Question
Manuel ate 1/3 of the crackers and his brother ate 1/4 of the crackers. If there were 5 total how many did they eat
Answers
GPT-4o mini
To find out how many crackers Manuel and his brother ate, we start by determining how many crackers each of them ate from the total of 5 crackers.
1. **Manuel's share:**
He ate \( \frac{1}{3} \) of the crackers:
\[
\text{Crackers eaten by Manuel} = \frac{1}{3} \times 5 = \frac{5}{3} \approx 1.67
\]
2. **Brother's share:**
He ate \( \frac{1}{4} \) of the crackers:
\[
\text{Crackers eaten by brother} = \frac{1}{4} \times 5 = \frac{5}{4} = 1.25
\]
Now, we add the fractions of the crackers they both ate:
- First, we need a common denominator to add \( \frac{5}{3} \) and \( \frac{5}{4} \). The least common multiple of 3 and 4 is 12.
3. **Convert to a common denominator:**
\[
\frac{5}{3} = \frac{5 \times 4}{3 \times 4} = \frac{20}{12}
\]
\[
\frac{5}{4} = \frac{5 \times 3}{4 \times 3} = \frac{15}{12}
\]
4. **Add the fractions:**
\[
\frac{20}{12} + \frac{15}{12} = \frac{35}{12}
\]
Now, we convert \( \frac{35}{12} \) back to a mixed number:
\[
\frac{35}{12} = 2 \frac{11}{12}
\]
Thus, together, Manuel and his brother ate \( 2 \frac{11}{12} \) crackers.
To summarize:
- Manuel ate approximately 1.67 crackers.
- His brother ate 1.25 crackers.
- Together, they ate \( 2 \frac{11}{12} \) crackers, which exceeds the total of 5 crackers available, indicating a mistake since both portions cannot be more than the whole amount.
To correct this miscalculation:
1. The total eaten by each according to the total is actually \( \frac{1}{3} + \frac{1}{4} \) of 5 must be effectively limited to only the total:
Instead,
Manuel's 1 cracker and brother's 1 cracker would have been sufficient based on the totals.
Final correct tally:
- Total eaten = 2 by both.
Thus to reiterate, rounded:
They collectively ate nearly 2 crackers according to needs without exceeding 5.
Please note:
If guaranteed ratios of more crackers than exist, flawed amounts, as portioning should not exceed total available; thus, an error exists in setup. Would need to clarify constraints.
1. **Manuel's share:**
He ate \( \frac{1}{3} \) of the crackers:
\[
\text{Crackers eaten by Manuel} = \frac{1}{3} \times 5 = \frac{5}{3} \approx 1.67
\]
2. **Brother's share:**
He ate \( \frac{1}{4} \) of the crackers:
\[
\text{Crackers eaten by brother} = \frac{1}{4} \times 5 = \frac{5}{4} = 1.25
\]
Now, we add the fractions of the crackers they both ate:
- First, we need a common denominator to add \( \frac{5}{3} \) and \( \frac{5}{4} \). The least common multiple of 3 and 4 is 12.
3. **Convert to a common denominator:**
\[
\frac{5}{3} = \frac{5 \times 4}{3 \times 4} = \frac{20}{12}
\]
\[
\frac{5}{4} = \frac{5 \times 3}{4 \times 3} = \frac{15}{12}
\]
4. **Add the fractions:**
\[
\frac{20}{12} + \frac{15}{12} = \frac{35}{12}
\]
Now, we convert \( \frac{35}{12} \) back to a mixed number:
\[
\frac{35}{12} = 2 \frac{11}{12}
\]
Thus, together, Manuel and his brother ate \( 2 \frac{11}{12} \) crackers.
To summarize:
- Manuel ate approximately 1.67 crackers.
- His brother ate 1.25 crackers.
- Together, they ate \( 2 \frac{11}{12} \) crackers, which exceeds the total of 5 crackers available, indicating a mistake since both portions cannot be more than the whole amount.
To correct this miscalculation:
1. The total eaten by each according to the total is actually \( \frac{1}{3} + \frac{1}{4} \) of 5 must be effectively limited to only the total:
Instead,
Manuel's 1 cracker and brother's 1 cracker would have been sufficient based on the totals.
Final correct tally:
- Total eaten = 2 by both.
Thus to reiterate, rounded:
They collectively ate nearly 2 crackers according to needs without exceeding 5.
Please note:
If guaranteed ratios of more crackers than exist, flawed amounts, as portioning should not exceed total available; thus, an error exists in setup. Would need to clarify constraints.