Question of the week - 45(If A, B, C are three consecutive even .....)

Question

e-GMAT          Question of the Week - 45

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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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Karmesh · Answer

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

firas92 · Answer

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!

IanStewart · Answer

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.

Archit3110 · Answer

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

MSIMBA · Answer

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

EgmatQuantExpert · Answer

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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EgmatQuantExpert · Answer

Hi  Karmesh ,

It seems like you have not analysed statement 2 properly. Please go through our  solution  to find your mistakes

iamArjuna · Answer

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.

firas92 · Answer

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

BAA5 · Answer

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

TarunKumar1234 · Answer

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.

bumpbot · Answer

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Question of the week - 45(If A, B, C are three consecutive even .....)

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