From 9.00 AM to 2.00 PM, the temperature rose at a constant rate from 21°C to 36°C. The temperature at noon was:
(a) 27°C
(b) 30°C
(c) 32°C
(d) 28.5°C
From 9.00 AM to 2.00 PM, the temperature rose at a constant rate from \(21^{\circ} \mathrm{C}\) to \(36^{\circ} \mathrm{C}\). What was the temperature at noon ? (a) \(27^{\circ} \mathrm{C}\) (b) \(30^{\circ} \mathrm{C}\) (c) \(32^{\circ} \mathrm{C}\) (d) \(28.5^{\circ} \mathrm{C}\)
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The temperature increased from \(21^{\circ}C\) to \(36^{\circ}C\) over a 5-hour period from 9:00 AM to 2:00 PM, which means the temperature rose \(36^{\circ}C – 21^{\circ}C = 15^{\circ}C\) in total.
To find the rate of increase per hour, divide the total temperature increase by the number of hours:
\[
\frac{15^{\circ}C}{5 \text{ hours}} = 3^{\circ}C/\text{hour}
\]
From 9:00 AM to noon (12:00 PM) is 3 hours. At a rate of \(3^{\circ}C/\text{hour}\), the temperature increase from 9:00 AM to noon would be:
\[
3 \text{ hours} \times 3^{\circ}C/\text{hour} = 9^{\circ}C
\]
Therefore, the temperature at noon, starting from \(21^{\circ}C\) at 9:00 AM, would be:
\[
21^{\circ}C + 9^{\circ}C = 30^{\circ}C
\]
So, the temperature at noon was \(30^{\circ}C\).
The correct answer is (b) \(30^{\circ} \mathrm{C}\).