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akurathi12
GMAT 1: 490 Q47 V13
Posts: 111
06 Jan 2019, 04:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
42% (02:28) correct
58%
(02:41)
wrong
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N=abc, N is a three digit number with hundreds digit a, tens digit b and units digit c. If a+b+c=20, what is the value of b?
(1) The product of digits a and c is 18.
(2) b/c=1 and b<a
(1) The product of digits a and c is 18.
(2) b/c=1 and b<a
06 Jan 2019, 06:59
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akurathi12
\(N = \left\langle {abc} \right\rangle \,\,\,\,\,\left( {a \ne 0} \right)\)
\(a + b + c = 20\,\,\,\left( * \right)\,\,\,\)
\(? = b\)
\(\left( 1 \right)\,\,ac = 18\,\,\,\,\mathop \Rightarrow \limits^{{\rm{all}}\,\,{\rm{possibilities}}} \,\,\,\left\{ \matrix{\\
\,\left( {a,b,c} \right) = \left( {2,9,9} \right)\,\,\, \Rightarrow \,\,\,? = 9 \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {9,9,2} \right)\,\,\, \Rightarrow \,\,\,? = 9 \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {3,11,6} \right)\,\,\,{\rm{impossible}} \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {6,11,3} \right)\,\,\,{\rm{impossible}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 9\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)
\(\left( 2 \right)\,\,a > b = c\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,a\,\,{\rm{even}}\,\,{\rm{ > }}\,\,\,{\rm{6}}\,\,\,\left( {a \le 6\,\,\,\, \Rightarrow \,\,\,b = c \ge 7} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {a,b,c} \right) = \left( {8,6,6} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)
The correct answer is therefore (D).
Obs.: in official questions, when the correct alternative choice is (D), we expect the unique answers (obtained in each statement alone) to be equal.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
06 Jan 2019, 07:03
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N=abc, N is a three digit number with hundreds digit a, tens digit b and units digit c. If a+b+c=20, what is the value of b?
(1) The product of digits a and c is 18.
a*c = 18
a,b,c are single digits
18 = 6*3 OR 9*2
if a = 6 and c= 3 then from a+b+c = 20, b = 11, which is not a single digit
So (6,3) is not possible
only possible solution is 9,2 for a,c
We are least bothered about the value of a, c as required value is b
a+b+c = 20
9+2+b = 30
b = 9
Sufficient
(2) b/c=1 and b<a
Since b,c are equal, their sum will be even (even if b=c = odd OR even)
As Even + even = even
odd + odd = even
possibilities of a is only even, as 2*a + b = 20 an even number
Since 10 is 2 digit number, and b < a, lets consider a =8, let b = c = x
8 + 2(x) = 20
b = c = x = 6
Next a = 6 (only even's possible from above explanation)
6 + 2(x) = 20
b= c = x = 7
b > a which is a contradiction, So a can only take 8, and in turn b,c can only be 6
So, b = 6
Sufficient
Option D is correct
(1) The product of digits a and c is 18.
a*c = 18
a,b,c are single digits
18 = 6*3 OR 9*2
if a = 6 and c= 3 then from a+b+c = 20, b = 11, which is not a single digit
So (6,3) is not possible
only possible solution is 9,2 for a,c
We are least bothered about the value of a, c as required value is b
a+b+c = 20
9+2+b = 30
b = 9
Sufficient
(2) b/c=1 and b<a
Since b,c are equal, their sum will be even (even if b=c = odd OR even)
As Even + even = even
odd + odd = even
possibilities of a is only even, as 2*a + b = 20 an even number
Since 10 is 2 digit number, and b < a, lets consider a =8, let b = c = x
8 + 2(x) = 20
b = c = x = 6
Next a = 6 (only even's possible from above explanation)
6 + 2(x) = 20
b= c = x = 7
b > a which is a contradiction, So a can only take 8, and in turn b,c can only be 6
So, b = 6
Sufficient
Option D is correct
