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21 Mar 2017, 04:49
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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:
95%
(hard)
Question Stats:
50% (02:41) correct
50%
(02:36)
wrong
based on 356
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If a, b, and c represent distinct digits, what is the value of the two-digit number ab?
(1) ab*ba = aca
(2) a*b < 10
(1) ab*ba = aca
(2) a*b < 10
11 Jan 2018, 12:29
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Statement 1:
\(ab\) * \(ba\) = \(aca\)
=> \((10a+b)(10b+a)\) = \(100a + 10c + a\)
=> \(100ab + 10a^2 + 10b^2 + ab\) = \(100a + 10c + a\)
=> \(100ab + (a^2 + b^2)10 + ab\) = \(100a + 10c + a\)
now comparing units digit on both the sides
\(ab\)= a => \(a= 0\) OR \(b = 1\) => but a cannot be 0, if it were, then aca would be 2-digit number => b = 1
now comparing tens digit on both the sides
\(a^2 + b^2 = c\)
\(a^2 = c - b^2\)
\(a^2 = c - 1\)
\(a\) = \(\sqrt{c - 1}\)
only for c = 1, 2, 5, \(\sqrt{c - 1}\) yields integer
if c = 1, a = 0 (which is not possible)
if c = 2, a = 1 (but b is already 1) => not possible
so only c = 5, a = 2, possible
so number ab = 21 => sufficient
Statement 2: a * b < 10
lot of combinations ab = {14, 21, 24} not sufficient
(A)
\(ab\) * \(ba\) = \(aca\)
=> \((10a+b)(10b+a)\) = \(100a + 10c + a\)
=> \(100ab + 10a^2 + 10b^2 + ab\) = \(100a + 10c + a\)
=> \(100ab + (a^2 + b^2)10 + ab\) = \(100a + 10c + a\)
now comparing units digit on both the sides
\(ab\)= a => \(a= 0\) OR \(b = 1\) => but a cannot be 0, if it were, then aca would be 2-digit number => b = 1
now comparing tens digit on both the sides
\(a^2 + b^2 = c\)
\(a^2 = c - b^2\)
\(a^2 = c - 1\)
\(a\) = \(\sqrt{c - 1}\)
only for c = 1, 2, 5, \(\sqrt{c - 1}\) yields integer
if c = 1, a = 0 (which is not possible)
if c = 2, a = 1 (but b is already 1) => not possible
so only c = 5, a = 2, possible
so number ab = 21 => sufficient
Statement 2: a * b < 10
lot of combinations ab = {14, 21, 24} not sufficient
(A)
General Discussion
21 Mar 2017, 11:49
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Bunuel
(1) b^2=c
if b =-2 , c= 4
if b= -3 , c=9
a could be 5 or 6
not suff
(2)again same as (1)
insuff
combining same as above
insuff
Ans E
