A temperature 20^\circC is the same as
A
36oF
B
68oF
C
11.1oF
D
43.1oF
E
25oF
Verified Correct Answer: B.
Physical Principles & Numerical Solution
212^\circF ... 100^\circC
\(\theta\)^\circF ... 20^\circC
32^\circF ... 0^\circC
\(\frac{\theta - 32}{212 - 32}\) = \(\frac{20 - 0}{100 - 0}\)
180 = 5(\(\theta\) - 32)
5\(\theta\) = 180 + 160 = 340
\(\theta\) = 68^\circF
Source: Myschool Past Questions Archive (Record #29265)