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Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
Updated on: 26 Jun 2019, 01:52
Updated on: 26 Jun 2019, 01:52
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Expert Reply
Show timer
00:00
A
B
C
D
E
Difficulty:
95%
(hard)
Question Stats:
37% (03:02) correct
63%
(03:22)
wrong
based on 260
sessions
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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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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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
21 Jun 2019, 07:02
21 Jun 2019, 07:02
9
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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!
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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
21 Jun 2019, 06:46
21 Jun 2019, 06:46
1
Bookmarks
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
Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
21 Jun 2019, 07:34
21 Jun 2019, 07:34
2
Kudos
2
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Expert Reply
Assuming the arrangement of the letters on the number line can be trusted, then the average of C, B and A is just the median, B, because C, B and A are consecutive even integers and so are equally spaced. Similarly the average of D, E and F is just E. Since those two sets are equal in size, the average of all six numbers will be the average of B and E, so will be (B+E)/2. So if we can find B+E we can answer the question.
Statement 1 gives us two possibilities (A = -2, D = 15, or A = -6, D = 5) which give different values for B+E.
For Statement 2, P = B and Q = E, so Statement 2 just says |B| = |E| - 2. Since B < 0 from the number line, |B| = -B and since E > 0, |E| = E. So Statement 2 tells us
-B = E - 2
2 = E + B
and the average of the six numbers must be (E+B)/2 = 2/2 = 1.
Statement 1 gives us two possibilities (A = -2, D = 15, or A = -6, D = 5) which give different values for B+E.
For Statement 2, P = B and Q = E, so Statement 2 just says |B| = |E| - 2. Since B < 0 from the number line, |B| = -B and since E > 0, |E| = E. So Statement 2 tells us
-B = E - 2
2 = E + B
and the average of the six numbers must be (E+B)/2 = 2/2 = 1.
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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
21 Jun 2019, 08:46
21 Jun 2019, 08:46
good question
#1
A * D = -30
A= 2,6,
D= 15,5,
insufficient
#2
|P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F
in this case we can say that P=B and Q=E
also B=E
so we can say 2=E+B
so avg of 6 no. ; E+B/2 = 2/2 ; 1
IMO B
#1
A * D = -30
A= 2,6,
D= 15,5,
insufficient
#2
|P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F
in this case we can say that P=B and Q=E
also B=E
so we can say 2=E+B
so avg of 6 no. ; E+B/2 = 2/2 ; 1
IMO B
EgmatQuantExpert wrote:
e-GMAT Question of the Week
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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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
22 Jun 2019, 20:13
22 Jun 2019, 20:13
Statement 1 :
A*D = -30
2 possibilities - A(-6) * B(5) or A(-2) * B(15)
NOT SUFFICIENT.
Statement 2:
|P| = |Q| - 2
P = -(A+B+C)/3
Q = (D+E+F)/3
A+B+C = D+E+F -6
A + B+ C +D + E + F = 6
So, B
A*D = -30
2 possibilities - A(-6) * B(5) or A(-2) * B(15)
NOT SUFFICIENT.
Statement 2:
|P| = |Q| - 2
P = -(A+B+C)/3
Q = (D+E+F)/3
A+B+C = D+E+F -6
A + B+ C +D + E + F = 6
So, B
Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
26 Jun 2019, 01:49
26 Jun 2019, 01:49
1
Kudos
Expert Reply
Solution
Given:
- • A, B, C are three consecutive even integers, and all of them are negative integers
• D, E, F are three consecutive multiples of 5, and all of them are positive integers
To find:
- • The average of all six integers, \(\frac{(A + B+ C + D + E + F)}{6}\)
Analyzing Statement 1:
“A * D = -30”
- • We know that D is a multiple of 5.
- o So, let’s write -30 as product of two integers, such that one of them is a positive multiple of 5.
o 5 * -6 & 15 * -2 are the only two possible ways of doing it
Case 1: A = -2 and D = 15
- • Implies, average of six numbers = \(\frac{(-2 - 4 – 6 + 15 + 20 + 25)}{6} = \frac{48}{6} = 8\)
Case 2: A = -6 and D = 5
- • Implies, average of six numbers = \(\frac{(-6 - 8 – 10 + 5 + 10 + 15)}{6} = \frac{6}{6} = 1\)
Since, we are not able to find a unique answer from this step, Statement 1 is not sufficient.
Analyzing Statement 2:
“|P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F”
- • Implies, \(P = \frac{(A + B + C)}{3}\)
- o Thus, (A + B + C) = 3P
• \(Q = \frac{(D + E + F)}{3}\)
- o Thus, (D + E + F) = 3Q
• So, average of six numbers = (A + B+ C + D + E + F)/6 = (3P + 3Q)/6 = (P + Q)/2 ……………….. (1)
And, we know |P|= |Q| - 2
- • From this, we can say that |P|=-P, since P < 0, and |Q|= Q, since Q > 0.
- o Thus, -P = Q – 2
o Implies, P + Q = 2
• Substituting this value in equation (1), we get,
- o Average of six integers = \(\frac{2}{2} = 1\)
Since, we are getting a unique value from this step, Statement 2 is sufficient.
Hence, the correct answer is Option B.
