Question

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation

`C = (5F - 160)/9`

1. If the temperature is 86°F, what is the temperature in Celsius? 
2. If the temperature is 35°C, what is the temperature in Fahrenheit?
3. If the temperature is 0°C what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
4. What is the numerical value of the temperature which is same in both the scales?

Answer 1

Given relation is, `C = (5F - 160)/9`  ...(i)

⇒ 9C = 5F – 160 

⇒ 5F = 9C + 160

⇒ `F = (9C + 160)/5`  ...(ii)

i. Given, F =  86°F, then from equation (i), we get

`C = (5 xx 86 - 160)/9`

= `(430 - 160)/9`

= `270/9`

= 30°C

ii. Given, C = 35°C, then from equation (ii), we get

`F = (9 xx 35 + 160)/5`

= `(315 + 160)/5`

= `475/5`

= 95°F

iii. Given, C = 0°C, then from equation (ii), we get

`F = (9 xx 0 + 160)/5`

= `160/5`

= 32°F

F = 0°F, then from equation (i), we get

`C = (5 xx 0 - 160)/9`

= `(-160)/9`

= `(- 160/9)^circC`

iv. By given condition, C = F

Put this value in equation (i), we get

`C = (5C - 160)/9`

⇒ 9C = 5C – 160

⇒ 9C – 5C = –160

⇒ 4C = –160

⇒ `C = (-160)/4`

⇒ C = –40 = F 

∴ The numerical value of the temperature is –40.