Question

Two candles A and B of the same height are lighted at the same instant. A is consumed in $4h$ while B in $3h$. Assume each candle burns at a constant rate. In how many hours after being lighted was A twice the height of B?
A. $1h48\min $
B. $2h24\min $
C. $2h18\min $
D. $1h58\min $

Answer 1

Hint: To obtain the answer of the given question we will use the length of the candles which is the same. Firstly we will find the burning rate of candles by dividing the length of the candles by the time taken by them to burn. Then we will find the length of the candle after time $t$ and then use the comparison given from an equation and solve it to get the desired answer.

Complete step by step answer:
It is given that the length of the candles is the same.
Let length of each candle is $l$.
Now as we known candle A take $4h$ to burn so burning rate of candle A is:
$\dfrac{l}{4}$
Next the candle B take $3h$ to burn so burning rate of candle B is:
$\dfrac{l}{3}$
Next let after $t$ hours A candle is twice the height of B candle.
So let length of candle A after $t$ hours is ${{l}_{1}}$ which will be:
${{l}_{1}}=l-\dfrac{l}{4}\times t$……$\left( 1 \right)$
Let length of candle B after $t$ hours is ${{l}_{2}}$ which will be:
${{l}_{2}}=l-\dfrac{l}{3}\times t$……$\left( 2 \right)$
As given after $t$ hours A candle is twice the height of B candle so we can write ${{l}_{1}}=2{{l}_{2}}$ so using equation (1) and (2) we get,
$\begin{align}
  & l-\dfrac{l}{4}\times t=2\left( l-\dfrac{l}{3}\times t \right) \\
 & \Rightarrow l-\dfrac{lt}{4}=2l-\dfrac{2lt}{3} \\
 & \Rightarrow \dfrac{2lt}{3}-\dfrac{lt}{4}=2l-l \\
 & \Rightarrow \dfrac{8lt-3lt}{12}=l \\
\end{align}$
Take $l$ common from both side and simplify for $t$ as follows:
$\begin{align}
  & \Rightarrow \dfrac{5lt}{12}=l \\
 & \Rightarrow l\left( \dfrac{5t}{12} \right)=l \\
 & \Rightarrow \dfrac{5t}{12}=1 \\
 & \Rightarrow t=\dfrac{12}{5} \\
\end{align}$
$t=2\dfrac{2}{5}$
Hence we got time as $2\dfrac{2}{5}h$ we will change it in hour and minute by multiplying the fraction by 60 as follows:
$\begin{align}
  & \Rightarrow \dfrac{2}{5}\times 60 \\
 & \Rightarrow 2\times 12 \\
 & \Rightarrow 24\min \\
\end{align}$
So we got the time as $2h24\min $

So, the correct answer is “Option B”.

Note: To solve such a question always assume the things given especially if the value is the same for two conditions or in this case two candles given which is the length of the candles. Then secondly try to form an equation so that the value we need can be obtained from it.