inequalities with mod ds

IMO E

1)| st | = t
t>o
s=1
says nothing else. t may or may not be <s

2)| st | = 7
if st is positive
t=7/s, again s may or may not be >t
ex: t=50, s=7/50
t=1/2,s=14
when st is negative
-st=7
t<0
not sufficient
combined

t>0
s=1

t may be>1
may be<1

not sufficient

would go with C

stment 1 and 2 are not sufficient,

together, s=1, t>0 from stmnt 1
from stmnt 2 : | st | = 7
=> |t|=7 , t>0
=>t=7

=> s(1) is always less than t(7)

I will go with A.

ern5231 wrote:Given that S>0. Is S>T?
(1) | st | = t
(2) | st | = 7

from 1

t is 0 and s anything or s = 1 (s is +ve given ) and t is +ve or -ve

from 2
s,t could be anything

both

t = 7 or -7thus s = 1...insuff

E

so what is the answer finally ??

is it C or E?

It's C

When both statements are combined, T MUST BE POSITIVE. The left side of the equation is an absolute value, which is positive. Therefore, T unmodified must be positive.

With |ST| = T, we know that S is 1 b/c the passage already states that S > 0. Therefore, S cannot be -1.

We know that T unmodified is +7 because the absolute value on the left side is positive.

We now know that T = 7 and S = 1, so S is not greater than T.

Even I feel it should be C but it's surprising that the Answer given is A! Can any of the instructors help us out?

ern5231 wrote:Even I feel it should be C but it's surprising that the Answer given is A! Can any of the instructors help us out?

The answer cannot be A simply because here are 2 conflicting solutions that would satisfy (1)
Either you missed something in conditions / problem formulation or the OA is wrong.

1) S=1, T=7 hence S<T
2) S=1, T=0.5, hence S>T

both satisfy |ST|=T

imo OA is correct

as with statement 1 we can find only one possible value as S=1 and T=0 in order to consider it true which implies==> s>t and hence suff

Statement 2
value of s can be 7/2,7/3,7/4
and corresponding value of t can be 2,3,4

and condition S>t is not proved and hence not suff

yes, i agree that A is the answer.

For example,
if s=1 t=7, s is not greater than t
if s=1 t=1, s is not greater than t

I cannot find any value of s and t that satisfies the condition 1, and gives s > t.

And, second condition is not sufficient as :
if s=1 t=7, then s is not greater than t
if s=1 t=-7, then s is greater than t

xcusemeplz2009 wrote:imo OA is correct
as with statement 1 we can find only one possible value as S=1 and T=0 in order to consider it true which implies==> s>t and hence suff

AND

acenikk wrote:I cannot find any value of s and t that satisfies the condition 1, and gives s > t

s=1, t=7 ??
s=1, t=0.565 ??

acenikk and xcusemeplz2009:

if S=1 and T=0.5 then | 1x0.5 | =0.5 and S>T
if S=1 and T=1 then | 1x1 | =1 and S is NOT > T

so how is A the answer?

ok seriously...some of the explainations here are truly puzzling!!

Stmt 1 says:

|st|=t

If: s=1 (Since s>0) and t = 1, it satisfies. And this means: S=T and therefore: S not greater than t

Hence: 1 alone not sufficient

Stmt 2 says:

|st|=7

Here: S>0 so t can be +ve or -ve. Hence not enough

Combining:

t=7

So answer is C

right???

Given that S>0. Is S>T?
(1) | st | = t
t is obviously not negative. We know that s is positive, hence we can drop absolute value. We have s*t = t <=> {t = 0 AND s >0} {t > 0 AND s = 1}

INSUFF

(2) | st | = 7
1 * 7 = |1 * -7| = 7

INSUFF

Combine both. t cannot be 0, hence t > 0 and s = 1 (from A). s = 1 -> t = 7

Answer is C