using smart numbers, it is not a lot faster but easier to imagine



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There are three vessels of equal capacity .Vessel A contains Milk and
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28 May 2016, 16:33
m=portion of milk in solution m=(4/7+2/3+3/5)/3➡193/315 315193=122 122/315=portion of water in solution ratio of milk to water=193/122



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Re: There are three vessels of equal capacity .Vessel A contains Milk and
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