X, Y and Z are three positive numbers. Further, X = 0.AAAAAA...., Y =

Question

X, Y and Z are three positive numbers. Further, X = 0.AAAAAA...., Y = 0.ABABABAB...., and X + Y = Z/33. Which of the following is one possible value of B?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 7

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GMATinsight · Answer

X = 0.AAAAAA.... = \frac{A}{9}

Y = 0.ABABABAB.... = \frac{AB}{99}

X + Y = \frac{A}{9} + \frac{AB}{99} = \frac{11A+AB}{99} = \frac{Z}{33}

ALso, AB = 10A + B and A, B and Z are single Digit numbers

i.e.  11A+(10A + B) = 3Z

i.e.  B = 3Z - 21A 
if A = 1, then Z could be 7 and B will be 0
if A = 1, then Z could be 8 and  B will be 3 
if A = 1, then Z could be 9 and B will be 6

Answer: Option B

BPranav7 · Answer

Bunuel : 
I worked backwards to solve this.. 3/33=1/11=0.090909.. 
I'm not sure it is the correct conceptual approach... A little help here please!!

arushi118 · Answer

10X=A.AAA....
X=0.AAAA
Subtract these 2 eqns to get: 9X = A -> X = A/9

Similarly,
100Y=AB.ABABABAB....
Y=0.ABABAB...
99Y = AB -> which is actually 10A + B

Now substitute value of X and Y in X + Y = Z/33:
you will get:
(21A+B)/99 = Z/33
B = 3(Z-7A) -> So B can be a multiple of 3.

Hence, answer is B.

X, Y and Z are three positive numbers. Further, X = 0.AAAAAA...., Y =

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X, Y and Z are three positive numbers. Further, X = 0.AAAAAA...., Y =

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