N = abc, N with hundreds digit a, tens digit b and units digit c. If a

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N = abc, N 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

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GMATinsight · Answer

Given: a + b + c = 20

Question : what is the value of b?

Statement 1 : a*c = 18

since a, b c are all single digit positive integer (Being digits of number N)
a and c could be {2, 9} or {3, 6}

If {a, c} = {2, 9} then b = 20-(2+9) = 9 POSSIBLE
If {a, c} = {3, 6} then b = 20-(3+6) = 11 NOT POSSIBLE (cause b is a single digit number)

i.e. b = 9
SUFFICIENT

Statement 2 : b/c = 1 and b > a 
b = c > a and a+b+c=20
Case 1: b = c = 6 and a = 8 NOT POSSIBLE
Case 2: b = c = 7 and a = 6 POSSIBLE (cause 7+7+6 = 20)
Case 3: b = c = 8 and a = 4 POSSIBLE (cause 8+8+4 = 20)
Case 4: b = c = 9 and a = 2 POSSIBLE

Different possible values of b hence

NOT SUFFICIENT

Answer: Option A

rajatchopra1994 · Answer

Statement 1: The product of digits a and c is 18.
a*c = 18; as 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.
An only possible solution is 9,2 for a and c
a+b+c = 20
9+2+b = 30
b = 9
Sufficient

Statement 2: b/c=1 and b<a
c can’t be zero so we can safely divide by c
.: b= c and b> a , c > a
we can have 4+8+8 =20 ,here b=8
Also , 6+7+7 =20 ,here b=7
Not sufficient

IMO-A

freedom128 · Answer

a+b+c=20 and a,b,c are between 0 to 9

(1) a.c = 18 
We rule out (a,b, c)=(3,11,6),(1,1,18). The only possible (a,b,c) is (2,9,9) and b=9
SUFFICIENT

(2) b=c and b>a
Possible (a,b,c) is (6,7,7), (4,8,8), (2,9,9). We don't know which one. 
NOT SUFFICIENT

FINAL ANSWER IS (A)

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Staphyk · Answer

N = abc, N with hundreds digit a, tens digit b and units digit c. If a + b + c = 20, what is the value of b?
Constraints : a+b+c=20
Maximum digits for (a,b,c) = 9
Value of b= ?

(1)The product of digits a and c is 18. 
a•c = 18 
We have four possibilities (3•6),(6•3),(2•9),and (9•2)
However (3•6) and (6•3) is ruled out since digit b can’t be 11 ,so a=2/9 and c=9/2

Now, a+b+c = 2+9+9 = 20 ,b=9
Again, a+b+c =9+9+2 =20, b=9
(Sufficient)

(2) b/c = 1 and b > a 
Here since amax. and bmax.is 9 
c can’t be zero so we can safely divide by c
.: b= c and b> a , c > a
we can have 4+8+8 =20 ,here b=8
Also , 6+7+7 =20 ,here b=7
(Not sufficient)

Hit that A

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Archit3110 · Answer

a+b+c = 20 
possible ; 299, 488 , 389, 578,677,776,866, so on....

#1The product of digits a and c is 18.
possible 2*9 ; 18 ; value of b would be 9 only
sufficient
#2
b/c = 1 and b > a
299,677, insufficient
IMO A ;

N = abc, N 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

eakabuah · Answer

We are to find digits a,b, and c such that a+b+c=20.
Note that we are not given that a,b, and c are distinct.

Statement 1: the product of digits a and c is 18
Sufficient. Because the possible combinations are 2*9=18 and 3*6=18.
But 3+6=9, implying we a and c can never be 3 and 6 since there is no digit that can be added to 9 to equal 20.
However, 2+9=9, implying the third digit is also 9. Hence we can conclusively say that b=9.

Statement 2: b/c=1 and b>a
Statement says that b=c and that b>a
We can have more than one value of b=c and b>a whereby a+b+c=20
For instance b=c=9, this implies a=2 and the condition of statement 2 is satisfied. In addition, b=c=8, and a=4 satisfies statement 2 and a+b+c= 4+8+8=20.

Statement 2 is therefore not sufficient.

The answer is A.

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exc4libur · Answer

A is a digit from 1 to 9
B,C are digits from 0 to 9

a+b+c=20

(1) The product of digits a and c is 18.   sufic

ac=18 (both are digits): f(18)=18-1, 9-2, 6-3.
a,c=9,2; b=20-9-2=9
a,c=6,3; b=20-6-3=20-9=11>9=invalid

(2) b/c = 1 and b > a   insufic

b/c=1, b=c, b>a; 2b+a=20, b≥6,7,8,9, a=8,6,4,2

Ans (A)

shameekv1989 · Answer

N = abc, N 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

Given - a+b+c = 20

from 1) product of a and c is 18 -> only possible values of a,c pair can be 2,9 or 3,6

Now let's check for these numbers by plugging in the given statement (1) 
-> 2+b+9 = 20 -> b = 9 -> valid
-> 3+b+6 = 20 -> b = 11 -> Not valid

Hence sufficient

from 2) b=c and b > a -> check for values such that a + b + c = 20, b=c and b>a -> (4 + 8 + 8 = 20) and (2 + 9 + 9 = 20) -> 2 values of b -  hence not sufficient

Correct answer - A

unraveled · Answer

N = abc, N 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.
ac = 18 
If a = 2, c = 9 then b = 9
No other values of a and c satisfy the condition.

SUFFICIENT.

(2) b/c = 1 and b > a
 \frac{7}{7}  = 1 then a = 6

\frac{8}{8}  = 1 then a = 4

INSUFFICIENT.

Answer A.

lacktutor · Answer

N = abc, N with hundreds digit a, tens digit b and units digit c. If a + b + c = 20, what is the value of b?
--> a,b and c are the positive integers from 1 to 9, inclusive. ( None of them can be zero)

( Statement1 ): The product of digits a and c is 18. (a*c=18)

--> 2*9=18 --> 2+9+b= 20 --> b=9
--> 3*6=18--> 3+6+ b=20 --> b≠ 11 (b must be any positive integer from 1 to 9.)
Sufficient

( Statement2 ):  \frac{b}{c }= 1  and b > a
--> b=c and b > a

case1: b=c=8 --> a=4 --> 4+8+8=20 --> b=8
case2: b=c=9--> a =2 --> 2+9+9= 20 --> b =9
Insufficient

The answer is A.

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