Some part of a 50% solution of acid was replaced

[bhumika.k.shah]

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 C

Can #s be plugged ? if so , then how ?

[ajith]

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 C

Can #s be plugged ? if so , then how ?

if x fraction of a 50% acid solution is replaced with an equal amount of 30% acid the resulting concentration is

(1-x)*50 + x*30 [Since 1-x parts of 50% acid is remaining and x parts of 30% acid is added)

which is given as 40

solving x = 0.5 = 1/2

Plugging the answers- well, yes

but we use the same formula as above and since it involves only 1 factor (x) it might be easier to solve than to plug

[bhumika.k.shah]

I am sowree but i dint quite understand...
could u break it down step by step or explain using #s...As i am quite comfortable using #s than equations

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 C

Can #s be plugged ? if so , then how ?

if x fraction of a 50% acid solution is replaced with an equal amount of 30% acid the resulting concentration is

(1-x)*50 + x*30 [Since 1-x parts of 50% acid is remaining and x parts of 30% acid is added)

which is given as 40

solving x = 0.5 = 1/2

Plugging the answers- well, yes

but we use the same formula as above and since it involves only 1 factor (x) it might be easier to solve than to plug

[ajith]

I am sowree but i dint quite understand...
could u break it down step by step or explain using #s...As i am quite comfortable using #s than equations

A 50% acid is acid diluted with water - 50% pure concentrated acid and rest water. In other words, 100 ml of the 50% acid contains 50 ml of 100%concentrated acid.

Similarly a 30% acid contains 30% acid and rest water. 100 ml of the 30% acid contains 30ml of 100%concentrated acid.

Plugging the answers .....

A.

Now if take 1/5 out of first mix - 80 ml of 50% acid remains and 40 ml of it is pure concentrated acid
if you replace it with 20 ml of 30% acid which contains 6 ml of pure concentrated acid

Now the the mixture contains 40+6 = 46 ml of concentrated acid in 100 ml hence is a 46% acid. but, you wanted to obtain 40% acid hence A is not the answer

B. 75 ml of 50% acid contains - 37.5ml of pure acid
25 ml of 30% acid contains - 7.5 ml of pure acid
Mixture contains - 45ml of pure acid in 100 ml hence 45%

C. 50 ml of 50% acid - 25 ml
50 ml of 30% acid - 15 ml
Mixture contains - 40 ml of pure acid in 100 ml hence 40%. The answer you are looking for....