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There are x numbers in list L, where x is a positive integer

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There are x numbers in list L, where x is a positive integer  [#permalink]

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New post Updated on: 14 Jan 2014, 05:15
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There are x numbers in list L, where x is a positive integer, and there are y numbers in list M, where y is a positive integer. The average (arithmetic mean) of the numbers in list L is p, and the average of all the numbers in both lists L and M is q. Which of the following expressions is the average of the numbers in list M?

A. (qy - px) / x
B. (qy - px) / y
C. (p + q)x / y
D. [(q - p)x - py] / (x + y)
E. [(q - p)x + qy] / y

OE
Sum = (Average × Number of terms)
Average of all (x + y) numbers in both lists L and M = q
Sum of all numbers in both lists L and M = q(x + y)
Average of x numbers in list L = p, sum of numbers in list L = px
Sum of all (x + y) numbers in both lists L and M = q(x + y)
Sum of x numbers in list L = px
If subtract sum of x numbers in list L from sum of (x + y) numbers in both lists L and M, we will be left with sum of y numbers in list M. → Sum of y numbers in list M = [q(x + y) – px]
Average of y numbers in list M = [q(x + y) – px] / y
Rewriting numerator q(x + y) − px → q(x + y) − px = (qx + qy − px = qx − px + qy) = (q − p)x + qy
→ q(x + y) − px = [(q − p)x + qy]
M = [(q − p)x + qy] / y

Originally posted by goodyear2013 on 14 Jan 2014, 05:05.
Last edited by Bunuel on 14 Jan 2014, 05:15, edited 1 time in total.
Edited the question.
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Re: There are x numbers in list L, where x is a positive integer  [#permalink]

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New post 14 Jan 2014, 05:23
3
goodyear2013 wrote:
There are x numbers in list L, where x is a positive integer, and there are y numbers in list M, where y is a positive integer. The average (arithmetic mean) of the numbers in list L is p, and the average of all the numbers in both lists L and M is q. Which of the following expressions is the average of the numbers in list M?

A. (qy - px) / x
B. (qy - px) / y
C. (p + q)x / y
D. [(q - p)x - py] / (x + y)
E. [(q - p)x + qy] / y

OE
Sum = (Average × Number of terms)
Average of all (x + y) numbers in both lists L and M = q
Sum of all numbers in both lists L and M = q(x + y)
Average of x numbers in list L = p, sum of numbers in list L = px
Sum of all (x + y) numbers in both lists L and M = q(x + y)
Sum of x numbers in list L = px
If subtract sum of x numbers in list L from sum of (x + y) numbers in both lists L and M, we will be left with sum of y numbers in list M. → Sum of y numbers in list M = [q(x + y) – px]
Average of y numbers in list M = [q(x + y) – px] / y
Rewriting numerator q(x + y) − px → q(x + y) − px = (qx + qy − px = qx − px + qy) = (q − p)x + qy
→ q(x + y) − px = [(q − p)x + qy]
M = [(q − p)x + qy] / y


There are x numbers in list L and the average is p --> the sum of the numbers in list L is (average)*(number of terms) = \(px\).

There are x+y numbers in lists L and M and the average is q --> the sum of the numbers in lists L and M is (average)*(number of terms) = \(q(x+y)\).

The above implies that the sum of the numbers in list M is \(q(x+y) - px=(q-p)x+qy\).

Therefore the average of list M is (sum)/(number of terms) = \(\frac{(q-p)x+qy}{y}\).

Answer: E.
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Re: There are x numbers in list L, where x is a positive integer  [#permalink]

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New post 12 Feb 2015, 13:40
Hi All,

While this question certainly looks "crazy", it can be solved rather easily by TESTing VALUES, staying organized and looking for logical patterns.

We're told that there are 2 lists of numbers (List L and List M), but we're NOT told ANYTHING about the lists. We can make them as simple of complex as we choose. I choose to make them SIMPLE....

List L: {2, 2} Average = 2, number of terms = 2
List M: {3, 3, 3} Average = 3, number of terms = 3

The prompt assigns a number of variables to different terms:
X = # of terms in List L
Y = # of terms in List M
P = Average of List L
Q = Average of ALL the terms in Lists L and M.

Using the above 2 lists as reference:
X = 2
Y = 3
P = 2
Q = 13/5 = 2.6

We're asked for the AVERAGE of LIST M. Before you jump into the answer choices, think about what you're asked for....The AVERAGE of List M.....Since list M has 3 terms, we will eventually divide a sum by 3. Now, look at the answer choices.....which answers do NOT do that.... You can eliminate Answer A (it divides by 2) and Answer D (it divides by 5). Now, we just have to check 3 answers; we're looking for an answer that is the AVERAGE of List M, so we're looking for the number 3.

Answer B: (QY-PX)/Y = (7.8 - 4)/3 = 3.8/3 This is clearly too small. Eliminate B.

Answer C: (P+Q)X/Y = (4.6)(2)/3 = 9.2/3 This is too big. Eliminate C.

Answer E is all that's left, but I'll check it just to be sure....

Answer E: [(Q-P)X +QY]/Y = [(0.6)(2) +7.8]/3 = 9/3 = 3. This is a MATCH.

Final Answer:

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Re: There are x numbers in list L, where x is a positive integer  [#permalink]

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New post 03 Apr 2017, 00:44
Total for list L= xp
Total for List M=ay (Let 'a' be the average here)
Total average for two lists= q
So the total sum for two lists= q(x+y)

q(x+y)- xp= ay
a= [x(q-p)+qy]/ y
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Re: There are x numbers in list L, where x is a positive integer  [#permalink]

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New post 06 Apr 2017, 09:37
goodyear2013 wrote:
There are x numbers in list L, where x is a positive integer, and there are y numbers in list M, where y is a positive integer. The average (arithmetic mean) of the numbers in list L is p, and the average of all the numbers in both lists L and M is q. Which of the following expressions is the average of the numbers in list M?

A. (qy - px) / x
B. (qy - px) / y
C. (p + q)x / y
D. [(q - p)x - py] / (x + y)
E. [(q - p)x + qy] / y


Let’s use the average formula: average = sum/number. An equivalent equation is (average) x (number) = sum.

Since the average of the numbers in list L is p and there are x numbers in L, the sum of the numbers in L is px. Since the average of all of the numbers in both lists L and M is q and there are x + y numbers in both lists, the sum of all of the numbers in the two lists is q(x + y). Thus, the sum of the numbers in list M is
q(x + y) - px, and since there are y numbers in M, the average of the numbers in M is:

[q(x + y) - px]/y

(qx + qy - px)/y

(qx - px + qy)/y

[x(q - p) + qy]/y

[(q - p)x + qy]/y

Answer: E
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Re: There are x numbers in list L, where x is a positive integer &nbs [#permalink] 06 Apr 2017, 09:37
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