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Exercise 2.3 · Q4

Q.The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C)=9C5+32t(C) = \dfrac{9C}{5} + 32. Find

(i) t(0)t(0)
(ii) t(28)t(28)
(iii) t(−10)t(-10)
(iv) The value of CC, when t(C)=212t(C) = 212.
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The function t(C)=9C5+32t(C) = \frac{9C}{5} + 32 is a linear conversion from Celsius to Fahrenheit. We evaluate it by substituting the given Celsius values and solving for CC when the output is 212212. The results are: (i) 3232,

(ii) 3325\frac{332}{5} or 66.466.4,

(iii) 1414,

(iv) C=100C = 100.

This problem is about linear function evaluation — the simplest and most powerful idea in algebra. A linear function like t(C)=9C5+32t(C) = \frac{9C}{5} + 32 tells you: take any input CC, multiply it by a fixed rate (95\frac{9}{5}), then add a constant (3232). That constant is the starting point when C=0C = 0, and the rate tells you how fast the output changes per unit change in input.

Here, the function is the real-world Celsius-to-Fahrenheit conversion. Every time you plug in a Celsius temperature, you get the corresponding Fahrenheit temperature. And if you know the Fahrenheit value, you can reverse the process to find the Celsius value.

Let’s work through each part.

  1. Find t(0)t(0) Substitute C=0C = 0 into the formula:

t(0)=9×05+32=0+32=32t(0) = \frac{9 \times 0}{5} + 32 = 0 + 32 = 32

This is the freezing point of water in Fahrenheit — a familiar landmark.

  1. Find t(28)t(28) Substitute C=28C = 28:

t(28)=9×285+32=2525+32t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32

Write 3232 as 1605\frac{160}{5} to combine:

2525+1605=4125\frac{252}{5} + \frac{160}{5} = \frac{412}{5}

As a decimal, that’s 82.4∘F82.4^\circ\text{F}. So t(28)=4125t(28) = \frac{412}{5} or 82.482.4.

  1. Find t(−10)t(-10) Substitute C=−10C = -10:

t(−10)=9×(−10)5+32=−905+32=−18+32=14t(-10) = \frac{9 \times (-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14

So t(−10)=14∘Ft(-10) = 14^\circ\text{F} — a cold but not freezing day.

  1. Find CC when t(C)=212t(C) = 212 Here we know the output and need the input. Set the function equal to 212212:

9C5+32=212\frac{9C}{5} + 32 = 212

Subtract 3232 from both sides: …

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