nalinnair wrote:
\(3r\leq{4s + 5}\)
\(|s|\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20
B. –5
C. 0
D. 5
E. 20
To get the range of r, we need the value of s.
But what we have is the range for s. Let's evaluate it:
\(|s|\leq{5}\)
This means \(-5 \leq s \leq 5\)
Check at the extremes.
s = -5 gives \(3r\leq{4*-5 + 5}\) so we get \(r\leq{-5}\)s = 5 gives \(3r\leq{4*5 + 5}\) so we get \(r\leq{8.33}\)
Note that any intermediate value of s will ensure that r is less than 8.33. The maximum value that s can take is 5 and corresponding to that, the max value r can take is 8.33.
Hence r can never be 20
Answer (E)
I am not able to understand the highlighted part . isnt r>=-5