Data Sufficiency

Given that R is positive three-digit integer, what is the hundreds digit of R?

1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.

[spoiler]OA : C[/spoiler]

Source : 800score.com

Statement 1:

Hundreds digit of 3R is 8.

THink of it this way 3*(some 3 digit numbers) = X8XX

Possible Cases;

3R can be between 801-897 (R between 267-299

(or) 3R can be between 1800 - 1899 (R between 600 - 633)

(or) 3R can be between 2802-2898 ( R between 934 - 966)

Hundreds place can be 2 or 6 or 9

Statement 2:

R+1 results in a number with hundreds digit 9.

The number can be 899 -> R+1 = 900

(or) 901 -> R+1 = 902 and so on.

the hundreds digit can be 8 (or) 9

Combined, we have only cases of 934-966 possible. So hundreds digit will be 9. C IMO

Given that R is positive three-digit integer, what is the hundreds digit of R?
Let R = ABC => R = 100A + 10B + C

1. The hundreds digit of 3R is 8

If the hundreds digit of 3R is 8, R can be 266<R<300 or 599<R<634 or 933<R<967 and the hundreds digit of R can be 2 or 5 or 9

2. (R+1) results in a number with the hundreds digit of 9.

R = 899 or R = 9xx i.e. The hundreds digit of R can be 8 or 9

From 1 and 2

The hundreds digit is 9, Sufficient!

shankar.ashwin wrote:Statement 1:

Hundreds digit of 3R is 8.

THink of it this way 3*(some 3 digit numbers) = X8XX

Possible Cases;

3R can be between 801-897 (R between 267-299

(or) 3R can be between 1800 - 1899 (R between 600 - 633)

(or) 3R can be between 2802-2898 ( R between 934 - 966)
Hundreds place can be 2 or 6 or 9

Statement 2:

R+1 results in a number with hundreds digit 9.

The number can be 899 -> R+1 = 900

(or) 901 -> R+1 = 902 and so on.

the hundreds digit can be 8 (or) 9

Combined, we have only cases of 934-966 possible. So hundreds digit will be 9. C IMO

How did you so quickly find out the ranges? It took quite sometime for me to get that.

gmatIntent wrote:
How did you so quickly find out the ranges? It took quite sometime for me to get that.

I am guessing you did it the same way I did, you're given R is a 3 digit number.

If R=1000, 3R = 3000.

So we are looking at options < 3000

Possible cases, 800-899 | 1800-1899 | 2800-2899

After this, we just need to divide the range by 3, since R is an integer chose 3R such that its a multiple of 3 (801 and not 800 and so on)
This is one of those questions in which you look at options and know its a C/E question, so immediately start using both statements together you can do it faster.

shankar.ashwin wrote:

gmatIntent wrote:
How did you so quickly find out the ranges? It took quite sometime for me to get that.

I am guessing you did it the same way I did, you're given R is a 3 digit number.

If R=1000, 3R = 3000.

So we are looking at options < 3000

Possible cases, 800-899 | 1800-1899 | 2800-2899

After this, we just need to divide the range by 3, since R is an integer chose 3R such that its a multiple of 3 (801 and not 800 and so on)
This is one of those questions in which you look at options and know its a C/E question, so immediately start using both statements together you can do it faster.

Thanks