Question

Manuel ate $\frac{1}{3}$ of the cracker on a plate.

His brother ate $\frac{1}{4}$ of the cracker.

There were $5$ crackers left on the plate.

How many cracker were there to begin with?

Answer 1

Finding the number of crackers:

Given,

Manuel ate $\frac{1}{3}$ of the cracker on a plate

His brother ate $\frac{1}{4}$ of the cracker

Remaining crackers were $5$

Let the total number of crackers on the plate be $x$

Then, Manuel ate $\frac{x}{3}$ crackers on the plate.

His brother then ate, $\frac{x-{\displaystyle \frac{x}{3}}}{4}$ crackers and after that $5$ crackers were left on the plate.

We can frame an equation for the remaining crackers as,

$$\begin{gathered} x-\left(\frac{x}{3}+\frac{{\displaystyle \frac{2x}{3}}}{4}\right)=5 \\ \Rightarrow x-\left(\frac{x}{3}+\frac{x}{6}\right)=5 \\ \Rightarrow x-\frac{x}{2}=5 \\ \Rightarrow x=10 \end{gathered}$$

Hence, in the beginning, there were $10$ crackers on the plate.