Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
a 1/5
b 1/4
c 1/2
d 3/4
e 4/5
OA: After sometime
Mixtures
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- neelgandham
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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a
result, 40% solution of acid was obtained, what part of the original solution was replaced?
Let us complete forget that a part of a 50% solution of acid is replaced and assume that X ml of a 50% solution of acid and Y amount of 30% solution of acid are mixed to form a 40% solution of acid.
Using allegations:
Quantity of the 50% acid solution/Quantity of the 30% acid solution = X/Y = (%strength of resultant - % strength of the 30% solution)/(% strength of the 50% solution - strength of resultant)
= (40 - 30)/(50 - 40) = 1:1
So, One part of 50% acid solution is mixed with one part of 30% acid solution. i.e. In two parts of 50% acid solution, one part is replaced with one part of 30% acid solution. So, the answer to this question is 1/2.
result, 40% solution of acid was obtained, what part of the original solution was replaced?
Let us complete forget that a part of a 50% solution of acid is replaced and assume that X ml of a 50% solution of acid and Y amount of 30% solution of acid are mixed to form a 40% solution of acid.
Using allegations:
Quantity of the 50% acid solution/Quantity of the 30% acid solution = X/Y = (%strength of resultant - % strength of the 30% solution)/(% strength of the 50% solution - strength of resultant)
= (40 - 30)/(50 - 40) = 1:1
So, One part of 50% acid solution is mixed with one part of 30% acid solution. i.e. In two parts of 50% acid solution, one part is replaced with one part of 30% acid solution. So, the answer to this question is 1/2.
Anil Gandham
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We can see that 40% is in middle of of 50% and 30% , in weighted avg. when both the coefficients are same (a*50% + a*30%)2 mean would fall exactly in between the two numbers i.e. a*40%.
Therefore equal quantities of 30% and 50% solution must have been mixed, and to make it equal 1/2 of the qty. of original mixture must have been removed.
Therefore equal quantities of 30% and 50% solution must have been mixed, and to make it equal 1/2 of the qty. of original mixture must have been removed.
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Intuitive Approach:shrey2287 wrote:Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?
a 1/5
b 1/4
c 1/2
d 3/4
e 4/5
OA: After sometime
Just observe that the concentration of the final solution (40%) is exactly at the middle of the concentrations of the solutions mixed together (50% and 30%).
Hence, exactly same amount of both solutions were mixed together, i.e. 1/2 of the original solution was replaced.
The correct answer is D.
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Algebraic Approach:
Say, x part of 50% solution was replaced with x part of 30% solution.
So, (1 - x)*50 + x*30 = 40
--> 50 - 50x + 30x = 40
--> 20x = 10
--> x = 1/2
The correct answer is D.
Say, x part of 50% solution was replaced with x part of 30% solution.
So, (1 - x)*50 + x*30 = 40
--> 50 - 50x + 30x = 40
--> 20x = 10
--> x = 1/2
The correct answer is D.
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