Answer: B
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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
30 Jun 2019, 07:56
30 Jun 2019, 07:56
Expert Reply
Karmesh wrote:
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
Hi Karmesh,
It seems like you have not analysed statement 2 properly. Please go through our solution to find your mistakes
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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
04 Aug 2019, 01:46
04 Aug 2019, 01:46
Please correct me if i am wrong, but everyone who has provided solution to this question is assuming value of D, E and F to be negative and value of A, B and C to be positive. Sure we are given A*D=-30 , but there is no mention about A, B, C being positive and D,E and F being negative. It is just written that they are three consecutive even integers and three consecutive multiples of 5. Multiples of 5 can be negative value also. So according to st1: A can have values 2, -2, 6 and -6. And D can have 15, -15, 5 and -5.
Please help me with this.
Please help me with this.
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Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
04 Aug 2019, 02:00
04 Aug 2019, 02:00
iamArjuna wrote:
Please correct me if i am wrong, but everyone who has provided solution to this question is assuming value of D, E and F to be negative and value of A, B and C to be positive. Sure we are given A*D=-30 , but there is no mention about A, B, C being positive and D,E and F being negative. It is just written that they are three consecutive even integers and three consecutive multiples of 5. Multiples of 5 can be negative value also. So according to st1: A can have values 2, -2, 6 and -6. And D can have 15, -15, 5 and -5.
Please help me with this.
Please help me with this.
iamArjuna
Its actually the other way around. A,B and C are negative and D, E and F are positive.
We get this from the number line that is given.
In the number line, everything to the left of 0 is less than 0 or negative and everything to the right of 0 is greater than 0 or positive.
Hope this helps!
Posted from my mobile device
_________________
Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
03 Nov 2021, 07:51
03 Nov 2021, 07:51
EgmatQuantExpert wrote:
Solution
Given:
- • A, B, C are three consecutive even integers, and all of them are negative integers
• D, E, F are three consecutive multiples of 5, and all of them are positive integers
To find:
- • The average of all six integers, \(\frac{(A + B+ C + D + E + F)}{6}\)
Analyzing Statement 1:
“A * D = -30”
- • We know that D is a multiple of 5.
- o So, let’s write -30 as product of two integers, such that one of them is a positive multiple of 5.
o 5 * -6 & 15 * -2 are the only two possible ways of doing it
Case 1: A = -2 and D = 15
- • Implies, average of six numbers = \(\frac{(-2 - 4 – 6 + 15 + 20 + 25)}{6} = \frac{48}{6} = 8\)
Case 2: A = -6 and D = 5
- • Implies, average of six numbers = \(\frac{(-6 - 8 – 10 + 5 + 10 + 15)}{6} = \frac{6}{6} = 1\)
Since, we are not able to find a unique answer from this step, Statement 1 is not sufficient.
Analyzing Statement 2:
“|P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F”
- • Implies, \(P = \frac{(A + B + C)}{3}\)
- o Thus, (A + B + C) = 3P
• \(Q = \frac{(D + E + F)}{3}\)
- o Thus, (D + E + F) = 3Q
• So, average of six numbers = (A + B+ C + D + E + F)/6 = (3P + 3Q)/6 = (P + Q)/2 ……………….. (1)
And, we know |P|= |Q| - 2
- • From this, we can say that |P|=-P, since P < 0, and |Q|= Q, since Q > 0.
- o Thus, -P = Q – 2
o Implies, P + Q = 2
• Substituting this value in equation (1), we get,
- o Average of six integers = \(\frac{2}{2} = 1\)
Since, we are getting a unique value from this step, Statement 2 is sufficient.
Hence, the correct answer is Option B.
Answer: B
Hello,
Could you help me understand this
If we open the modulus |P| = |Q| -2; we know P is negative and Q is positive, Example Q is 8: Could the value of P be -6 Only or could it be -10 too based on the way we open the modulus?
I am sorry it may be a silly doubt
Re: Question of the week - 45(If A, B, C are three consecutive even .....)
[#permalink]
03 Nov 2021, 11:56
03 Nov 2021, 11:56
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?
This question uses famous trap of GMAT, as you can get the required solution by equation itself. You don't need to solve it.
Stat1: A * D = -30
A = -2 and D = 15 or, A = -6, D = 5, as we don't know any relation between A and D. Not sufficient
Stat2: |P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F
Since, P <0 so, |P| = -P and As Q >0, |Q| = Q
So, (-A-B-C)/3 = (D+E+F)/3 -2 or, (A+B+C+D+E+F)/6 = 1. Sufficient
So, I think B.
This question uses famous trap of GMAT, as you can get the required solution by equation itself. You don't need to solve it.
Stat1: A * D = -30
A = -2 and D = 15 or, A = -6, D = 5, as we don't know any relation between A and D. Not sufficient
Stat2: |P|= |Q| - 2, where P is the average of A, B and C, and Q is the average of D, E and F
Since, P <0 so, |P| = -P and As Q >0, |Q| = Q
So, (-A-B-C)/3 = (D+E+F)/3 -2 or, (A+B+C+D+E+F)/6 = 1. Sufficient
So, I think B.
Re: Question of the week - 45(If A, B, C are three consecutive even .....)
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30 Jul 2023, 03:14
30 Jul 2023, 03:14
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Re: Question of the week - 45(If A, B, C are three consecutive even .....) [#permalink]
30 Jul 2023, 03:14
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