Question

Manuel ate 1/3 of the crackers on a plate 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
12
7
17
24

Answer 1

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

Manuel ate \( \frac{1}{3}x \) of the crackers, and his brother ate \( \frac{1}{4}x \) of the crackers. Together, they ate:

\[
\frac{1}{3}x + \frac{1}{4}x
\]

To combine these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12, so we can express the fractions with a common denominator:

\[
\frac{1}{3}x = \frac{4}{12}x \quad \text{and} \quad \frac{1}{4}x = \frac{3}{12}x
\]

Now, adding them together:

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

Thus, the total crackers eaten is \( \frac{7}{12}x \). According to the problem, there were 5 crackers left, which means:

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

Simplifying the left side:

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

Now, we have:

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

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

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

Therefore, the total number of crackers on the plate initially was:

\[
\boxed{12}
\]