N=abc, N is a three digit number with hundreds digit a, tens digit b

\(N = \left\langle {abc} \right\rangle \,\,\,\,\,\left( {a \ne 0} \right)\)

\(a + b + c = 20\,\,\,\left( * \right)\,\,\,\)

\(? = b\)

\(\left( 1 \right)\,\,ac = 18\,\,\,\,\mathop \Rightarrow \limits^{{\rm{all}}\,\,{\rm{possibilities}}} \,\,\,\left\{ \matrix{\\
\,\left( {a,b,c} \right) = \left( {2,9,9} \right)\,\,\, \Rightarrow \,\,\,? = 9 \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {9,9,2} \right)\,\,\, \Rightarrow \,\,\,? = 9 \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {3,11,6} \right)\,\,\,{\rm{impossible}} \hfill \cr \\
\,\left( {a,b,c} \right) = \left( {6,11,3} \right)\,\,\,{\rm{impossible}} \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 9\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)

\(\left( 2 \right)\,\,a > b = c\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,a\,\,{\rm{even}}\,\,{\rm{ > }}\,\,\,{\rm{6}}\,\,\,\left( {a \le 6\,\,\,\, \Rightarrow \,\,\,b = c \ge 7} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {a,b,c} \right) = \left( {8,6,6} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)


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.
_________________

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

From 2 options we are getting two different values for B..is it right approach?

I have made the following observation in my previous post:

Obs.: in official questions, when the correct alternative choice is (D), we expect the unique answers (obtained in each statement alone) to be equal.

Reason for the observation: it was not the case here.

Regards,
Fabio.
_________________

Give 100a+10b+c = 20 , to find 10b ?

Statement 1 : (a)*(c)=18, hence none of them is zero.
Possible Values :
2*9 = 18 ---> B can be 9 since a + b + c = 20
9*2 = 18---> B can be 9 since a + b + c = 20

Ruled out options : - (Since range of values for any figure can lie between 0 and 9 )
3*6 = 18---> B can be 11 since a + b + c = 20
6*3 = 18---> B can be 11 since a + b + c = 20

Hence statement 1 is alone sufficient. (B=9)

Statement 2: -
Give B/C =1 implies B=C additionally, its given b<a
From the above information we can infer that,
possible values of A , B , C can be
A + B + C =20
9 5.5 5.5
8 6 6
7 6.5 6.5
6 7 7
abc have to be an integer and given that a > b
Hence only possible option ---> a = 8 and B & C 6 .
Alone Sufficient

Hence, answer "D"