Each number in each of lists L and M above is positive.

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Each number in each of lists L and M above is positive.

by DivyaD » Fri Jan 18, 2019 4:20 am
List L: r, s, t, u, v
List M: w, x, y, z
Each number in each of lists L and M above is positive. The ratio of the average (arithmetic mean) of the numbers in list L to the average of the numbers in list M is 3 to 7. What is the ratio of the average of all the numbers in both lists L and M to the average of the numbers in list L ?

a) $$\frac{10}{7}$$
b) $$\frac{43}{27}$$
c) $$\frac{9}{5}$$
d) $$\frac{23}{12}$$
e) $$\frac{10}{3}$$

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by [email protected] » Fri Jan 18, 2019 5:43 pm
DivyaD wrote:List L: r, s, t, u, v
List M: w, x, y, z
Each number in each of lists L and M above is positive. The ratio of the average (arithmetic mean) of the numbers in list L to the average of the numbers in list M is 3 to 7. What is the ratio of the average of all the numbers in both lists L and M to the average of the numbers in list L ?

$$a) \frac{10}{7}\,\,\,\,\,\,b) \frac{43}{27}\,\,\,\,\,\,c) \frac{9}{5}\,\,\,\,\,\,d) \frac{23}{12}\,\,\,\,\,\,e) \frac{10}{3}$$
$${{\,{\mu _L}\,} \over {\,{\mu _M}\,}} = {3 \over 7}\,\,\,\,\,\,;\,\,\,\,\,\,\,? = {{\,{\mu _{L \oplus M}}\,} \over {{\mu _L}}}$$
$${\rm{Take}}\,\,\left\{ \matrix{ \,L = \left\{ {3,3,3,3,3} \right\} \hfill \cr \,M = \left\{ {7,7,7,7} \right\} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,{\mu _{L \oplus M}} = \mu \left( {\left\{ {3,3,3,3,3,7,7,7,7} \right\}} \right) = {{43} \over 9}$$
$$?\,\, = \,\,{{\,\,{{43} \over 9}\,\,} \over 3}\,\, = \,\,{{43} \over {27}}$$

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by [email protected] » Fri Jan 25, 2019 2:54 am
DivyaD wrote:List L: r, s, t, u, v
List M: w, x, y, z

Each number in each of lists L and M above is positive. The ratio of the average (arithmetic mean) of the numbers in list L to the average of the numbers in list M is 3 to 7. What is the ratio of the average of all the numbers in both lists L and M to the average of the numbers in list L ?

A. 10/7
B. 43/27
C. 9/5
D. 23/12
E. 10/3
Given that the ratio of the average (arithmetic mean) of the numbers in list L to the average of the numbers in list M is 3 to 7, let's take the average of list L = 3x and the average of list M = 7x.

Thus, the sum of 7 numbers in list (M + L) = 4*(7x) + 5*(3x) = 28x + 15x = 43x

Thus, average of 7 numbers in list (M + L) = 43x/9;

Thus, the ratio of the average of all the numbers in both lists L and M to the average of the numbers in list L = (43x/9) / (3x) = 43/27

The correct answer: B

Hope this helps!

-Jay
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