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Updated on: 26 Jun 2019, 01:52
Originally posted by EgmatQuantExpert on 21 Jun 2019, 06:08.
Last edited by EgmatQuantExpert on 26 Jun 2019, 01:52, edited 2 times in total.
Last edited by EgmatQuantExpert on 26 Jun 2019, 01:52, edited 2 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (03:02) correct
63%
(03:23)
wrong
based on 262
sessions
History
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Time
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Not Attempted Yet
e-GMAT Question of the Week - 45
If A, B, C are three consecutive even integers, and D, E, F are three consecutive multiples of 5, then what is the average of all six numbers?
If A, B, C are three consecutive even integers, and D, E, F are three consecutive multiples of 5, then what is the average of all six numbers?
- (1) A * D = -30
(2) |P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F
- A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
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21 Jun 2019, 07:02
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From (1)
\((A,D) = (-2,15)\) or \((-6, 5)\) ------------ Not sufficient
From (2)
Since \(\frac{(A+B+C)}{3} < 0\) or \(P<0, |P|=-P\)
\(|P| = \frac{(-A-B-C)}{3}\)
Similarly, \(Q>0\) and so \(|Q|=Q\)
\(|Q| = \frac{(D+E+F)}{3}\)
=> \(\frac{(-A-B-C)}{3} = \frac{(D+E+F)}{3} - 2\)
=> \(-A-B-C = D+E+F-6\)
=> \(A+B+C+D+E+F = 6\) and therefore \(\frac{(A+B+C+D+E+F)}{6} = 1\)
Sufficient
Answer is (B)
Hit Kudos if this helped!
\((A,D) = (-2,15)\) or \((-6, 5)\) ------------ Not sufficient
From (2)
Since \(\frac{(A+B+C)}{3} < 0\) or \(P<0, |P|=-P\)
\(|P| = \frac{(-A-B-C)}{3}\)
Similarly, \(Q>0\) and so \(|Q|=Q\)
\(|Q| = \frac{(D+E+F)}{3}\)
=> \(\frac{(-A-B-C)}{3} = \frac{(D+E+F)}{3} - 2\)
=> \(-A-B-C = D+E+F-6\)
=> \(A+B+C+D+E+F = 6\) and therefore \(\frac{(A+B+C+D+E+F)}{6} = 1\)
Sufficient
Answer is (B)
Hit Kudos if this helped!
General Discussion
21 Jun 2019, 06:46
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To find : (A+B+C+D+E+F)/6
Statement 1 :
A*D = -30
There can be 2 possibilities - A(-6) * B(5) or A(-2) * B(15)
Therefore, NOT SUFFICIENT.
Statement 2:
|P| = |Q| - 2
P = -(A+B+C)/3
Q = (D+E+F)/3
On solving, we get A+B+C = D+E+F -6
If we substitute this in (A+B+C+D+E+F)/6 we still don't get a value. Therefore INSUFFICIENT.
1+2
Only 5, -6 will satisfy statement 2. Therefore we have specific values for A & D, which also give us all the other values required.
So, C
Statement 1 :
A*D = -30
There can be 2 possibilities - A(-6) * B(5) or A(-2) * B(15)
Therefore, NOT SUFFICIENT.
Statement 2:
|P| = |Q| - 2
P = -(A+B+C)/3
Q = (D+E+F)/3
On solving, we get A+B+C = D+E+F -6
If we substitute this in (A+B+C+D+E+F)/6 we still don't get a value. Therefore INSUFFICIENT.
1+2
Only 5, -6 will satisfy statement 2. Therefore we have specific values for A & D, which also give us all the other values required.
So, C
