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) \frac{1}{5}
(B) \frac{1}{4}
(C) \frac{1}{2}
(D) \frac{3}{4}
(E) \frac{4}{5}


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Can sum1 give a simpler explanation?

A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1


I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

The correct answer is E.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?

I understand the alegbraic approach here but can this question be solved using more basic math principles?

If a solution goes from 50% to 40% in a mixture with 30% of some proportion, doesnt logic dictate that the mixture be 50/50?

i understand this approach doesnt work with many other mixture problems, but in this case where equal amounts are used and it is asking about proportions and not the overall amounts used, wouldnt it work? thoughts?
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You came up with the answer 0 for 4:1.
Now if you choose red:green=8:2 (which is equivalent to the ratio 4:1), you get an answer(2/10*1/9) which is different from 0. So its E

A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1


I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

The correct answer is E.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?

Addressing your problem - your intent is right, however what you have to realise is that you are only provided with the ratios in the quesion and not specefic values. It is easy assume that 20 out of 100 or 2 out of 10 chips are green, which is not the case here.

Option 1 says - 20% of chips are green. Now this info is clearly not enough as we don't know total number of chips. So we can't say whether number of green chips are 20 or 200 or 200000. So choice 1 is clearly insufficient.

Now if you get my point above, you'd realise that Option 2 is just a re-wording of the option 1, i.e. if the ratio is 4:1, then green chips are certainly 20% of the total.

I hope you understand why D is wrong. D is incorrect since we do not have values and only have ratios. If the question had asked us to pick just one chip, D wld have been correct.

Cheers!

This one is cake. Answer C.
Can be solved on 30 sec. It is just becasue data is very simple.

50% solution of acid and 30% solution of acid = 40% solution of acid -- Straight average of both acid is required. Hence, 50-50 of both. Answer C. Hope it helps.

Cheers!

P.S. - This is applicable for all the questions in which final concentration is becoming average concentration of 2 solutions which are getting mixed.
10% solution of acid and 30% solution of acid = 20% solution of acid
20% solution of acid and 40% solution of acid = 30% solution of acid
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you can easily solve this question by just seeing that 40% is also in the mid of the two numbers and to be in the middle you must have an equal amount of both.
Caution though that this will only work in this situation where it is perfectly in the middle of the two numbers... it would be harder to tell if the numbers were larger and you couldnt see the middle number of the two.
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What is this mgmat method that you speak of.. could you elaborate. might be useful elsewhere.

I solved this question by following method without being sure if it was true, but it gave a correct answer and took some times ...

0.5x - a0.5x + a0.3x = 0.4x (1)

a is the same amount of solutions replaced

from (1) we have: a0.2x = 0.1x
thus, a = 1/2

and the answer is C