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Miscellaneous Exercise · Q11

Q.A solution is to be kept between 68∘68^\circ F and 77∘77^\circ F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F=95C+32F = \dfrac{9}{5}C + 32?

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The problem asks for the Celsius range corresponding to a Fahrenheit range of 68∘68^\circF to 77∘77^\circF. Using the linear conversion F=95C+32F = \frac{9}{5}C + 32, we solve for CC at each endpoint and find the range is 20∘20^\circC to 25∘25^\circC.

The key here is that the conversion formula is a linear function — it’s a straight line. That means the relationship between Fahrenheit and Celsius is one-to-one and order-preserving: if FF increases, CC increases too. So the lowest Fahrenheit corresponds to the lowest Celsius, and the highest Fahrenheit to the highest Celsius. No trickiness with flipping inequalities.

We are given:

F=95C+32F = \frac{9}{5}C + 32

We want the Celsius range when 68≤F≤7768 \leq F \leq 77.

  1. Solve for CC in terms of FF. Subtract 32 from both sides:

F−32=95CF - 32 = \frac{9}{5}C

Multiply both sides by 59\frac{5}{9}:

C=59(F−32)C = \frac{5}{9}(F - 32)

  1. Find CC at the lower bound F=68F = 68.

C=59(68−32)=59×36=5×4=20C = \frac{5}{9}(68 - 32) = \frac{5}{9} \times 36 = 5 \times 4 = 20

  1. Find CC at the upper bound F=77F = 77.

C=59(77−32)=59×45=5×5=25C = \frac{5}{9}(77 - 32) = \frac{5}{9} \times 45 = 5 \times 5 = 25

  1. State the range. Since the function is increasing (the coefficient 59\frac{5}{9} is positive), the Celsius values increase as Fahrenheit increases. So the range is: 20∘C≤C≤25∘C20^\circ \text{C} \leq C \leq 25^\circ \text{C} …

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