Question

According to an article published in a science magazine the speed of ants in a particular place is affected by the temperature of that place. If the equation

s = \frac{t - 20}{5}

shows the relation between the temperature of a place and the speed of ants where temperature,

t

is in '

° C

' and speed,

s

is in

cm / s

, then draw the graph between the speed of ants and temperature using

t = 25 °, 30 °, 35 °, 40 °

. What is the temperature of the place where the speed of ants is

2.5 cm

per second?

Accepted Answer

The given equation is $s = \frac{t - 20}{5}$ , where the temperature $t$ is in $° C$ and speed, $s$ is in $cm / s$ . We are given to draw the graph between the speed of ants and temperature using $t = 25 °, 30 °, 35 °, 40 °$ .

First, we shall find the different values of $s$ for $t = 25 °,_{} 30 °, 35 °, 40 °$ .

For, $t = 25 °$ , we have, $s = \frac{t - 20}{5} = \frac{25 - 20}{5} = \frac{5}{5} = 1 cm / s$ .

For, $t = 30 °$ , we have, $s = \frac{t - 20}{5} = \frac{30 - 20}{5} = \frac{10}{5} = 2 cm / s$

For, $t = 35 °$ , we have, $s = \frac{t - 20}{5} = \frac{35 - 20}{5} = \frac{15}{5} = 3 cm / s$

For, $t - 40 °$ , we have, $s = \frac{t - 20}{5} = \frac{40 - 20}{5} = \frac{20}{5} = 4 cm / s$

Now, we shall draw the graph.

Here, the different temperatures of the place will be considered as the independent variables $(X)$ and these will be represented on the $X$ -axis. The speeds of the ant are considered the dependent variables $(Y)$ and these will be represented on the $Y$ -axis.

We shall write the values of both variable in $(X, Y)$ table as shown below.

The temperature $t$ (in $° C$ ) $(X)$ $25$ $30$ $35$ $40$

The speed of the ant (in $cm / s$ ) $(Y)$ $1$ $2$ $3$ $4$

The scale chosen on the $X$ -axis is: $1$ unit $= 5 ° C$

and, the scale chosen on the $Y$ -axis is: $1$ unit $= 1 cm / s$ .

Now, we plot the points $(X_{1}, Y_{1}) = (25, 1), (X_{2}, Y_{2}) = (30, 2), (X_{3}, Y_{3}) = (35, 3)$ and $(X_{4}, Y_{4}) = (40, 4)$ on the graph paper and draw a line connecting the points. The line is the required graph as shown below.

Now, substituting the value of $s = 2.5 cm / s$ in $s = \frac{t - 20}{5}$ , we get,

$s = \frac{t - 20}{5}$

$\Rightarrow 2.5 = \frac{t - 20}{5}$

$\Rightarrow t - 20 = 2.5 \times 5$

$\Rightarrow t = 12.5 + 20$

$\Rightarrow t = 32.5 ° C$

Therefore, the required temperature of the place is $32.5 ° C$ .

Chhattisgarh Board Solutions for Exercise 8: Exercise–1

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 8: Exercise–1

Questions from Chhattisgarh Board Solutions for Exercise 8: Exercise–1 with Hints & Solutions

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