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sherxon
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There were 12 litres of sulphuric acid in a bottle 07 Sep 2021, 04:08
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61% (02:34) correct 39% (02:47) wrong based on 236 sessions

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There were 12 litres of sulphuric acid in a bottle. N litres of acid is replaced with water. Then again, N litres of new fluid is replaced with water. In the end, the share of acid in the fluid has decreased to 25%. Find N.

A. 3
B. 4
C. 5
D. 6
E. 8
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D
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DashingNightmare2
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There were 12 litres of sulphuric acid in a bottle 08 Sep 2021, 12:25
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sherxon
There were 12 litres of sulphuric acid in a bottle. N litres of acid is replaced with water. Then again, N litres of new fluid is replaced with water. In the end, the share of acid in the fluid has decreased to 25%. Find N.

A. 3
B. 4
C. 5
D. 6
E. 8
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share of acid in the fluid has decreased to 25%
so new share of sulphuric acid =25% of 12 =3
sulphuric acid is reduced by 9 in 2 time swap
taking values,
if we replace 1st time by 6 => sulphuric acid = 6 water =6
next swap of 6 litres will remove sulphuric acid by 3 and water by 3
sulphuric acid = 3 water = 9

Hence D
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There were 12 litres of sulphuric acid in a bottle 11 Sep 2021, 22:46
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sherxon
There were 12 litres of sulphuric acid in a bottle. N litres of acid is replaced with water. Then again, N litres of new fluid is replaced with water. In the end, the share of acid in the fluid has decreased to 25%. Find N.

A. 3
B. 4
C. 5
D. 6
E. 8
Show more

Alt solution

N liters of acid is replaced and the process is repeated two times, and the final ratio acid to solution becomes 25/100

{12-n/12}^2 = 25/100, solving it we get N = 6
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There were 12 litres of sulphuric acid in a bottle 12 Sep 2021, 03:13
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N from 12 is replaced with water twice

12*N/12*N/12 = 3


N = 6
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There were 12 litres of sulphuric acid in a bottle 12 Sep 2021, 13:48
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Original: 100% concentration of acid, 12 liters

Remove: N lifers, 100% concentration of acid

Replace with: N liters, 0% concentration of acid

Intermediate MIX: 12 liters broken down into:

(12 - N) / (12) = concentration of acid

(N) / (12) = concentration of water


2nd step, we remove and replace N more liters:

Original: (N - 12) / (N) concentration acid, 12 liters

Remove: N liters that has an acid concentration of (N - 12) / (N)

Add: N liters, 0% concentration of acid

FiNAL Mix: 12 liters, 25% concentration of acid


We can set up the equation starting from the intermediate step to the Final Mix


(Amount of acid in intermediate mix) - (amount of acid removed) + (0 amount of acid added) = amount of acid in FINAL Mix

Plugging in the above numbers:

(12 - N / 12) * (12)

- (12 - N / 12) * (N)

+ 0% * (N)

____________

25% * (12) or 3


—multiply both sides by 12 to remove the DEN in the fractions—-

(Also of course 0% of N is zero and irrelevant to the calculation)

(12 - N) * (12) - (12 - N) * (N) = (3) * (12)

Distributing the factors and multiplying we end up with the Quadratic:

(N)^2 - 24(N) + 108 = 0

—-factoring the quadratic——

(N - 6) (N - 18) = 0

N can not be 18 liters because that is more than the original amount in the bottle (12)

Thus:

N = 6 liters

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There were 12 litres of sulphuric acid in a bottle 05 Aug 2024, 11:09
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Acid, which is equal to 12 liters initially is decaying exponantially to become equal to 3 liters in two steps.

Removing N of 12 from 12 and then N of (12-N) from (12-N) is equal to say: 12-N/12 - N/(12-N/12). We can translate this process of "12 liters becoming equal to 3 liters in two steps" into decaying 12 liters by a certain ratio "r" which is equal to "subtracting N"

12*r*r = 12*r^2 = 3 means r = 1/2

Thus:

First step is 12*1/2 must be equal to 12-N --> 12*1/2 = 12-N --> solve and get N=6

IMO D­
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