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

Question

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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KarishmaB · Accepted Answer

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/01 ... -mixtures/

pratikdas007 · Answer

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 ,

/   = 2/3

Solve for K which is K = 4

So initial volume of liquid A = 4K = 16 litres

Hope this clarifies

amit2k9 · Answer

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.

hussi9 · Answer

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

gracie · Answer

let x=liters in jar
.2x-.2(10)+10=.6x
x=20 liters
(.8)(20)=16 liters of liquid A in jar initially

SVaidyaraman · Answer

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

suramya26 · Answer

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

sobby · Answer

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.

sobby · Answer

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.

mvictor · Answer

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.

ScottTargetTestPrep · Answer

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

kanwar08 · Answer

KarishmaB

Can we solve this via the scale method?

How I approach Scale:
1. Find Individual Elements beings mixed and pick only 1 component (eg. A or B in this Example)
2. Find Earlier Concentration, Mixed Concentraion and Average

Initial Conc of B: 1/5 i.e. 20%. Mixed with 100% B solution. The weighted avg being 60% (After mixture.

So,

20------60-----100

The ratio of both mixtures will be 1:1 i.e. Earlier solution will be 10L

Where am I going wrong here? Can't get the answer as 20 this way

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=84638 Harsha GMAT-Club-Forum-sacv1o5c.png

KarishmaB · Answer

You are correct in your solution. The error you made - answer what is asked, not what you obtained.

Question:  How many liters of liquid A was present in mixture initially?

Yes, 10 liters of initial mixture was mixed with 10 liters of B but don't forget that 10 liters of initial mixture was first removed and then B was added. This means that the volume of initial mixture to begin with was 20 liters, not 10 liters.

kanwar08 · Answer

Thanks for responding to my query.

I understand that we needed to find the concentration of the solution(earlier) and then later get the quantity of A out of it.

When using scale, when I arrived at the conclusion that their weights are in ratio 1:1, does that mean 1/2 of the total will be 100% solution of B in the mixture and 1/2 of total be the earlier replaced sol'n.

So, the answer I got (10L) represents just how much of earlier solution was replaced with the 100% B soln. I.e 10 L which represents 50% of the current mixture, which is 20 L.

So, the earlier total must have been 20L as there is only replacement (no removal)

Is my understanding correct?

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

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