Question

Select the correct statement.
Alex covers $2\frac{1}{3}$ mi on the first day.

Second day: He covers $1\frac{2}{3}$ times as much as he did on the first day.

Third day: He covers $1\frac{2}{5}$ times as much as he did on the first day.

Statement $1$: He covers more distance on the second day as compared to the third day.

Statement $2$: He covers less distance on the second day as compared to the third day.

• A. Statement $2$ is correct
• B. Both statements are incorrect.
• C. Both statements are correct.
• D. Statement $1$ is correct.

Answer 1

The correct option is A Statement $2$ is correct

Detailed step-by-step solution:
On the first day, Alex covers $2\frac{1}{3}\text{mi}$.

On the second day, he covers $1\frac{2}{3}$ times as much as he did on the first day, i.e., $1\frac{2}{3}\times 2\frac{1}{3}\text{mi}$.

Distance covered on the second day = $1\frac{2}{3}\times 2\frac{1}{3}\text{mi}=\frac{5}{3}\times \frac{7}{3}=\frac{35}{9}\text{mi}$

On the third day, he covers $1\frac{2}{5}$ times as much as he did on the first day, i.e., $1\frac{2}{5}\times 2\frac{1}{3}\text{mi}$.

Distance covered on second day = $1\frac{2}{5}\times 2\frac{1}{3}\text{mi}=\frac{7}{5}\times \frac{7}{3}=\frac{49}{15}\text{mi}$

So, on both the second and third days, he covers the same distance, i.e., $\frac{49}{15}\text{mi}$

Both fractions are unlike fractions.
Make both fractions like fractions.

$\frac{35}{9}=\frac{35\times 15}{9\times 15}=\frac{525}{135},\frac{49\times 9}{15\times 9}=\frac{441}{135}$

$\frac{441}{135}<\frac{525}{135}$

He covers less distance on the second day as compared to the third day.
So, statement $2$ is correct.

Hence, option C is correct.