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10 Jan 2017, 04:03
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:19) correct
37%
(01:12)
wrong
based on 212
sessions
History
Date
Time
Result
Not Attempted Yet
Let a and b be digits. What is the sum b.aaa + a.bbb?
(1) a + b = 10
(2) a = b
(1) a + b = 10
(2) a = b
10 Jan 2017, 06:34
Kudos
Bookmarks
Bunuel
\(b.aaa + a.bbb = \frac{baaa}{1000} + \frac{abbb}{1000} = \frac{1000b + 100a + 10 a + a + 1000a + 100b + 10b + b}{1000} = \\
\\
= a+b + \frac{a + b}{10} + \frac{a + b}{100} + \frac{a + b}{1000}\)
(1) a + b = 10
10 + 1 + 0.1 + 0.01 = 11.11 Sufficient.
(2) a = b
\(2a + \frac{2a}{10} + \frac{2a}{100} + \frac{2a}{1000}\). No information about a or b (1.111 + 1.111 or 2.222 + 2.222 etc.). Insufficient.
Answer A
General Discussion
11 Jan 2017, 00:20
Kudos
Bookmarks
Bunuel
Digits are from 0 to 9.
As per statement 1:
a + b = 10
various combinations are 1+9, 2+8, 3+6 etc.
So if a = 1 and b = 9 we have 1.999 + 9.111 = 11.11
a = 2 and b = 8 we have 2.888 + 8.222 = 11.11 etc (various combinations gives the same value . Sufficient
As per statement 2:
a = b
There may various combinations that will yield different sums of a & b
1.111 + 1.111 = 2.222
2.222 + 2.222 = 4.444
Not Sufficient.
(A)
