Question

Manuel ate 1/3 of the crackers on a plate and his brother ate 1/4 of the crackers. There were 5 crackers left on the plate. How many crackers were on the plate to begin with

Answer 1

Let \( x \) be the total number of crackers on the plate initially.

According to the problem, Manuel ate \( \frac{1}{3}x \) and his brother ate \( \frac{1}{4}x \). After they ate the crackers, there were 5 crackers left. We can set up the equation as follows:

\[
x - \left(\frac{1}{3}x + \frac{1}{4}x\right) = 5
\]

To combine the fractions, we first find a common denominator for \( \frac{1}{3} \) and \( \frac{1}{4} \). The least common multiple of 3 and 4 is 12. We can rewrite the fractions:

\[
\frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12}
\]

Now substitute these into the equation:

\[
x - \left(\frac{4}{12}x + \frac{3}{12}x\right) = 5
\]

Combine the fractions inside the parentheses:

\[
\frac{4}{12}x + \frac{3}{12}x = \frac{7}{12}x
\]

So we have:

\[
x - \frac{7}{12}x = 5
\]

This can be expressed as:

\[
\frac{12}{12}x - \frac{7}{12}x = 5
\]

This simplifies to:

\[
\frac{5}{12}x = 5
\]

To solve for \( x \), multiply both sides by \( \frac{12}{5} \):

\[
x = 5 \cdot \frac{12}{5} = 12
\]

Thus, the total number of crackers on the plate to begin with was \(\boxed{12}\).

To verify, if there were 12 crackers initially:

- Manuel ate \( \frac{1}{3} \times 12 = 4 \) crackers.
- His brother ate \( \frac{1}{4} \times 12 = 3 \) crackers.

Together, they ate \( 4 + 3 = 7 \) crackers, which leaves:

\[
12 - 7 = 5
\]

This confirms that the initial number of crackers is indeed \(\boxed{12}\).