During a laboratory experiment, a liquid cooling solution's temperature dropped by a total… — ACT

During a laboratory experiment, a liquid cooling solution's temperature dropped by a total of $14.4^\circ\text{C}$ over a duration of $t$ minutes. If the temperature decreased at a constant rate of $1.8^\circ\text{C}$ per minute, which of the following equations can be solved to find $t$ ?

$1.8t = 14.4$ Answer
B
$\frac{t}{1.8} = 14.4$
C
$t - 1.8 = 14.4$
D
$t + 1.8 = 14.4$
E
$14.4t = 1.8$

Answer

The equation

1.8t = 14.4

correctly models the situation.

Because the cooling rate is constant at

1.8^\circ\text{C}

per minute, the total temperature change over

t

minutes is found by multiplying the rate by time (

1.8 \cdot t

). Setting this equal to the total drop of

14.4^\circ\text{C}

gives

1.8t = 14.4

Step-by-Step Solution

Identify the given quantities and their relationship.

Rate of change =

1.8^\circ\text{C/min}

, Time =

t

minutes, Total change =

14.4^\circ\text{C}

The total change in a constant rate scenario is the product of the rate and the time duration.

Set up the one-step algebraic equation.

\text{Rate} \times \text{Time} = \text{Total Change} \implies 1.8 \times t = 14.4

Translating the verbal relationship into an algebraic equation yields

1.8t = 14.4

Key Concept

Translating constant rate scenarios into one-step linear multiplication equations

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