Ex.1.3 Q6 Integers Solution - NCERT Maths Class 7

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A certain freezing process requires that room temperature be lowered from \(40^\circ \rm C\)  at the rate of \(5^\circ{\rm C}\)  every hour. What will be the temperature \(10\) hours after the process begins?

 Video Solution
Ex 1.3 | Question 6

Text Solution


Present temperature of room \(=\) \(40^\circ \rm C\)

Decrease in temperature after every hour \(= 5^\circ{\rm C}\)

Temperature of room after \(10\) hours

\[\begin{align}&=40^{\circ} \mathrm{c}+10 \times(-5)^{\circ} \mathrm{c}\\&=40^{\circ} \mathrm{c}-50^{\circ} \mathrm{c}\\&=-10^{\circ} \mathrm{c}\end{align}\]

Thus, the room temperature after \(10\) hours is \(–10^\circ{\rm C},\) after the process begins.

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