# The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/5 + 32. Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212

**Solution:**

The given function is t (C) = 9C/5 + 32

**(i)** t (0) = = (9 x 0) / 5 + 32 = 0 + 32 = 32

**(ii**) t (28) = (9 x 28) / 5 + 32

= (252 + 160) / 5

= 412 / 5 = 82.4

**(iii)** t (- 10) = 9 x (- 10) + 32

= 9 x (- 2) + 32

= - 18 + 32 = 14

**(iv)** It is given that

t (C) = 212

⇒ 9C/5 + 32 = 212

⇒ 9C/5 = 212 - 32

⇒ C = (180 x 5)/9

= 100

Thus, the value of ‘t’, when t (C) = 212 is 100

NCERT Solutions Class 11 Maths Chapter 2 Exercise 2.3 Question 4

## The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/5 + 32. Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212

**Summary:**

A function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is given. We have found that the value of ‘t’, when t (C) = 212 is 100