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Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app

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Bunuel
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Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   03 Jun 2018, 12:40
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Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   03 Jun 2018, 13:00
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Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%


40 oz total
- 30% apple juice = 12 oz
- 70% seltzer = 28 oz

We are looking for (seltzer oz)/(new total oz)

Since the amount of seltzer doesn't change it's still 28 oz.

The new total oz is just the old total + 10 ---> 40 + 10 ---> 50 oz

seltzer/new total = 28/50 = 56/100 = 56%

Answer: D
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   03 Jun 2018, 13:19
Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%


let x=% of seltzer in mixture
.30*40+10=(1-x)*50
x=56%
D
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   03 Jun 2018, 13:45
Apple= 40×30%=12
Seltzer=40-12=28
According to question,

10 ounces apple added...so total apple 12+10=22

So, Total Apple + Seltzer=50

% of Seltzer in the mix = (28/50)×100=56% :-) :-) :-)
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   23 Jun 2018, 08:00
Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%



let the percentage of seltzer in the final mixture be x %

40/10 = (0-x)/(x-70)
4x - 280 = -x
x = 280/5
x = 56 %

Link for the method in the signature.
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Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   23 Jun 2018, 18:18
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Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%

1) Find the number of ounces of seltzer now
2) Find new mixture's total ounces of liquid. Ignore the apple juice percent.
3) Find percent seltzer in new mixture

1) How many ounces of seltzer now?
40 ounces of liquid are 30% apple juice, so 70% are seltzer. Number of ounces of seltzer:
\((.7*40)=28\) ounces seltzer

2) Jane pours 10 more ounces of apple juice into the mix.
Total ounces of liquid now: \(40oz. + 10oz.=50\) ounces*

3) New mixture is what percent seltzer?
Number of seltzer ounces has not changed. Percent seltzer:

\(\frac{SeltzerOz}{TotalOz}=\frac{28}{50}=(\frac{56}{100}*100) = 56\) percent

Answer D

*We don't care about the fact that the new ounces of liquid are apple juice. If we were tracking on apple juice percent, we would care. We are tracking on seltzer percent. No seltzer is added. All we need is the new total volume of liquid to find percent seltzer in new mix.
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Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   18 May 2019, 10:50
generis wrote:
Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%

1) Find the number of ounces of seltzer now
2) Find new mixture's total ounces of liquid. Ignore the apple juice percent.
3) Find percent seltzer in new mixture

1) How many ounces of seltzer now?
40 ounces of liquid are 30% apple juice, so 70% are seltzer. Number of ounces of seltzer:
\((.7*40)=28\) ounces seltzer

2) Jane pours 10 more ounces of apple juice into the mix.
Total ounces of liquid now: \(40oz. + 10oz.=50\) ounces*

3) New mixture is what percent seltzer?
Number of seltzer ounces has not changed. Percent seltzer:

\(\frac{SeltzerOz}{TotalOz}=\frac{28}{50}=(\frac{56}{100}*100) = 56\) percent

Answer D

*We don't care about the fact that the new ounces of liquid are apple juice. If we were tracking on apple juice percent, we would care. We are tracking on seltzer percent. No seltzer is added. All we need is the new total volume of liquid to find percent seltzer in new mix.


I miss read the question so badly. This happens when someone immersed too much in Sentence Correction. I thought only seltzer is 30% apple juice as "that " was modifying seltzer. :-D . And, therefore, I was not able to make sense of the question. :cool:

Height of smartness.
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   18 May 2019, 16:30
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AkshdeepS wrote:
generis wrote:
Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%

1) Find the number of ounces of seltzer now
2) Find new mixture's total ounces of liquid. Ignore the apple juice percent.
3) Find percent seltzer in new mixture

1) How many ounces of seltzer now?
40 ounces of liquid are 30% apple juice, so 70% are seltzer. Number of ounces of seltzer:
\((.7*40)=28\) ounces seltzer

2) Jane pours 10 more ounces of apple juice into the mix.
Total ounces of liquid now: \(40oz. + 10oz.=50\) ounces*

3) New mixture is what percent seltzer?
Number of seltzer ounces has not changed. Percent seltzer:

\(\frac{SeltzerOz}{TotalOz}=\frac{28}{50}=(\frac{56}{100}*100) = 56\) percent

Answer D

*We don't care about the fact that the new ounces of liquid are apple juice. If we were tracking on apple juice percent, we would care. We are tracking on seltzer percent. No seltzer is added. All we need is the new total volume of liquid to find percent seltzer in new mix.

I miss read the question so badly. This happens when someone immersed too much in Sentence Correction. I thought only seltzer is 30% apple juice as "that " was modifying seltzer. :-D . And, therefore, I was not able to make sense of the question. :cool:

Height of smartness.

AkshdeepS , you quoted me. The post is humorous (thanks for the laugh). And I can't resist. (Easy mistake, BTW.)
Quote:
. . . as "that " was modifying seltzer.

"That" [is 30%] modifies mixture, not seltzer. :grin:
That can be separated from the noun it modifies by essential modifiers such as the prepositional phrase "of apple juice and seltzer."

You're plenty smart. Maybe just a little tired. :)
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink] New post   18 May 2019, 17:17
Bunuel wrote:
Jane has a 40-ounce mixture of apple juice and seltzer that is 30% apple juice. If she pours 10 more ounces of apple juice into the mixture, what percent of the mixture will be seltzer?

(A) 33%
(B) 44%
(C) 50%
(D) 56%
(E) 67%


Originally, A:S :: 12:28
Now after adding 10 A we have 50 ounce in total,

Required answer=> (28/50) i.e. 56%
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Re: Jane has a 40-ounce mixture of apple juice and seltzer that is 30% app [#permalink]
18 May 2019, 17:17
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