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A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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23 May 2011, 07:27
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A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially. A) 12 B) 15 C) 16 D) 20 E) 25 *************************** IIM CAT Level Question ***************************
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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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01 Nov 2015, 20:06
hussi9 wrote: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially.
A) 12 B) 15 C) 16 D) 20 E) 25
*************************** IIM CAT Level Question *************************** Let the volume of the initial total mixture was V. When 10L of mixture is removed the volume becomes (V  10). The concentration stays 4:1. Next step, when you add 10 litres of B, the amount of A does not change. So Amount of A in original mix = Amount of A in final mix C1 * V1 = C2 * V2 (4/5)*(V  10) = (2/5)*V 1  10/V = 1/2 V = 20 So volume of A in initial mixture was (4/5) * 20 = 16 litres Answer (C) Suggest you to check out this post on replacement: http://www.veritasprep.com/blog/2012/01 ... mixtures/
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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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23 May 2011, 08:11
10 litres of mixture that is replaced will contain 8 litres of A and 2 litres of B (as A:B = 4:1)
Let the initial volume of the mixture be 4K + 1K = 5K
So by condition ,
[ 4K8 ]/ [ K2+10 ] = 2/3
Solve for K which is K = 4
So initial volume of liquid A = 4K = 16 litres
Hope this clarifies




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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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24 May 2011, 02:09
new ratio (a8)/(b2+10) = 2/3 gives 3a2b = 40. a/b old ratio = 4:1 3*(4x) 2 *(x) = 40 gives x=4 thus a = 4*4 = 16. Yes please post more questions on Mixtures,Time work and distance,geometry and probability.



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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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25 May 2011, 02:28
I give simple solution Ratios of A:B becomes from 4:1 >3:2 From the mixture some amount is removed and replaced with solution B. This means A is not replaced. Volume of A become from 4>2 i.e. 50 % You cannot only remove 50 % of solution but 50 % of entire solution is removed. => this 50% corespondents to 10 L that is removed. Hence total volume of mixture is 20 L Volume of A intitailly will be 20 (4/5) = 16 L
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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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01 Nov 2015, 16:10
let x=liters in jar .2x.2(10)+10=.6x x=20 liters (.8)(20)=16 liters of liquid A in jar initially



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A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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01 Nov 2015, 20:44
Let s1 be the volume of the initial mixture. 2/3= (4/5)*s1  (4/5) *10) / (1/5)*s1  (1/5) *10 + 10 s1=20L Volume of A in s1=(4/5)*20 = 16L
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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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17 Oct 2016, 21:11
HEY I HAVE USED A DIFFERENT APPROACH BUT NOT ABLE TO FIGURE OUT WHAT I HAVE DONE WRONG
INITIALLY A:B IS 4:1 i.e A=80% and B=20% 10L of the solution was replaced by B hence the amount of A remains constant Therefore, by the formula a%of b=b%of a 80%of X=20% of(X+10) by this I get X=10 A=80%of X OR A=80/100(10)=8Litres
Kindly tell me where I am wrong



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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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17 Oct 2016, 21:29
hussi9 wrote: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially.
A) 12 B) 15 C) 16 D) 20 E) 25
*************************** IIM CAT Level Question *************************** Let A= 4x and B = x now we removed 10L of mixture , so some part of A and some part of b were removed . and we know the initial ratio is 4:1 , so in 10L we will have =4x+x =10 x=2 8L of A and 2L of B. According to question: 4x8(removed)/x2(removed)+10(add 10 L)=2:3 x=4. initial mixture:4x=4*4 =16.



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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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17 Oct 2016, 21:35
suramya26 wrote: HEY I HAVE USED A DIFFERENT APPROACH BUT NOT ABLE TO FIGURE OUT WHAT I HAVE DONE WRONG
INITIALLY A:B IS 4:1 i.e A=80% and B=20% 10L of the solution was replaced by B hence the amount of A remains constant Therefore, by the formula a%of b=b%of a 80%of X=20% of(X+10) by this I get X=10 A=80%of X OR A=80/100(10)=8Litres
Kindly tell me where I am wrong Hi suramya26, You considered only addition of 10 L in the mixture but question says that we are replacing 10 L of mixture , first remove 10 L and then add.



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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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11 Nov 2016, 15:03
hussi9 wrote: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially.
A) 12 B) 15 C) 16 D) 20 E) 25
*************************** IIM CAT Level Question *************************** If we have a mixture in ratio 4:1, it must be true that we should look for a multiple of 4 as an answer. B and E are right away out. since we need more than 10 liters, we could have had: 8 liters of A and 2 liters of B  since we don't have 8, let's go further... 12 liters of A and 3 liters of B > removing 10 liters > means remove 8 of A and 2 of B > new result is 4:11  nope. 16 liters of A and 4 liters of B > removing 10 liters > means remove 8 of A and 2 of B > we have 8 liters of A and 12 of B. or 8:12 or 4:6 or 2:3  satisfies the condition. answer must be C.



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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of
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27 Mar 2018, 10:35
hussi9 wrote: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially.
A) 12 B) 15 C) 16 D) 20 E) 25 We can let the amount of liquid B, in liters, originally in the mixture = m; thus the amount of liquid A, in liters, originally in the mixture = 4m. We can let the amount of liquid B in the mixture being replaced = x, thus the amount of liquid A in the mixture being replaced = 4x. So we have 4x + x = 10 5x = 10 x = 2 Thus of the 10 L of the mixture that is replaced with liquid B, 8 L is liquid A and 2 L is liquid B in the original mixture. We see that the amount of liquid A has a net loss of 8 L, whereas the amount of liquid B has a net gain of 10  2 = 8 L. We can create the following equation to reflect this change and the new ratio: (4m  8)/(m + 8) = 2/3 3(4m  8) = 2(m + 8) 12m  24 = 2m + 16 10m = 40 m = 4 Thus the amount of liquid A originally in the mixture is 4(4) = 16 L. Answer: C
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