Case Study II

A few countries such as the USA officially use Fahrenheit as a unit for measuring temperature. Other countries prefer Celsius over Fahrenheit. The two different scales are related by the linear equation, $\dfrac{F - 32}{9} = \dfrac{C}{5}$. A scientist wants to store an experimental solution between a temperature range of 68°F and 77°F.

Based on the above information, answer the following questions:

1. The algebraic representation of the given information in degree Celsius is :

1. 68 < $\dfrac{5}{9}C$ + 32 ≤ 77, C ∈ R

2. 68 ≤ $\dfrac{5}{9}C$ − 32 ≤ 77, C ∈ R

3. 68 ≤ $\dfrac{9}{5}C$ − 32 < 77, C ∈ R

4. 68 < $\dfrac{9}{5}C$ + 32 < 77, C ∈ R

2. The solution set for the temperature in degree Celsius is:

1. {C ∈ R : 18 < C < 23}

2. {C ∈ R : 20 < C < 25}

3. {C ∈ R : 22 < C < 27}

4. {C ∈ R : 25 < C < 30}

3. What is the range of the temperature in degree Celsius?

1. between 20°C and 25°C

2. between 25°C and 30°C

3. between 18°C and 23°C

4. between 22°C and 27°C

4. Which of the following is the graphical representation of the temperature in degree Celsius?

a.

b.

c.

d.

5. If the minimum temperature that can be maintained in a particular refrigerator is 0°C, what is the possible temperature range of the refrigerator on a Fahrenheit scale?

1. F > $\dfrac{160}{9}$

2. F < $\dfrac{160}{9}$

3. F > 32

4. F < 32

1. Given,

$\Rightarrow \dfrac{F - 32}{9} = \dfrac{C}{5} \\[1em] \Rightarrow F - 32 = \dfrac{9}{5}C \\[1em] \Rightarrow F = \dfrac{9}{5}C + 32.$

Given,

A scientist wants to store an experimental solution between a temperature range of 68°F and 77°F.

$\therefore 68 \lt \dfrac{9}{5}C + 32 \lt 77$

Hence, Option 4 is the correct option.

2. Solving,

⇒ 68 < $\dfrac{9}{5}C$ + 32 < 77

Solving L.H.S of inequation,

⇒ 68 < $\dfrac{9}{5}C$ + 32

⇒ $\dfrac{9}{5}C$ + 32 > 68

⇒ $\dfrac{9}{5}C$ > 68 - 32

⇒ $\dfrac{9}{5}C$ > 36

⇒ 9C > 36 × 5

⇒ 9C > 180

⇒ C > $\dfrac{180}{9}$

⇒ C > 20 ………(1)

Solving R.H.S of inequation,

⇒ $\dfrac{9}{5}C$ + 32 < 77

⇒ $\dfrac{9}{5}C$ < 77 - 32

⇒ $\dfrac{9}{5}C$ < 45

⇒ 9C < 45 × 5

⇒ 9C < 225

⇒ C < $\dfrac{225}{9}$

⇒ C < 25 ……..(2)

From (1) and (2) we get,

∴ 20 < C < 25

Since C ∈ R

Solution set = {C ∈ R : 20 < C < 25 }

Hence, Option 2 is the correct option.

3. The solution set for the temperature in degree Celsius is: {C ∈ R : 20 < C < 25}

Hence, Option 1 is the correct option.

4. Solution set = {C ∈ R : 20 < C < 25}

Hence, Option a is the correct option.

5. Given,

Minimum temperature that can be maintained = 0°C

⇒ F = $\dfrac{9}{5}C + 32$

Minimum temperature that can be maintained in term of Fahrenheit scale :

⇒ F = $\dfrac{9}{5} \times (0) + 32$ = 32.

⇒ F > 32

Hence, Option 3 is the correct option.