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

(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.

Solution

• 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

• The average of all six integers, \(\frac{(A + B+ C + D + E + F)}{6}\)

• 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

• Implies, average of six numbers = \(\frac{(-2 - 4 – 6 + 15 + 20 + 25)}{6} = \frac{48}{6} = 8\)

• Implies, average of six numbers = \(\frac{(-6 - 8 – 10 + 5 + 10 + 15)}{6} = \frac{6}{6} = 1\)

• 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)

• 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 = \(\frac{2}{2} = 1\)