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Updated on: 14 Jan 2014, 05:15
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.
Last edited by Bunuel on 14 Jan 2014, 05:15, edited 1 time in total.
Edited the question.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
77% (02:36) correct
23%
(02:45)
wrong
based on 408
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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
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
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
14 Jan 2014, 05:23
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goodyear2013
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.
12 Feb 2015, 13:40
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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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
