HPengineer wrote:If Carmen had 12 more tapes she would have twice as many tapes as Rafael. Does Carmen have fewer tapes then Rafael?

S1. Rafael has more then 5 tapes
S2. Carmen has fewer than 12 tapes



I got this right however i think it was by accident. Any one care to share their approach with me and so we can compare?

Take C + 12 = 2 R from stem. C will always be EVEN as C and R are whole numbers. Then try answering if C < R.

S1. If R > 5, then C + 12 is at least 12 as C and R are whole numbers. When R is 6, C is 0, making C < R as true, but when R is 12, C = R. So insufficient

S2. If C < 12, then it can at most be 10 to ensure C and R are whole numbers. When C is 10, R is 11, making C < R as true, which remains as true for all smaller possibilities for C. [spoiler]So sufficient

B[/spoiler]

also got B
the number of tapes Carmen-c
the number of tapes Rafael-r
we are given
c+12=2r
r=c/2+6
and asked if r>c, or
(c/2+6)-c>0
c+12-2c>0
is c<12
(1) insuffiicent, here to answer definitely we need to prove that r<12, but given that r>5
(2) c<12 suffiicent

Ok heres how i solved it... Im wondering if its dangerous to solve similar probems with out inequalities...

S1. Rafael has more then 5 tapes

C + 12 = 2R so plug 6 in for R and solve for C which = 0 so less tapes then Rafael

Now try a larger number say 20 for R and you will get C = 38 which is more tapes then Rafaeal therefore S1 is Insuff.



S2. Carmen has fewer than 12 tapes

Here i simply plugged in 11 for C and solved for R

11+12= 2R or 23/2 = R so R = 11.5 which is greater then C so statement is sufficient.

What troubled me here is that i did not get a integer for R... How can you have half a tape? Shouldnt this be a hidden constraint problem since were dealing with items?

@HPengineer:

i solved it in the same way......