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.
On the number line
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Ian Stewart
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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.
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.
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gmat009
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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.Ian Stewart wrote: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.
I am not understanding this " In case a), however, r < t < -s, so r will not be equal to -s "
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Rashmi1804
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@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!!
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!!
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Rashmi1804
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Let's transfer it to the VERBAL forum for further discussions...Rashmi1804 wrote:hey....whats wrong with the sentence ?????
Just in case & if.......redundant ?? :roll:
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sashish007
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@Rashmi1804: Can you elaborate on case ii please?Rashmi1804 wrote: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!!
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
Ashish
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