From a 100 litre container having milk and water in the ratio 8 : 7, 1

Question

From a 100 litre container having milk and water in the ratio 8 : 7, 10 litres of solution is removed and replaced with 20 litres of milk. Again 20 litres of solution is removed and replaced with 30 litres of milk. Find the amount of milk in the solution now.

A. 112.5
B. 105.6
C. 93.7
D. 87.5
E. 85.6

A. 112.5 
B. 105.6 
C. 93.7 
D. 87.5 
E. 85.6

KarishmaB · Accepted Answer

First, let's find the final concentration of water in the solution: Cf = Ci * (Vi/Vf) = (7/15) * (90/110) * (90/120) = 7*9/5*11*4 Now the volume of solution is 120 lts Amount of water = (63/20*11) * 120 = 6*63/11 = 34.4 lts Amount of milk = 120 - 34.4 = 85.6 lts Answer (E) Check this post for more on replacement: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

Tapesh03 · Answer

IMO - E
Milk to water ratio is 100 liters of solution is 8:7, which means 53.33 liters of milk and rest is water
10 liters of solution is removed, which means 10% of milk is removed because total solution is 100 liters. 
Milk left in the solution after 10% of milk is 53.33-5.33 - 48 liters
20 liters of milk added to the solution, so total amount of milk now is 48+20=68 liters and total amount of solution now is 110 liters
Now, 20 liters of solution is removed, which is 18.18% of total solution
Milk left in the solution after taking 18.18% of milk is 55.64 liters
30 liters of milk added to the solution, so total amount of milk now is 55.64+30=85.64 liters

ScottTargetTestPrep · Answer

Let’s first determine the amount of milk and water originally in the container first:

8x + 7x = 100

15x = 100

x = 100/15 = 20/3

So originally, 8(20/3) = 160/3 liters are milk and 7(20/3) = 140/3 liters are water.

When 10 liters of the solution is removed, the amount of milk and water removed are proportional to their weight. So 16/3 liters of milk and 14/3 liters of water are removed (notice that 10 is 1/10th of 100). With 20 liters of milk added back in the container, there are now 160/3 - 16/3 + 20 = 144/3 + 60/3 = 204/3 liters of milk and 140/3 - 14/3 = 126/3 liters of water and the ratio of milk to water is:

(240/3)/(126/3) = 204/126 = 34/21

Now, when 20 liters of the solution is removed, the amount of milk and water removed are proportional to their weight. Furthermore, the amount of milk and water we have left are also proportional to their weight. So we have (notice that now the amount of solution is 100 - 10 + 20 - 20 = 90):

34y + 21y = 90

55y = 90

y = 90/55 = 18/11

When 30 liters of milk are added back in the container, there are now 34(18/11) + 30 = 612/11 + 330/11 = 942/11 ≈ 85.6 liters of milk in the container.

Answer: E

CrackverbalGMAT · Answer

This is one of those rare sums, where the amount being removed and replaced is different and therefore the total volume of the solution is also changing.

When doing Iterations, we use the equation  \frac{Final }{ Initial}  =  (1 - \frac{b }{ a})^n

Final = The Final Ratio of that substance who's   concentration is being reduced  .

Initial = The Initial Ratio of that substance who's   concentration is being reduced  .

b = Amount of liquid (  who's concentration is increasing  ) is added back

a = Final Volume in the container   after the replacement   of amount b.

n = number of iterations, or the number of times the process is repeated.

Here since both b and a are changing, we convert the above equation to the form

\frac{Final }{ Initial}  =  (1 - \frac{b_1 }{ a_1})  *  (1 - \frac{b_2 }{ a_2})

In the question, the concentration of milk is increasing and that of water is decreasing

Initial Ratio of water = 7/15

Amount replaced the first time,  b_1  = 20 L

Total Volume after replacement for the first time,  a_1  = 100 - 10 + 20 = 110

Amount replaced the second time,  b_2  = 30 L

Total Volume after replacement for the second time,  a_2  = 110 - 20 + 30 = 120

Substituting these values we get

\frac{Final}{7/15}  =  (1 - \frac{20 }{ 110})  *  (1 - \frac{30 }{ 120})

Final Ratio =  \frac{7}{15}  *  \frac{90}{110}  *  \frac{90}{120}  =  \frac{56700 }{ 198000}  =  \frac{63}{220}

Final Amount of water in 120 L =  \frac{63}{220}  * 120 = 34.4 L

Therefore Final Amount of Milk = 120 - 34.4 = 85.6 L

Option E

Arun Kumar

Therefore Amount of milk in 120 L solution which has 28.64 L water = 100 - 28.64 = 91.36 L

SairamRamo · Answer

Initial Mix
Total lit - 100
Milk - 100*(8/15)
water - 100*(7/15)

after 1st replacement
total lit - 100-10+20 = 110
Milk - (90*(8/15))+20 =68
water - 90*(7/15) = 42

after 2nd replacement
total lit - 110-20+30 = 120
Milk - (80*(68/110))+30 = 89.45
water - 80*(42/110) = 30.54

Can any point out what's wrong in my approach here ?
 Bunuel

abhee84 · Answer

red highlighted 80 should be 90. 110-20 = 90.

jaideepjs8 · Answer

Mam, why did you not do it like this:
Water left behind = (7/15) x 100 x (90/110) x (90/120) = 28.6 l
Therefore, final amount of milk = 120 - 28.6 = 91.4 l
Because the amount of water will only reduce in the both transactions. So it would stay same even if the solution is 100 l or 120 l. There's no increase in water happening so why you calculated water w.r.t. 120 l ?

cskarlapudi · Answer

Too few kudos to this approach.

A little more explanation here: The concentration of water is to be taken to start with since no water is being added later, making this approach of calculating the final concentration using multiplying factors feasible.

From a 100 litre container having milk and water in the ratio 8 : 7, 1

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