I solved it this way -
assume r = 100x + 10y + z ....equation 1
The question is asking is you can find y.

From hint 1 & eqn. 1,
r/10 = 10x + y +(maybe some fraction). All the hint gives you here is the ten's place of r/10 which gives you x, not y.

From hint 2 & eqn.1,
10r = 1000x + 100y + 10z. -- which can otherwise be written as the number xyz0. The hint gives you the hundred's place of 10r , which happens to be y, which is what you want.

So, the answer is B.

I should admit that I couldn't crack it in under 2 minutes, but now I discovered how I can, for the next time!

no, plugging the number to finish this in 1 minute

Thank you very much!!!

metalhead wrote:I solved it this way -
assume r = 100x + 10y + z ....equation 1
The question is asking is you can find y.

From hint 1 & eqn. 1,
r/10 = 10x + y +(maybe some fraction). All the hint gives you here is the ten's place of r/10 which gives you x, not y.

From hint 2 & eqn.1,
10r = 1000x + 100y + 10z. -- which can otherwise be written as the number xyz0. The hint gives you the hundred's place of 10r , which happens to be y, which is what you want.

So, the answer is B.

I should admit that I couldn't crack it in under 2 minutes, but now I discovered how I can, for the next time!

Could you please explain in detail

adorendra wrote:What is the tens digit of positive intiger r?
1. The tens digit of r/10= 3
2. The hundreds digit of 10r = 6


OA B

let r=300
s1) r/10=30 tenth dig is 3----> tenth dig of (300) is 0
let r=340
s10 R/10=34 TENNTH DIG IS 3---------> tenth dig of r(340) is 4
hence insuff

s2) let R=60
10R=600......tenth dig of r is 6

r=660
10* r=6600 (100th place is 6 satisfying s2 and 10th dig of r is 6)

let r=6660(100th dig is 6 for 10r.......and 10th dig is 6)

logic is since multiplication s with 10 , the 10th place digit of original no. will be same as 100th place dig no. of 10r
which for this case is 6
hence suff

B is the ans