Hong swam kilometers against the current in the same amount of time it took him to swim kilometers 16 with the current. The rate of the current was 1 kilometer per hour. How fast would Hong swim if there were no current?

Hong swam 8 kilometers against the current in the same amount of time it took him to swim kilometers 16 with the current. The rate of the current was 1 kilometer per hour. How fast would Hong swim if there were no current?

Let x represent Hang's swimming rate.

Hang's rate upstream would be: [tex]x-1[/tex]

Hang's rate downstream would be: [tex]x+1[/tex]

[tex]\text{Time}=\frac{\text{Distance}}{\text{Rate}}[/tex]

[tex]\text{Time taken downstream}=\frac{16}{x+1}[/tex]

[tex]\text{Time taken against current}=\frac{8}{x-1}[/tex]

Since downstream and upstream times are equal, so we can equate both equation as:

[tex]\frac{16}{x+1}=\frac{8}{x-1}[/tex]

Cross multiply:

[tex]16(x-1)=8(x+1)[/tex]

[tex]16x-16=8x+8[/tex]

[tex]16x-8x-16=8x-8x+8[/tex]

[tex]8x-16=8[/tex]

[tex]8x-16+16=8+16[/tex]

[tex]8x=24\\\\\frac{8x}{8}=\frac{24}{8}\\\\x=3[/tex]

Therefore, Hong would swim at a rate of 3 kilometers per hour, if there were no current.