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# A certain quantity of 40% concentration solution is replaced

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Intern
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A certain quantity of 40% concentration solution is replaced  [#permalink]

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21 Jan 2015, 08:11
1
5
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Difficulty:

65% (hard)

Question Stats:

58% (01:41) correct 42% (01:49) wrong based on 209 sessions

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A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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21 Jan 2015, 09:04
1
25% - - - - - - - - - 35% - - - - - - 40%

From 25 to 35 =10

From 35 to 40 = 5

So the ratio is 10/5 = 2 to 1

B it is
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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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21 Jan 2015, 19:23
1
Hi notrandom,

Mixture questions can often be solved by using the Weighted Average Formula. Here's how you can set that up:

From the wording in the prompt, we're essentially mixing two liquids:

Liquid X is 40% concentrate
Liquid Y is 25% concentrate

We're supposed to mix some of each and end up with a solution that is 35% concentrate. The prompt doesn't give us specific amounts of liquids to work with and the question itself asks for the ratio of the two liquids, so we could even TEST VALUES if doing it that way made it easier for you. For now though, here's the Algebra:

X = # of ounces of liquid X
Y = # of ounces of liquid Y

(.4X + .25Y) / (X + Y) = .35

.4X + .25Y = .35X + .35Y
.05X = .1Y
5X = 10Y

From here, you have to be CAREFUL to answer the question that is ASKED.

5X = 10Y

Y/X = 5/10 = 1/2

For every 1 ounce of liquid Y (the replacement liquid), we had 2 ounces of liquid X (the original liquid).

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A certain quantity of 40% concentration solution is replaced  [#permalink]

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21 Jan 2015, 23:05
2
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notrandom wrote:
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

This replacement question is nothing but a mixture problem with an extra step. Essentially, a certain amount of 40% solution is mixed with 25% solution to get 35% solution. The ratio of the amounts added will be given by:

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (25 - 35)/(35 - 40) = 2/1

So 40% solution amount:25% solution amount = 2:1 i.e. of total 3 parts of solution, 2 parts is 40% solution and 1 part is 25% solution.
So 1 part of the 40% solution was replaced and 2 parts of the 40% solution were not replaced.

For an explanation of the formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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22 Jan 2015, 00:02

Using allegation method:

40 .................... 25

........... 35 ............

35-25 ..................... 40-35
=10 ...................... = 5

Answer$$= \frac{5}{10} = \frac{1}{2}$$
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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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31 Jul 2016, 20:39
VeritasPrepKarishma wrote:
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

This replacement question is nothing but a mixture problem with an extra step. Essentially, a certain amount of 40% solution is mixed with 25% solution to get 35% solution. The ratio of the amounts added will be given by:

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (25 - 35)/(35 - 40) = 2/1

So 40% solution amount:25% solution amount = 2:1 i.e. of total 3 parts of solution, 2 parts is 40% solution and 1 part is 25% solution.
So 1 part of the 40% solution was replaced and 2 parts of the 40% solution were not replaced.

For an explanation of the formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/

Quote:
Hi Karishma,
Here the question says, the amount of solution that was replaced to the amount of solution that was not replaced.
So, when 40 % of the solution is replaced, should we not consider about 60 % of the solution, which was not replaced?

Regards,
Yosita

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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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31 Jul 2016, 21:14
1
yosita18 wrote:
VeritasPrepKarishma wrote:
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

This replacement question is nothing but a mixture problem with an extra step. Essentially, a certain amount of 40% solution is mixed with 25% solution to get 35% solution. The ratio of the amounts added will be given by:

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (25 - 35)/(35 - 40) = 2/1

So 40% solution amount:25% solution amount = 2:1 i.e. of total 3 parts of solution, 2 parts is 40% solution and 1 part is 25% solution.
So 1 part of the 40% solution was replaced and 2 parts of the 40% solution were not replaced.

For an explanation of the formula, check: http://www.veritasprep.com/blog/2011/03 ... -averages/

Quote:
Hi Karishma,
Here the question says, the amount of solution that was replaced to the amount of solution that was not replaced.
So, when 40 % of the solution is replaced, should we not consider about 60 % of the solution, which was not replaced?

Regards,
Yosita

We are given that a CERTAIN QUANTITY (we don't know how much) of a 40% solution (the concentration of spirit/sugar/... in the solution is 40%. So you have 40 ml spirit in 100 ml solution) is mixed with 25% concentration solution (the concentration of spirit/sugar/... in the solution is 25%).
The final solution has concentration of 35% (the concentration of spirit/sugar/... in the solution is 35%).
Now review the solution. Does it make sense?
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A certain quantity of 40% concentration solution is replaced  [#permalink]

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Updated on: 01 Jun 2018, 19:56
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

let x=amount of solution replaced
.4-.4x+.25x=.35
x=1/3
1-x=2/3=amount of solution not repaced
1/3:2/3=1:2
B

Originally posted by gracie on 01 Aug 2016, 12:36.
Last edited by gracie on 01 Jun 2018, 19:56, edited 1 time in total.
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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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31 May 2018, 15:12
notrandom wrote:
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?

A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1

We let m = the amount of solution with 40% concentration and n = the amount of solution with 25% concentration.

Originally we have 0.4m concentration. Since the amount n is removed from m, 0.4n concentration is removed. Since the original amount n is replaced a new amount n, the total amount is still m, however, the 0.4n concentration that is removed is only replaced with 0.25n concentration. Thus we have:

(0.4m - 0.4n + 0.25n)/(m - n + n) = 0.35

(0.4m - 0.15n)/m = 0.35

0.4m - 0.15n = 0.35m

0.05m = 0.15n

5m = 15n

m = 3n

Since m = 3n and amount n is replaced, thus amount 2n is not replaced. Therefore, the ratio is 1/2.

Alternate Solution:

Let x = the original amount of 40% concentration and y = the amount that was taken out and replaced. Note that we started with x liters, and we ended with the same amount (because it was replaced).

Thus, we start with x liters of 40% solution, we take out y liters of 40% solution, and then add back y liters of 25% solution. The result is x liters of 35% solution. Converting the percents to decimals, we can create the equation:

0.40x – 0.40y + 0.25y = 0.35x

.05x = 0.15y

5x = 15y

x = 3y

The question is to determine the ratio of the solution that was replaced to the amount of solution that was not replaced. The amount of solution that was replaced is y, and the amount not replaced is (x – y). Thus, by substituting 3y for x, the ratio can be expressed as:

y/(x – y) = y/(3y – y) = y/2y = 1/2

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Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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31 May 2018, 20:30
Let x be the quantity of 40% solution that was replaced by 25% solution.

The new concentration is now 35%

35% includes 'x' quantity of 25% solution and remaining quantity of 40% solution.

Remaining quantity = total quantity - x
where total quantity = 1 (since in fractions the total is 1)

Remaining quantity = 1-x

Equation: X*25% + (1-X)*40% = 35%
X = 5/15 = 1/3

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Joined: 02 Apr 2018
Posts: 20
Re: A certain quantity of 40% concentration solution is replaced  [#permalink]

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01 Jun 2018, 09:06
I will try solving this by using alligation:

40% 25%
35%
10 5

So if you mix 40% and 25% in 10:5 or rather 2:1 ratio then you get 35% solution.
Now, replaced to not replaced is indeed 1:2.
Re: A certain quantity of 40% concentration solution is replaced &nbs [#permalink] 01 Jun 2018, 09:06
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