IMO B
s1) x+y=9
bth can have any values
s2) 52x+58y=492
only one set of value satisfy this condition
i.e x=5 and y=4
hence suff....
source pls kindly give the name of book .how many DS questions are there in the book and what is difficulty level

It is from the Veritas Prep Data Sufficiency workbook. Over a hundred problems.

How do you quickly determine:

s2) 52x+58y=492
only one set of value satisfy this condition
i.e x=5 and y=4

This is where I struggled.

this was something called recognizing the pattern....
in my first attempt when i faced another version of same concept i went for C
but then when i understood the logic i made it a point in my flash card, which goes something like this
" in DS having two variable just find out whether they are int or not (this will either stated explicitly or implied as in this case) , typical statements will be (s1)x+y=some number and s2) Ax+by=$some value; since x and y are things such as some fruit ,items like pen book ....etc and they can not have a fractional value then just go for s2) in order to find the combintaion use s1 for limiting the values as here x+y will be 9 only and becoz x and y are int we have (1,8);(2,7);(3,6);(4,5)(5,4)(6,3)(7,2)(9,1)
using all these value in 2nd eqn you will come to know that only one set satisfy......

these are my findings

HTH

thephoenix wrote:this was something called recognizing the pattern....
in my first attempt when i faced another version of same concept i went for C
but then when i understood the logic i made it a point in my flash card, which goes something like this
" in DS having two variable just find out whether they are int or not (this will either stated explicitly or implied as in this case) , typical statements will be (s1)x+y=some number and s2) Ax+by=$some value; since x and y are things such as some fruit ,items like pen book ....etc and they can not have a fractional value then just go for s2) in order to find the combintaion use s1 for limiting the values as here x+y will be 9 only and becoz x and y are int we have (1,8);(2,7);(3,6);(4,5)(5,4)(6,3)(7,2)(9,1)
using all these value in 2nd eqn you will come to know that only one set satisfy......

these are my findings

HTH

Man, I went for C & now I realised, B is sufficient. But we are so used 2 use 2 variable equations& tend to get "addicted" to that!

gmatmachoman wrote: Man, I went for C & now I realised, B is sufficient. But we are so used 2 use 2 variable equations& tend to get "addicted" to that!

Gmat is a game of recognizing the pattern , if we devlop this quality then half of battle is won without any extra effort

A much better explination than the book, thank you!!!