A laboratory is testing a new steroid on mice.

Question

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...

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...

mromrell · Accepted Answer

https://i.imgur.com/isnQmgRm.png

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

abhimahna · Answer

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.

damybox · Answer

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

Abhishek009 · Answer

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

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

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

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}

msk0657 · Answer

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.

shamim2k14 · Answer

Should be C. Calculate the distance from each end to the mean and then flip the ratios.

Posted from my mobile device

ProfX · Answer

I could not figure it out, but if I have to guess, I will go with E

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ProfX · Answer

Aah you are very smart

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rohit8865 · Answer

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

Likhitha30chowdary · Answer

26.8a + 19.2b = 22.4(a+b)

or a/b = 8/11. Hence, answer is C.

ScottTargetTestPrep · Answer

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.

bumpbot · Answer

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A laboratory is testing a new steroid on mice.

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