Rahul takes a solution which contains 100 percent pure water and repla

Question

Rahul takes a solution which contains 100 percent pure water and replaces 40 percent of it with baking soda. He repeats this process two more times and stops. What is the ratio of soda to water in the final solution?

A. 41:9
B. 98:27
C. 18:7
D. 16:9
E. 14:9

A. 41:9
B. 98:27
C. 18:7
D. 16:9
E. 14:9

abhinavchitre13 · Answer

It should be 98/27!! Answer is B

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divyadna · Answer

Amount of water left in final solution/Amount of water in initial solution= {1-40/100}^3
Amount of water in final solution=21.6 
Amount of soda= 100-21.6=78.4
78.4/21.6= 98/27
Option B

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Can,Will · Answer

KarishmaB  
can you solve this with V2=V1(C1/C2)^n
I am having hard time implementing the formula
values for V1,V2,C1,C2

SaquibHGMATWhiz · Answer

Solution:

Water is 100 percent in the solution and 40% is taken out 3 times 
 This means final volume of water  =100(1-\frac{40}{100})^3 
 ⇒100(\frac{60}{100})^3 
 ⇒100	imes \frac{6	imes 6	imes 6}{1000} 
 ⇒21.6  
 Amount of soda  =100-21.6=78.4  
 Ratio  =\frac{78.4}{21.6}=\frac{98}{27}

Hence the right answer is    Option B

KarishmaB · Answer

C1 - Initial Concentration (after removing solution but before addition of soda ) = 100% water
V1 - Initial Volume (after removing solution but before addition of soda) = 60 ml
C2 - Final Concentration
V2 - Final Volume (after addition) - 100 ml

C2 = C1 * (V1/V2)^3

C2 = 100 *(60/100)^3 = 21.6% water = 216/1000 water
Ratio of water: soda = 216:784 = 27:98
which means Ratio of soda:water = 98:27

Answer (B)

ScottTargetTestPrep · Answer

After each application, the amount of water in the solution will reduce by 40 percent, which means that after three applications, the amount of water in the solution will be (1 - 0.4)^3 = 0.6^3 = 0.216 = 21.6% of the initial amount. Thus, the remaining 100 - 21.6 = 78.4% will be baking soda. The ratio of soda to water in the final solution will be 78.4/21.6 = 784/216 = 392/108 = 196/54 = 98/27.

Answer: B

shaurya_gmat · Answer

KarishmaB  ,  SaquibHGMATWhiz

Can either of you help me understand how the following equation is valid -
 100(1−40/100)^3

I got the answer, but my approach was-

Initial 100 units of water
after first switch -
60 water, 40 BS
After second switch -
60 - 3*8  
36 - 2*8 + 40

I repeated this one more time to get the answer.

How can we directly reduce the amount of water using the equation above ? As per this the only content that is being switched/removed is water.

KarishmaB · Answer

Replacement is discussed here:  https://anaprep.com/arithmetic-replacement-in-mixtures/

If doesn't matter what initial concentration you begin with 100%, 80% or 30%; it is all the same. The formula deals with the addition aspect of the replacement.

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Rahul takes a solution which contains 100 percent pure water and repla

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