At breakfast, the outside temperature was ${{35}^{\circ }}C.$ At lunch time, it had gone up by ${{5}^{\circ }}C$, and then at dinner time it had gone down by ${{2}^{\circ }}C$. What was the temperature at dinner time?
(a). ${{32}^{\circ }}C$
(b). ${{39}^{\circ }}C$
(c). ${{38}^{\circ }}C$
(d). ${{40}^{\circ }}C$

Hint: In this question we need to find the outside temperature by simple operations like addition and subtraction. We need to remember, the outside temperature in the afternoon rises and during night or in the evening decreases.

Complete step-by-step answer:
We have the outside temperature as 35 degree Celsius, and during lunch time which is afternoon the temperature goes up by 5 degree Celsius and then at dinner time in the night the temperature goes down by 2 degree Celsius.
Let’s consider the outside temperature at breakfast be x, during lunch time be y and during dinner time be z.
$x={{35}^{\circ }}C$
Now, when the temperature goes up in the afternoon during lunch time by 5 degree Celsius, the outside temperature is
$\begin{align}
  & y={{35}^{\circ }}C+{{5}^{\circ }}C \\
 & ={{40}^{\circ }}C
\end{align}$
After evening, during dinner time at night the temperature goes down by 2 degree Celsius, the outside temperature is
$\begin{align}
  & z={{40}^{\circ }}C-{{2}^{\circ }}C \\
 & ={{38}^{\circ }}C
\end{align}$
Hence, the outside temperature during dinner time is ${{38}^{\circ }}C$.
Correct answer is option (c)

Note: Here, in the first step, use the temperature at breakfast and the temperature it increases in the afternoon to calculate the outside temperature during lunch time. In the second step, use the temperature during lunch time and the temperature it goes down in the night or evening to find the temperature during dinner time. Also, if the answer is asked in Kelvin, simply add the final answer to 273.15 K.