# At 3:00 P.M., Carrie and Chad had already made 46 candles. By 5:00 p.m., the total had reached...

## Question:

At 3:00 P.M., Carrie and Chad had already made 46 candles. By 5:00 p.m., the total had reached 100 candles.

Assuming a constant production rate, at what time did they make their 82nd candle?

## Algebraic Equation Word Problems:

An algebraic equation is formed when a relation between any variable is equated to a constant. For example, {eq}4x +3y = 1{/eq} is an algebraic equation where {eq}x{/eq} and {eq}y{/eq} are variables and {eq}1{/eq} is the equivalent constant of the relation shown in the equation.

In solving word problems, the relationship between variables is usually stated in the problem and we need to form an algebraic equation out of it. Assuming a base variable and determining other unknowns in its term also helps in simplifying the equation.

## Answer and Explanation:

In {eq}2{/eq} hours or {eq}120{/eq} minutes, between 3:00 PM to 5:00 PM, a total of {eq}(100 - 46) =54 {/eq} candles were made.

Rate of production of candles:

$$\begin{align} \rm Rate\ of\ production &=\rm \dfrac{number\ of\ candles}{Time} \\[0.2cm] &=\rm \dfrac{54}{120} \\[0.2cm] &=\rm \dfrac{9}{20}\ candles\ per\ minutes \end{align} $$

{eq}82\rm nd{/eq} candle is {eq}(82 - 46) \rm th{/eq} candle from 3:00 PM, or equal to {eq}36\rm th {/eq}.

Time taken for {eq}82 \rm nd{/eq} candle is:

$$\begin{align} \rm Time&= \rm \dfrac{Number\ of\ candles}{rate\ of\ production} \\[0.2cm] &= \dfrac{36}{\dfrac{9}{20}} \\[0.2cm] &= \dfrac{36*20}{9} \\[0.2cm] &= 4*20 \\[0.2cm] &=\rm 80\ minutes \end{align} \\ $$

They make their {eq}82\rm nd{/eq} candle after {eq}\rm 1\ hour\ 20\ minutes {/eq} at **4:20 PM**.

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