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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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A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of  [#permalink]

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New post 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


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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  [#permalink]

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New post 01 Nov 2015, 20:06
6
5
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  [#permalink]

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New post 23 May 2011, 08:11
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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 ,

[ 4K-8 ]/ [ K-2+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  [#permalink]

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New post 24 May 2011, 02:09
2
new ratio
(a-8)/(b-2+10) = 2/3
gives 3a-2b = 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  [#permalink]

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New post 25 May 2011, 02:28
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1
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  [#permalink]

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New post 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  [#permalink]

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New post 01 Nov 2015, 20:44
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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:
4x-8(removed)/x-2(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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 27 Mar 2018, 10:35
1
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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Re: A Jar contains a mixture of A & B in the ratio 4:1. When 10 L of   [#permalink] 31 Mar 2020, 14:10

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