On the number line

[s_raizada]

Thanks!

[durgesh79]

i think it should be B

statement 2 : distance between t and r is t-r
distance between t and -s is t+s

both are same, t-r = t+s => r = -s

0 is in middle of two points.

[s_raizada]

OA is C

[Ian Stewart]

Note first that we can rephrase the question. It's asking: is r = -s ?

The answer is C. The second statement is not sufficient, because s (and t) might be positive or negative. From statement 2 alone, we have two possibilities:

a) -s is to the right of t (which can happen if s < 0)
b) -s is to the left of t

In case b), r = -s. In case a), however, r < t < -s, so r will not be equal to -s (and if someone wants an example with numbers, note that Statement 2 will be true if t = -1, s = -2, and r = -4, but 0 is not halfway between s and r).

Together the two statements are sufficient, because they guarantee that only case b) above is possible.

[durgesh79]

Thanks for the explaination.

[s_raizada]

Excellent explanation!!!!!!!!!!!!!

[gmat009]

Note first that we can rephrase the question. It's asking: is r = -s ?

The answer is C. The second statement is not sufficient, because s (and t) might be positive or negative. From statement 2 alone, we have two possibilities:

a) -s is to the right of t (which can happen if s < 0)
b) -s is to the left of t

In case b), r = -s. In case a), however, r < t < -s, so r will not be equal to -s (and if someone wants an example with numbers, note that Statement 2 will be true if t = -1, s = -2, and r = -4, but 0 is not halfway between s and r).

Together the two statements are sufficient, because they guarantee that only case b) above is possible.

Can someone plz. explain this. In both a & b cases I am calculating distance as t-r=t-(-s) and regardless of whether -s is to the right or left of t getting equal distance.
I am not understanding this " In case a), however, r < t < -s, so r will not be equal to -s "

[Rashmi1804]

@gmat009: Just in case if you havent already gotten it.....

Consider stmt 2) distance between t and r is equal to distance between t and -s
two cases are possible from this hint:

case i:
if t:r =t:-s and -S IS ON THE LEFT HAND SIDE OF T ==> <------r(-s)---0-----s----t------> i.e, r= -s in which case, 0 lies between r and s.
eg: <--------(-2)---0----(2)-----4--->

case ii
If t:r =t:-s and -S IS ON THE RIGHT HAND SIDE OF T ==>
<---------r-------s---t---0-----(-s)-------> i.e, s , r and t are negative. and so 0 doesnt lie in between r and s.

eg: <-----(r=-4)--------(s=-2)----(t=-1)----0---------(-s=2)---->

even in this case dist between t:r= -1-(-4) = 3 and -s:t =2-(-1) = 3, though r and -s are not equal. So we need stmt:1 to know where S is.

HTH!!

[Rashmi1804]

hey....whats wrong with the sentence ?????

Just in case & if.......redundant ?? :roll:

[sanju09]

hey....whats wrong with the sentence ?????

Just in case & if.......redundant ?? :roll:

Let's transfer it to the VERBAL forum for further discussions...

[ssuarezo]

@gmat009: Just in case if you havent already gotten it.....

HTH!!

Thanks for the explanation
Silvia

[sashish007]

case ii
If t:r =t:-s and -S IS ON THE RIGHT HAND SIDE OF T ==>
<---------r-------s---t---0-----(-s)-------> i.e, s , r and t are negative. and so 0 doesnt lie in between r and s.

eg: <-----(r=-4)--------(s=-2)----(t=-1)----0---------(-s=2)---->

even in this case dist between t:r= -1-(-4) = 3 and -s:t =2-(-1) = 3, though r and -s are not equal. So we need stmt:1 to know where S is.

HTH!!

@Rashmi1804: Can you elaborate on case ii please?

I accounted for
[2] as |t-r| = |t-(-s)| which will yield 2 cases,
.........t-r = t+s | t-r = -(t+s)
...............-r=s | t-r = -t-s
......................| t= (r-s)/2

thanks