1) The tens digit of r/10 is 3

r=320 r/10=32
r=300 r/10=30
Both r/10 have a tens digit of 3
But the two r have different tens digit: 2 and 0

Insufficient

2) The hundreds digit of 10r is 6
When an integer is multiplyed by 10 the original tens digit becomes its hundreds digit.
Sufficient.

California4jx wrote:any exaplantion .. got this right but not very clear - Thanks in advance !

I go with B.

Statement 1. r/10 = 3 doesnt tell us anything abt whar r can be ...hence insufficient.

statement 2. 10r= 6 .. which is sufficient..

if u think carefully u wll realise that any integer if u multiply by 10.. for that matter any powers of 10( 10,100,1000) the digits will move one place to the left.. for example

63.. here units place is 3 and tens place is 6 .. when mulitplied by 10 becomes 630. here 3 has moved one place to the left and has taken tens place and 6 has taken hundredths place...

when statement says"The hundreds digit of 10r is 6" we know for sure that in the original value of r 6 would have been in the tens place.

which is the Answer.

Hope it helps.. do let me know if u have any doubts..

Hi california4jx,

I dont have a short answer , just a analysis:

the question says r - a positive integer

now option 1 says: r/10 which could be 10^n (a) + 10^(n-1)(b)......m(10) +x

where a,n,m x are all + integers

option one says m=3 , and also that r is a factor of 10 which means the units digit in r=0 and the tens digit is x , which we dont know...Insufficient

Option 2 on the other hand says :

if r= 10^n........m(100)+n(10)+x
therefore 10r=10^(n+1) ........m(1000)+n(100)+10(x)+ 0(units place)

the hundreds place here is n =6 , which is the tens digit of r, Sufficient

therefore B

Hope it makes sense !

terrific ! - thanks very much - you guys rock !

Similar to OG 12 Problem 167