Question

Bill's record-breaking jump was 1 1/8 times longer than the previous best in the jump competition. If Bill's distance was 5 2/5 m, what was the old record

Answer 1

To find the old record, we can set up the relationship based on the information given. Let \( x \) be the previous best distance in the jump competition. According to the problem, Bill's distance was \( 1 \frac{1}{8} \) times longer than the previous record, which can be expressed as:

\[
\text{Bill's distance} = \frac{9}{8} x
\]

We know that Bill's distance is \( 5 \frac{2}{5} \) m. We can convert this mixed number into an improper fraction:

\[
5 \frac{2}{5} = \frac{5 \times 5 + 2}{5} = \frac{25 + 2}{5} = \frac{27}{5}
\]

Now we can set up the equation:

\[
\frac{27}{5} = \frac{9}{8} x
\]

To solve for \( x \), we can multiply both sides by \( \frac{8}{9} \):

\[
x = \frac{27}{5} \times \frac{8}{9}
\]

Next, we can simplify the right side:

\[
x = \frac{27 \times 8}{5 \times 9} = \frac{216}{45}
\]

We can simplify \( \frac{216}{45} \) by finding the greatest common divisor (GCD) of 216 and 45:

1. The GCD of 216 and 45 is 9.
2. Now, divide the numerator and denominator by their GCD:

\[
x = \frac{216 \div 9}{45 \div 9} = \frac{24}{5}
\]

We can convert \( \frac{24}{5} \) back into a mixed number:

\[
\frac{24}{5} = 4 \frac{4}{5}
\]

Thus, the old record was:

\[
\boxed{4 \frac{4}{5} \text{ m}}
\]