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A laboratory is testing a new steroid on mice. The average

tagged by: BTGmoderatorLU

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A

B

C

D

E

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Source: Veritas Prep

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

The OA is C.

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BTGmoderatorLU wrote:
Source: Veritas Prep

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

The OA is C.
Say,

the average weight of a mouse that has been treated with the steroid = x = 26.4;
the average weight of a mouse that has NOT been treated with the steroid = y = 19.2; and
the average weight of all the mice = z = 22.4

Then,

The ratio of mice that have been treated to mice that have not been treated = (z - y) / (x - z) = (22.4 - 19.2) / (26.8 - 22.4) = 8/11

Hope this helps!

-Jay
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BTGmoderatorLU wrote:
Source: Veritas Prep

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
Perfect opportunity for the alligation, a nice technique included in our course!

$? = T:N\,\,\,\,\left( {{\text{see}}\,\,{\text{image}}\,\,{\text{attached}}} \right)$
$\frac{T}{{total}} = \frac{{22.4 - 19.2}}{{26.8 - 19.2}} = \frac{{3.2 \cdot \boxed{10}}}{{7.6 \cdot \boxed{10}}} = \frac{8}{{19}}$

We are done, but if you have not got the answer yet, have a look at the k technique...

$?\,\,\,:\,\,\,\left\{ \begin{gathered} T = 8k \hfill \\ {\text{total}} = 19k \hfill \\ \end{gathered} \right.\,\,\,\,\,\left( {k > 0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 11k\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\, = \,\,\,8:11\,\,\,\,\,$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Let the number of mice treated with steroid = x
Let the number of mice NOT treated with steroid = y

As per questions, we are given

28.6x + 19.2y = 22.4(x+y)
or
x/y = 8/11.

I hope this help! Regards!

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