
Two temperature scales A and B are related by $\dfrac{{A - 42}}{{110}} = \dfrac{{B - 72}}{{220.}}$. At which temperature two scales have the same reading?
A) $ - 42^\circ $
B) $ - 72^\circ $
C) $ + 12^\circ $
D) $ - 40^\circ $
Answer
586.2k+ views
Hint: Two temperatures, namely A and B, should have the same reading, so it means temperature A and temperature B have the same value. We can use this result in the above relation given in the question.
Complete step by step answer:
Since both the temperatures have the same reading, therefore we can use the result $A = B$.
Using given relation in question, $\dfrac{{A - 42}}{{110}} = \dfrac{{B - 72}}{{220.}}$
Putting value of B as A in above relation (because, $A = B$ )
$\dfrac{{A - 42}}{{110}} = \dfrac{{A - 72}}{{220}}$
On further solving, we get,
$2 \times \left( {A - 42} \right) = A - 72$
On solving this, we get,
$2A - A = 84^\circ - 72^\circ $
On further solving, we get,
$A = + 12^\circ $
So, we get our answer as $A = + 12^\circ $.
So option (C) is correct.
Note: Since we got $A = + 12^\circ $, it means the two temperature scales, named as A and B which are related by the given relation in question will give the same reading whenever we will get A and B both as positive $12$ degrees.
Complete step by step answer:
Since both the temperatures have the same reading, therefore we can use the result $A = B$.
Using given relation in question, $\dfrac{{A - 42}}{{110}} = \dfrac{{B - 72}}{{220.}}$
Putting value of B as A in above relation (because, $A = B$ )
$\dfrac{{A - 42}}{{110}} = \dfrac{{A - 72}}{{220}}$
On further solving, we get,
$2 \times \left( {A - 42} \right) = A - 72$
On solving this, we get,
$2A - A = 84^\circ - 72^\circ $
On further solving, we get,
$A = + 12^\circ $
So, we get our answer as $A = + 12^\circ $.
So option (C) is correct.
Note: Since we got $A = + 12^\circ $, it means the two temperature scales, named as A and B which are related by the given relation in question will give the same reading whenever we will get A and B both as positive $12$ degrees.
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