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A laboratory is testing a new steroid on mice.

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theironhand
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A laboratory is testing a new steroid on mice. [#permalink] New post   24 Sep 2016, 16:49
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A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9

I couldn't understand the solution explanation on Veritas Prep. Please explain how you solve it...
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C
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A laboratory is testing a new steroid on mice. [#permalink] New post   30 Sep 2016, 07:59
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Anytime you deal with averages it help to draw it out like this. All you have to do is get the difference between the average and the numbers on each side.
32:44
Those differences simplify down to a ratio.

8:11. Answer C
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Re: A laboratory is testing a new steroid on mice. [#permalink] New post   24 Sep 2016, 22:20
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theironhand wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9

I couldn't understand the solution explanation on Veritas Prep. Please explain how you solve it...


Simple question.

Let the number of mice treated with steroid = x

Let the number of mice NOT treated with steroid = y

As per question, we are given

26.8x + 19.2y = 22.4(x+y)

or x/y = 8/11. Hence, answer is C.
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A laboratory is testing a new steroid on mice. [#permalink] New post   29 Sep 2016, 03:30
theironhand wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9

I couldn't understand the solution explanation on Veritas Prep. Please explain how you solve it...





let's call the number mice Treated T, and the number of mice non treated N

\(\frac{(26.9T+19.2N)}{N+T}= 22.4\) =====> \((26.5-22.4)T=(22.4-19.2)N\)

===> \(4.4T=3.2N\)

\(\frac{T}{N}=\frac{8}{11}\)

answer C
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Re: A laboratory is testing a new steroid on mice. [#permalink] New post   30 Sep 2016, 08:43
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theironhand wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9

I couldn't understand the solution explanation on Veritas Prep. Please explain how you solve it...


Quote:
The average weight of a mouse that has been treated with the steroid is 26.8 grams

Let \(S\) mice be treated with Steroid
Total weight of mouse treated with steroid is \(26.8 S\)

Quote:
the average weight of a mouse that has not been treated with the steroid is 19.2 grams.

Let \(N\) mice not be treated with Steroid
Total weight of mouse not treated with steroid is \(19.2 N\)

Quote:
the average weight of all mice at the laboratory is 22.4 grams,


Total weight of mouse ( not treated + treated ) with steroid is \(22.4 ( N + S )\)


So, \(22.4 ( N + S )\) = \(26.8 S\) \(+\) \(19.2 N\)

Or, \(22.4N + 22.4S\) = \(26.8 S\) \(+\) \(19.2 N\)

Or, \(22.4N\) \(-\) \(19.2 N\) = \(26.8 S\) \(- 22.4S\)

Or, \(3.2N\) = \(4.4S\)

Or, \(8N\) = \(11S\)

Quote:
what is the ratio of mice that have been treated to mice that have not been treated?


So, we need to calculate the ratio of \(S:N\)

As, \(11S\) = \(8N\)

So, \(\frac{S}{N} = \frac{8}{11}\)

Hence the correct answer is \((C) \frac{8}{11}\)
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Re: A laboratory is testing a new steroid on mice. The average weight of a [#permalink] New post   09 Oct 2016, 03:10
Bunuel wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A. 8:13
B. 3:4
C. 8:11
D. 8:9
E. 7:9



Let x be treated with steroid and the y be non steroid

26.8x + 19.2y / x+y = 22.4

4.4x = 3.2y => x/y = 3.2/4.4 = 8/11.

IMO option C.
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A laboratory is testing a new steroid on mice. The average weight of a [#permalink] New post   09 Oct 2016, 05:07
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Should be C. Calculate the distance from each end to the mean and then flip the ratios.

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Re: A laboratory is testing a new steroid on mice. The average weight of a [#permalink] New post   09 Oct 2016, 05:24
I could not figure it out, but if I have to guess, I will go with E :)


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Re: A laboratory is testing a new steroid on mice. The average weight of a [#permalink] New post   09 Oct 2016, 05:25
shamim2k14 wrote:
Should be C. Calculate the distance from each end to the mean and then flip the ratios.

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Aah you are very smart


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Re: A laboratory is testing a new steroid on mice. [#permalink] New post   09 Oct 2016, 19:54
theironhand wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9

I couldn't understand the solution explanation on Veritas Prep. Please explain how you solve it...

This could also be done with weighted avg. formula suggested by karishma

w1/w2=(A2-Avg)/(Avg-A1)
A1=26.8
A2=19.2
Avg.=22.4
substituting w1/w2=3.2/4.4=8/11

Ans C
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Re: A laboratory is testing a new steroid on mice. [#permalink] New post   23 May 2018, 00:47
26.8a + 19.2b = 22.4(a+b)

or a/b = 8/11. Hence, answer is C.
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A laboratory is testing a new steroid on mice. [#permalink] New post   24 May 2018, 18:01
Expert Reply
theironhand wrote:
A laboratory is testing a new steroid on mice. The average weight of a mouse that has been treated with the steroid is 26.8 grams and the average weight of a mouse that has not been treated with the steroid is 19.2 grams. If the average weight of all mice at the laboratory is 22.4 grams, what is the ratio of mice that have been treated to mice that have not been treated?

A) 8:13

B) 3:4

C) 8:11

D) 8:9

E) 7:9


We can let m = the number of mice that have been treated and n = the number of mice that have not been treated.

(26.8m + 19.2n)/(m + n) = 22.4

26.8m + 19.2n = 22.4m + 22.4n

4.4m = 3.2n

44m = 32n

m/n = 32/44

m/n = 8/11

Answer: C

Too bad the topic is animal testing. There has to be a way to use the same math with a more positive topic.
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Re: A laboratory is testing a new steroid on mice. [#permalink] New post   04 Jun 2023, 07:23
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