The answer is c right?
The question states that the sum of the digits is 11.
Let the digits be a, b and c
i.e a+b+c=11
Statement one says that b=9
So, a+b=2
So, the number can be 290 or 191 Not sufficient
Statement 2 says the c=1
So a+b=10
So there are many options like 821, 281,731,...... Not sufficient
Combining both we get
191 as the answer
hence C is the answer
hope this helps
Whats the OA?

Hey,

OA is different...it's A. I made the same mistake myself going for C.

Although, there is a typo in the OA, this is how I interepreted it.

Since m is a 3-digit positive number, its 3 digits can be labeled d1, for the hundreds place, d2 for the tens place, and d3 for the digit in the ones place. Additionally, it is known that d1 + d2 + d3 = 11, and that d3 can only be 1, 3, 5, 7, or 9, since m is an odd number.

(1) if d1 is 9, then d2 + d3 = 2. Since d3 must be odd, the only combination of d2 and d3 that satisfies this condition is d2=1 and d3=1. so m is 911; suff.

(2) if d2=1, then d1 + d3 = 10. since d3 must be odd, d3 could be 1,3,5, or 7. since there can be a few different possibilities for d3, there is not enough info to identify a single specific value for d1, d3, or thus m; not suff.

Robinmrtha wrote:The answer is c right?
The question states that the sum of the digits is 11.
Let the digits be a, b and c
i.e a+b+c=11
Statement one says that b=9

(1) says that the hundreds digit is 9, not the tens digit.

Using your notation, this means that a=9, not b=9.

So, our number is:

9bc.

We know that c is odd, so the smallest possible value of c is 1. Since the digits sum to 11, the only possibility is 911: sufficient.

The answer is actually A. Not C. Here's how: let the number be abc where a is the hundreds digit, b is the tens digit, and c is the units digit. Then,

a + b + c = 11

and

c = 1, 3, 5, 7, 9.

Now, 1) tells us that a = 9, so b + c = 2. But since both b and c are positive, c has to be 1. So, this leaves b = 1, with m = 911.

We can't do the same for 2). So, the answer is A.