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A marine biologist measured one fish that was 3 2/4 feet long and a second fish that was 2/3 of a foot long. How much longer wa...Question
A marine biologist measured one fish that was 6 1/2 feet long and a second fish that was 1 2/3 feet long. How much longer was the first fish?
Write your answer as a fraction or as a whole or mixed number.
feet
Write your answer as a fraction or as a whole or mixed number.
feet
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Answered by
GPT-4o mini
To find out how much longer the first fish is than the second fish, we need to subtract the length of the second fish from the length of the first fish.
The lengths of the fish are:
- First fish: \( 6 \frac{1}{2} \) feet
- Second fish: \( 1 \frac{2}{3} \) feet
First, we convert these mixed numbers into improper fractions.
For the first fish:
\[
6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{6 \times 2 + 1}{2} = \frac{12 + 1}{2} = \frac{13}{2}
\]
For the second fish:
\[
1 \frac{2}{3} = 1 + \frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3}
\]
Now we have:
- First fish: \( \frac{13}{2} \)
- Second fish: \( \frac{5}{3} \)
Next, we need to subtract \( \frac{5}{3} \) from \( \frac{13}{2} \). To do this, we need a common denominator. The least common multiple of 2 and 3 is 6.
Now we convert each fraction:
\[
\frac{13}{2} = \frac{13 \times 3}{2 \times 3} = \frac{39}{6}
\]
\[
\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}
\]
Now we can perform the subtraction:
\[
\frac{39}{6} - \frac{10}{6} = \frac{39 - 10}{6} = \frac{29}{6}
\]
Now we can convert \( \frac{29}{6} \) back to a mixed number:
\[
29 \div 6 = 4 \quad \text{(with a remainder of 5)}
\]
This means:
\[
\frac{29}{6} = 4 \frac{5}{6}
\]
Thus, the first fish is
\[
\boxed{4 \frac{5}{6}}
\]
feet longer than the second fish.
The lengths of the fish are:
- First fish: \( 6 \frac{1}{2} \) feet
- Second fish: \( 1 \frac{2}{3} \) feet
First, we convert these mixed numbers into improper fractions.
For the first fish:
\[
6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{6 \times 2 + 1}{2} = \frac{12 + 1}{2} = \frac{13}{2}
\]
For the second fish:
\[
1 \frac{2}{3} = 1 + \frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3}
\]
Now we have:
- First fish: \( \frac{13}{2} \)
- Second fish: \( \frac{5}{3} \)
Next, we need to subtract \( \frac{5}{3} \) from \( \frac{13}{2} \). To do this, we need a common denominator. The least common multiple of 2 and 3 is 6.
Now we convert each fraction:
\[
\frac{13}{2} = \frac{13 \times 3}{2 \times 3} = \frac{39}{6}
\]
\[
\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}
\]
Now we can perform the subtraction:
\[
\frac{39}{6} - \frac{10}{6} = \frac{39 - 10}{6} = \frac{29}{6}
\]
Now we can convert \( \frac{29}{6} \) back to a mixed number:
\[
29 \div 6 = 4 \quad \text{(with a remainder of 5)}
\]
This means:
\[
\frac{29}{6} = 4 \frac{5}{6}
\]
Thus, the first fish is
\[
\boxed{4 \frac{5}{6}}
\]
feet longer than the second fish.
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