Bunuel wrote:
If q, s, and t are all different numbers, is q < s < t ?
(1) t - q = |t - s| + |s - q|
Notice that the right hand side is positive (it's the sum of two absolute values, so two non-negative values, in fact, in our case two positive values, since we know that the variables are distinct). Thus the left hand side must also be positive, which means that t > q. So, we can have 3 cases for s:
a. ---s---q-------t-------
In this case \(s < q < t\):
\(t - s > 0\) and \(s - q < 0\), which would mean that \(|t - s| = t -s\) and \(|s - q| = -(s - q)\) (recall that |x| = x when x > 0 and x = -x when x <= 0).
So, \(|t - s| + |s - q| = (t -s) - (s - q) = t - 2s + q\).
So, in his case we'd have \(t - q = t - 2s + q\) or \(q=s\). But we are told that q, s, and t are all different numbers, so this case is out.
b. -------q---s---t-------
In this case \(q < s < t\):
\(t - s > 0\) and \(s - q > 0\), which would mean that \(|t - s| = t -s\) and \(|s - q| = s - q\). So, \(|t - s| + |s - q| = (t -s) + (s - q) = t - q\).
This matches the info given in the statement.
c. -------q-------t---s---
In this case \(q < t < s\):
\(t - s < 0\) and \(s - q > 0\), which would mean that \(|t - s| = -(t -s)\) and \(|s - q| = s - q\). So, \(|t - s| + |s - q| = -(t -s) + (s - q) = -t + 2s - q\).
So, in his case we'd have \(t - q = -t + 2s - q\) or \(t=s\). But we are told that q, s, and t are all different numbers, so this case is out.
Only q < s < t case is possible. Sufficient.
(2) t > q. Not sufficient.
Answer: A.
Bunnuel I tried it in this way . Is it right or wrong
given If q, s, and t are all different numbers, is q < s < t ?
or is 0<s-q<t-q?
Since i am doing the same operation on all the parts on inequality it doesn't violates any rule i suppose
ST 1 t - q = |t - s| + |s - q|
Here t-q is positive ( sum of two mod's) and also greater than s-q as something is been added to |s - q|
Now this equation can be solved in two ways only either same sign or opposite sign of mods
same sign leads t-q= t-s +s-q lhs=rhs therefore 0<s-q<t-q
different signs lead to t-q= t-s -s+q ,,,,,, not possible
or t-q = -t+s+s-q ,,,,,, again lhs is not equal to rhs . this means that the case s-q<0<t-q doesnt holds
Hence st 1 is sufficient
st 2 clearly not suficient hence a